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diff --git a/.gitattributes b/.gitattributes new file mode 100644 index 0000000..6833f05 --- /dev/null +++ b/.gitattributes @@ -0,0 +1,3 @@ +* text=auto +*.txt text +*.md text diff --git a/13693-pdf.pdf b/13693-pdf.pdf Binary files differnew file mode 100644 index 0000000..b4bb016 --- /dev/null +++ b/13693-pdf.pdf diff --git a/13693-pdf.zip~ b/13693-pdf.zip~ Binary files differnew file mode 100644 index 0000000..10c17cb --- /dev/null +++ b/13693-pdf.zip~ diff --git a/13693-t.zip~ b/13693-t.zip~ Binary files differnew file mode 100644 index 0000000..452074c --- /dev/null +++ b/13693-t.zip~ diff --git a/13693-t/13693-t.tex b/13693-t/13693-t.tex new file mode 100644 index 0000000..b41dd37 --- /dev/null +++ b/13693-t/13693-t.tex @@ -0,0 +1,4682 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of The Theory of Numbers, by Robert D. Carmichael +% % +% This eBook is for the use of anyone anywhere in the United States and % +% most other parts of the world at no cost and with almost no restrictions% +% whatsoever. You may copy it, give it away or re-use it under the terms % +% of the Project Gutenberg License included with this eBook or online at % +% www.gutenberg.org. If you are not located in the United States, you % +% will have to check the laws of the country where you are located before % +% using this eBook. % +% % +% % +% Title: The Theory of Numbers % +% % +% Author: Robert D. Carmichael % +% % +% Release Date: April 8, 2013 [EBook #13693] % +% Revised: January 9, 2924 % +% % +% Language: English % +% % +% Character set encoding: TeX % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{13693} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: Additional mathematical symbols. Required. %% +%% %% +%% makeidx: Indexing. Required. %% +%% %% +%% PDF pages: 88 %% +%% PDF page size: US Letter (8.5 x 11in) %% +%% %% +%% Summary of log file: %% +%% * Two overfull hboxes (~2.6pt and ~15.2pt too wide). %% +%% %% +%% Command block: %% +%% %% +%% pdflatex x2 %% +%% makeindex %% +%% pdflatex %% +%% %% +%% %% +%% April 2013: pglatex. %% +%% Compile this project with: %% +%% pdflatex 13693-t.tex ..... TWO times %% +%% makeindex 13693-t.idx %% +%% pdflatex 13693-t.tex %% +%% %% +%% pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) %% +%% %% +%% December 2021: okrick. %% +%% MiKTeX Console 4.3, Windows 10 Home %% +%% TeXworks 0.6.6 used to generate PDF output. %% +%% %% +%% %% +%% January 2024: okrick. %% +%% MiKTeX Console 4.9, Windows 10 Home %% +%% TeXworks 0.6.7 used to generate PDF output. %% +%% %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\listfiles +\documentclass[oneside]{book} +\usepackage[latin1]{inputenc} +\usepackage[reqno]{amsmath} +\usepackage{amssymb} +\usepackage{makeidx} +\makeindex +\begin{document} + +\thispagestyle{empty} +\small +\begin{verbatim} +The Project Gutenberg EBook of The Theory of Numbers, by Robert D. Carmichael + +This eBook is for the use of anyone anywhere in the United States and +most other parts of the world at no cost and with almost no restrictions +whatsoever. You may copy it, give it away or re-use it under the terms +of the Project Gutenberg License included with this eBook or online at +www.gutenberg.org. If you are not located in the United States, you +will have to check the laws of the country where you are located before +using this eBook. + + +Title: The Theory of Numbers + +Author: Robert D. Carmichael + +Release Date: October 10, 2003 [eBook #13693] +Revised Date: December 23, 2021 + +Language: English + +Character set encoding: TeX + +*** START OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +Produced by David Starner, Joshua Hutchinson, John Hagerson. +Revised by Richard Tonsing +\end{verbatim} +\normalsize +\newpage + +\frontmatter + +\begin{center} +\noindent \Large MATHEMATICAL MONOGRAPHS \\ + +\bigskip \footnotesize{EDITED BY} \\ +\normalsize \textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD.} \\ + +\bigskip\bigskip \huge +No. 13. + +\bigskip\bigskip \huge THE THEORY \\ +\bigskip\small \textsc{of} \\ +\bigskip\huge NUMBERS \\ + +\bigskip\bigskip\footnotesize\textsc{by} \\ +\bigskip\large ROBERT D. CARMICHAEL, \\ +\footnotesize\textsc{Associate Professor of Mathematics in Indiana +University} + +\bigskip\bigskip\normalsize NEW YORK: \\ +\medskip JOHN WILEY \& SONS. \\ +\medskip \textsc{London: CHAPMAN \& HALL, Limited.} \\ +\medskip 1914. + +\bigskip\bigskip +\tiny \textsc{Copyright 1914} \\ +\textsc{by} \\ +ROBERT D. CARMICHAEL. \\ +\medskip \textsc{the scientific press} \\ +\textsc{robert drummond and company} \\ +\textsc{brooklyn, n.~y.} +\end{center} + +\bigskip\bigskip +\scriptsize \noindent \textsc{Transcriber's Note:} \emph{I did my +best to recreate the index.} \normalsize + +\newpage + +\fbox{\parbox{11cm}{ +\begin{center} +\textbf{MATHEMATICAL MONOGRAPHS.} \\ +\small\textsc{edited by}\normalsize \\ +\textbf{Mansfield Merriman and Robert S. Woodward.} \\ +\footnotesize \textbf{Octavo. Cloth. \$1.00 each.} \\ + +\bigskip \textbf{No. 1. History of Modern Mathematics.} \\ +By \textsc{David Eugene Smith.} + +\smallskip \textbf{No. 2. Synthetic Projective Geometry.} \\ +By \textsc{George Bruce Halsted.} + +\smallskip \textbf{No. 3. Determinants.} \\ +By \textsc{Laenas Gifford Weld.} + +\smallskip \textbf{No. 4. Hyperbolic Functions.} \\ +By \textsc{James McMahon.} + +\smallskip \textbf{No. 5. Harmonic Functions.} \\ +By \textsc{William E. Byerly.} + +\smallskip \textbf{No. 6. Grassmann's Space Analysis.} \\ +By \textsc{Edward W. Hyde.} + +\smallskip \textbf{No. 7. Probability and Theory of Errors.} \\ +By \textsc{Robert S. Woodward.} + +\smallskip \textbf{No. 8. Vector Analysis and Quaternions.} \\ +By \textsc{Alexander Macfarlane.} + +\smallskip \textbf{No. 9. Differential Equations.} \\ +By \textsc{William Woolsey Johnson.} + +\smallskip \textbf{No. 10. The Solution of Equations.} \\ +By \textsc{Mansfield Merriman.} + +\smallskip \textbf{No. 11. Functions of a Complex Variable.} \\ +By \textsc{Thomas S. Fiske.} + +\smallskip \textbf{No. 12. The Theory of Relativity.} \\ +By \textsc{Robert D. Carmichael.} + +\smallskip \textbf{No. 13. The Theory of Numbers.} \\ +By \textsc{Robert D. Carmichael.} \normalsize + +\bigskip \small PUBLISHED BY \\ +\smallskip \textbf{JOHN WILEY \& SONS, Inc., NEW YORK. \\ +CHAPMAN \& HALL, Limited, LONDON.} +\end{center}}} + +\chapter{Editors' Preface.} + +The volume called Higher Mathematics, the third edition of which was +published in 1900, contained eleven chapters by eleven authors, each +chapter being independent of the others, but all supposing the +reader to have at least a mathematical training equivalent to that +given in classical and engineering colleges. The publication of that +volume was discontinued in 1906, and the chapters have since been +issued in separate Monographs, they being generally enlarged by +additional articles or appendices which either amplify the former +presentation or record recent advances. This plan of publication was +arranged in order to meet the demand of teachers and the convenience +of classes, and it was also thought that it would prove advantageous +to readers in special lines of mathematical literature. + +It is the intention of the publishers and editors to add other +monographs to the series from time to time, if the demand seems to +warrant it. Among the topics which are under consideration are those +of elliptic functions, the theory of quantics, the group theory, the +calculus of variations, and non-Euclidean geometry; possibly also +monographs on branches of astronomy, mechanics, and mathematical +physics may be included. It is the hope of the editors that this +Series of Monographs may tend to promote mathematical study and +research over a wider field than that which the former volume has +occupied. + +\chapter{Preface} + +The purpose of this little book is to give the reader a convenient +introduction to the theory of numbers, one of the most extensive and +most elegant disciplines in the whole body of mathematics. The +arrangement of the material is as follows: The first five chapters +are devoted to the development of those elements which are essential +to any study of the subject. The sixth and last chapter is intended +to give the reader some indication of the direction of further study +with a brief account of the nature of the material in each of the +topics suggested. The treatment throughout is made as brief as is +possible consistent with clearness and is confined entirely to +fundamental matters. This is done because it is believed that in +this way the book may best be made to serve its purpose as an +introduction to the theory of numbers. + +Numerous problems are supplied throughout the text. These have been +selected with great care so as to serve as excellent exercises for +the student's introductory training in the methods of number theory +and to afford at the same time a further collection of useful +results. The exercises marked with a star are more difficult than +the others; they will doubtless appeal to the best students. + +Finally, I should add that this book is made up from the material +used by me in lectures in Indiana University during the past two +years; and the selection of matter, especially of exercises, has +been based on the experience gained in this way. + +\hfill \textsc{R.~D.\ Carmichael.} + +\tableofcontents + +%% CHAPTER I. ELEMENTARY PROPERTIES OF INTEGERS +%% 1. Fundamental Notions and Laws +%% 2. Definition of Divisibility. The Unit +%% 3. Prime Numbers. The Sieve of Eratosthenes +%% 4. The Number of Primes is Infinite +%% 5. The Fundamental Theorem of Euclid +%% 6. Divisibility by a Prime Number +%% 7. The Unique Factorization Theorem +%% 8. The Divisors of an Integer +%% 9. The Greatest Common Factor of Two or More Integers +%% 10. The Least Common Multiple of Two or More Integers +%% 11. Scales of Notation +%% 12. Highest Power of a Prime $p$ Contained in $n!$ +%% 13. Remarks Concerning Prime Numbers +%% +%% CHAPTER II. ON THE INDICATOR OF AN INTEGER +%% 14. Definition. Indicator of a Prime Power +%% 15. The Indicator of a Product +%% 16. The Indicator of Any Positive Integer +%% 17. Sum of the Indicators of the Divisors of a Number +%% +%% CHAPTER III. ELEMENTARY PROPERTIES OF CONGRUENCES +%% 18. Congruences Modulo $m$ +%% 19. Solutions of Congruences by Trial +%% 20. Properties of Congruences Relative to Division +%% 21. Congruences with a Prime Modulus +%% 22. Linear Congruences +%% +%% CHAPTER IV. THE THEOREMS OF FERMAT AND WILSON +%% 23. Fermat's General Theorem +%% 24. Euler's Proof of the Simple Fermat Theorem +%% 25. Wilson's Theorem +%% 26. The Converse of Wilson's Theorem +%% 27. Impossibility of $1\cdot 2\cdot 3\cdot \ldots \cdot +%% \overline{n-1}+1=n^k, n>5$ +%% 28. Extension of Fermat's Theorem +%% 29. On the Converse of Fermat's Simple Theorem +%% 30. Application of Previous Results to Linear Congruences +%% 31. Application of the Preceding Results to the Theory of +%% Quadratic Residues +%% +%% CHAPTER V. PRIMITIVE ROOTS MODULO $m$ +%% 32. Exponent of an Integer Modulo $m$ +%% 33. Another Proof of Fermat's General Theorem +%% 34. Definition of Primitive Roots +%% 35. Primitive Roots Modulo $p$ +%% 36. Primitive Roots Modulo $p^\alpha$, $p$ an Odd Prime +%% 37. Primitive Roots Modulo $2p^\alpha$, $p$ an Odd Prime +%% 38. Recapitulation +%% 39. Primitive $\lambda$-Roots +%% +%% CHAPTER VI. OTHER TOPICS +%% 40. Introduction +%% 41. Theory of Quadratic Residues +%% 42. Galois Imaginaries +%% 43. Arithmetic Forms +%% 44. Analytical Theory of Numbers +%% 45. Diophantine Equations +%% 46. Pythagorean Triangles +%% 47. The Equation $x^n+y^n = z^n$ + +\mainmatter + +\chapter{ELEMENTARY PROPERTIES OF INTEGERS} +\section{Fundamental Notions and Laws}\label{s1}% +\index{Fundamental notions} + +In the present chapter we are concerned primarily with certain +elementary properties of the positive integers 1, 2, 3, 4, \ldots It +will sometimes be convenient, when no confusion can arise, to employ +the word \emph{integer} or the word \emph{number} in the sense of +positive integer. + +We shall suppose that the integers are already defined, either by +the process of counting or otherwise. We assume further that the +meaning of the terms \emph{greater, less, equal, sum, difference, +product} is known. + +From the ideas and definitions thus assumed to be known follow +immediately the theorems: +\begin{table}[h] +\begin{tabular}{rl} + I.\ & The sum of any two integers is an integer. \\ + II.\ & The difference of any two integers is an integer. \\ + III.\ & The product of any two integers is an integer. +\end{tabular} +\end{table} + +Other fundamental theorems, which we take without proof, are +embodied in the following formulas: +\begin{table}[h] +\begin{tabular}{rrcl} + IV.\ & $a + b$ & = & $b + a$. \\ + V.\ & $a \times b$ & = & $b \times a$. \\ + VI.\ & $(a + b) + c$ & = & $a + (b + c)$. \\ + VII.\ & $(a \times b) \times c$ & = & $a \times (b \times c)$. \\ +VIII.\ & $a \times (b + c)$ & = & $a \times b + a \times c$. +\end{tabular} +\end{table} +Here $a$, $b$, $c$ denote any positive integers. + +\newpage +These formulas are equivalent in order to the following five +theorems: addition is commutative; multiplication is commutative; +addition is associative; multiplication is associative; +multiplication is distributive with respect to addition. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove the following relations: +\begin{align*} + 1 + 2 + 3 \ldots + n &= \frac{n(n+1)}{2} \\ + 1 + 3 + 5 + \ldots + (2n - 1) &= n^2, \\ +1^3 + 2^3 + 3^3 + \ldots + n^3 &= \left(\frac{n(n+1)}{2}\right)^2 + = (1+2+\ldots+n)^2. +\end{align*} + +\item[2.] Find the sum of each of the following series: +\begin{align*} +1^2 + 2^2 + 3^2 + &\ldots + n^2, \\ +1^2 + 3^2 + 5^2 + &\ldots + (2n - 1)^2, \\ +1^3 + 3^3 + 5^3 + &\ldots + (2n - 1)^3. +\end{align*} + +\item[3.] Discover and establish the law suggested by the equations +$1^2 = 0 + 1$, $2^2 = 1 + 3$, $3^2 = 3 + 6$, $4^2 = 6 + 10$, +$\ldots$; by the equations $1 = 1^3$, $3 + 5 = 2^3$, $7 + 9 + 11 = +3^3$, $13 + 15 + 17 + 19 = 4^3$, $\ldots$. +\end{enumerate} \normalsize + +\section{Definition of Divisibility. The Unit}\label{s2}% +\index{Divisibility}\index{Unit} + +\textsc{Definitions.} An integer $a$ is said to be divisible by an +integer $b$ if there exists an integer $c$ such that $a = bc$. It is +clear from this definition that $a$ is also divisible by $c$. The +integers $b$ and $c$ are said to be divisors or factors of $a$; and +$a$ is said to be a multiple of $b$ or of $c$. The process of +finding two integers $b$ and $c$ such that $bc$ is equal to a given +integer $a$ is called the process of resolving $a$ into factors or +of factoring $a$; and $a$ is said to be resolved into factors or to +be factored. + +We have the following fundamental theorems: + +\smallskip I.~\emph{If $b$ is a divisor of $a$ and $c$ is a divisor +of $b$, then $c$ is a divisor of $a$.} + +Since $b$ is a divisor of a there exists an integer $\beta$ such +that $a = b\beta$. Since $c$ is a divisor of $b$ there exists an +integer $\gamma$ such that $b = c\gamma$. Substituting this value of +$b$ in the equation $a = b\gamma$ we have $a = c\gamma\beta$. But +from theorem III of \S~\ref{s1} it follows that $\gamma\beta$ is an +integer; hence, $c$ is a divisor of $a$, as was to be proved. + +\smallskip II.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the sum of $a$ and $b$.} + +From the hypothesis of the theorem it follows that integers $\alpha$ +and $\beta$ exist such that +\begin{gather*} +a = c\alpha,\quad b = c\beta. \\ +\intertext{Adding, we have} +a + b = c\alpha + c\beta = c(\alpha + \beta) = c\delta, +\end{gather*} +where $\delta$ is an integer. Hence, $c$ is a divisor of $a+b$. + +\smallskip III.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the difference of $a$ and $b$.} + +The proof is analogous to that of the preceding theorem. + +\smallskip \textsc{Definitions.} If $a$ and $b$ are both divisible +by $c$, then $c$ is said to be a common divisor or a common factor +of $a$ and $b$. Every two integers have the common factor 1. The +greatest integer which divides both $a$ and $b$ is called the +greatest common divisor of $a$ and $b$. More generally, we define in +a similar way a common divisor and the greatest common divisor of +$n$ integers $a_1$, $a_2$, $\ldots$, $a_n$.\index{Common!divisors} + +\smallskip \textsc{Definitions.} If an integer $a$ is a multiple of +each of two or more integers it is called a common multiple of these +integers. The product of any set of integers is a common multiple of +the set. The least integer which is a multiple of each of two or +more integers is called their least common multiple.% +\index{Common!multiples} + +It is evident that the integer $1$ is a divisor of every integer and +that it is the only integer which has this property. It is called +the unit. + +\smallskip \textsc{Definition.} Two or more integers which have no +common factor except $1$ are said to be prime to each other or to be +relatively prime.\index{Relatively prime} + +\smallskip \textsc{Definition.} If a set of integers is such that no +two of them have a common divisor besides $1$ they are said to be +prime each to each.\index{Prime each to each} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove that $n^3 - n$ is divisible by $6$ for every +positive integer $n$. + +\item[2.] If the product of four consecutive integers is increased by +$1$ the result is a square number. + +\item[3.] Show that $2^{4n + 2} + 1$ has a factor different from itself +and $1$ when $n$ is a positive integer. +\end{enumerate} \normalsize + +\section{Prime Numbers. The Sieve of Eratosthenes}\label{s3}% +\index{Eratosthenes}\index{Sieve of Eratosthenes} + +\textsc{Definition.} If an integer $p$ is different from 1 and has +no divisor except itself and 1 it is said to be a prime number or to +be a prime. + +\smallskip \textsc{Definition.} An integer which has at least one +divisor other than itself and 1 is said to be a composite number or +to be composite. + +All integers are thus divided into three classes: +\begin{table}[h] +\begin{tabular}{rl} +1.\ & The unit; \\ +2.\ & Prime numbers; \\ +3.\ & Composite numbers. +\end{tabular} +\end{table}\index{Composite numbers}\index{Prime numbers} + +We have seen that the first class contains only a single number. The +third class evidently contains an infinitude of numbers; for, it +contains all the numbers $2^2, 2^3, 2^4, \ldots$ In the next section +we shall show that the second class also contains an infinitude of +numbers. We shall now show that every number of the third class +contains one of the second class as a factor, by proving the +following theorem: + +\smallskip I.~\emph{Every integer greater than 1 has a prime factor.} + +Let $m$ be any integer which is greater than 1. We have to show that +it has a prime factor. If $m$ is prime there is the prime factor $m$ +itself. If $m$ is not prime we have +\begin{equation*} +m = m_1 m_2 +\end{equation*} +where $m_1$ and $m_2$ are positive integers both of which are less +than $m$. If either $m_1$ or $m_2$ is prime we have thus obtained a +prime factor of $m$. If neither of these numbers is prime, then +write +\begin{equation*} +m_1 = m'_1 m'_2,\quad m'_1 > 1, m'_2 > 1. +\end{equation*} +Both $m'_1$ and $m'_2$ are factors of $m$ and each of them is less +than $m_1$. Either we have not found in $m'_1$ or $m'_2$ a prime +factor of $m$ or the process can be continued by separating one of +these numbers into factors. Since for any given $m$ there is +evidently only a finite number of such steps possible, it is clear +that we must finally arrive at a prime factor of $m$. From this +conclusion, the theorem follows immediately. + +Eratosthenes has given a useful means of finding the prime numbers +which are less than any given integer $m$. It may be described as +follows: + +Every prime except 2 is odd. Hence if we write down every odd number +from 3 up to $m$ we shall have in the list every prime less than $m$ +except 2. Now 3 is prime. Leave it in the list; but beginning to +count from 3 strike out every third number in the list. Thus every +number divisible by 3, except 3 itself, is cancelled. Then begin +from 5 and cancel every fifth number. Then begin from the next +uncancelled number, namely 7, and strike out every seventh number. +Then begin from the next uncancelled number, namely 11, and strike +out every eleventh number. Proceed in this way up to $m$. The +uncancelled numbers remaining will be the odd primes not greater +than $m$. + +It is obvious that this process of cancellation need not be carried +altogether so far as indicated; for if $p$ is a prime greater than +$\sqrt{m}$, the cancellation of any $p^\text{th}$ number from $p$ +will be merely a repetition of cancellations effected by means of +another factor smaller than $p$, as one may see by the use of the +following theorem. + +\smallskip II.~\emph{An integer $m$ is prime if it has no prime +factor equal or less than $I$, where $I$ is the greatest integer +whose square is equal to or less than $m$.} + +Since $m$ has no prime factor less than $I$, it follows from theorem +I that it has no factor but unity less than $I$. Hence, if $m$ is +not prime it must be the product of two numbers each greater than +$I$; and hence it must be equal to or greater than $(I+1)^2$. This +contradicts the hypothesis on $I$; and hence we conclude that $m$ is +prime. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] By means of the method of Eratosthenes determine the primes +less than 200. +\end{enumerate} +\normalsize + +\section{The Number of Primes is Infinite}\label{s4}% +\index{Prime numbers} + +I.~\emph{The number of primes is infinite.} + +We shall prove this theorem by supposing that the number of primes +is not infinite and showing that this leads to a contradiction. If +the number of primes is not infinite there is a greatest prime +number, which we shall denote by $p$. Then form the number +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p + 1. +\end{equation*} +Now by theorem 1 of \S~\ref{s3} $N$ has a prime divisor $q$. But +every non-unit divisor of $N$ is obviously greater than $p$. Hence +$q$ is greater than $p$, in contradiction to the conclusion that $p$ +is the greatest prime. Thus the proof of the theorem is complete. + +In a similar way we may prove the following theorem: + +\smallskip II.~\emph{Among the integers of the arithmetic +progression $5$, $11$, $17$, $23$, $\ldots$, there is an infinite +number of primes.} + +If the number of primes in this sequence is not infinite there is a +greatest prime number in the sequence; supposing that this greatest +prime number exists we shall denote it by $p$. Then the number $N$, +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p-1, +\end{equation*} +is not divisible by any number less than or equal to $p$. This +number $N$, which is of the form $6n - 1$, has a prime factor. If +this factor is of the form $6k - 1$ we have already reached a +contradiction, and our theorem is proved. If the prime is of the +form $6k_1 + 1$ the complementary factor is of the form $6k_2 - 1$. +Every prime factor of $6k_2 - 1$ is greater than $p$. Hence we may +treat $6k_2 - 1$ as we did $6n - 1$, and with a like result. Hence +we must ultimately reach a prime factor of the form $6k_3 - 1$; for, +otherwise, we should have $6n - 1$ expressed as a product of prime +factors all of the form $6t + 1$---a result which is clearly +impossible. Hence we must in any case reach a contradiction of the +hypothesis. Thus the theorem is proved. + +The preceding results are special cases of the following more +general theorem: + +\smallskip III.~\emph{Among the integers of the arithmetic +progression $a$, $a + d$, $a + 2d$, $a + 3d$, $\ldots$, there is an +infinite number of +primes, provided that $a$ and $d$ are relatively prime.}% +\index{Arithmetic progression} + +For the special case given in theorem II we have an elementary +proof; but for the general theorem the proof is difficult. We shall +not give it here. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Prove that there is an infinite number of primes of the +form $4n - 1$. + +\item[2.] Show that an odd prime number can be represented as the +difference of two squares in one and in only one way. + +\item[3.] The expression $m^p - n^p$, in which $m$ and $n$ are integers +and $p$ is a prime, is either prime to $p$ or is divisible by $p^2$. + +\item[4.] Prove that any prime number except $2$ and $3$ is of one of +the forms $6n + 1$, $6n - 1$. +\end{enumerate}\normalsize + +\section{The Fundamental Theorem of Euclid}\label{s5}% +\index{Euclid, Theorem of} + +\emph{If $a$ and $b$ are any two positive integers there exist +integers $q$ and $r$, $q\stackrel{=}{>} 0, 0 \leqq r < b$, such +that} +\begin{equation*} +a = qb + r. +\end{equation*} + +If $a$ is a multiple of $b$ the theorem is at once verified, $r$ +being in this case $0$. If $a$ is not a multiple of $b$ it must lie +between two consecutive multiples of $b$; that is, there exists a +$q$ such that +\begin{equation*} +qb < a < (q + 1)b. +\end{equation*} +Hence there is an integer $r$, $0 < r < b$, such that $a = qb + r$. +In case $b$ is greater than $a$ it is evident that $q = 0$ and $r = +a$. Thus the proof of the theorem is complete. + +\section{Divisibility by a Prime Number}\label{s6}\index{Prime numbers} + +I.~\emph{If $p$ is a prime number and $m$ is any integer, then $m$ +either is divisible by $p$ or is prime to $p$.} + +This theorem follows at once from the fact that the only divisors of +$p$ are $1$ and $p$. + +\smallskip II.~\emph{The product of two integers each less than a +given prime number $p$ is not divisible by $p$.} + +Let $a$ be a number which is less than $p$ and suppose that $b$ is a +number less than $p$ such that $ab$ is divisible by $p$, and let $b$ +be the least number for which $ab$ is so divisible. Evidently there +exists an integer $m$ such that +\begin{equation*} +mb < p < (m + 1)b. +\end{equation*} +Then $p - mb < b$. Since $ab$ is divisible by $p$ it is clear that +$mab$ is divisible by $p$; so is $ap$ also; and hence their +difference $ap - mab$, $=a(p - mb)$, is divisible by $p$. That is, +the product of $a$ by an integer less than $b$ is divisible by $p$, +contrary to the assumption that $b$ is the least integer such that +$ab$ is divisible by $p$. The assumption that the theorem is not +true has thus led to a contradiction; and thus the theorem is +proved. + +\smallskip III.~\emph{If neither of two integers is divisible by a +given prime number $p$ their product is not divisible by $p$.} + +Let $a$ and $b$ be two integers neither of which is divisible by the +prime $p$. According to the fundamental theorem of Euclid there +exist integers $m$, $n$, $\alpha$, $\beta$ such that +\begin{align*} +a &= mp + \alpha,& 0 &< \alpha < p, \\ +b &= np + \beta, & 0 &< \beta < p. +\end{align*} +Then +\begin{equation*} +ab = (mp + \alpha)(np + \beta) + = (mnp + \alpha + \beta)p + \alpha\beta. +\end{equation*} +If now we suppose $ab$ to be divisible by $p$ we have $\alpha\beta$ +divisible by $p$. This contradicts II, since $\alpha$ and $\beta$ +are less than $p$. Hence $ab$ is not divisible by $p$. + +By an application of this theorem to the continued product of +several factors, the following result is readily obtained: + +\smallskip IV.~\emph{If no one of several integers is divisible by a +given prime $p$ their product is not divisible by $p$.} + +\section{The Unique Factorization Theorem}\label{s7}% +\index{Factorization theorem}\index{Factors} + +I.~\emph{Every integer greater than unity can be represented in one +and in only one way as a product of prime numbers.} + +In the first place we shall show that it is always possible to +resolve a given integer $m$ greater than unity into prime factors by +a finite number of operations. In the proof of theorem I, +\S~\ref{s3}, we showed how to find a prime factor $p_1$ of $m$ by a +finite number of operations. Let us write +\begin{equation*} +m = p_1 m_1. +\end{equation*} +If $m_1$ is not unity we may now find a prime factor $p_2$ of $m_1$. +Then we may write +\begin{equation*} +m = p_1 m_1 = p_1 p_2 m_2. +\end{equation*} +If $m_2$ is not unity we may apply to it the same process as that +applied to $m_1$ and thus obtain a third prime factor of $m$. Since +$m_1 > m_2 > m_3 > \ldots$ it is clear that after a finite number of +operations we shall arrive at a decomposition of $m$ into prime +factors. Thus we shall have +\begin{equation*} +m = p_1 p_2 \ldots p_r +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_r$ are prime numbers. We have thus +proved the first part of our theorem, which says that the +decomposition of an integer (greater than unity) into prime factors +is always possible. + +Let us now suppose that we have also a decomposition of $m$ into +prime factors as follows: +\begin{gather*} +m = q_1 q_2 \ldots q_s. \\ +\intertext{Then we have} +p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_s. +\end{gather*} +Now $p_1$ divides the first member of this equation. Hence it also +divides the second member of the equation. But $p_1$ is prime; and +therefore by theorem IV of the preceding section we see that $p_1$ +divides some one of the factors $q$; we suppose that $p_1$ is a +factor of $q_1$. It must then be equal to $q_1$. Hence we have +\begin{equation*} +p_2 p_3 \ldots p_r = q_2 q_3 \ldots q_s. +\end{equation*} +By the same argument we prove that $p_2$ is equal to some $q$, say +$q_2$. Then we have +\begin{equation*} +p_3 p_4 \ldots p_r = q_3 q_4 \ldots q_s. +\end{equation*} +Evidently the process may be continued until one side of the +equation is reduced to $1$. The other side must also be reduced to +$1$ at the same time. Hence it follows that the two decompositions +of $m$ are in fact identical. + +This completes the proof of the theorem. + +\smallskip The result which we have thus demonstrated is easily the +most important theorem in the theory of integers. It can also be +stated in a different form more convenient for some purposes: + +\smallskip II.~\emph{Every non-unit positive integer $m$ can be +represented in one and in only one way in the form +\begin{equation*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_n$ are different primes and +$\alpha_1$, $\alpha_2$, $\ldots$, $\alpha_n$ are positive integers.}% +\index{Factors} + +This comes immediately from the preceding representation of $m$ in +the form $m = p_1 p_2 \ldots p_r$ by combining into a power of $p_1$ +all the primes which are equal to $p_1$. + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ and $b$ are relatively +prime integers and $c$ is divisible by both $a$ and $b$, then $c$ is +divisible by $ab$.} + +\smallskip \textsc{Corollary 2.}~\emph{If $a$ and $b$ are each prime +to $c$ then $ab$ is prime to $c$.} + +\smallskip \textsc{Corollary 3.}~\emph{If $a$ is prime to $c$ and +$ab$ is divisible by $c$, then $b$ is divisible by $c$.} + +\section{The Divisors of an Integer}\label{s8}% +\index{Divisors of a number|(}\index{Factors} + +The following theorem is an immediate corollary of the results in +the preceding section: + +I.~\emph{All the divisors of $m$, +\begin{gather*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}, \\ +\intertext{are of the form} +p_1^{\beta_1} p_2^{\beta_2} \ldots p_n^{\beta_n},\ + 0 \leqq \beta_i \leqq \alpha_i; +\end{gather*} +and every such number is a divisor of $m$.} + +From this it is clear that every divisor of $m$ is included once and +only once among the terms of the product +\begin{multline*} +(1 + p_1 + p_1^2 + \ldots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \ldots + + p_2^{\alpha_2}) \ldots \\ +(1 + p_n + p_n^2 + \ldots + p_n^{\alpha_n}), +\end{multline*} +when this product is expanded by multiplication. It is obvious that +the number of terms in the expansion is $(\alpha_1 + 1)(\alpha_2 + +1) \ldots (\alpha_n+1)$. Hence we have the theorem: + +\smallskip II.~\emph{The number of divisors of $m$ is} +$(\alpha_1 + 1) (\alpha_2 + 1) \ldots (\alpha_n+1)$. + +Again we have +\begin{equation*} +\prod_i(1 + p_i + p_i^2 + \ldots + p_i^{\alpha_i}) = + \prod_i\frac{p_i^{\alpha_i+1} - 1}{p_i - 1}. +\end{equation*} +Hence, + +\smallskip III.~\emph{The sum of the divisors of $m$ is} +\begin{equation*} +\frac{p_1^{\alpha_1 + 1} - 1}{p_1 - 1} \cdot + \frac{p_2^{\alpha_2 + 1} - 1}{p_2 - 1} \cdot + \ldots \cdot + \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}. +\end{equation*} + +In a similar manner we may prove the following theorem: + +\smallskip IV.~\emph{The sum of the $h^{th}$ powers of the divisors +of $m$ is} +\begin{equation*} +\frac{p_1^{h(\alpha_1 + 1)} - 1}{p_1^h - 1} \cdot + \ldots \cdot + \frac{p_n^{h(\alpha_n + 1)} - 1}{p_n^h - 1}. +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find numbers $x$ such that the sum of the divisors of $x$ +is a perfect square. + +\item[2.] Show that the sum of the divisors of each of the following +integers is twice the integer itself: 6, 28, 496, 8128, 33550336. +Find other integers $x$ such that the sum of the divisors of $x$ is +a multiple of $x$. + +\item[3.] Prove that the sum of two odd squares cannot be a square. + +\item[4.] Prove that the cube of any integer is the difference of the +squares of two integers. + +\item[5.] In order that a number shall be the sum of consecutive +integers, it is necessary and sufficient that it shall not be a +power of 2. + +\item[6.] Show that there exist no integers $x$ and $y$ (zero excluded) +such that $y^2 = 2x^2$. Hence, show that there does not exist a +rational fraction whose square is 2. + +\item[7.] The number $m = p_1^{\alpha_1} p_2^{\alpha_2} \cdots +p_n^{\alpha_n}$, where the $p$'s are different primes and the +$\alpha$'s are positive integers, may be separated into relatively +prime factors in $2^{n-1}$ different ways. + +\item[8.] The product of the divisors of $m$ is $\sqrt{m^v}$ where $v$ +is the number of divisors of $m$. +\end{enumerate} \normalsize\index{Divisors of a number|)} + +\section{The Greatest Common Factor of Two or More +Integers}\label{s9}% +\index{Common!divisors|(}\index{Factors}% +\index{Greatest common factor|(} + +Let $m$ and $n$ be two positive integers such that $m$ is greater +than $n$. Then, according to the fundamental theorem of Euclid, we +can form the set of equations +\begin{align*} +m &= qn + n_1, & 0 &< n_1 < n, \\ +n &= q_1 n_1 + n_2, & 0 &< n_2 < n_1, \\ +n_1 &= q_2 n_2 + n_3, & 0 &< n_3 < n_2, \\ +&\vdots \qquad \vdots &&\vdots \qquad \vdots \\ +n_{k - 2} &= q_{k - 1} n_{k-1} + n_k, & 0 &< n_k < n_{k - 1}, \\ +n_{k - 1} &= q _k n_k. & & +\end{align*} +If $m$ is a multiple of $n$ we write $n = n_0$, $k = 0$, in the +above equations. + +\smallskip \textsc{Definition.} The process of reckoning involved in +determining the above set of equations is called the Euclidian +Algorithm.\index{Euclidian algorithm} + +\smallskip I.~\emph{The number $n_k$ to which the Euclidian +algorithm leads is the greatest common divisor of $m$ and $n$.} + +In order to prove this theorem we have to show two things: + +1)~That $n_k$ is a divisor of both $m$ and $n$; + +2)~That the greatest common divisor $d$ of $m$ and $n$ is a divisor +of $n_k$. + +To prove the first statement we examine the above set of equations, +working from the last to the first. From the last equation we see +that $n_k$ is a divisor of $n_{k-1}$. Using this result we see that +the second member of next to the last equation is divisible by $n_k$ +Hence its first member $n_{k-2}$ must be divisible by $n_k$. +Proceeding in this way step by step we show that $n_2$ and $n_1$, +and finally that $n$ and $m$, are divisible by $n_k$. + +For the second part of the proof we employ the same set of equations +and work from the first one to the last one. Let $d$ be any common +divisor of $m$ and $n$. From the first equation we see that $d$ is a +divisor of $n_1$. Then from the second equation it follows that $d$ +is a divisor of $n_2$. Proceeding in this way we show finally that +$d$ is a divisor of $n_k$. Hence any common divisor, and in +particular the greatest common divisor, of $m$ and $n$ is a factor +of $n_k$. + +This completes the proof of the theorem. + +\smallskip \textsc{Corollary.} \emph{Every common divisor of $m$ and +$n$ is a factor of their greatest common divisor.} + +\smallskip II.~\emph{Any number $n_i$ in the above set of equations +is the difference of multiples of $m$ and $n$.} + +From the first equation we have +\begin{equation*} +n_i = m - qn +\end{equation*} +so that the theorem is true for $i = 1$. We shall suppose that the +theorem is true for every subscript up to $i - 1$ and prove it true +for the subscript $i$. Thus by hypothesis we have\footnote{If $i = +2$ we must replace $n_{i-2}$ by $n$.} +\begin{align*} +n_{i-2} &= \pm(\alpha_{i-2}m - \beta_{i-2}n ), \\ +n_{i-1} &= \mp(\alpha_{i-1}m - \beta_{i-1}n). +\intertext{Substituting in the equation} +n_i &= -q_{i-1}n_{n-1} + n_{i-2} \\ +\intertext{we have a result of the form} +n_i &= \pm (\alpha_i m - \beta_i n). +\end{align*} +From this we conclude at once to the truth of the theorem. + +Since $n_k$ is the greatest common divisor of $m$ and $n$, we have +as a corollary the following important theorem: + +\smallskip III.~\emph{If $d$ is the greatest common divisor of the +positive integers $m$ and $n$, then there exist positive integers +$\alpha$ and $\beta$ such that} +\begin{equation*} +\alpha m - \beta n = \pm d. +\end{equation*} + +If we consider the particular case in which $m$ and $n$ are +relatively prime, so that $d = 1$, we see that there exist positive +integers $\alpha$ and $\beta$ such that $\alpha m - \beta n = \pm +1$. Obviously, if $m$ and $n$ have a common divisor $d$, greater +than $1$, there do not exist integers $\alpha$ and $\beta$ +satisfying this relation; for, if so, $d$ would be a divisor of the +first member of the equation and not of the second. Thus we have the +following theorem: + +\smallskip IV.~\emph{A necessary and sufficient condition that $m$ +and $n$ are relatively prime is that there exist integers $\alpha$ +and $\beta$ such that $\alpha m - \beta n = \pm 1$.} + +The theory of the greatest common divisor of three or more numbers +is based directly on that of the greatest common divisor of two +numbers; consequently it does not require to be developed in detail. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] If $d$ is the greatest common divisor of $m$ and $n$, +then $m / d$ and $n / d$ are relatively prime. + +\item[2.] If $d$ is the greatest common divisor of $m$ and $n$ and +$k$ is prime to $n$, then $d$ is the greatest common divisor of $km$ +and $n$. + +\item[3.] The number of multiple of $b$ in the sequence $a, 2a, 3a, +\cdots, ba$ is equal to the greatest common divisor of $a$ and $b$. + +\item[4.] If the sum or the difference of two irreducible fractions is +an integer, the denominators of the fractions are equal. + +\item[5.] The algebraic sum of any number of irreducible fractions, +whose denominators are prime each to each, cannot be an integer. + +\item[6*.] The number of divisions to be effected in finding the +greatest common divisor of two numbers by the Euclidian algorithm +does not exceed five times the number of digits in the smaller +number (when this number is written in the usual scale of 10). +\end{enumerate}\normalsize% +\index{Common!divisors|)}\index{Greatest common factor|)} + +\section{The Least Common Multiple of Two or More +Integers}\label{s10}% +\index{Common!multiples|(}\index{Least common multiple|(} + +I.~\emph{The common multiples of two or more numbers are the +multiples of their least common multiple.} + +This may be readily proved by means of the unique factorization +theorem. The method is obvious. We shall, however, give a proof +independent of this theorem. + +Consider first the case of two numbers; denote them by $m$ and $n$ +and their greatest common divisor by $d$. Then we have +\begin{equation*} +m = d\mu, \quad n = d\nu, +\end{equation*} +where $\mu$ and $\nu$ are relatively prime +integers.\index{Common!divisors}\index{Greatest common factor} The +common multiples sought are multiples of $m$ and are all comprised +in the numbers $am=ad\mu$, where $a$ is any integer whatever. In +order that these numbers shall be multiples of $n$ it is necessary +and sufficient that $ad\mu$ shall be a multiple of $d\nu$; that is, +that $a\mu$ shall be a multiple of $\nu$; that is, that $a$ shall be +a multiple of $\nu$, since $\mu$ and $\nu$ are relatively prime. +Writing $a = \delta\nu$ we have as the multiples in question the set +$\delta d\mu\nu$ where $\delta$ is an arbitrary integer. This proves +the theorem for the case of two numbers; for $d\mu\nu$ is evidently +the least common multiple of $m$ and $n$. + +We shall now extend the proposition to any number of integers $m, n, +p, q,\ldots$. The multiples in question must be common multiples of +$m$ and $n$ and hence of their least common multiple $\mu$. Then the +multiples must be multiples of $\mu$ and $p$ and hence of their +least common multiple $\mu_1$. But $\mu_1$ is evidently the least +common multiple of $m, n, p$. Continuing in a similar manner we may +show that every multiple in question is a multiple of $\mu$, the +least common multiple of $m, n, p, q, \ldots$. And evidently every +such number is a multiple of each of the numbers $m, n, p, q, +\ldots$. + +Thus the proof of the theorem is complete. + +When the two integers $m$ and $n$ are relatively prime their +greatest common divisor is $1$ and their least common multiple is +their product. Again if $p$ is prime to both $m$ and $n$ it is prime +to their product $mn$; and hence the least common multiple of $m, n, +p$ is in this case $mnp$. Continuing in a similar manner we have the +theorem: + +\smallskip II.~\emph{The least common multiple of several integers, +prime each to each, is equal to their product.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] In order that a common multiple of $n$ numbers shall be +the least, it is necessary and sufficient that the quotients +obtained by dividing it successively by the numbers shall be +relatively prime. + +\item[2.] The product of $n$ numbers is equal to the product of +their least common multiple by the greatest common divisor of their +products $n - 1$ at a time. + +\item[3.] The least common multiple of $n$ numbers is equal to any +common multiple $M$ divided by the greatest common divisor of the +quotients obtained on dividing this common multiple by each of the +numbers. + +\item[4.] The product of $n$ numbers is equal to the product of their +greatest common divisor by the least common multiple of the products +of the numbers taken $n - 1$ at a time. +\end{enumerate} \normalsize% +\index{Common!multiples|)}\index{Least common multiple|)} + +\section{Scales of Notation}\label{s11}\index{Scales of notation|(} + +I.~\emph{If $m$ and $n$ are positive integers and $n > 1$, then $m$ +can be represented in terms of $n$ in one and in only one way in the +form} +\begin{gather*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1} n + a_h, \\ +\intertext{where} +a_0 \ne 0,\ 0 \leqq a_i < n, \quad i = 0, 1, 2, \ldots, h. +\end{gather*} + +That such a representation of $m$ exists is readily proved by means +of the fundamental theorem of Euclid. For we have +\begin{align*} +m &= n_0 n + a_h, & 0 &\leqq a_h < n, \\ +n_0 &= n_1n + a_{h-1}, & 0 &\leqq a_{h-1} < n, \\ +n_1 &= n_2 n + a_{h-2}, & 0 &\leqq a_{h-2} < n, \\ +\hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots & + \hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots \\ +n_{h-3} &= n_{h-2} n + a_2, & 0 &\leqq a_2 < n, \\ +n_{h-2} &= n_{h-1} n + a_1, & 0 &\leqq a_1 < n, \\ +n_{h-1} &= a_0, & 0 &< a_0 < n. +\end{align*} +If the value of $n_{h-1}$ given in the last of these equations is +substituted in the second last we have +\begin{equation*} +n_{h-2} = a_0n + a_1. +\end{equation*} +This with the preceding gives +\begin{equation*} +n_{h-3} = a_0 n^2 + a_1n + a_2. +\end{equation*} +Substituting from this in the preceding and continuing the process +we have finally +\begin{equation*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1}n + a_h, +\end{equation*} +a representation of $m$ in the form specified in the theorem. + +To prove that this representation is unique, we shall suppose that +$m$ has the representation +\begin{gather*} +m = b_0 n^k + b_1 n^{k-1} + \ldots + b_{k-1}n + b_k, \\ +\intertext{where} +b_0 \ne 0,\ 0 < b_i < n,\quad i=0, 1, 2, \ldots, k, \\ +\intertext{and show that the two representations are identical. We +have} +a_0 n^h + \ldots + a_{h-1} n + a_h = + b_0 n^k + \ldots + b_{k-1} n + b_k. +\intertext{Then} +a_0 n^h + \ldots + a_{h-1} n - + (b_0 n^k + \ldots + b_{k-1} n) = b_k - a_h. +\end{gather*} +The first member is divisible by $n$. Hence the second is also. But +the second member is less than $n$ in absolute value; and hence, in +order to be divisible by $n$, it must be zero. That is, $b_k = a_h$. +Dividing the equation through by $n$ and transposing we have +\begin{equation*} +a_0 n^{h-1} + \ldots + a_{h-2} n - (b_0 n^{k-1} + \ldots + + b_{k-2} n) + = b_{k-1} - a_{h-1}. +\end{equation*} +It may now be seen that $b_{k-1} = a_{h-1}$. It is evident that this +process may be continued until either the $a$'s are all eliminated +from the equation or the $b$'s are all eliminated. But it is obvious +that when one of these sets is eliminated the other is also. Hence, +$h = k$. Also, every $a$ equals the $b$ which multiplies the same +power of $n$ as the corresponding $a$. That is, the two +representations of $m$ are identical. Hence the representation in +the theorem is unique. + +From this theorem it follows as a special case that any positive +integer can be represented in one and in only one way in the scale +of 10; that is, in the familiar Hindoo notation. It can also be +represented in one and in only one way in any other scale. Thus +\begin{equation*} +120759 = 1 \cdot 7^6 + 0 \cdot 7^5 + 1 \cdot 7^4 + 2 \cdot 7^3 + + 0 \cdot 7^2 + 3 \cdot 7^1 + 2. +\end{equation*} +Or, using a subscript to denote the scale of notation, this may be +written +\begin{equation*} +(120759)_{10} = (1012032)_7. +\end{equation*} + +For the case in which $n$ (of theorem I) is equal to 2, the only +possible values for the $a$'s are 0 and 1. Hence we have at once the +following theorem: + +II.~\emph{Any positive integer can be represented in one and in only +one way as a sum of different powers of 2.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Any positive integer can be represented as an aggregate of +different powers of $3$, the terms in the aggregate being combined +by the signs $+$ and $-$ appropriately chosen. + +\item[2.] Let $m$ and $n$ be two positive integers of which $n$ is the +smaller and suppose that $2^k \leq n < 2^{k+1}$. By means of the +representation of $m$ and $n$ in the scale of 2 prove that the +number of divisions to be effected in finding the greatest common +divisor of $m$ and $n$ by the Euclidian algorithm does not exceed +$2k$. +\end{enumerate}\normalsize\index{Scales of notation|)} + +\section{Highest Power of a Prime $p$ Contained in $n!$.}\label{s12}% +\index{Highest power of \emph{p} in \emph{n}"!|(} + +Let $n$ be any positive integer and $p$ any prime number not greater +than $n$. We inquire as to what is the highest power $p^\nu$ of the +prime $p$ contained in $n!$. + +In solving this problem we shall find it convenient to employ the +notation +\begin{equation*} +\left [ \frac{r}{s} \right ] +\end{equation*} to denote the greatest integer $\alpha$ such that +$\alpha s \leq r$. With this notation it is evident that we have +\begin{gather} +\left [ + \frac{\left [ \frac{n}{p} \right ]} + {p} +\right ] = \left [ \frac{n}{p^2} \right ]; \tag{1} \\ +\intertext{and more generally} +\left [ + \frac{\left [ \frac{n}{p^i} \right ]} + {p^j} +\right ] = \left [ \frac{n}{p^{i+j}} \right ]. \notag +\end{gather} + +If now we use $H\{x\}$ to denote the index of the highest power of +$p$ contained in an integer $x$, it is clear that we have +\begin{gather*} +H\{n!\} = + H \left \{ p \cdot 2p \cdot 3p \ldots + \left [ \frac{n}{p} \right ] p \right \}, \\ +\intertext{since only multiples of $p$ contain the factor $p$. +Hence} +H\{n!\} = + \left [ \frac{n}{p} \right ] + + H \left \{ 1 \cdot 2 \ldots \left [ \frac{n}{p} \right ] + \right \}. +\end{gather*} +Applying the same process to the $H$-function in the second member +and remembering relation (1) it is easy to see that we have +\begin{align*} +H\{n!\} &= \left[ \frac{n}{p} \right] + + H\left\{ p \cdot 2p \cdot \ldots \cdot + \left[ \frac{n}{p^2} \right]p\right\} \\ + &= \left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + + H \left\{\cdot 1 \cdot 2 \cdot 3 + \ldots \left[ \frac{n}{p^2} \right] \right\}. \\ +\intertext{Continuing the process we have finally} +H\{n1\} &= \left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] + + \ldots, +\end{align*} +the series on the right containing evidently only a finite number of +terms different from zero. Thus we have the theorem: + +\smallskip I.~\emph{The index of the highest power of a prime $p$ +contained in $n!$ is} +\begin{gather*} +\left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + + \left[ \frac{n}{p^3} \right] + \ldots. +\end{gather*} + +The theorem just obtained may be written in a different form, more +convenient for certain of its applications. Let $n$ be expressed in +the scale of $p$ in the form +\begin{gather*} +n = a_0p^h + a_1p^{h-1} + \ldots + a_{h-1}p + a_h, \\ +\intertext{where} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h. +\end{gather*} +Then evidently +\begin{align*} +\left[ \frac{n}{p} \right] &= a_0p^{h-1} + a_1p^{h-2} + \ldots + + a_{h-2}p + a_{h-1}, \\ +\left[ \frac{n}{p^2} \right] &= a_0p^{h-2} + a_1p^{h-3} + \ldots + + a_{h-2}, \\ +.\ \ .\ \ .\ \ .\ \ &.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ +.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ . +\end{align*} +Adding these equations member by member and combining the second +members in columns as written, we have +\begin{align*} +\left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] &+ + \left[ \frac{n}{p^3} \right] + \ldots \\ +&= \sum_{i=0}^h \frac{a_i(p^{h-i} - 1)}{p - 1} \\ +&= \frac{a_0p^h + a_1p^{h-1} + \ldots + a_h - + (a_0 + a_1 + \ldots + a_h)}{p-1} \\ +&= \frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{align*} +Comparing this result with theorem I we have the following theorem: + +\smallskip II.~\emph{If $n$ is represented in the scale of $p$ in +the form +\begin{gather*} +n = a_0 p^h + a_1 p^{h-1} + \ldots + a_h, \\ +\intertext{where $p$ is prime and} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h, \\ +\intertext{then the index of the highest power of $p$ contained in +$n!$ is} +\frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{gather*}} + +Note the simple form of the theorem for the case $p = 2$; in this +case the denominator $p - 1$ is unity. + +We shall make a single application of these theorems by proving the +following theorem: + +\smallskip III.~\emph{If $n$, $\alpha$, $\beta$, $\ldots$, $\lambda$ +are any positive integers such that $n = \alpha + \beta + \ldots + +\lambda$, then +\begin{equation} +\frac{n!}{\alpha! \beta! \ldots \lambda!} \tag{A} +\end{equation} +is an integer.} + +Let $p$ be any prime factor of the denominator of the fraction (A). +To prove the theorem it is sufficient to show that the index of the +highest power of $p$ contained in the numerator is at least as great +as the index of the highest power of $p$ contained in the +denominator. This index for the denominator is the sum of the +expressions +\begin{equation} + \left . + \begin{gathered} + \left [ \frac{\alpha}{p} \right ] + + \left [ \frac{\alpha}{p^2} \right ] + + \left [ \frac{\alpha}{p^3} \right ] + + \ldots \\ + \left [ \frac{\beta}{p} \right ] + + \left [ \frac{\beta}{p^2} \right ] + + \left [ \frac{\beta}{p^3} \right ] + + \ldots \\ + \vdots \\ + \left [ \frac{\lambda}{p} \right ] + + \left [ \frac{\lambda}{p^2} \right ] + + \left [ \frac{\lambda}{p^3} \right ] + + \ldots + \end{gathered} + \right \} \tag{B} +\end{equation} + +The corresponding index for the numerator is +\begin{equation} +\left [ \frac{n}{p} \right ] + +\left [ \frac{n}{p^2} \right ] + +\left [ \frac{n}{p^3} \right ] + +\ldots \tag{C} +\end{equation} +But, since $n = \alpha + \beta + \ldots + \lambda$, it is evident +that +\begin{equation*} + \left [ \frac{n}{p^r} \right ] \stackrel{=}{>} + \left [ \frac{\alpha}{p^r} \right ] + + \left [ \frac{\beta}{p^r} \right ] + + \ldots + + \left [ \frac{\lambda}{p^r} \right ]. +\end{equation*} +From this and the expressions in (B) and (C) it follows that the +index of the highest power of any prime $p$ in the numerator of (A) +is equal to or greater than the index of the highest power of p +contained in its denominator. The theorem now follows at once. + +\smallskip \textsc{Corollary.}~\emph{The product of $n$ consecutive +integers is divisible by $n!$.} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the highest power of 2 contained in 1000! is +$2^{994}$; in 1900! is $2^{1893}$. Show that the highest power of 7 +contained in 10000! is $7^{1665}$. + +\item[2.] Find the highest power of 72 contained in 1000! + +\item[3.] Show that 1000! ends with 249 zeros. + +\item[4.] Show that there is no number $n$ such that $3^7$ is the +highest power of 3 contained in $n!$. + +\item[5.] Find the smallest number $n$ such that the highest power +of 5 contained in $n!$ is $5^{31}$. What other numbers have the same +property? + +\item[6.] If $n = rs$, $r$ and $s$ being positive integers, show that +$n!$ is divisible by $(r!)^s$; by $(s!)^r$; by the least common +multiple of $(r!)^s$ and $(s!)^r$. + +\item[7.] If $n = \alpha + \beta + pq + rs$, where $\alpha, \beta, p, +q, r, s$, are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha ! \beta ! (q!)^p (s!)^r. +\end{equation*} + +\item[8.] When $m$ and $n$ are two relatively prime positive integers +the quotient +\begin{equation*} +Q = \frac{(m + n + 1)!} {m! n!} +\end{equation*} +as an integer. + +\item[9*.] If $m$ and $n$ are positive integers, then each of the +quotients +\begin{equation*} +Q = \frac{(mn)!} {n! (m!)^n},\quad +Q = \frac{(2m)! (2n)!} {m! n! (m+n)!}, +\end{equation*} +is an integer. Generalize to $k$ integers $m, n, p, \ldots$. + +\item[10*.] If $n = \alpha + \beta + pq + rs$ where $\alpha, \beta, +p, q, r, s$ are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha! \beta! r! p! (q!)^p (s!)^r. +\end{equation*} + +\item[11*.] Show that +\begin{equation*} +\frac{(rst)!} {t! (s!)^t (r!)^{st}}, +\end{equation*} is an integer ($r, s, t$ being positive integers). +Generalize to the case of $n$ integers $r, s, t, u, \ldots$. +\end{enumerate}\normalsize% +\index{Highest power of \emph{p} in \emph{n}"!|)} + +\section{Remarks Concerning Prime Numbers}\label{s13}% +\index{Prime numbers|(} + +We have seen that the number of primes is infinite. But the integers +which have actually been identified as prime are finite in number. +Moreover, the question as to whether a large number, as for instance +$2^{257}-1$, is prime is in general very difficult to answer. Among +the large primes actually identified as such are the following: +\begin{equation*} +2^{61}-1, \quad 2^{75} \cdot 5+1, \quad 2^{89}-1, \quad 2^{127}-1. +\end{equation*} + +\emph{No analytical expression for the representation of prime +numbers has yet been discovered.} Fermat believed, though he +confessed that he was unable to prove, that he had found such an +analytical expression in +\begin{equation*} +2^{2^n} + 1. +\end{equation*} +Euler showed the error of this opinion by finding that $641$ is a +factor of this number for the case when $n = 5$.% +\index{Euler}\index{Fermat} + +The subject of prime numbers is in general one of exceeding +difficulty. In fact it is an easy matter to propose problems about +prime numbers which no one has been able to solve. Some of the +simplest of these are the following: + +\begin{enumerate} +\item Is there an infinite number of pairs of primes differing by +2? +\item Is every even number (other than 2) the sum of two primes or +the sum of a prime and the unit? +\item Is every even number the difference of two primes or the +difference of 1 and a prime number? +\item To find a prime number greater than a given prime. +\item To find the prime number which follows a given prime. +\item To find the number of primes not greater than a given number. +\item To compute directly the $n^\text{th}$ prime number, when $n$ +is given. +\end{enumerate}\index{Prime numbers|)} + +\chapter{ON THE INDICATOR OF AN INTEGER}% +\index{Indicator|(} + +\section{Definition. Indicator of a Prime Power}\label{s14}% +\index{Indicator!of a prime power} + +\emph{Definition.} If $m$ is any given positive integer the number +of positive integers not greater than $m$ and prime to it is called +the indicator of $m$. It is usually denoted by $\phi(m)$, and is +sometimes called Euler's $\phi$-function of $m$.% +\index{Euler's!$\phi$-function}\index{$\phi(m)$} More rarely, it has +been given the name of totient of $m$.\index{Totient} + +As examples we have +\begin{equation*} +\phi(1) = 1,\ \phi(2) = 1,\ \phi(3) = 2,\ \phi(4) = 2. +\end{equation*} + +If $p$ is a prime number it is obvious that +\begin{equation*} +\phi(p) = p - 1; +\end{equation*} +for each of the integers 1, 2, 3, $\ldots$, $p-1$ is prime to $p$. + +Instead of taking $m = p$ let us assume that $m = p^\alpha$, where +$\alpha$ is a positive integer, and seek the value of +$\phi(p^\alpha)$. Obviously, every number of the set 1, 2, 3, +$\ldots$, $p^\alpha$ either is divisible by $p$ or is prime to +$p^\alpha$. The number of integers in the set divisible by $p$ is +$p^{\alpha - 1}$. Hence $p^\alpha-p^{\alpha-1}$ of them are prime to +$p$. Hence $\phi(p^\alpha) = p^\alpha-p^{\alpha-1}$. Therefore + +\emph{If $p$ is any prime number and $\alpha$ is any positive +integer, then} +\begin{equation*} +\phi(p^\alpha) = p^\alpha \left ( 1 - \frac{1}{p} \right ). +\end{equation*} + +\section{The Indicator of a Product}\label{s15}% +\index{Indicator!of a product|(} + +I.~\emph{If $\mu$ and $\nu$ are any two relatively prime positive +integers, then} +\begin{equation*} +\phi(\mu\nu) = \phi(\mu) \phi(\nu). +\end{equation*} + +In order to prove this theorem let us write all the integers up to +$\mu\nu$ in a rectangular array as follows: +\footnotesize\begin{equation} + \left . + \begin{aligned} + 1 && 2 && 3 && + \ldots && h && \ldots && \mu \\ + \mu + 1 && \mu + 2 && \mu + 3 && + \ldots && \mu + h && \ldots && 2\mu \\ + 2 \mu + 1 && 2 \mu + 2 && 2 \mu + 3 && + \ldots && 2 \mu + h && \ldots && 3\mu \\ + \vdots && \vdots && \vdots && + && \vdots && && \vdots \\ + (\nu - 1)\mu + 1 && (\nu - 1)\mu + 2 && (\nu - 1)\mu + 3 && + \ldots && (\nu - 1)\mu + h && \ldots && \nu\mu \\ + \end{aligned} + \right \} \tag{A} +\end{equation}\normalsize + +If a number $h$ in the first line of this array has a factor in +common with $\mu$ then every number in the same column with $h$ has +a factor in common with $\mu$. On the other hand if $h$ is prime to +$\mu$, so is every number in the column with $h$ at the top. But the +number of integers in the first row prime to $\mu$ is $\phi(\mu)$. +Hence the number of columns containing integers prime to $\mu$ is +$\phi(\mu)$ and every integer in these columns is prime to $\mu$. + +Let us now consider what numbers in one of these columns are prime +to $\nu$; for instance, the column with $h$ at the top. We wish to +determine how many integers of the set +\begin{gather*} +h,\ \mu + h,\ 2\mu + h,\ \ldots,\ (\nu - 1)\mu + h \\ +\intertext{are prime to $\nu$. Write} +s\mu + h = q_s\nu + r_s +\end{gather*} where s ranges over the numbers $s = 0,\ 1,\ 2,\ +\ldots,\ \nu - 1$ and $0\leqq r_s < \nu$. Clearly $s\mu + h$ is or +is not prime to $\nu$ according as $r_s$ is or is not prime to +$\nu$. Our problem is then reduced to that of determining how many +of the quantities $r_s$ are prime to $\nu$. + +First let us notice that all the numbers $r_s$ are different; for, +if $r_s = r_t$ then from +\begin{equation*} +s\mu + h = q_s\nu + r_s,\quad t\mu + h = q_t\nu + r_t, +\end{equation*} +we have by subtraction that $(s-t)\mu$ is divisible by $\nu$. But +$\mu$ is prime to $\nu$ and $s$ and $t$ are each less than $\nu$. +Hence $(s-t)\mu$ can be a multiple of $\nu$ only by being zero; that +is, $s$ must equal $t$. Hence no two of the remainders $r_s$ can be +equal. + +Now the remainders $r_s$ are $\nu$ in number, are all zero or +positive, each is less than $\nu$, and they are all distinct. Hence +they are in some order the numbers 0, 1, 2, $\ldots$, $\nu-1$. The +number of integers in this set prime to $\nu$ is evidently +$\phi(\nu)$. + +Hence it follows that in any column of the array (A) in which the +numbers are prime to $\mu$ there are just $\phi(\nu)$ numbers which +are prime to $\nu$. That is, in this column there are just +$\phi(\nu)$ numbers which are prime to $\mu\nu$. But there are +$\phi(\mu)$ such columns. Hence the number of integers in the array +(A) prime to $\mu\nu$ is $\phi(\mu)\phi(\nu)$. + +But from the definition of the $\phi$-function it follows that the +number of integers in the array (A) prime to $\mu\nu$ is +$\phi(\mu\nu).$ Hence, +\begin{equation*} +\phi(\mu\nu) = \phi(\mu)\phi(\nu), +\end{equation*} which is the theorem to be proved. + +\smallskip \textsc{Corollary.}~\emph{In the series of $n$ +consecutive terms of an arithmetical progression the common +difference of which is prime to $n$, the number of terms prime to +$n$ is $\phi(n)$.} + +From theorem I we have readily the following more general result: + +\smallskip II.~\emph{If $m_1, m_2, \ldots, m_k$ are $k$ positive +integers which are prime each to each, then} +\begin{equation*} +\phi(m_1 m_2 \ldots m_k) = \phi(m_1) \phi(m_2) \ldots \phi(m_k). +\end{equation*}\index{Indicator!of a product|)} + +\section{The Indicator of any Positive Integer}\label{s16}% +\index{Indicator!of any integer|(} + +From the results of \S\S \ref{s14} and \ref{s15} we have an +immediate proof of the following fundamental theorem: + +\emph{If $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}$ +where $p_1, p_2, \ldots, p_n$ are different primes and $\alpha_1, +\alpha_2, \ldots, \alpha_n$ are positive integers, then} +\begin{equation*} +\phi(m) = m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{equation*} + +For, +\begin{align*} +\phi(m) &= \phi(p_1^{\alpha_1}) \phi(p_2^{\alpha_2}) \ldots + \phi(p_n^{\alpha_n}) \\ + &= p_1^{\alpha_1} \left ( 1-\frac{1}{p_1} \right ) + p_2^{\alpha_2} \left ( 1-\frac{1}{p_2} \right ) + \ldots + p_n^{\alpha_n} \left ( 1-\frac{1}{p_n} \right ) \\ + &= m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{align*} + +On account of the great importance of this theorem we shall give a +second demonstration of it. + +It is clear that the number of integers less than $m$ and divisible +by $p_1$ is +\begin{gather*} +\frac{m}{p_1}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_2$ is} +\frac{m}{p_2}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_1 p_2$ is} +\frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and divisible +by either $p_1$ or $p_2$ is} +\frac{m}{p_1} + \frac{m}{p_2} - \frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and prime to +$p_1 p_2$ is} +m - \frac{m}{p_1} - \frac{m}{p_2} + \frac{m}{p_1 p_2} = + m \left ( 1-\frac{1}{p_1} \right ) \left ( 1-\frac{1}{p_2} \right ). +\end{gather*} + +We shall now show that if the number of integers less than $m$ and +prime to $p_1 p_2 \ldots p_i$, where $i$ is less than $n$, is +\begin{gather*} +m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_i} \right ), \\ +\intertext{then the number of integers less than $m$ and prime to +$p_1 p_2 \ldots p_i p_{i+1}$ is} + m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_{i+1}} \right ). +\end{gather*} +From this our theorem will follow at once by induction. + +From our hypothesis it follows that the number of integers less than +$m$ and divisible by at least one of the primes $p_1$, $p_2$, +$\ldots$, $p_i$ is +\begin{gather} +m - + m \left (1 - \frac{1}{p_1}\right ) + \ldots + \left (1 - \frac{1}{p_i}\right ), \notag \\ +\intertext{or} +\sum \frac{m}{p_1} - \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - \ldots, \tag{A} +\end{gather} +where the summation in each case runs over all numbers of the type +indicated, the subscripts of the $p$'s being equal to or less than +$i$. + +Let us consider the integers less than $m$ and having the factor +$p_{i+1}$ but not having any of the factors $p_1$, $p_2$, $\ldots$, +$p_i$. Their number is +\begin{gather} +\frac{m}{p_{i+1}} - + \frac{1}{p_{i+1}} \left \{ + \sum \frac{m}{p_1} - + \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - + \ldots + \right \}, \tag{B} +\end{gather} +where the summation signs have the same significance as before. For +the number in question is evidently $\frac{m}{p_{i+1}}$ \emph{minus} +the number of integers not greater than $\frac{m}{p_{i+1}}$ and +divisible by at least one of the primes $p_1$, $p_2$, $\ldots$, +$p_i$. + +If we add (A) and (B) we have the number of integers less than $m$ +and divisible by one at least of the numbers $p_1$, $p_2$, $\ldots$, +$p_{i+1}$. Hence the number of integers less than $m$ and prime to +$p_1$, $p_2$, $\ldots$, $p_{i+1}$ is +\begin{gather*} +m - + \sum \frac{m}{p_1} + + \sum \frac{m}{p_1 p_2} - + \sum \frac{m}{p_1 p_2 p_3} + + \ldots, \\ +\intertext{where now in the summations the subscripts run from 1 to +$i+1$. This number is clearly equal to} +m + \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_{i+1}} \right ). +\end{gather*} +From this result, as we have seen above, our theorem follows at once +by induction.\index{Indicator!of any integer|)} + +\section{Sum of the Indicators of the Divisors of a Number}% +\label{s17} + +We shall first prove the following lemma: + +\smallskip \emph{Lemma. If $d$ is any divisor of $m$ and $m = nd$, +the number of integers not greater than $m$ which have with $m$ the +greatest common divisor $d$ is $\phi(n)$.} + +Every integer not greater than $m$ and having the divisor $d$ is +contained in the set $d$, $2d$, $3d$, $\ldots$, $nd$. The number of +these integers which have with $m$ the greatest common divisor $d$ +is evidently the same as the number of integers of the set 1, 2, +$\ldots$, $n$ which are prime to $\frac{m}{d}$, or $n$; for $\alpha +d$ and $n$ have or have not the greatest common divisor $d$ +according as $\alpha$ is or is not prime to $\frac{m}{d}=n$. Hence +the number in question is $\phi(n)$. + +From this lemma follows readily the proof of the following theorem: + +\smallskip \emph{If $d_1$, $d_2$, $\ldots$, $d_r$ are the different +divisors of $m$, then} +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = m. +\end{equation*} + +Let us define integers $m_1$, $m_2$, $\ldots$, $m_r$ by the +relations +\begin{equation*} +m = d_1 m_1 = d_2 m_2 = \ldots = d_r m_r. +\end{equation*} +Now consider the set of $m$ positive integers not greater than $m$, +and classify them as follows into $r$ classes. Place in the first +class those integers of the set which have with $m$ the greatest +common divisor $m_1$; their number is $\phi(d_1)$, as may be seen +from the lemma. Place in the second class those integers of the set +which have with $m$ the greatest common divisor $m_2$; their number +is $\phi(d_2)$. Proceeding in this way throughout, we place finally +in the last class those integers of the set which have with $m$ the +greatest common divisor $m_r$; their number is $\phi(d_r)$. It is +evident that every integer in the set falls into one and into just +one of these $r$ classes. Hence the total number $m$ of integers in +the set is $\phi(d_1) + \phi(d_r) + \ldots + \phi(d_r)$. From this +the theorem follows immediately. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the indicator of any integer greater than $2$ +is even. + +\item[2.] Prove that the number of irreducible fractions not greater +than $1$ and with denominator equal to $n$ is $\phi(n)$. + +\item[3.] Prove that the number of irreducible fractions not greater +than $1$ and with denominators not greater than $n$ is +\begin{equation*} +\phi(1) + \phi(2) + \phi(3) + \ldots + \phi(n). +\end{equation*} + +\item[4.] Show that the sum of the integers less than $n$ and prime to +$n$ is $\frac{1}{2} n \phi(n)$ if $n > 1$. + +\item[5.] Find ten values of $x$ such that $\phi(x) = 24$. + +\item[6.] Find seventeen values of $x$ such that $\phi(x) = 72$. + +\item[7.] Find three values of $n$ for which there is no $x$ satisfying +the equation $\phi(x) = 2n$. + +\item[8.] Show that if the equation +\begin{equation*} +\phi(x) = n +\end{equation*} +has one solution it always has a second solution, $n$ being given +and $x$ being the unknown. + +\item[9.] Prove that all the solutions of the equation +\begin{equation*} +\phi(x) = 4n - 2, n > 1, +\end{equation*} +are of the form $p^\alpha$ and $2p^\alpha$, where $p$ is a prime of +the form $4s-1$. + +\item[10.] How many integers prime to $n$ are there in the set +\begin{enumerate} +\item $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots, n(n+1)$? +\item $1 \cdot 2 \cdot 3, 2 \cdot 3 \cdot 4, + 3 \cdot 4 \cdot 5, \ldots, n(n+1)(n+2)$? +\item $\frac{1 \cdot 2}{2}, \frac{2 \cdot 3}{2}, + \frac{3 \cdot 4}{2}, \ldots, \frac{n(n+1)}{2}$? +\item $\frac{1 \cdot 2 \cdot 3}{6}, + \frac{2 \cdot 3 \cdot 4}{6}, + \frac{3 \cdot 4 \cdot 5}{6}, + \ldots, + \frac{n(n+1)(n+2)}{6}$? +\end{enumerate} + +\item[11*.] Find a method for determining all the solutions of the +equation +\begin{equation*} +\phi(x) = n, +\end{equation*} +where $n$ is given and $x$ is to be sought. + +\item[12*.] A number theory function $\phi(n)$ is defined for every +positive integer $n$; and for every such number $n$ it satisfies the +relation +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = n, +\end{equation*} +where $d_1, d_2, \ldots, d_r$ are the divisors of $n$. From this +property alone show that +\begin{equation*} +\phi(n) = n \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_k} \right ), +\end{equation*} +where $p_1, p_2, \ldots, p_k$ are the different prime factors of +$n$. \end{enumerate} \normalsize\index{Indicator|)} + +\chapter{ELEMENTARY PROPERTIES OF CONGRUENCES}% +\index{Congruences|(} + +\section{Congruences Modulo $m$}\label{s18} + +\textsc{Definitions.} If $a$ and $b$ are any two integers, positive +or zero or negative, whose difference is divisible by $m$, $a$ and +$b$ are said to be congruent modulo $m$, or congruent for the +modulus $m$, or congruent according to the modulus $m$. Each of +the numbers $a$ and $b$ is said to be a residue of the other.% +\index{Residue} + +\smallskip To express the relation thus defined we may write +\begin{equation*} +a = b + cm, +\end{equation*} +where $c$ is an integer (positive or zero or negative). It is more +convenient, however, to use a special notation due to Gauss, and to +write +\begin{equation*} +a \equiv b \mod m, +\end{equation*} +an expression which is read $a$ is congruent to $b$ modulo $m$, or +$a$ is congruent to $b$ for the modulus $m$, or $a$ is congruent to +$b$ according to the modulus $m$.\index{Gauss} This notation has the +advantage that it involves only the quantities which are essential +to the idea involved, whereas in the preceding expression we had the +irrelevant integer $c$. The Gaussian notation is of great value and +convenience in the study of the theory of divisibility. In the +present chapter we develop some of the fundamental elementary +properties of congruences. It will be seen that many theorems +concerning equations are likewise true of congruences with fixed +modulus; and it is this analogy with equations which gives +congruences (as such) one of their chief claims to attention. + +As immediate consequences of our definitions we have the following +fundamental theorems: + +\smallskip I.~\emph{If} $a\equiv c \mod m$, $b\equiv c\mod m$, +\emph{then} $a\equiv b\mod m$; \noindent \emph{that is, for a given +modulus, numbers congruent to the same number are congruent to each +other.} + +For, by hypothesis, $a - c = c_1 m$, $b - c = c_2 m$, where $c_1$ +and $c_2$ are integers. Then by subtraction we have $a - b = (c_1 - +c_2) m$; whence $a \equiv b \mod m$. + +\smallskip II.~\emph{If} $a \equiv b \mod m$, $\alpha \equiv +\beta \mod m$, \emph{then} $a \pm \alpha \equiv b \pm \beta \mod m$; +\emph{that is, congruences with the same modulus may be added or +subtracted member by member.} + +For, by hypothesis, $a - b = c_1 m$, $\alpha - \beta = c_2 m$; +whence $(a \pm \alpha) - (b \pm \beta) = (c_1 \pm c_2)m$. Hence $a +\pm \alpha = b \pm \beta \mod m$. + +\smallskip III.~\emph{If} $a = b \mod m$, \emph{then} +$ca = cb \mod m$, \emph{$c$ being any integer whatever.} + +The proof is obvious and need not be stated. + +\smallskip IV.~\emph{If} $a \equiv b \mod m$, +$\alpha \equiv \beta \mod m$, \emph{then} $a \alpha \equiv b \beta +\mod m$; \emph{that is, two congruences with the same modulus may be +multiplied member by member.} + +For, we have $a = b + c_1 m$, $\alpha = \beta + c_2 m$. Multiplying +these equations member by member we have $a \alpha = b \beta + m (b +c_2 + \beta c_1 + c_1 c_2 m)$. Hence $a \alpha \equiv b \beta \mod +m$. + +\smallskip A repeated use of this theorem gives the following +result: + +\smallskip V.~\emph{If} $a \equiv b \mod m$, \emph{then} +$a^n \equiv b^n \mod m$ \emph{where $n$ is any positive integer.} + +\smallskip As a corollary of theorems II, III and V we have the +following more general result: + +\smallskip VI.~\emph{If $f(x)$ denotes any polynomial in $x$ with +coefficients which are integers (positive or zero or negative) and +if further $a\equiv b \bmod m$, then} +\begin{equation*} +f(a) \equiv f(b) \bmod m. +\end{equation*} + +\section{Solutions of Congruences by Trial}\label{s19}% +\index{Congruences!Solution by trial|(} + +Let $f(x)$ be any polynomial in $x$ with coefficients which are +integers (positive or negative or zero). Then if $x$ and $c$ are any +two integers it follows from the last theorem of the preceding +section that +\begin{gather*} +f(x) \equiv f(x + cm) \bmod m. \tag{1} \\ +\intertext{Hence if $a$ is any value of $x$ for which the +congruence} +f(x)\equiv 0\bmod m. \tag{2} +\end{gather*} +is satisfied, then the congruence is also satisfied for $x = \alpha ++ cm$, where $c$ is any integer whatever. The numbers $\alpha + cm$ +are said to form a \emph{solution} (or to be a \emph{root}) of the +congruence, $c$ being a variable integer. Any one of the integers +$\alpha + cm$ may be taken as the representative of the solution. We +shall often speak of one of these numbers as the solution itself. + +Among the integers in a solution of the congruence (2) there is +evidently one which is positive and not greater than $m$. Hence all +solutions of a congruence of the type (2) may be found by trial, a +substitution of each of the numbers $1, 2, \ldots, m$ being made for +$x$. It is clear also that $m$ is the maximum number of solutions +which (2) can have whatever be the function $f(x)$. By means of an +example it is easy to show that this maximum number of solutions is +not always possessed by a congruence; in fact, it is not even +necessary that the congruence have a solution at all. + +This is illustrated by the example +\begin{equation*} +x^2 - 3 \equiv 0 \bmod 5. +\end{equation*} +In order to show that no solution is possible it is necessary to +make trial only of the values $1, 2, 3, 4, 5$ for $x$. A direct +substitution verifies the conclusion that none of them satisfies the +congruence; and hence that the congruence has no solution at all. + +On the other hand the congruence +\begin{equation*} +x^5 - x \equiv 0 \bmod 5 +\end{equation*} +has the solutions $x = 1, 2, 3, 4, 5$ as one readily verifies; that +is, this congruence has five solutions---the maximum number possible +in accordance with the results obtained above. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $(a + b)^p \equiv a^p + b^p \bmod p$ +where $a$ and $b$ are any integers and $p$ is any prime. + +\item[2.] From the preceding result prove that +$\alpha^p \equiv \alpha \bmod p$ for every integer $\alpha$. + +\item[3.] Find all the solutions of each of the congruences $x^{11} +\equiv x \bmod 11, x^{10} \equiv 1 \bmod 11, x^{5} \equiv 1 \bmod +11$. +\end{enumerate} \normalsize\index{Congruences!Solution by trial|)} + +\section{Properties of Congruences Relative to Division}\label{s20} + +The properties of congruences relative to addition, subtraction and +multiplication are entirely analogous to the properties of algebraic +equations. But the properties relative to division are essentially +different. These we shall now give. + +\smallskip I.~\emph{If two numbers are congruent modulo $m$ they are +congruent modulo $d$, where $d$ is any divisor of $m$.} + +For, from $a \equiv b \bmod m$, we have $a = b + cm = b + c'd$. +Hence $a\equiv b \bmod d$. + +\smallskip II.~\emph{If two numbers are congruent for different +moduli they are congruent for a modulus which is the least common +multiple of the given moduli.} + +The proof is obvious, since the difference of the given numbers is +divisible by each of the moduli. + +\smallskip III.~\emph{When the two members of a congruence are +multiples of an integer $c$ prime to the modulus, each member of the +congruence may be divided by $c$.} + +For, if $ca \equiv cb \bmod m$ then $ca - cb$ is divisible by $m$. +Since $c$ is prime to $m$ it follows that $a - b$ is divisible by +$m$. Hence $a\equiv b \bmod m$. + +\smallskip IV.~\emph{If the two members of a congruence are +divisible by an integer $c$, having with the modulus the greatest +common divisor $\delta$, one obtains a congruence equivalent to the +given congruence by dividing the two members by $c$ and the modulus +by $\delta$.} + +By hypothesis $ac \equiv bc \bmod m,\quad c = \delta c_1,\quad m = +\delta m_1$. Hence $c(a-b)$ is divisible by $m$. A necessary and +sufficient condition for this is evidently that $c_1(a-b)$ is +divisible by $m_1$. This leads at once to the desired result. + +\section{Congruences with a Prime Modulus}\label{s21}% +\index{Congruences!with prime modulus|(} + +\emph{The congruence\footnote{The sign $\not\equiv$ is read \emph{is +not congruent to}.}} +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod p, + \quad a_0 \not\equiv 0 \bmod p +\end{equation*} +\emph{where $p$ is a prime number and the $a$'s are any integers, +has not more than $n$ solutions.} + +Denote the first member of this congruence by $f(x)$ so that the +congruence may be written +\begin{gather} +f(x) \equiv 0 \bmod p \tag{1} \\ +\intertext{Suppose that $a$ is a root of the congruence, so that} +f(a) \equiv 0 \bmod p. \notag \\ +\intertext{Then we have} f(x) +\equiv f(x) - f(a) \bmod p. \notag \\ +\intertext{But, from algebra, $f(x) - f(a)$ is divisible by $x - a$. +Let $(x-a)^{\alpha}$ be the highest power of $x - a$ contained in +$f(x) - f(a)$. Then we may write} +f(x) - f(a) = (x - a)^{\alpha} f_1(x), \tag{2} \\ +\intertext{where $f_1(x)$ is evidently a polynomial with integral +coefficients. Hence we have} +f(x) \equiv (x - a)^{\alpha} f_1(x) \bmod p. \tag{3} +\end{gather} +We shall say that $a$ occurs $\alpha$ times as a solution of (1); or +that the congruence has $\alpha$ solutions each equal to $a$. + +Now suppose that congruence (1) has a root $b$ such that +$b\not\equiv a \bmod p$. Then from (3) we have +\begin{gather*} +f(b) \equiv (b-a)^{\alpha}f_1(b) \bmod p. \\ +\intertext{But} +f(b)\equiv 0 \bmod p,\quad (b-a)^{\alpha} \not\equiv 0 \bmod p. \\ +\intertext{Hence, since $p$ is a prime number, we must have} +f_1(b)\equiv 0 \bmod p. +\end{gather*} + +By an argument similar to that just used above, we may show that +$f_1(x) - f_1(b)$ may be written in the form +\begin{gather*} +f_1(x) - f_1(b) = (x-b)^{\beta}f_2(x), \\ +\intertext{where $\beta$ is some positive integer. Then we have} +f(x) \equiv (x-a)^{\alpha}(x-b)^{\beta}f_2(x) \bmod p. +\end{gather*} + +Now this process can be continued until either all the solutions of +(1) are exhausted or the second member is a product of linear +factors multiplied by the integer $a_0$. In the former case there +will be fewer than $n$ solutions of (1), so that our theorem is true +for this case. In the other case we have +\begin{equation*} +f(x) \equiv a_0(x-a)^{\alpha}(x-b)^{\beta} + \ldots (x-l)^{\lambda} \bmod p. +\end{equation*} +We have now $n$ solutions of (1): $a$ counted $\alpha$ times, $b$ +counted $\beta$ times, \ldots, $l$ counted $\lambda$ times; $\alpha ++ \beta + \ldots +\lambda = n$. + +Now let $\eta$ be any solution of (1). Then +\begin{equation*} +f(\eta) \equiv a_0(\eta-a)^{\alpha}(\eta-b)^{\beta} \ldots + (\eta-l)^{\lambda} \equiv 0 \bmod p. +\end{equation*} +Since $p$ is prime it follows now that some one of the factors +$\eta-a, \eta-b, \ldots, \eta-l$ is divisible by $p$. Hence $\eta$ +coincides with one of the solutions $a, b, c, \ldots, l$. That is, +(1) can have only the $n$ solutions already found. + +This completes the proof of the theorem. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Construct a congruence of the form +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod m, \quad + a_0 \not\equiv 0 \bmod m, +\end{equation*} +having more than $n$ solutions and thus show that the limitation to +a prime modulus in the theorem of this section is essential. + +\item[2.] Prove that +\begin{equation*} +x^6-1 \equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) \bmod 7 +\end{equation*} +for every integer $x$. + +\item[3.] How many solutions has the congruence $x^5 \equiv 1 \bmod +11$? the congruence $x^5\equiv 2 \bmod 11$? +\end{enumerate}\normalsize\index{Congruences!with prime modulus|)} + +\section{Linear Congruences}\label{s22}% +\index{Congruences!Linear|(} + +From the theorem of the preceding section it follows that the +congruence +\begin{equation*} +ax \equiv c \bmod p,\quad a \not\equiv 0 \bmod p, +\end{equation*} +where $p$ is a prime number, has not more than one solution. In this +section we shall prove that it always has a solution. More +generally, we shall consider the congruence +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +where $m$ is any integer. The discussion will be broken up into +parts for convenience in the proofs. + +\smallskip I.~\emph{The congruence} +\begin{equation} +ax \equiv 1 \bmod m, \tag{1} +\end{equation} +\emph{in which a and m are relatively prime, has one and only one +solution.} + +The question as to the existence and number of the solutions of (1) +is equivalent to the question as to the existence and number of +integer pairs $x, y$ satisfying the equation, +\begin{equation} +ax - my = 1, \tag{2} +\end{equation} +the integers $x$ being incongruent modulo $m$. Since $a$ and $m$ are +relatively prime it follows from theorem IV of \S~\ref{s9} that +there exists a solution of equation (2). Let $x = \alpha$ and $y = +\beta$ be a particular solution of (2) and let $x = \bar{\alpha}$ +and $y = \bar{\beta}$ be any solution of (2). Then we have +\begin{gather*} +a\alpha-m\beta = 1, \\ +a \bar{\alpha} - m\bar{\beta} = 1; \\ +\intertext{whence} +a(\alpha - \bar{\alpha}) - m(\beta - \bar{\beta}) = 0. +\end{gather*} +Hence $\alpha-\bar{\alpha}$ is divisible by $m$, since $a$ and $m$ +are relatively prime. That is, $a \equiv \bar{\alpha} \mod m$. Hence +$\alpha$ and $\bar{\alpha}$ are representatives of the same solution +of (1). Hence (1) has one and only one solution, as was to be +proved. + +\smallskip II.~\emph{The solution $x = \alpha$ of the congruence +$ax \equiv 1 \mod m$, in which $a$ and $m$ are relatively prime, is +prime to $m$.} + +For, if $a\alpha - 1$ is divisible by $m$, $\alpha$ is divisible by +no factor of $m$ except $1$. + +\smallskip III.~\emph{The congruence} +\begin{equation} +ax \equiv c \mod m \tag{3} +\end{equation} +\emph{in which $a$ and $m$ and also $c$ and $m$ are relatively +prime, has one and only one solution.} + +Let $x = \gamma$ be the unique solution of the congruence $cx = 1 +\mod m$. Then we have $a\gamma x \equiv c\gamma \equiv 1 \mod m$. +Now, by I we see that there is one and only one solution of the +congruence $a\gamma x \equiv 1 \mod m$; and from this the theorem +follows at once. + +Suppose now that $a$ is prime to $m$ but that $c$ and $m$ have the +greatest common divisor $\delta$ which is different from 1. Then it +is easy to see that any solution $x$ of the congruence $ax \equiv c +\mod m$ must be divisible by $\delta$. The question of the existence +of solutions of the congruence $ax \equiv c \bmod m$ is then +equivalent to the question of the existence of solutions of the +congruence +\begin{equation*} +a \frac{x}{\delta} \equiv \frac{c}{\delta} \bmod \frac{m}{\delta}, +\end{equation*} +where $\frac{x}{\delta}$ is the unknown integer. From III it follows +that this congruence has a unique solution $\frac{x}{\delta} = +\alpha$. Hence the congruence $ax \equiv c \bmod m$ has the unique +solution $x = \delta\alpha$. Thus we have the following theorem: + +\smallskip IV.~\emph{The congruence $ax \equiv c \bmod m$, in which +$a$ and $m$ are relatively prime, has one and only one solution.} + + +\smallskip\textsc{Corollary.}~\emph{The congruence +$ax \equiv c \bmod p$, $a \not\equiv 0 \bmod p$, where $p$ is a +prime number, has one and only one solution.} + +It remains to examine the case of the congruence $ax =c \bmod m$ in +which $a$ and $m$ have the greatest common divisor $d$. It is +evident that there is no solution unless $c$ also contains this +divisor $d$. Then let us suppose that $a = \alpha d$, $c = \gamma +d$, $m = \mu d$. Then for every $x$ such that $ax = c \bmod m$ we +have $\alpha x = \gamma \bmod \mu$; and conversely every $x$ +satisfying the latter congruence also satisfies the former. Now +$\alpha x = \gamma \bmod \mu$, has only one solution. Let $\beta$ be +a non-negative number less than $\mu$, which satisfies the +congruence $\alpha x = \gamma \bmod \mu$. All integers which satisfy +this congruence are then of the form $\beta + \mu\nu$, where $\nu$ +is an integer. Hence all integers satisfying the congruence $ax = c +\bmod m$ are of the form $\beta + \mu\nu$; and every such integer is +a representative of a solution of this congruence. It is clear that +the numbers +\begin{equation} +\beta,\ \beta + \mu,\ \beta + 2\mu,\ \ldots,\ \beta + (d-1)\mu +\tag{A} +\end{equation} +are incongruent modulo $m$ while every integer of the form $\beta + +\mu\nu$ is congruent modulo $m$ to a number of the set (A). Hence +the congruence $ax = c \bmod m$ has the $d$ solutions (A). + +This leads us to an important theorem which includes all the other +theorems of this section as special cases. It may be stated as +follows: + +\smallskip V.~\emph{Let} +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +\emph{be any linear congruence and let $a$ and $m$ have the greatest +common divisor $d (d \geq 1)$. Then a necessary and sufficient +condition for the existence of solutions of the congruence is that +$c$ be divisible by $d$. If this condition is satisfied the +congruence has just $d$ solutions, and all the solutions are +congruent modulo $m / d$.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find the remainder when $2^{40}$ is divided by $31$; when +$2^{43}$ is divided by $31$. + +\item[2.] Show that $2^{2^5}+1$ has the factor $641$. + +\item[3.] Prove that a number is a multiple of $9$ if and only if the +sum of its digits is a multiple of $9$. + +\item[4.] Prove that a number is a multiple of $11$ if and only if the +sum of the digits in the odd numbered places diminished by the sum +of the digits in the even numbered places is a multiple of $11$. +\end{enumerate} \normalsize% +\index{Congruences|)}\index{Congruences!Linear|)} + +\chapter{THE THEOREMS OF FERMAT AND WILSON} + +\section{Fermat's General Theorem}\label{s23}% +\index{Fermat's!General Theorem} + +Let $m$ be any positive integer and let +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)} \tag{A} +\end{equation} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$. Let $a$ be any integer prime to $m$ and form the set +of integers +\begin{equation} +aa_1,\ aa_2,\ \ldots,\ aa_{\phi(m)} \tag{B} +\end{equation} +No number $aa_i$ of the set (B) is congruent to a number $aa_j$, +unless $j = i;$ for, from +\begin{equation*} +aa_i \equiv aa_j \bmod m +\end{equation*} +we have $a_i \equiv a_j \bmod m$; whence $a_i = a_j$ since both +$a_i$ and $a_j$ are positive and not greater than $m$. Therefore $j += i$. Furthermore, every number of the set (B) is congruent to some +number of the set (A). Hence we have congruences of the form +\begin{align*} +aa_1 & \equiv a_{i_1} \bmod m, \\ +aa_2 & \equiv a_{i_2} \bmod m, \\ + & \vdots \\ +aa_{\phi(m)} & \equiv a_{i_{\phi(m)}} \bmod m. +\end{align*} +No two numbers in the second members are equal, since $aa_i +\not\equiv aa_j$ unless $i= j$. Hence the numbers $a_{i_1},\ +a_{i_2},\ \ldots,\ a_{i_{\phi(m)}}$ are the numbers $a_1,\ a_2,\ +\ldots,\ a_{\phi(m)}$ in some order. Therefore, if we multiply the +above system of congruences together member by member and divide +each member of the resulting congruence by $a_1\cdot a_2\ldots +a_{\phi(m)}$ (which is prime to $m$), we have +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} +This result is known as Fermat's general theorem.% +\index{Fermat's!general theorem} It may be stated as follows: + +\emph{If $m$ is any positive integer and $a$ is any integer prime to +$m$, then} +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ is any integer +not divisible by a prime number $p$, then} +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +\smallskip \textsc{Corollary 2.}~\emph{If $p$ is any prime number +and $a$ is any integer, then} +\begin{equation*} +a^p \equiv a \bmod p. +\end{equation*} + +\section{Euler's Proof of the Simple Fermat Theorem}\label{s24}% +\index{Euler}\index{Fermat}\index{Fermat's!Simple Theorem} + +The theorem of Cor.\ 1, \S~\ref{s23}, is often spoken of as the +simple Fermat theorem. It was first announced by Fermat in 1679, but +without proof. The first proof of it was given by Euler in 1736. +This proof may be stated as follows: + +From the Binomial Theorem it follows readily that +\begin{gather*} +(a+1)^p \equiv a^p + 1 \bmod p \\ +\intertext{since} +\frac{p!}{r!(p-r)!}, \quad 0 < r < p, \\ +\intertext{is obviously divisible by $p$. Subtracting $a + 1$ from +each side of the foregoing congruence, we have} +(a+1)^p - (a+1) \equiv a^p - a \bmod p. +\end{gather*} +Hence if $a^p - a$ is divisible by $p$, so is $(a + 1)^p - (a + 1)$. +But $1^p - 1$ is divisible by $p$. Hence $2^p - 2$ is divisible by +$p$; and then $3^p - 3$; and so on. Therefore, in general, we have +\begin{equation*} +a^p \equiv a \mod p. +\end{equation*} +If $a$ is prime to $p$ this gives $a^{p-1} \equiv 1 \mod p$, as was +to be proved. + +If instead of the Binomial Theorem one employs the Polynomial +Theorem, an even simpler proof is obtained. For, from the latter +theorem, we have readily +\begin{gather*} +(\alpha_1 + \alpha_2 + \ldots + \alpha_a)^p \equiv + \alpha_1^p + \alpha_2^p + \ldots + \alpha_a^p \mod p. \\ +\intertext{Putting $\alpha_1 = \alpha_2 = \ldots = \alpha_a = 1$ we +have} +a^p\equiv a \mod p, +\end{gather*} +from which the theorem follows as before. + +\section{Wilson's Theorem}\label{s25}\index{Wilson's theorem|(} + +From the simple Fermat theorem it follows that the congruence +\begin{gather*} +x^{p-1} \equiv 1\mod p \\ +\intertext{has the $p-1$ solutions $1$, $2$, $3$, $\ldots$, $p-1$. +Hence from the discussion in \S \ref{s21} it follows that} +x^{p-1}-1 \equiv (x-1)(x-2)\ldots(x-\overline{p-1}) \mod p, \\ +\intertext{this relation being satisfied for every value of $x$. +Putting $x = 0$ we have} +(-1) = (-1)^{p - 1}\cdot 1\cdot 2\cdot 3 \ldots + \overline{p-1}\mod p. \\ +\intertext{If $p$ is an odd prime this leads to the congruence} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} + 1 = 0 \mod p. +\end{gather*} +Now for $p = 2$ this congruence is evidently satisfied. Hence +we have the Wilson theorem: + +\smallskip \emph{Every prime number $p$ satisfies the relation} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} + 1 \equiv 0 \mod p. +\end{equation*} + +An interesting proof of this theorem on wholly different principles +may be given. Let $p$ points be distributed at equal intervals on +the circumference of a circle. The whole number of $p$-gons which +can be formed by joining up these $p$ points in every possible order +is evidently +\begin{equation*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1; +\end{equation*} +for the first vertex can be chosen in $p$ ways, the second in $p - +1$ ways, $\ldots$, the $(p-1)^{\mathrm{th}}$ in two ways, and the +last in one way; and in counting up thus we have evidently counted +each polygon $2p$ times, once for each vertex and for each direction +from the vertex around the polygon. Of the total number of polygons +$\frac{1}{2}(p-1)$ are regular (convex or stellated) so that a +revolution through $\frac{360^\circ}{p}$ brings each of these into +coincidence with its former position. The number of remaining +$p$-gons must be divisible by $p$; for with each such $p$-gon we may +associate the $p-1$ $p$-gons which can be obtained from it by +rotating it through successive angles of $\frac{360^\circ}{p}$. That +is, +\begin{gather*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - + \frac 12 (p-1) \equiv 0 \bmod p. \\ +\intertext{Hence} +(p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - p + 1 \equiv 0 \bmod p; \\ +\intertext{and from this it follows that} +1 \cdot 2 \ldots \overline{p-1} + 1 \equiv 0 \bmod p, \\ +\end{gather*} +as was to be proved. + +\section{The Converse of Wilson's Theorem}\label{s26} + +Wilson's theorem is noteworthy in that its converse is also true. +The converse may be stated as follows: + +\smallskip \emph{Every integer $n$ such that the congruence} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n +\end{equation*} +\emph{is satisfied is a prime number.} + +For, if $n$ is not prime, there is some divisor $d$ of $n$ different +from $1$ and less than $n$. For such a $d$ we have $1 \cdot 2 \cdot +3 \ldots \overline{n-1} \equiv 0 \bmod d$; so that $1 \cdot 2 \ldots +\overline{n-1}+1 \not\equiv 0 \bmod d$; and hence $1 \cdot 2 \ldots +\overline{n-1}+1 \equiv 0 \bmod n$. Since this contradicts our +hypothesis the truth of the theorem follows. + +\smallskip Wilson's theorem and its converse may be combined into +the following elegant theorem: + +\smallskip \emph{A necessary and sufficient condition that an +integer $n$ is prime is that} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n. +\end{equation*}\index{Prime numbers} + +Theoretically this furnishes a complete and elegant test as to +whether a given number is prime. But, practically, the labor of +applying it is so great that it is useless for verifying large +primes. + +\section{Impossibility of $1 \cdot 2 \cdot 3 \cdots +\overline{n-1} + 1 = n^k$ for $n > 5$.}\label{s27} + +In this section we shall prove the following theorem: + +\emph{There exists no integer $k$ for which the equation} +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-1} + 1 = n^k +\end{equation*} +is true when $n$ is greater than $5$. + +If $n$ contains a divisor $d$ different from $1$ and $n$, the +equation is obviously false; for the second member is divisible by +$d$ while the first is not. Hence we need to prove the theorem only +for primes $n$. + +Transposing $1$ to the second member and dividing by $n - 1$ we have +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-2} = n^{k-1} + n^{k-2} + + \ldots + n+1. +\end{equation*} +If $n>5$ the product on the left contains both the factor $2$ and +the factor $\frac{1}{2} (n-1)$; that is, the first member contains +the factor $n - 1$. But the second member does not contain this +factor, since for $n = 1$ the expression $n^{k-1} + \ldots n + 1$ is +equal to $k \neq 0$. Hence the theorem follows at once. + +\section{Extension of Fermat's Theorem}\label{s28}% +\index{Fermat's!theorem extended|(} + +The object of this section is to extend Fermat's general theorem and +incidentally to give a new proof of it. We shall base this proof on +the simple Fermat theorem, of which we have already given a simple +independent proof. This theorem asserts that for every prime $p$ and +integer $a$ not divisible by $p$, we have the congruence +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +Then let us write +\begin{gather} +a^{p-1} = 1 + hp. \tag{1} \\ +\intertext{Raising each member of this equation to the +$p^{\text{th}}$ power we may write the result in the form} +a^{p(p-1)} = 1 + h_1p^2. \tag{2} \\ +\intertext{where $h_1$ is an integer. Hence} +a^{p(p-1)} \equiv 1 \bmod p^2. \notag \\ +\intertext{By raising each member of (2) to the $p^{\text{th}}$ +power we can readily show that} +a^{p^2(p-1)} \equiv 1 \bmod p^3. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{p^{\alpha - 1}(p-1)} \equiv 1 \bmod p^{\alpha}. \notag \\ +\intertext{where $\alpha$ is a positive integer; that is,} +a^{\phi(p^{\alpha})} \equiv 1 \bmod p^{\alpha}. \notag +\end{gather} + +For the special case when $p$ is 2 this result can be extended. For +this case (1) becomes +\begin{gather} +a = 1 + 2h. \notag \\ +\intertext{Squaring we have} +a^2 = 1 + 4h(h+1). \notag \\ +\intertext{Hence,} +a^2 = 1+8h_1, \tag{3} \\ +\intertext{where $h_1$ is an integer. Therefore} +a^2 \equiv 1 \bmod 2^3. \notag \\ +\intertext{Squaring (3) we have} +a^{2^2} = 1 + 2^4h_2; \notag \\ +\intertext{or} +a^{2^2} \equiv 1 \bmod 2^4. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{2^{\alpha-2}} \equiv 1 \bmod 2^{\alpha} \notag \\ +\intertext{if $\alpha > 2$. That is,} +a^{\frac{1}{2}\phi(2^{\alpha})} \equiv 1 \bmod 2^{\alpha} + \text{ if } a > 2. +\end{gather} + +Now in terms of the $\phi$-function let us define a new function +$\lambda(m)$ as follows: +\begin{align*} +\lambda(2^{\alpha}) &= \phi(2^{\alpha}) \text{ if $a = 0, 1, 2$;} \\ +\lambda(2^{\alpha}) &= \frac{1}{2}\phi(2^{\alpha}) + \text{ if $a > 2$;} \\ +\lambda(p^{\alpha}) &= \phi(p^{\alpha}) + \text{ if $p$ is an odd prime;} \\ +\lambda(2^{\alpha} p_1^{\alpha_1} p_2^{\alpha_2} \cdots + p_n^{\alpha_n}) &= M, +\end{align*} +where $M$ is the least common multiple of +\begin{equation*} + \lambda(2^{\alpha}), + \lambda(p_1^{\alpha_1}), + \lambda(p_2^{\alpha_2}), \ldots, \lambda(p_n^{\alpha_n}), +\end{equation*} +$2, p_1, p_2, \ldots, p_n$ being different primes.% +\index{$\lambda(m)$} + +Denote by $m$ the number +\begin{equation*} +m = 2^{\alpha}p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_n^{\alpha_n}. +\end{equation*} +Let $a$ be any number prime to $m$. From our preceding results we +have +\begin{align*} +a^{\lambda(2^{\alpha})} &\equiv 1 \bmod 2^{\alpha}, \\ +a^{\lambda(p_1^{\alpha_1})} &\equiv 1 \bmod p_1^{\alpha_1},\\ +a^{\lambda(p_2^{\alpha_2})} &\equiv 1 \bmod p_2^{\alpha_2}, \\ +\ldots \\ +a^{\lambda(p_n^{\alpha_n})} &\equiv 1 \bmod p_2^{\alpha_n}. \\ +\end{align*} + +Now any one of these congruences remains true if both of its members +are raised to the same positive integral power, whatever that power +may be. Then let us raise both members of the first congruence to +the power $\frac{\lambda(m)}{\lambda(2^\alpha)}$; both members of +the second congruence to the power +$\frac{\lambda(m)}{\lambda(p_1^{\alpha_1})}$; $\ldots$; both members +of the last congruence to the power +$\frac{\lambda(m)}{\lambda(p_n^{\alpha_n})}$. Then we have +\begin{align*} +a^{\lambda(m)} &\equiv 1 \mod 2^\alpha, \\ +a^{\lambda(m)} &\equiv 1 \mod p_1^{\alpha_1}, \\ +\ldots \ldots \\ +a^{\lambda(m)} &\equiv 1 \mod p_n^{\alpha_n}. \\ +\intertext{From these congruences we have immediately} +a^{\lambda(m)} &\equiv 1 \mod m. +\end{align*} + +We may state this result in full in the following theorem: + +\smallskip \emph{If $a$ and $m$ are any two relatively prime positive +integers, the congruence} +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +\emph{is satisfied.} + +As an excellent example to show the possible difference between the +exponent $\lambda(m)$ in this theorem and the exponent $\phi(m)$ in +Fermat's general theorem, let us take +\begin{gather*} +m = 2^6 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 19 + \cdot 37 \cdot 73. \\ +\intertext{Here} +\lambda(m) = 2^4 \cdot 3^2, \quad \phi(m) = 2^{31} \cdot 3^{10}. +\end{gather*} + +In a later chapter we shall show that there is no exponent $\nu$ +less than $\lambda(m)$ for which the congruence +\begin{equation*} +a^\nu = 1 \mod m +\end{equation*} +is verified for every integer $a$ prime to $m$. + +From our theorem, as stated above, Fermat's general theorem follows +as a corollary, since $\lambda(m)$ is obviously a factor of +$\phi(m)$, +\begin{equation*} +\phi(m) = \phi(2^\alpha) \phi(p_1^{\alpha_1}) \ldots + \phi(p_n^{\alpha_n}). +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $a^{16} \equiv 1 \bmod 16320$, for every $a$ +which is prime to $16320$. + +\item[2.] Show that $a^{12} \equiv 1 \bmod 65520$, for every $a$ which +is prime to $65520$. + +\item[3*.] Find one or more composite numbers $P$ such that +\begin{equation*} +a^{P-1} \equiv 1 \bmod P +\end{equation*} +for every a prime to $P$. (Compare this problem with the next +section.) \end{enumerate} \normalsize% +\index{Fermat's!theorem extended|)} + +\section{On the Converse of Fermat's Simple Theorem}\label{s29}% +\index{Fermat's!Simple Theorem} + +The fact that the converse of Wilson's theorem is a true proposition +leads one naturally to inquire whether the converse of Fermat's +simple theorem is true. Thus, we may ask the question: Does the +existence of the congruence $2^{n-1} \equiv 1 \bmod n$ require that +$n$ be a prime number? The Chinese answered this question in the +affirmative and the answer passed unchallenged among them for many +years. An example is sufficient to show that the theorem is not +true. We shall show that +\begin{equation*} +2^{340} \equiv 1 \bmod 341 +\end{equation*} +although $341 = 11 \cdot 31$, is not a prime number. Now $2^{10}-1 = +3 \cdot 11 \cdot 31$. Hence $2^{10} \equiv 1 \bmod 341$. Hence +$2^{340} \equiv 1 \bmod 341$. From this it follows that the direct +converse of Fermat's theorem is not true. The following theorem, +however, which is a converse with an extended hypothesis, is readily +proved. + +\smallskip \emph{If there exists an integer $a$ such that} +\begin{equation*} +a^{n-1} \equiv 1 \bmod n +\end{equation*} +\emph{and if further there does not exist an integer $\nu$ less than +$n - 1$ such that} +\begin{equation*} +a^{\nu} \equiv 1 \bmod n, +\end{equation*} +\emph{then the integer $n$ is a prime number.} + +For, if $n$ is not prime, $\phi(n) < n - 1$. Then for $\nu = +\phi(n)$ we have $a^{\nu} \equiv 1 \bmod n$, contrary to the +hypothesis of the theorem. + +\section{Application of Previous Results to Linear +Congruences}\label{s30}% +\index{Congruences!Linear} + +The theorems of the present chapter afford us a ready means of +writing down a solution of the congruence +\begin{equation} +ax \equiv c \bmod m. \tag{1} +\end{equation} +We shall consider only the case in which $a$ and $m$ are relatively +prime, since the general case is easily reducible to this one, as we +saw in the preceding chapter. + +Since $a$ and $m$ are relatively prime we have the congruences +\begin{gather*} +a^{\lambda(m)} \equiv 1,\quad a^{\phi(m)} \equiv 1 \bmod m. \\ +\intertext{Hence either of the numbers $x$,} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +is a representative of the solution of (1). Hence the following +theorem: + +\smallskip \emph{If} +\begin{gather*} +ax \equiv c \bmod m \\ +\intertext{\emph{is any linear congruence in which $a$ and $m$ are +relatively prime, then either of the numbers $x$,}} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +\emph{is a representative of the solution of the congruence.} + +The former representative of the solution is the more convenient of +the two, since the power of $a$ is in general much less in this case +than in the other. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] Find a solution of $7x \equiv 1 \bmod 2^6 \cdot 3 \cdot 5 \cdot +17.$ Note the greater facility in applying the first of the above +representatives of the solution rather than the second. +\end{enumerate} \normalsize + +\section{Application of the Preceding Results to the Theory +of Quadratic Residues}\label{s31}\index{Quadratic residues|(} + +In this section we shall apply the preceding results of this chapter +to the problem of finding the solutions of congruences of the form +\begin{equation*} +\alpha z^2 + \beta z + \gamma \equiv 0 \mod \mu +\end{equation*} +where $\alpha, \beta, \gamma, \mu$ are integers. These are called +quadratic congruences. + +The problem of the solution of the quadratic congruence (1) can be +reduced to that of the solution of a simpler form of congruence as +follows: Congruence (1) is evidently equivalent to the congruence +\begin{gather} +4\alpha^2 z^2 + 4\alpha\beta z + 4\alpha\gamma \equiv + 0 \mod 4\alpha\mu. \tag{1} \\ +\intertext{But this may be written in the form} +(2\alpha z + \beta)^2 \equiv \beta^2 - 4\alpha\gamma + \mod 4\alpha\mu. \notag \\ +\intertext{Now if we put} +2\alpha z + \beta\equiv x \mod 4\alpha\mu \tag{2} \\ +\intertext{and} +\beta^2 - 4\alpha\gamma = a,\quad 4\alpha\mu = m, \notag \\ +\intertext{we have} +x^2 \equiv a\mod m. \tag{3} +\end{gather} +We have thus reduced the problem of solving the general congruence +(1) to that of solving the binomial congruence (3) and the linear +congruence (2). The solution of the latter may be effected by means +of the results of \S \ref{s30}. We shall therefore confine ourselves +now to a study of congruence (3). We shall make a further limitation +by assuming that $a$ and $m$ are relatively prime, since it is +obvious that the more general case is readily reducible to this one. + +The example +\begin{equation*} +x^2 \equiv 3 \mod 5 +\end{equation*} +shows at once that the congruence (3) does not always have a +solution. First of all, then, it is necessary to find out in what +cases (3) has a solution. Before taking up the question it will be +convenient to introduce some definitions. + +\smallskip\textsc{Definitions.} An integer $a$ is said to be a +quadratic residue modulo $m$ or a quadratic non-residue modulo $m$ +according as the congruence +\begin{equation*} +x^2 = a \mod m +\end{equation*} +has or has not a solution. We shall confine our attention to the +case when $m > 2$.\index{Residue} + +We shall now prove the following theorem: + +\smallskip I.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\lambda(m)} \equiv 1 \mod m. +\end{equation*} + +Suppose that the congruence $x^2 \equiv a \mod m$ has the solution $x = +\alpha$. Then $\alpha^2 \equiv a \mod m$. Hence +\begin{equation*} +\alpha^{\lambda(m)} \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +Since $a$ is prime to $m$ it is clear from $\alpha^2 \equiv a \mod +m$ that $\alpha$ is prime to $m$. Hence $\alpha^{\lambda(m)} \equiv 1 +\mod m$. Therefore we have +\begin{equation*} +1 \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +That is, this is a necessary condition in order that $a$ shall be a +quadratic residue modulo $m$. + +In a similar way one may prove the following theorem: + +\smallskip II.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\phi(m)} \equiv 1 \mod m. +\end{equation*} + +When $m$ is a prime number $p$ each of the above results takes the +following form: If $a$ is prime to $p$ and is a quadratic residue +modulo $p$, then +\begin{equation*} +a^{\frac{1}{2}(p-1)} \equiv 1 \mod p. +\end{equation*} +We shall now prove the following more complete theorem, without the +use of I or II. + +\smallskip III.~\emph{If $p$ is an odd prime number and $a$ is an +integer not divisible by $p$, then $a$ is a quadratic residue or a +quadratic non-residue modulo $p$ according as} +\begin{equation*} +a^{\tfrac{1}{2}(p-1)} \equiv +1 \quad \text{or} \quad +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p. +\end{equation*} + +This is called Euler's criterion.\index{Euler's!criterion} + +Given a number $a$, not divisible by $p$, we have to determine +whether or not the congruence +\begin{gather} +x^2 \equiv a \bmod p \notag \\ +\intertext{has a solution. Let $r$ be any number of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 \tag{A} \\ +\intertext{and consider the congruence} +rx \equiv a \bmod p. +\end{gather} +This has always one and just one solution $x$ equal to a number $s$ +of the set (A). Two cases can arise: either for every $r$ of the set +(A) the corresponding $s$ is different from $r$ or for some $r$ of +the set (A) the corresponding $s$ is equal to $r$. The former is the +case when $a$ is a quadratic non-residue modulo $p$; the latter is +the case when $a$ is a quadratic residue modulo $p$. We consider the +two cases separately. + +In the first case the numbers of the set (A) go in pairs such that +the product of the numbers in the pair is congruent to a modulo $p$. +Hence, taking the product of all $\tfrac{1}{2}(p - 1)$ pairs, we +have +\begin{align*} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &\equiv + +a^{\tfrac{1}{2}(p-1)} \bmod p. \\ +\intertext{But} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &= -1 \bmod p. \\ +\intertext{Hence} +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p, +\end{align*} +whence the truth of one part of the theorem. + +In the other case, namely that in which some $r$ and corresponding +$s$ are equal, we have for this $r$ +\begin{gather*} +r^{2} \equiv a \bmod p \\ +\intertext{and} +(p - r)^{2} \equiv a \bmod p. +\end{gather*} +Since $x^{2} \equiv a \bmod p$ has at most two solutions it follows +that all the integers in the set (A) except $r$ and $p - r$ fall in +pairs such that the product of the numbers in each pair is congruent +to a modulo $p$. Hence, taking the product of all these pairs, which +are $\frac{1}{2}(p - 1) - 1$ in number, and multiplying by $r(p-r)$ +we have +\begin{align*} +1 \cdot 2 \cdot 3 \cdots \overline{p -1} + &\equiv (p - r) r a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -r^{2} a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a^{\frac{1}{2}(p -1)} \bmod p. \\ +\intertext{Since $1 \cdot 2 \cdot 3 \cdots \overline{p - 1} \equiv -1 +\bmod p$ we have} +a^{\frac{1}{2}(p -1)} &\equiv + 1 \bmod p +\end{align*} +whence the truth of another part of the theorem. + +Thus the proof of the entire theorem is complete.% +\index{Quadratic residues|)}\index{Wilson's theorem|)} + +\chapter{PRIMITIVE ROOTS MODULO $m$.} + +\section{Exponent of an Integer Modulo $m$}\label{s32}% +\index{Exponent of an integer|(}\index{Primitive roots|(} + +Let +\begin{equation*} +a_{1},\ a_{2},\ \cdots,\ a_{\phi(m)} \tag{A} +\end{equation*} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$; and let $a$ denote any integer of the set (A). Now any +positive integral power of $a$ is prime to $m$ and hence is +congruent modulo $m$ to a number of the set (A). Hence, among all +the powers of a there must be two, say $a^{n}$ and $a^{\nu}$, $n > +\nu$, which, are congruent to the same integer of the set (A). These +two powers are then congruent to each other; that is, +\begin{equation*} +a^{n} \equiv a^{\nu} \bmod m +\end{equation*} +Since $a^{\nu}$ is prime to $m$ the members of this congruence may +be divided by $a^{\nu}$. Thus we have +\begin{equation*} +a^{n - \nu} \equiv 1 \bmod m. +\end{equation*} +That is, among the powers of $a$ there is one at least which is +congruent to $1$ modulo $m$. + +\smallskip Now, in the set of all powers of $a$ which are congruent +to $1$ modulo $m$ there is one in which the exponent is less than in +any other of the set. Let the exponent of this power be $d$, so that +$a^{d}$ is the lowest power of $a$ such that +\begin{equation} +a^{d} \equiv 1 \bmod m. \tag{1} +\end{equation} + +We shall now show that if $a^{\alpha} \equiv 1 \bmod m$, then +$\alpha$ is a multiple of $d$. Let us write +\begin{gather} +\alpha = d\delta + \beta, \quad 0 \leqq \beta < d. \notag \\ +\intertext{Then} +a^{\alpha} \equiv 1 \bmod m, \tag{2} \\ +a^{d\delta} \equiv 1 \bmod m, \tag{3} \\ +\intertext{the last congruence being obtained by raising (1) to the +power $\delta$. From (3) we have} +a^{d\delta + \beta} \equiv a^{\beta} \bmod m; \notag \\ +\intertext{or} +a^{\beta}\equiv 1 \bmod m. \notag +\end{gather} +Hence $\beta = 0$, for otherwise $d$ is not the exponent of the +lowest power of $a$ which is congruent to 1 modulo $m$. Hence $d$ is +a divisor of $\alpha$. + +\smallskip These results may be stated as follows: + +\smallskip I.~\emph{If $m$ is any integer and $a$ is any integer +prime to $m$, then there exists an integer $d$ such that} +\begin{gather*} +a^d\equiv 1 \bmod m \\ +\intertext{\emph{while there is no integer $\beta$ less than $d$ for +which}} +a^\beta\equiv 1 \bmod m. \\ +\intertext{\emph{Further, a necessary and sufficient condition +that}} +a^\nu \equiv 1 \bmod m +\end{gather*} +\emph{is that $\nu$ is a multiple of $d$.} + +\smallskip \textsc{Definition.} The integer $d$ which is thus +uniquely determined when the two relatively prime integers $a$ and +$m$ are given is called the exponent of $a$ modulo $m$. Also, $d$ is +said to be the exponent to which $a$ belongs modulo $m$. + +Now, in every case we have +\begin{equation*} +a^{\phi(m)} \equiv 1,\quad a^{\lambda(m)} \equiv 1 \bmod m, +\end{equation*} +if $a$ and $m$ are relatively prime. Hence from the preceding +theorem we have at once the following: + +\smallskip II.~\textit{The exponent $d$ to which $a$ belongs modulo +$m$ is a divisor of both $\phi(m)$ and $\lambda(m)$.}% +\index{Exponent of an integer|)} + +\section{Another Proof of Fermat's General Theorem}\label{s33} + +In this section we shall give an independent proof of the theorem +that the exponent $d$ of $a$ modulo $m$ is a divisor of $\phi(m)$; +from this result we have obviously a new proof of Fermat's theorem +itself. + +We retain the notation of the preceding section. We shall first +prove the following theorem: + +\smallskip I.~\textit{The numbers} +\begin{equation} +1,\ a,\ a^2,\ \ldots,\ a^{d-1} \tag{A} +\end{equation} +\textit{are incongruent each to each modulo $m$.} + +For, if $a^\alpha \equiv a^\beta \bmod m$, where $0 \leqq \alpha < +d$ and $0 \leqq \beta < d$, $\alpha > \beta$, we have +$a^{\alpha-\beta} \equiv 1 \bmod m$, so that $d$ is not the exponent +to which $a$ belongs modulo $m$, contrary to hypothesis. + +\smallskip Now any number of the set (A) is congruent to some number +of the set +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)}. \tag{B} +\end{equation} +Let us undertake to separate the numbers (B) into classes after the +following manner: Let the first class consist of the numbers +\begin{equation} +\alpha_1,\ \alpha_2,\ \ldots,\ \alpha_{d-1}, \tag{I} +\end{equation} +where $\alpha_i$ is the number of the set (B) to which $a^i$ is +congruent modulo $m$. + +If the class (I) does not contain all the numbers of the set (B), +let $a_i$ be any number of the set (B) not contained in (I) and form +the following set of numbers: +\begin{equation} +\alpha_0 a_i,\ \alpha_1 a_i,\ \alpha_2 a_i,\ \ldots,\ + \alpha_{d-1}a_i. \tag{II'} +\end{equation} +We shall now show that no number of this set is congruent to a +number of class (I). For, if so, we should have a congruence of the +form +\begin{gather*} +a_i \alpha_j \equiv \alpha_k \bmod m; \\ +\intertext{hence} +a_i a^j \equiv a^k \bmod m, \\ +\intertext{so that} +a_i a^d \equiv a^{k+d-j} \bmod m; \\ +\intertext{or} +a_i \equiv a^{k+d-j} \bmod m, +\end{gather*} +so that $a_i$ would belong to the set (I) contrary to hypothesis. + +Now the numbers of the set (II$'$) are all congruent to numbers of +the set (B); and no two are congruent to the same number of this +set. For, if so, we should have two numbers of (II') congruent; that +is, $\alpha_k a_i \equiv \alpha_j a_i \bmod m,$ or $\alpha_k \equiv +\alpha_j \bmod m;$ and this we have seen to be impossible. + +Now let the numbers of the set (B) to which the numbers of the set +(II') are congruent be in order the following: +\begin{equation} +\beta_0,\ \beta_1,\ \beta_2,\ \ldots,\ \beta_{d-1}. \tag{II} +\end{equation} +These numbers constitute our class (II). + +If classes (I) and (II) do not contain all the numbers of the set +(B), let $a_j$ be a number of the set ($B$) not contained in either +of the classes (I) and (II): and form the set of numbers +\begin{equation} +\alpha_0 a_j,\ \alpha_1 a_j,\ \alpha_2 a_j,\ \ldots,\ + \alpha_{d-1} a_j. \tag{III'} +\end{equation} +Just as in the preceding case it may be shown that no number of this +set is congruent to a number of class (I) and that the numbers of +(III') are incongruent each to each. We shall also show that no +number of (III') is congruent to a number of class (II). For, if so, +we should have $\alpha_k a_j \equiv \beta_l \bmod m$. Hence $a^k a_j +\equiv a^l a_i \bmod m$; or $a_j \equiv a^{l+d-k} a_i \bmod m$, from +which it follows that $a_j$ is of class (II), contrary to +hypothesis. + +Now let the numbers of the set (B) to which the numbers of the set +(III') are congruent be in order the following: +\begin{equation} +\gamma_0,\ \gamma_1,\ \gamma_2,\ \ldots,\ \gamma_{d-1}. \tag{III} +\end{equation} +These numbers form our class (III). + +It is now evident that the process may be continued until all the +numbers of the set (B) have been separated into classes, each class +containing $d$ integers, thus: +\begin{equation*} +\begin{matrix} +(\text{I}) & \alpha_0, & \alpha_1, & \alpha_2, + & \ldots, & \alpha_{d-1}, \\ +(\text{II}) & \beta_0, & \beta_1, & \beta_2, + & \ldots, & \beta_{d-1}, \\ +(\text{III}) & \gamma_0, & \gamma_1, & \gamma_2, + & \ldots, & \gamma_{d-1}, \\ +&\hdotsfor{5} \\ +(\quad ) & \lambda_0, & \lambda_1, & \lambda_2, + & \ldots, & \lambda_{d-1}. +\end{matrix} +\end{equation*} +The set (B), which consists of $\phi(m)$ integers, has thus been +separated into classes, each class containing $d$ integers. Hence we +conclude that $d$ is a divisor of $\phi(m)$. Thus we have a second +proof of the theorem: + +\smallskip II.~\emph{If $a$ and $m$ are any two relatively prime +integers and $d$ is the exponent to which $a$ belongs modulo $m$, +then $d$ is a divisor of $\phi(m)$.} + +In our classification of the numbers (B) into the rectangular array +above we have proved much more than theorem II; in fact, theorem II +is to be regarded as one only of the consequences of the more +general result contained in the array. + +If we raise each member of the congruence +\begin{equation*} +a^d \equiv 1 \bmod m +\end{equation*} +to the (integral) power $\phi(m)/d$, the preceding theorem leads +immediately to an independent proof of Fermat's general theorem. + +\section{Definition of Primitive Roots}\label{s34} + +\textsc{Definition.} Let $a$ and $m$ be two relatively prime +integers. If the exponent to which $a$ belongs modulo $m$ is +$\phi(m)$, $a$ is said to be a primitive root modulo $m$ (or a +primitive root of $m$). + +In a previous chapter we saw that the congruence +\begin{equation*} +a^{\lambda(m)} \equiv 1 \bmod m +\end{equation*} +is verified by every pair of relatively prime integers $a$ and $m$. +Hence, primitive roots can exist only for such a modulus $m$ as +satisfies the equation +\begin{equation*} +\phi(m) = \lambda(m). \tag{1} +\end{equation*} +We shall show later that this is also sufficient for the existence +of primitive roots. + +From the relation which exists in general between the +$\phi$-function and the $\lambda$-function in virtue of the +definition of the latter, it follows that (1) can be satisfied only +when $m$ is a prime power or is twice an odd prime power. + +Suppose first that $m$ is a power of $2$, say $m = 2^\alpha$. Then +(1) is satisfied only if $\alpha = 0,\ 1,\ 2$. For $\alpha = 0$ or +$1$, $1$ itself is a primitive root. For $\alpha = 2$, $3$ is a +primitive root. We have therefore left to examine only the cases +\begin{equation*} +m = p^\alpha,\quad m = 2p^\alpha +\end{equation*} +where $p$ is an odd prime number. The detailed study of these cases +follows in the next sections. + +\section{Primitive roots modulo $p$.}\label{s35} + +We have seen that if $p$ is a prime number and $d$ is the exponent +to which $a$ belongs modulo $p$, then $d$ is a divisor of $\phi(p) = +p - 1$. Now, let +\begin{gather*} +d_1,\ d_2,\ d_3,\ \ldots,\ d_r \\ +\intertext{be all the divisors of $p-1$ and let $\psi(d_i)$ denote +the number of integers of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 +\end{gather*} +which belong to the exponent $d_i$. If there is no integer of the +set belonging to this exponent, then $\psi(d_i) = 0$. + +Evidently every integer of the set belongs to some one and only one +of the exponents $d_1, d_2, \ldots, d_r$. Hence we have the relation +\begin{gather} +\psi(d_1) + \psi(d_2) + \ldots + \psi(d_r) = p-1. \tag{1} \\ +\intertext{But} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = p-1. \tag{2} \\ +\intertext{If then we can show that} +\psi(d_i) \leqq \phi(d_i) \tag{3} \\ +\intertext{for $i = 1, 2, \ldots, r$, it will follow from a +comparison of (1) and (2) that} +\psi(d_i) = \phi(d_i). \notag +\end{gather} +Accordingly, we shall examine into the truth of (3). + +Now the congruence +\begin{equation} +x^{d_i} \equiv 1 \mod p \tag{4} +\end{equation} +has not more than $d_i$ roots. If no root of this congruence belongs +to the exponent $d_i$, then $\psi(d_i) = 0$ and therefore in this +case we have $\psi(d_i) < \phi(d_i)$. On the other hand if $a$ is a +root of (4) belonging to the exponent $d_i$, then +\begin{equation} +a, a^2, a^3, \ldots, a^{d_i} \tag{5} +\end{equation} +are a set of $d_i$ incongruent roots of (4); and hence they are the +complete set of roots of (4). + +But it is easy to see that $a^k$ does or does not belong to the +exponent $d_i$ according as $k$ is or is not prime to $d_i$; for, if +$a^k$ belongs to the exponent $t$, then $t$ is the least integer +such that $kt$ is a multiple of $d_i$. Consequently the number of +roots in the set (5) belonging to the exponent $d_i$ is $\phi(d_i)$. +That is, in this case $\psi(d_i) = \phi(d_i)$. Hence in general +$\psi(d_i) \leqq \phi(d_i)$. Therefore from (1) and (2) we conclude +that +\begin{equation*} +\psi(d_i) = \phi(d_i), \quad i = 1,\ 2,\ \ldots,\ r. +\end{equation*} +The result thus obtained may be stated in the form of the following +theorem: + +\smallskip I.~\emph{If $p$ is a prime number and $d$ is any divisor +of $p-1$, then the number of integers belonging to the exponent $d$ +modulo $p$ is $\phi(d)$.} + +In particular: + +\smallskip II.~\emph{There exist primitive roots modulo $p$ and their +number is $\phi(p-1)$.} + +\section{Primitive Roots Modulo $p^\alpha$, $p$ an Odd +Prime}\label{s36} + +In proving that there exist primitive roots modulo $p^\alpha$, where +$p$ is an odd prime and $\alpha > 1$, we shall need the following +theorem: + +I.~\emph{There always exists a primitive root $\gamma$ modulo $p$ +for which $\gamma^{p-1} - 1$ is not divisible by $p^2$.} + +Let $g$ be any primitive root modulo $p$. If $g^{p-1}-1$ is not +divisible by $p^2$ our theorem is verified. Then suppose that +$g^{p-1}-1$ is divisible by $p^2$, so that we have +\begin{gather*} +g^{p-1}-1 = kp^2 \\ +\intertext{where $k$ is an integer. Then put} +\gamma \equiv g + xp \\ +\intertext{where $x$ is an integer. Then $\gamma \equiv g \mod p$, and +hence} +\gamma^h \equiv g^h \mod p; +\end{gather*} +whence we conclude that $\gamma$ is a primitive root modulo $p$. But +\begin{align*} +\gamma^{p-1}-1 &= + g^{p-1} - 1 + \frac{p-1}{1!}g^{p-2}xp + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p^2 + \ldots \\ + &= p\left(kp + \frac{p-1}{1!}g^{p-2}x + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p + \ldots\right). +\end{align*} +Hence +\begin{equation*} +\gamma^{p-1}-1 \equiv p(-g^{p-2}x) \mod p^2. +\end{equation*} +Therefore it is evident that $x$ can be so chosen that +$\gamma^{p-1}-1$ is not divisible by $p^2$. Hence there exists a +primitive root $\gamma$ modulo $p$ such that $\gamma^{p-1}-1$ is not +divisible by $p^2$. Q.~E.~D. + +\smallskip We shall now prove that this integer $\gamma$ is a +primitive root modulo $p^\alpha$, where $\alpha$ is any positive +integer. + +If +\begin{equation*} +\gamma^k \equiv 1\mod p, +\end{equation*} +then $k$ is a multiple of $p-1$, since $\gamma$ is a primitive root +modulo $p$. Hence, if +\begin{equation*} +\gamma^k \equiv 1 \mod p^\alpha, +\end{equation*} +then $k$ is a multiple of $p-1$. + +Now, write +\begin{equation*} +\gamma^{p-1} = 1 + hp. +\end{equation*} +Since $\gamma^{p-1}-1$ is not divisible by $p^2$, it follows that $h$ +is prime to $p$. If we raise each member of this equation to the +power $\beta p^{\alpha-2}$, $\alpha \stackrel{=}{>}2$, we have +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} = + 1 + \beta p^{\alpha-1}h + p^\alpha I, +\end{equation*} +where $I$ is an integer. Then if +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} \equiv 1 \mod p^\alpha, +\end{equation*} +$\beta$ must be divisible by $p$. Therefore the exponent of the +lowest power of $\gamma$ which is congruent to $1$ modulo $p^\alpha$ +is divisible by $p^{\alpha-1}$. But we have seen that this exponent +is also divisible by $p-1$. Hence the exponent of $\gamma$ modulo +$p^\alpha$ is $p^{\alpha-1}(p-1)$ since $\phi(p^\alpha) = +p^{\alpha-1}(p-1)$. That is, $\gamma$ is a primitive root modulo +$p^\alpha$. + +It is easy to see that no two numbers of the set +\begin{equation} +\gamma, \gamma^2, \gamma^3, \ldots, \gamma^{p^{\alpha-1}(p-1)} +\tag{A} +\end{equation} +are congruent modulo $p^\alpha$; for, if so, $\gamma$ would belong +modulo $p^\alpha$ to an exponent less than $p^{\alpha-1}(p-1)$ and +would therefore not be a primitive root modulo $p^\alpha$. Now every +number in the set (A) is prime to $p^\alpha$; their number is +$\phi(p^\alpha) = p^{\alpha -1}(p-1)$. Hence the numbers of the set +(A) are congruent in some order to the numbers of the set (B): +\begin{equation} +a_1,\ a_2,\ a_3,\ \ldots ,\ a_{p^{\alpha-1}(p-1)}, \tag{B} +\end{equation} +where the integers (B) are the positive integers less than +$p^\alpha$ and prime to $p^\alpha$. + +But any number of the set (B) is a solution of the congruence +\begin{equation} +x^{p^{\alpha-1} (p-1)} \equiv 1 \bmod p^\alpha. \tag{1} +\end{equation} +Further, every solution of this congruence is prime to $p^\alpha$. +Hence the integers (B) are a complete set of solutions of (1). +Therefore the integers (A) are a complete set of solutions of (1). +But it is easy to see that an integer $\gamma^k$ of the set (A) is +or is not a primitive root modulo $p^\alpha$ according as $k$ is or +is not prime to $p^{\alpha-1} (p-1)$. Hence the number of primitive +roots modulo $p^\alpha$ is $\phi \{p^{\alpha-1} (p-1) \}.$ + +The results thus obtained may be stated as follows: + +\smallskip II.~\emph{If $p$ is any odd prime number and $\alpha$ is +any positive integer, then there exist primitive roots modulo +$p^\alpha$ and their number is $\phi \{ \phi(p^\alpha) \}$}. + +\section{Primitive Roots Modulo $2p^\alpha$, $p$ an Odd +Prime}\label{s37} + +In this section we shall prove the following theorem: + +\emph{If $p$ is any odd prime number and $\alpha$ is any positive +integer, then there exist primitive roots modulo $2p^\alpha$ and +their number is $\phi \{\phi(2 p^{\alpha} )\}.$} + +Since $2 p^\alpha$ is even it follows that every primitive root +modulo $2 p^\alpha$ is an odd number. Any odd primitive root modulo +$p^\alpha$ is obviously a primitive root modulo $2p^\alpha$. Again, +if $\gamma$ is an even primitive root modulo $p^\alpha$ then $\gamma ++ p^\alpha$ is a primitive root modulo $2 p^\alpha$. It is evident +that these two classes contain (without repetition) all the +primitive roots modulo $2 p^\alpha$. Hence the theorem follows as +stated above. + +\section{Recapitulation}\label{s38} + +The results which we have obtained in \S\S \ref{s34}--\ref{s37} +inclusive may be gathered into the following theorem: + +\emph{In order that there shall exist primitive roots modulo $m$, it +is necessary and sufficient that $m$ shall have one of the values} +\begin{equation*} +m = 1, 2, 4, p^\alpha, 2p^\alpha +\end{equation*} +\emph{where $p$ is an odd prime and $\alpha$ is a positive integer.} + +\emph{If $m$ has one of these values then the number of primitive +roots modulo $m$ is $\phi\{\phi(m)\}$.} + +\section{Primitive $\lambda$-roots}\label{s39}% +\index{Primitive roots!$\lambda$-roots|(} + +In the preceding sections of this chapter we have developed the +theory of primitive roots in the way in which it is usually +presented. But if one approaches the subject from a more general +point of view the results which may be obtained are more general and +at the same time more elegant. It is our purpose in this section to +develop the more general theory. + +\smallskip We have seen that if $a$ and $m$ are any two relatively +prime positive integers, then +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +Consequently there is no integer belonging modulo $m$ to an exponent +greater than $\lambda(m)$. It is natural to enquire if there are any +integers $a$ which belong to the exponent $\lambda(m)$. It turns out +that the question is to be answered in the affirmative, as we shall +show. Accordingly, we introduce the following definition: + +\smallskip \textsc{Definition.} If $a^{\lambda(m)}$ is the lowest +power of $a$ which is congruent to $1$ modulo $m$, $a$ is said to be +a primitive $\lambda$-root modulo $m$. We shall also say that it is +a primitive $\lambda$-root of the congruence $x^{\lambda(m)} = 1 +\mod m$. To distinguish we may speak of the usual primitive root as +a primitive $\phi$-root modulo $m$.% +\index{Primitive roots!$\phi$-roots} + +From the theory of primitive $\phi$-roots already developed it +follows that primitive $\lambda$-roots always exist when $m$ is a +power of any odd prime, and also when $m = 1,\ 2,\ 4$; for, for such +values of $m$ we have $\lambda(m) = \phi(m)$. + +We shall next show that primitive $\lambda$-roots exist when $m = +2^{\alpha}$, $a > 2$, by showing that 5 is such a root. It is +necessary and sufficient to prove that $5$ belongs modulo +$2^{\alpha}$ to the exponent $2^{\alpha-2} = \lambda(2^{\alpha})$. +Let $d$ be the exponent to which $5$ belongs modulo $2^{\alpha}$. +Then from theorem II of \S \ref{s32} it follows that $d$ is a +divisor of $2^{\alpha-2} = \lambda(2^{\alpha})$. Hence if $d$ is +different from $2^{\alpha-2}$ it is $2^{\alpha-3}$ or is a divisor +of $2^{\alpha-3}$. Hence if we can show that $5^{2^{\alpha-3}}$ is +not congruent to $1$ modulo $2^{\alpha}$ we will have proved that +$5$ belongs to the exponent $2^{\alpha-2}$. But, clearly, +\begin{gather*} +5^{2^{\alpha-3}} = (1+2^2)^{2^{\alpha-3}} + = 1+2^{\alpha-1}+ I\cdot 2^{\alpha}, \\ +\intertext{where $I$ is an integer. Hence} +5^{2^{\alpha-3}} \not\equiv 1 \bmod 2^{\alpha}. +\end{gather*} +Hence 5 belongs modulo $2^{\alpha}$ to the exponent +$\lambda(2^{\alpha})$. + +By means of these special results we are now in position to prove +readily the following general theorem which includes them as special +cases: + +\smallskip I.~\emph{For every congruence of the form} +\begin{gather*} +x^{\lambda(m)} \equiv 1 \bmod m +\end{gather*} +\emph{a solution $g$ exists which is a primitive $\lambda$-root, and +for any such solution $g$ there are $\phi\{\lambda(m)\}$ primitive +roots congruent to powers of $g$.} + +If any primitive $\lambda$-root $g$ exists, $g^\nu$ is or is not a +primitive $\lambda$-root according as $\nu$ is or is not prime to +$\lambda(m)$; and therefore the number of primitive $\lambda$-roots +which are congruent to powers of any such root $g$ is +$\phi\{\lambda(m)\}$. + +The existence of a primitive $\lambda$-root in every case may easily +be shown by induction. In case $m$ is a power of a prime the theorem +has already been established. We will suppose that it is true when +$m$ is the product of powers of $r$ different primes and show that +it is true when $m$ is the product of powers of $r+1$ different +primes; from this will follow the theorem in general. + +Put $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_r^{\alpha_r} +p_{r+1}^{\alpha_{r+1}}, \quad n = p_1^{\alpha_1} p_2^{\alpha_2} +\ldots p_r^{\alpha_r}$, and let $h$ be a primitive $\lambda$-root of +\begin{gather} +x^{\lambda(n)} \equiv 1 \mod n. \tag{1} \\ +\intertext{Then} +h + ny \notag +\end{gather} +is a form of the same root if $y$ is an integer. + +Likewise, if $c$ is any primitive $\lambda$-root of +\begin{equation} +x^\lambda(p_{r+1}^{\alpha_{r+1}}) + \equiv 1 \mod p_{r+1}^{\alpha_{r+1}} \tag{2} +\end{equation} +a form of this root is +\begin{equation*} +c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +where $z$ is any integer. + +Now, if $y$ and $z$ can be chosen so that +\begin{equation*} +h+ny = c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +the number in either member of this equation will be a common +primitive $\lambda$-root of congruences (1) and (2); that is, a +common primitive $\lambda$-root of the two congruences may always be +obtained provided that the equation +\begin{equation*} +p_1^{\alpha_1} \ldots p_r^{\alpha_r}y - p_{r+1}^{\alpha_{r+1}}z = c-h +\end{equation*} +has always a solution in which $y$ and $z$ are integers. That this +equation has such a solution follows readily from theorem III of \S +\ref{s9}; for, if $c-h$ is replaced by $1$, the new equation has a +solution $\bar{y}$, $\bar{z}$; and therefore for $y$ and $z$ we may +take $y = \bar{y}(c-h)$, $z = \bar{z}(c-h)$. + +Now let $g$ be a common primitive $\lambda$-root of congruences (1) +and (2) and write +\begin{equation*} +g^\nu \equiv 1 \mod m, +\end{equation*} +where $\nu$ is to be the smallest exponent for which the congruence +is true. Since $g$ is a primitive $\lambda$-root of (1) $\nu$ is a +multiple of $\lambda(p_1^{\alpha_1} \ldots p_r^{\alpha_r})$. Since +$g$ is a primitive $\lambda$-root of (2) $\nu$ is a multiple of +$\lambda\left(p_{r+1}^{\alpha_{r+1}} \right)$. Hence it is a +multiple of $\lambda(m)$. But $g^{\lambda(m)} \equiv 1 \bmod m$; +therefore $\nu = \lambda(m)$. That is, $g$ is a primitive +$\lambda$-root modulo $m$. + +The theorem as stated now follows at once by induction. + +\smallskip There is nothing in the preceding argument to indicate +that the primitive $\lambda$-roots modulo $m$ are all in a single +set obtained by taking powers of some root $g$; in fact it is not in +general true when $m$ contains more than one prime factor. + +By taking powers of a primitive $\lambda$-root $g$ modulo $m$ one +obtains $\phi\{\lambda(m)\}$ different primitive $\lambda$-roots +modulo $m$. It is evident that if $\gamma$ is any one of these +primitive $\lambda$-roots, then the same set is obtained again by +taking the powers of $\gamma$. We may say then that the set thus +obtained is the set belonging to $g$. + +\smallskip II.~\emph{If $\lambda(m)>2$ the product of the +$\phi\{\lambda(m)\}$ primitive $\lambda$-roots in the set belonging +to any primitive $\lambda$-root $g$ is congruent to $1$ modulo $m$.} + +These primitive $\lambda$-roots are +\begin{gather*} +g,\ g^{c_1},\ g^{c_2},\ \ldots,\ g^{c_\mu} \\ +\intertext{where} +1,\ c_1,\ c_2,\ \ldots,\ c_\mu \\ +\end{gather*} +are the integers less than $\lambda(m)$ and prime to $\lambda(m)$. +If any one of these is $c$ another is $\lambda(m)-c$, since +$\lambda(m) > 2$. Hence +\begin{gather*} +1 + c_1 + c_2 + \ldots + c_\mu \equiv 0 \bmod \lambda(m). \\ +\intertext{Therefore} +g^{1 + c_1 + c_2 + \ldots + c_\mu} \equiv 1 \bmod m. +\end{gather*} +From this the theorem follows. + +\smallskip \textsc{Corollary.}\emph{The product of all the +primitive $\lambda$-roots modulo $m$ is congruent to $1$ modulo $m$ +when $\lambda(m) > 2$.}\index{Primitive roots!$\lambda$-roots|)} + +\begin{center} +EXERCISES +\end{center} + +\small\begin{enumerate} +\item[1.] If $x_1$ is the largest value of $x$ satisfying the equation +$\lambda(x) = a$, where $a$ is a given integer, then any solution +$x_2$ of the equation is a factor of $x_1$. + +\item[2*.] Obtain an effective rule for solving the equation +$\lambda(x) = a$. + +\item[3*.] Obtain an effective rule for solving the equation +$\phi(x) = a$. + +\item[4.] A necessary and sufficient condition that $a^{P-1} \equiv 1 +\mod P$ for every integer $a$ prime to $P$ is that $P \equiv 1 \mod +\lambda(P)$. + +\item[5.] If $a^{P-1} \equiv 1\mod P$ for every a prime to $P$, then +(1) $P$ does not contain a square factor other than $1$, (2) $P$ +either is prime or contains at least three different prime factors. + +\item[6.] Let $p$ be a prime number. If $a$ is a root of the congruence +$x^q \equiv 1 \mod p$ and $\alpha$ is a root of the congruence +$x^\delta\equiv 1 \mod p$, then $a\alpha$ is a root of the +congruence $x^{d\delta}\equiv 1 \mod p$. If $a$ is a primitive root +of the first congruence and $\alpha$ of the second and if $d$ and +$\delta$ are relatively prime, then $a\alpha$ is a primitive root of +the congruence $x^{d\delta} \equiv 1\mod p$. +\end{enumerate} \normalsize\index{Primitive roots|)} + +\chapter{OTHER TOPICS} + +\section{Introduction}\label{s40} + +The theory of numbers is a vast discipline and no single volume can +adequately treat of it in all of its phases. A short book can serve +only as an introduction; but where the field is so vast such an +introduction is much needed. That is the end which the present +volume is intended to serve; and it will best accomplish this end +if, in addition to the detailed theory already developed, some +account is given of the various directions in which the matter might +be carried further. + +To do even this properly it is necessary to limit the number of +subjects considered. Consequently we shall at once lay aside many +topics of interest which would find a place in an exhaustive +treatise. We shall say nothing, for instance, about the vast domain +of algebraic numbers, even though this is one of the most +fascinating subjects in the whole field of +mathematics.\index{Algebraic numbers} Consequently, we shall not +refer to any of the extensive theory connected with the division of +the circle into equal parts.\index{Circle, Division of} Again, we +shall leave unmentioned many topics connected with the theory of +positive integers; such, for instance, is the frequency of prime +numbers in the ordered system of integers---a subject which contains +in itself an extensive and elegant theory.\index{Prime numbers} + +In \S\S \ref{s41}--\ref{s44} we shall speak briefly of each of the +following topics: theory of quadratic residues, Galois imaginaries, +arithmetic forms, analytical theory of numbers. Each of these alone +would require a considerable volume for its proper development. All +that we can do is to indicate the nature of the problem in each case +and in some cases to give a few of the fundamental results. + +In the remaining three sections we shall give a brief introduction +to the theory of Diophantine equations, developing some of the more +elementary properties of certain special cases. We shall carry this +far enough to indicate the nature of the problem connected with the +now famous Last Theorem of Fermat. The earlier sections of this +chapter are not required as a preliminary to reading this latter +part. + +\section{Theory of Quadratic Residues}\label{s41}% +\index{Quadratic residues|(} + +Let $a$ and $m$ be any two relatively prime integers. In \S +\ref{s31} we agreed to say that $a$ is a quadratic residue modulo +$m$ or a quadratic non-residue modulo $m$ according as the +congruence +\begin{equation*} +x^2 \equiv a \bmod m +\end{equation*} +has or has not a solution. We saw that if $m$ is chosen equal to an +odd prime number $p$, then $a$ is a quadratic residue modulo $p$ or +a quadratic non-residue modulo $p$ according as +\begin{equation*} +a^{\frac{1}{2} (p-1)} \equiv 1\quad \mathrm{or}\quad + a^{\frac{1}{2} (p-1)} \equiv -1 \bmod p. +\end{equation*} +This is known as Euler's criterion.\index{Euler's!criterion} + +It is convenient to employ the Legendre symbol +\begin{equation*} +\left( \frac{a}{p} \right ) +\end{equation*} +to denote the quadratic character of $a$ with respect to $p$.% +\index{Legendre symbol} This symbol is to have the value $+1$ or the +value $-1$ according as $a$ is a quadratic residue modulo $p$ or a +quadratic non-residue modulo $p$. We shall now derive some of the +fundamental properties of this symbol, understanding always that the +numbers in the numerator and the denominator are relatively prime. + +From the definition of quadratic residues and non-residues it is +obvious that +\begin{equation} +\left ( \frac{a}{p} \right ) = \left ( \frac{b}{p} \right ) + \quad \text{if}\quad a \equiv b \bmod p. \tag{1} +\end{equation} + +It is easy to prove in general that +\begin{equation} +\left ( \frac{a}{p} \right ) \left ( \frac{b}{p} \right ) = + \left (\frac {ab}{p} \right ). \tag{2} +\end{equation} +This comes readily from Euler's criterion. We have to consider the +three cases +\begin{align*} +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=+1; & +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=-1; \\ +&& \left( \frac{a}{p} \right ) &=-1,& + \left( \frac{b}{p} \right ) &=-1. +\end{align*} +The method will be sufficiently illustrated by the treatment +of the last case. Here we have +\begin{gather*} +a^{\frac 12 (p-1)}\equiv -1 \bmod p,\quad + b^{\frac 12 (p-1)}\equiv -1 \bmod p. \\ +\intertext{Multiplying these two congruences together member by +member we have} +(ab)^{\frac 12 (p-1)} \equiv 1 \bmod p, \\ +\intertext{whence} +\left( \frac {ab}{p} \right ) = 1 = + \left( \frac ap \right ) \left( \frac bp \right ), +\end{gather*} +as was to be proved. + +If $m$ is any number prime to $p$ and we write $m$ as the product of +factors +\begin{equation*} +m = \epsilon \cdot 2^\alpha \cdot q' q'' q''' \cdots +\end{equation*} +where $q',\ q'',\ q''',\ \ldots$ are odd primes, $\alpha$ is zero or +a positive integer and $\epsilon$ is $+1$ or $-1$ according as $m$ +is positive or negative, we have +\begin{equation} +\left( \frac{m}{p} \right ) = +\left( \frac{\epsilon}{p} \right ) +\left( \frac{2}{p} \right ) ^\alpha +\left( \frac{q'}{p} \right ) +\left( \frac{q''}{p} \right ) +\left( \frac{q'''}{p} \right ) \ldots, \tag{3} +\end{equation} +as one shows easily by repeated application of relation (2). +Obviously, +\begin{equation*} +\left( \frac{1}{p} \right ) = 1. +\end{equation*} +Hence, it follows from (3) that we can readily determine the +quadratic character of $m$ with respect to the odd prime $p$, that +is, the value of +\begin{equation*} +\left( \frac{m}{p} \right ), +\end{equation*} +provided that we know the value of each of the expressions +\begin{equation} +\left( \frac{-1}{p} \right ),\quad + \left( \frac{2}{p} \right ),\quad + \left( \frac{q}{p} \right ),\tag{4} +\end{equation} +where $q$ is an odd prime. + +The first of these can be evaluated at once by means of Euler's +criterion; for, we have +\begin{gather*} +\left( \frac{-1}{p} \right ) \equiv + (-1)^{\frac{1}{2} (p-1)} \bmod p \\ +\intertext{and hence} +\left( \frac{-1}{p} \right ) = (-1)^{\frac{1}{2} (p-1)}. +\end{gather*} +Thus we have the following result: The number $-1$ is a quadratic +residue of every prime number of the form $4k + 1$ and a quadratic +non-residue of every prime number of the form $4k + 3$. + +The value of the second symbol in (4) is given by the formula +\begin{equation*} +\left( \frac{2}{p} \right ) = (-1)^{\frac{1}{8} (p^2 -1)}. +\end{equation*} +The theorem contained in this equation may be stated in the +following words: The number $2$ is a quadratic residue of every +prime number of either of the forms $8k + 1, 8k + 7$; it is a +quadratic non-residue of every prime number of either of the forms +$8k + 3, 8k + 5$. + +The proof of this result is not so immediate as that of the +preceding one. To evaluate the third expression in (4) is still more +difficult. We shall omit the demonstration in both of these cases. +For the latter we have the very elegant relation +\begin{equation*} +\left( \frac{p}{q} \right ) \left( \frac{q}{p} \right ) = + (-1)^{\frac{1}{4}(p-1)(q-1)}. +\end{equation*} +This equation states the law which connects the quadratic character +of $q$ with respect to $p$ with the quadratic character of $p$ with +respect to $q$. It is known as the Law of Quadratic Reciprocity. +About fifty proofs of it have been given. Its history has been a +very interesting one; see Bachmann's Niedere Zablentheorie, Teil I, +pp.\ 180--318, especially pp.\ 200--206.\index{Bachmann}% +\index{Law of quadratic reciprocity}\index{Quadratic reciprocity} + +For a further account of this beautiful and interesting subject we +refer the reader to Bachmann, loc.\ cit., and to the memoirs to +which this author gives reference.\index{Quadratic residues|)} + +\section{Galois Imaginaries}\label{s42}% +\index{Galois imaginaries}\index{Imaginaries of Galois} + +If one is working in the domain of real numbers the equation +\begin{equation*} +x^2 + 1 = 0 +\end{equation*} +has no solution; for there is no real number whose square is $-1$. +If, however, one enlarges the ``number system'' so as to include not +only all real numbers but all complex numbers as well, then it is +true that every algebraic equation has a root. It is on account of +the existence of this theorem for the enlarged domain that much of +the general theory of algebra takes the elegant form in which we +know it. + +The question naturally arises as to whether we can make a similar +extension in the case of congruences. The congruence +\begin{equation*} +x^2 = 3 \bmod 5 +\end{equation*} +has no solution, if we employ the term solution in the sense in +which we have so far used it. But we may if we choose introduce an +imaginary quantity, or mark, $j$ such that +\begin{equation*} +j^2 \equiv 3 \bmod 5, +\end{equation*} +just as in connection with the equation $x^2 + 1 = 0$ we would +introduce the symbol $i$ having the property expressed by the +equation +\begin{equation*} +i^2 = -1. +\end{equation*} + +It is found to be possible to introduce in this way a general set of +imaginaries satisfying congruences with prime moduli; and the new +quantities or marks have the property of combining according to the +laws of algebra. + +The quantities so introduced are called Galois imaginaries. + +We cannot go into a development of the important theory which is +introduced in this way. We shall be content with indicating two +directions in which it leads. + +In the first place there is the general Galois field theory which is +of fundamental importance in the study of certain finite groups. It +may be developed from the point of view indicated here. An excellent +exposition, along somewhat different lines, is to be found in +Dickson's \emph{Linear Groups with an Exposition of the Galois Field +Theory.}\index{Dickson} + +Again, the whole matter may be looked upon from the geometric point +of view. In this way we are led to the general theory of finite +geometries, that is, geometries in which there is only a finite +number of points. For a development of the ideas which arise here +see Veblen and Young's \emph{Projective Geometry} and the memoir by +Veblen and Bussey in the Transactions of the American Mathematical +Society, vol.\ 7, pp.\ 241--259.\index{Bussey}\index{Veblen}% +\index{Young} + +\section{Arithmetic Forms}\label{s43}% +\index{Arithmetic forms|(}\index{Forms|(} + +The simplest arithmetic form is $ax + b$ where $a$ and $b$ are fixed +integers different from zero and $x$ is a variable integer. By +varying $x$ in this case we have the terms of an arithmetic +progression. We have already referred to Dirichlet's celebrated +theorem which asserts that the form $ax + b$ has an infinite number +of prime values if only $a$ and $b$ are relatively +prime.\index{Dirichlet} This is an illustration of one type of +theorem connected with arithmetic forms in general, namely, those in +which it is asserted that numbers of a given form have in addition a +given property.\index{Prime numbers} + +Another type of theorem is illustrated by a result stated in \S +\ref{s41}, provided that we look at that result in the proper way. +We saw that the number $2$ is a quadratic residue of every prime of +either of the forms $8k + 1$ and $8k + 7$ and a quadratic +non-residue of every prime of either of the forms $8k + 3$ and $8k + +5$. We may state that result as follows: A given prime number of +either of the forms $8k + 1$ and $8k + 7$ is a divisor of some +number of the form $x^2 - 2$, where $x$ is an integer; no prime +number of either of the forms $8k + 3$ and $8k + 5$ is a divisor of +a number of the form $x^2 - 2$, where $x$ is an integer. + +The result just stated is a theorem in a discipline of vast extent, +namely, the theory of quadratic forms. Here a large number of +questions arise among which are the following: What numbers can be +represented in a given form? What is the character of the divisors +of a given form? As a special case of the first we have the question +as to what numbers can be represented as the sum of three squares. +To this category belong also the following two theorems: Every +positive integer is the sum of four squares of integers; every prime +number of the form $4n + 1$ may be represented (and in only one way) +as the sum of two squares.\index{Prime numbers} + +For an extended development of the theory of quadratic forms we +refer the reader to Bachmann's Arithmetik der Quadratischen Formen +of which the first part has appeared in a volume of nearly seven +hundred pages.\index{Bachmann} + +It is clear that one may further extend the theory of arithmetic +forms by investigating the properties of those of the third and +higher degrees. Naturally the development of this subject has not +been carried so far as that of quadratic forms; but there is a +considerable number of memoirs devoted to various parts of this +extensive field, and especially to the consideration of various +special forms. + +Probably the most interesting of these special forms are the +following: +\begin{equation*} +\alpha^n + \beta^n , \quad + \frac{\alpha^n - \beta^n}{\alpha - \beta} = + \alpha^{n-1} + \alpha^{n-2} \beta + \cdots + \beta^{n-1}, +\end{equation*} +where $\alpha$ and $\beta$ are relatively prime integers, or, more +generally, where $\alpha$ and $\beta$ are the roots of the quadratic +equation $x^2 - ux + v = 0$ where $u$ and $v$ are relatively prime +integers. A development of the theory of these forms has been given +by the present author in a memoir published in 1913 in the Annals of +Mathematics, vol.\ 13, pp.\ 30--70.% +\index{Arithmetic forms|)}\index{Carmichael}\index{Forms|)}% +\index{Quadratic forms} + +\section{Analytical theory of numbers}\label{s44}% +\index{Analytical theory of numbers|(} + +Let us consider the function +\begin{equation*} +P(x) = \frac{1}{\prod_{k=0}^\infty (1-x^{2^k} )} , \quad + |x|\leqq \rho < 1. +\end{equation*} +It is clear that we have +\begin{align*} +P(x) = \prod_{k=0}^\infty \frac{1}{(1-x^{2^k} )} &= + \prod_{k=0}^\infty + ( 1 + x^{2k} + x^{2\cdot 2^k} + x^{3\cdot 2^k} + \cdots ) \\ +&= \sum_{s=0}^\infty G(s) x^s, +\end{align*} +where $G(0) = 1$ and $G(s)$ (for $s$ greater than $0$) is the number +of ways in which the positive integer $s$ may be separated into like +or distinct summands each of which is a power of $2$. + +We have readily +\begin{equation*} +(1-x)\sum_{s=0}^\infty G(s) x^s = (1-x)P(x) = P(x^2) = + \sum_{s=0}^\infty x^{2^s}; +\end{equation*} +whence +\begin{equation} +G(2s + 1) = G(2s) = G(2s - 1) + G(s), \tag{A} +\end{equation} +as one readily verifies by equating coefficients of like powers of +$x$. From this we have in particular +\begin{gather*} +G(0) = 1, \quad G(1) = 1, \quad G(2) = 2, \quad G(3) = 2, \\ +G(4) = 4, \quad G(5) = 4, \quad G(6) = 6, \quad G(7) = 6. +\end{gather*} +Thus in (A) we have recurrence relations by means of which we may +readily reckon out the values of the number theoretic function +$G(s)$. Thus we may determine the number of ways in which a given +positive integer $s$ may be represented as a sum of powers of $2$. + +We have given this example as an elementary illustration of the +analytical theory of numbers, that is, of that part of the theory of +numbers in which one employs (as above) the theory of a continuous +variable or some analogous theory in order to derive properties of +sets of integers. This general subject has been developed in several +directions. For a systematic account of it the reader is referred to +Bachmann's Analytische Zahlentheorie.% +\index{Analytical theory of numbers|)}\index{Bachmann} + +\section{Diophantine equations}\label{s45}% +\index{Diophantine equations}\index{Equations!Diophantine} + +If $f(x, y, z, \ldots)$ is a polynomial in the variables $x, y, z, +\ldots$ with integral coefficients, then the equation +\begin{equation*} +f(x, y, z, \ldots) = 0 +\end{equation*} +is called a Diophantine equation when we look at it from the point +of view of determining the integers (or the positive integers) $x, +y, z, \ldots$ which satisfy it. Similarly, if we have several such +functions $f_i(x, y, z, \ldots)$, in number less than the number of +variables $x, y, z, \ldots$, then the set of equations +\begin{equation*} +f_i(x, y, z, \ldots) = 0,\quad i = i, 2, \ldots, +\end{equation*} +is said to be a Diophantine system of equations. Any set of integers +$x, y, z, \ldots$ which satisfies the equation [system] is said to +be a solution of the equation [system]. + +We may likewise define Diophantine inequalities by replacing the +sign of equality above by the sign of inequality. But little has +been done toward developing a theory of Diophantine inequalities. +Even for Diophantine equations the theory is in a rather fragmentary +state. + +In the next two sections we shall illustrate the nature of the ideas +and the methods of the theory of Diophantine equations by developing +some of the results for two important special cases. + +\section{Pythagorean triangles}\label{s46}% +\index{Pythagorean triangles|(} + +\textsc{Definitions.} If three positive integers $x, y, z$ satisfy +the relation +\begin{equation} +x^2 + y^2 = z^2 \tag{1} +\end{equation} +they are said to form a Pythagorean triangle or a numerical right +triangle; $z$ is called the hypotenuse of the triangle and $x$ and +$y$ are called its legs. The area of the triangle is said to be +$\frac{1}{2} xy$.\index{Triangles, Numerical} + +We shall determine the general form of the integers $x$, $y$, $z$, +such that equation (1) may be satisfied. Let us denote by $\nu$ the +greatest common divisor of $x$ and $y$ in a particular solution of +(1). Then $\nu$ is a divisor of $z$ and we may write +\begin{equation*} +x = \nu u, \quad y = \nu v,\quad z = \nu w. +\end{equation*} +Substituting these values in (1) and reducing we have +\begin{equation} +u^2 + v^2 = w^2, \tag{2} +\end{equation} +where $u, v, w$ are obviously prime each to each, since $u$ and $v$ +have the greatest common divisor $1$. + +Now an odd square is of the form $4k + 1$. Hence the sum of two odd +squares is divisible by $2$ but not by $4$; and therefore the sum of +two odd squares cannot be a square. Hence one of the numbers $u$, +$v$ is even. Suppose that $u$ is even and write equation (2) in the +form +\begin{equation} +u^2 = (w - v)(w + v). \tag{3} +\end{equation} +Every common divisor of $w - v$ and $w + v$ is a divisor of their +difference $2v$. Therefore, since $w$ and $v$ are relatively prime, +it follows that $2$ is the greatest common divisor of $w - v$ and $w ++ v$. Then from (3) we see that each of these numbers is twice a +square, so that we may write +\begin{equation*} +w - v = 2b^2,\quad w + v = 2a^2 +\end{equation*} +where $a$ and $b$ are relatively prime integers. From these two +equations and equation (3) we have +\begin{equation} +w = a^2 + b^2, \quad v = a^2 -b^2,\quad u = 2ab. \tag{4} +\end{equation} +Since $u$ and $v$ are relatively prime it is evident that one of the +numbers $a$, $b$ is even and the other odd. + +The forms of $u$, $v$, $w$ given in (4) are necessary in order that +(2) may be satisfied. A direct substitution in (2) shows that this +equation is indeed satisfied by these values. Hence we have in (4) +the general solution of (2) where $u$ is restricted to be even. A +similar solution would be obtained if $v$ were restricted to be +even. Therefore \emph{the general solution of (1) is +\begin{gather*} +x = 2\nu ab,\quad y = \nu (a^2 - b^2),\quad z = \nu (a^2 + b^2)\\ +\intertext{and} +x = 2\nu (a^2 - b^2 ),\quad y = 2\nu ab,\quad z = \nu (a^2 + b^2) +\end{gather*} +where $a$, $b$, $\nu$ are arbitrary integers except that $a$ and $b$ +are relatively prime and one of them is even and the other odd.} + +By means of this general solution of (1) we shall now prove the +following theorem: + +\smallskip I.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero, such that} +\begin{equation} +q^2 + n^2 = m^2 , \quad m^2 + n^2 = p^2. \tag{5} +\end{equation} + +It is obvious that an equivalent theorem is the following: + +\smallskip II.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero such that} +\begin{equation} +p^2 + q^2 = 2m^2, \quad p^2 - q^2 = 2n^2. \tag{6} +\end{equation} + +Obviously, we may without loss of generality take $m$, $n$, $p$, $q$ +to be positive; and this we do. + +The method of proof is to assume the existence of integers +satisfying equations (5) and (6) and to show that we are thus led to +a contradiction. The argument we give is an illustration of Fermat's +famous method of ``infinite descent.''% +\index{Descent, Infinite}\index{Fermat}\index{Infinite descent} + +If any two of the numbers $p$, $q$, $m$, $n$ have a common prime +factor $t$, it follows at once from (5) and (6) that all four of +them have this factor. For, consider an equation in (5) or in (6) in +which these two numbers occur; this equation contains a third +number, and it is readily seen that this third number is divisible +by $t$. Then from one of the equations containing the fourth number +it follows that this fourth number is divisible by $t$. Now let us +divide each equation of system (6) through by $t^2$; the resulting +system is of the same form as (6). If any two numbers in this +resulting system have a common prime factor $t_1$, we may divide +through by $t_1^2$; and so on. Hence if a pair of simultaneous +equations (6) exists then there exists a pair of equations of the +same form in which no two of the numbers $m$, $n$, $p$, $q$ have a +common factor other than unity. Let this system of equations be +\begin{equation} +p_1^2 + q_1^2 = 2m_1^2, \quad p_1^2 - q_1^2 = 2n_1^2. \tag{7} +\end{equation} + +From the first equation in (7) it follows that $p_1$ and $q_1$ are +both even or both odd; and, since they are relatively prime, it +follows that they are both odd. Evidently $p_1 > q_1$. Then we may +write +\begin{equation*} +p_1 = q_1 + 2\alpha, +\end{equation*} +where $\alpha$ is a positive integer. If we substitute this value of +$p_1$ in the first equation of (7), the result may readily be put in +the form +\begin{equation} +(q_1 + \alpha)^2 + a^2 = m_1^2. \tag{8} +\end{equation} +Since $q_1$ and $m_1$ have no common prime factor it is easy to see +from this equation that $\alpha$ is prime to both $q_1$ and $m_1$, +and hence that no two of the numbers $q_1 + \alpha, \alpha, m_1$ +have a common factor. + +Now we have seen that if $a$, $b$, $c$ are positive integers no two +of which have a common prime factor, while +\begin{equation*} +a^2 + b^2 = c^2, +\end{equation*} +then there exist relatively prime integers $r$ and $s$, $r > s$, +such that +\begin{gather} +c = r^2 + s^2,\quad a = 2rs,\quad b = r^2 - s^2 \notag \\ +\intertext{or} +c = r^2 + s^2,\quad a = r^2 - s^2,\quad b = 2rs. \notag \\ +\intertext{Hence from (8) we see that we may write} +q_1 + \alpha = 2rs,\quad \alpha = r^2 - s^2 \tag{9} \\ +\intertext{or} +q_1 + \alpha = r^2 - s^2, \alpha = 2rs. \tag{10} \\ +\intertext{In either case we have} +p_1^2 - q_1^2 = (p_1 - q_1)(p_1 + q_1) = + 2\alpha \cdot 2(q_1 + \alpha) = 8rs(r^2 - s^2). \notag \\ +\intertext{If we substitute in the second equation of (7) and divide +by 2 we have} 4rs(r^2 - s^2) = n_1^2. \notag +\end{gather} + +From this equation and the fact that $r$ and $s$ are relatively +prime it follows at once that $r$, $s$, $r^2 - s^2$ are all square +numbers; say, +\begin{gather} +r = u^2,\quad s = v^2,\quad r^2 - s^2 = w^2. \notag \\ +\intertext{Now $r - s$ and $r + s$ can have no common factor other +than 1 or 2; hence from} +w^2 = (r^2-s^2) = (r-s)(r+s) = (u^2-v^2)(u^2+v^2) \notag \\ +\intertext{we see that either} +u^2 + v^2 = 2w_1^2,\quad u^2 - v^2 = 2w_2^2 \tag{11} \\ +\intertext{or} +u^2 + v^2 = w_1^2,\quad u^2 - v^2 = w_2^2. \notag \\ +\intertext{And if it is the latter case which arises, then} +w_1^2 + w_2^2 = 2u^2,\quad w_1^2 - w_2^2 = 2v^2. \tag{12} +\end{gather} +Hence, assuming equations of the form (6) we are led either to +equations (11) or to equations (12); that is, we are led to new +equations of the form with which we started. Let us write the +equations thus: +\begin{equation} +p_2^2 + q_2^2 = 2m_2^2,\quad p_2^2 - q_2^2 = 2n_2^2; \tag{13} +\end{equation} +that is, system (13) is identical with that one of systems (11), +(12) which actually arises. + +Now from (9) and (10) and the relations $p_1 = q_1 + 2\alpha, r +> s$, we see that +\begin{gather*} +p_1 = 2rs + r^2 - s^2 > 2s^2 + r^2 - s^2 = + r^2 + s^2 = u^4 + v^4. \\ +\intertext{Hence $u < p_1$. Also,} +w_1^2 \leqq w^2 \leqq r+s < r^2 + s^2. +\end{gather*} +Hence $w_1 < p_1$. Since $u$ and $w_1$ are both less than $p_1$ it +follows that $p_2$ is less than $p_1$. Hence, obviously, $p_2 < p$. +Moreover, it is clear that all the numbers $p_2, q_2, m_2, n_2$ are +different from zero. + +From these results we have the following conclusion: If we assume a +system of the form (6) we are led to a new system (13) of the same +form; and in the new system $p_2$ is less than $p$. + +Now if we start with (13) and carry out a similar argument +we shall be led to a new system +\begin{gather*} +p_3^2 + q_3^2 = 2m_3^2,\quad p_3^2 - q_3^2 = 2n_3^2, +\end{gather*} +with the relation $p_3 < p_2$, starting from this last system we +shall be led to a new one of the same form, with a similar relation +of inequality; and so on \emph{ad infinitum.} But, since there is +only a finite number of positive integers less than the given +positive integer $p$ this is impossible. We are thus led to a +contradiction; whence we conclude at once to the truth of II and +likewise of I. + +By means of theorems I and II we may readily prove the following +theorem: + +\smallskip III.~\emph{The area of a numerical right triangle is +never a square number.} + +Let the sides and hypotenuse of a numerical right triangle be $u, v, +w$, respectively. The area of this triangle is $\frac{1}{2} uv$. If +we assume this to be a square number $t^2$ we shall have the +following simultaneous Diophantine equations +\begin{equation} +u^2 + v^2 = w^2,\quad uv = 2t^2. \tag{14} +\end{equation} +We shall prove our theorem by showing that the assumption of such a +system leads to a contradiction. + +If any two of the numbers $u, v, w$ have a common prime factor $p$ +then the remaining one also has this factor, as one sees readily +from the first equation in (14). From the second equation in (14) it +follows that $t$ also has the same factor. Then if we put $u = pu_1, +v = pv_1, w = pw_1, t = pt_1$, we have +\begin{equation*} +u_1^2 + v_1^2 = w_1^2,\quad u_1 v_1 = 2t_1^2, +\end{equation*} +a system of the same form as (14). It is clear that we may start +with this new system and proceed in the same manner as before, and +so on, until we arrive at a system +\begin{equation} +\bar{u}^2 + \bar{v}^2 = \bar{w}^2,\quad + \bar{u}\bar{v} = 2\bar{t}^2, \tag{15} +\end{equation} +where $\bar{u}$, $\bar{v}$, $\bar{w}$ are prime each to each. + +Now the general solution of the first equation (15) may be written +in one of the forms +\begin{gather*} +\bar{u} = 2ab,\quad \bar{v} = a^2 - b^2,\quad \bar{w} = a^2 + b^2 \\ +\bar{u} = a^2 - b^2,\quad \bar{v} = 2ab, \quad \bar{w} = a^2 + b^2. \\ +\intertext{Then from the second equation in (15) we have} +\bar{t}^2 = ab(a^2 - b^2 ) = ab(a-b)(a+b). +\end{gather*} +It is easy to see that no two of the numbers $a$, $b$, $a - b$, $a + +b$ in the last member of this equation have a common factor; for, if +so, $\bar{u}$ and $\bar{v}$ would have a common factor, contrary to +hypothesis. Hence each of these four numbers is a square. That is, +we have equations of the form +\begin{gather*} +a = m^2,\quad b = n^2,\quad a + b = p^2,\quad a - b = q^2; \\ +\intertext{whence} +m^2 - n^2 = q^2,\quad m^2 + n^2 = p^2. +\end{gather*} +But, according to theorem I, no such system of equations can exist. +That is, the assumption of equations (14) leads to a contradiction. +Hence the theorem follows as stated above.% +\index{Pythagorean triangles|)} + +\section{The Equation $x^n + y^n = z^n$.}\label{s47}% +\index{Equation $x^n + y^n = z^n$|(}\index{Fermat's!last theorem} + +The following theorem, which is commonly known as Fermat's Last +Theorem, was stated without proof by Fermat in the seventeenth +century: + +\smallskip\emph{If n is an integer greater than 2 there do not exist +integers x, y, z, all different from zero, such that} +\begin{equation} +x^n + y^n = z^n. \tag{1} +\end{equation} + +No general proof of this theorem has yet been given. For various +special values of $n$ the proof has been found; in particular, for +every value of $n$ not greater than 100. + +In the study of equation (1) it is convenient to make some +preliminary reductions. If there exists any particular solution of +(1) there exists also a solution in which $x$, $y$, $z$ are prime +each to each, as one may show readily by the method employed in the +first part of \S \ref{s46}. Hence in proving the impossibility of +equation (1) it is sufficient to treat only the case in which $x$, +$y$, $z$ are prime each to each. + +Again, since $n$ is greater than 2 it must contain the factor +4 or an odd prime factor $p$. If $n$ contains the factor $p$ we write +$n = mp$, whence we have +\begin{gather*} +(x^m)^p + (y^m)^p = (z^m)^p). \\ +\intertext{If $n$ contains the factor 4 we write $n = 4m$, whence we +have} +(x^m)^4 + (y^m)^4 = (z^m)^4. +\end{gather*} +From this we see that in order to prove the impossibility of (1) in +general it is sufficient to prove it for the special cases when $n$ +is 4 and when $n$ is an odd prime $p$. For the latter case the proof +has not been found. For the former case we give a proof below. The +theorem may be stated as follows: + +\smallskip I.~\emph{There are no integers $x, y, z$, all different +from zero, such that} +\begin{equation*} +x^4 + y^4 = z^4. +\end{equation*} + +This is obviously a special case of the more general theorem: + +\smallskip II.~\emph{There are no integers $p$, $q$, $\alpha$, all +different from zero, such that} +\begin{equation} +p^4 - q^4 = \alpha^2. \tag{2} +\end{equation} + +The latter theorem is readily proved by means of theorem III of \S +\ref{s46}. For, if we assume an equation of the form (2), we have +\begin{gather} +(p^4 - q^4)p^2 q^2 = p^2 q^2 \alpha^2. \tag{3} \\ +\intertext{But, obviously,} +(2p^2 q^2)^2 + (p^4 - q^4)^2 = (p^4 + q^4)^2. \tag{4} +\end{gather} +Now, from (3) we see that the numerical right triangle determined by +(4) has its area $p^2 q^2(p^4 - q^4)$ equal to the square number +$p^2 q^2 \alpha^2$. But this is impossible. Hence no equation of the +form (2) exists. + +\begin{center} +EXERCISES +\end{center} + +\begin{enumerate} +\item[1.] Show that the equation $\alpha^4 + 4\beta^4 = \gamma^2$ is +impossible in integers $\alpha$, $\beta$, $\gamma$ all of which are +different from zero. + +\item[2.] Show that the system $p^2 - q^2 = km^2$, $p^2 + q^2 = kn^2$ +impossible in integers $p$, $q$, $k$, $m$, $n$, all of which are +different from zero. + +\item[3*.] Show that neither of the equations $m^4 - 4n^4 = \pm t^2$ +is possible in integers $m$, $n$, $t$, all of which are different +from zero. + +\item[4*.] Prove that the area of a numerical right triangle is not +twice a square number. + +\item[5*.] Prove that the equation $m^4 + n^4 = \alpha^2$ is not +possible in integers $m$, $n$, $\alpha$ all of which are different +from zero. + +\item[6*.] In the numerical right triangle $a^2 + b^2 = c^2$, +not more than one of the numbers $a$, $b$, $c$ is a square. + +\item[7.] Prove that the equation $x^{2k} + y^{2k} = z^{2k}$ implies +an equation of the form $m^k + n^k = 2^{k-2} t^k$. + +\item[8.] Find the general solution in integers of the equation +$x^2 + 2y^2 = t^2$. + +\item[9.] Find the general solution in integers of the equation +$x^2 + y^2 = z^4$. + +\item[10.] Obtain solutions of each of the following Diophantine +equations: +\begin{align*} +x^3 + y^3 + z^3 &= 2t^3, \\ +x^3 + 2y^3 + 3z^3 &= t^3, \\ +x^4 + y^4 + 4z^4 &= t^4, \\ +x^4 + y^4 + z^4 &= 2t^4. +\end{align*} +\end{enumerate}\index{Equation $x^n + y^n = z^n$|)} + +\addcontentsline{toc}{chapter}{Index} +\printindex + + +\newpage +\chapter{PROJECT GUTENBERG "SMALL PRINT"} +\small +\pagenumbering{gobble} + +*** END OF THE PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +***** This file should be named 13693-pdf.pdf or 13693-pdf.zip ***** +This and all associated files of various formats will be found in: + https://www.gutenberg.org/1/3/6/9/13693/ + +Updated editions will replace the previous one--the old editions will +be renamed. + +Creating the works from print editions not protected by U.S. copyright +law means that no one owns a United States copyright in these works, +so the Foundation (and you!) can copy and distribute it in the +United States without permission and without paying copyright +royalties. 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Carmichael + +This eBook is for the use of anyone anywhere at no cost and with +almost no restrictions whatsoever. You may copy it, give it away or +re-use it under the terms of the Project Gutenberg License included +with this eBook or online at www.gutenberg.org + + +Title: The Theory of Numbers + +Author: Robert D. Carmichael + +Release Date: April 8, 2013 [EBook #13693] + +Language: English + +Character set encoding: TeX + +*** START OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +Produced by David Starner, Joshua Hutchinson, John Hagerson, +\end{verbatim} +\normalsize +\newpage + +\frontmatter + +\begin{center} +\noindent \Large MATHEMATICAL MONOGRAPHS \\ + +\bigskip \footnotesize{EDITED BY} \\ +\normalsize \textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD.} \\ + +\bigskip\bigskip \huge +No. 13. + +\bigskip\bigskip \huge THE THEORY \\ +\bigskip\small \textsc{of} \\ +\bigskip\huge NUMBERS \\ + +\bigskip\bigskip\footnotesize\textsc{by} \\ +\bigskip\large ROBERT D. CARMICHAEL, \\ +\footnotesize\textsc{Associate Professor of Mathematics in Indiana +University} + +\bigskip\bigskip\normalsize NEW YORK: \\ +\medskip JOHN WILEY \& SONS. \\ +\medskip \textsc{London: CHAPMAN \& HALL, Limited.} \\ +\medskip 1914. + +\bigskip\bigskip +\tiny \textsc{Copyright 1914} \\ +\textsc{by} \\ +ROBERT D. CARMICHAEL. \\ +\medskip \textsc{the scientific press} \\ +\textsc{robert drummond and company} \\ +\textsc{brooklyn, n.~y.} +\end{center} + +\bigskip\bigskip +\scriptsize \noindent \textsc{Transcriber's Note:} \emph{I did my +best to recreate the index.} \normalsize + +\newpage + +\fbox{\parbox{11cm}{ +\begin{center} +\textbf{MATHEMATICAL MONOGRAPHS.} \\ +\small\textsc{edited by}\normalsize \\ +\textbf{Mansfield Merriman and Robert S. Woodward.} \\ +\footnotesize \textbf{Octavo. Cloth. \$1.00 each.} \\ + +\bigskip \textbf{No. 1. History of Modern Mathematics.} \\ +By \textsc{David Eugene Smith.} + +\smallskip \textbf{No. 2. Synthetic Projective Geometry.} \\ +By \textsc{George Bruce Halsted.} + +\smallskip \textbf{No. 3. Determinants.} \\ +By \textsc{Laenas Gifford Weld.} + +\smallskip \textbf{No. 4. Hyperbolic Functions.} \\ +By \textsc{James McMahon.} + +\smallskip \textbf{No. 5. Harmonic Functions.} \\ +By \textsc{William E. Byerly.} + +\smallskip \textbf{No. 6. Grassmann's Space Analysis.} \\ +By \textsc{Edward W. Hyde.} + +\smallskip \textbf{No. 7. Probability and Theory of Errors.} \\ +By \textsc{Robert S. Woodward.} + +\smallskip \textbf{No. 8. Vector Analysis and Quaternions.} \\ +By \textsc{Alexander Macfarlane.} + +\smallskip \textbf{No. 9. Differential Equations.} \\ +By \textsc{William Woolsey Johnson.} + +\smallskip \textbf{No. 10. The Solution of Equations.} \\ +By \textsc{Mansfield Merriman.} + +\smallskip \textbf{No. 11. Functions of a Complex Variable.} \\ +By \textsc{Thomas S. Fiske.} + +\smallskip \textbf{No. 12. The Theory of Relativity.} \\ +By \textsc{Robert D. Carmichael.} + +\smallskip \textbf{No. 13. The Theory of Numbers.} \\ +By \textsc{Robert D. Carmichael.} \normalsize + +\bigskip \small PUBLISHED BY \\ +\smallskip \textbf{JOHN WILEY \& SONS, Inc., NEW YORK. \\ +CHAPMAN \& HALL, Limited, LONDON.} +\end{center}}} + +\chapter{Editors' Preface.} + +The volume called Higher Mathematics, the third edition of which was +published in 1900, contained eleven chapters by eleven authors, each +chapter being independent of the others, but all supposing the +reader to have at least a mathematical training equivalent to that +given in classical and engineering colleges. The publication of that +volume was discontinued in 1906, and the chapters have since been +issued in separate Monographs, they being generally enlarged by +additional articles or appendices which either amplify the former +presentation or record recent advances. This plan of publication was +arranged in order to meet the demand of teachers and the convenience +of classes, and it was also thought that it would prove advantageous +to readers in special lines of mathematical literature. + +It is the intention of the publishers and editors to add other +monographs to the series from time to time, if the demand seems to +warrant it. Among the topics which are under consideration are those +of elliptic functions, the theory of quantics, the group theory, the +calculus of variations, and non-Euclidean geometry; possibly also +monographs on branches of astronomy, mechanics, and mathematical +physics may be included. It is the hope of the editors that this +Series of Monographs may tend to promote mathematical study and +research over a wider field than that which the former volume has +occupied. + +\chapter{Preface} + +The purpose of this little book is to give the reader a convenient +introduction to the theory of numbers, one of the most extensive and +most elegant disciplines in the whole body of mathematics. The +arrangement of the material is as follows: The first five chapters +are devoted to the development of those elements which are essential +to any study of the subject. The sixth and last chapter is intended +to give the reader some indication of the direction of further study +with a brief account of the nature of the material in each of the +topics suggested. The treatment throughout is made as brief as is +possible consistent with clearness and is confined entirely to +fundamental matters. This is done because it is believed that in +this way the book may best be made to serve its purpose as an +introduction to the theory of numbers. + +Numerous problems are supplied throughout the text. These have been +selected with great care so as to serve as excellent exercises for +the student's introductory training in the methods of number theory +and to afford at the same time a further collection of useful +results. The exercises marked with a star are more difficult than +the others; they will doubtless appeal to the best students. + +Finally, I should add that this book is made up from the material +used by me in lectures in Indiana University during the past two +years; and the selection of matter, especially of exercises, has +been based on the experience gained in this way. + +\hfill \textsc{R.~D.\ Carmichael.} + +\tableofcontents + +%% CHAPTER I. ELEMENTARY PROPERTIES OF INTEGERS +%% 1. Fundamental Notions and Laws +%% 2. Definition of Divisibility. The Unit +%% 3. Prime Numbers. The Sieve of Eratosthenes +%% 4. The Number of Primes is Infinite +%% 5. The Fundamental Theorem of Euclid +%% 6. Divisibility by a Prime Number +%% 7. The Unique Factorization Theorem +%% 8. The Divisors of an Integer +%% 9. The Greatest Common Factor of Two or More Integers +%% 10. The Least Common Multiple of Two or More Integers +%% 11. Scales of Notation +%% 12. Highest Power of a Prime $p$ Contained in $n!$ +%% 13. Remarks Concerning Prime Numbers +%% +%% CHAPTER II. ON THE INDICATOR OF AN INTEGER +%% 14. Definition. Indicator of a Prime Power +%% 15. The Indicator of a Product +%% 16. The Indicator of Any Positive Integer +%% 17. Sum of the Indicators of the Divisors of a Number +%% +%% CHAPTER III. ELEMENTARY PROPERTIES OF CONGRUENCES +%% 18. Congruences Modulo $m$ +%% 19. Solutions of Congruences by Trial +%% 20. Properties of Congruences Relative to Division +%% 21. Congruences with a Prime Modulus +%% 22. Linear Congruences +%% +%% CHAPTER IV. THE THEOREMS OF FERMAT AND WILSON +%% 23. Fermat's General Theorem +%% 24. Euler's Proof of the Simple Fermat Theorem +%% 25. Wilson's Theorem +%% 26. The Converse of Wilson's Theorem +%% 27. Impossibility of $1\cdot 2\cdot 3\cdot \ldots \cdot +%% \overline{n-1}+1=n^k, n>5$ +%% 28. Extension of Fermat's Theorem +%% 29. On the Converse of Fermat's Simple Theorem +%% 30. Application of Previous Results to Linear Congruences +%% 31. Application of the Preceding Results to the Theory of +%% Quadratic Residues +%% +%% CHAPTER V. PRIMITIVE ROOTS MODULO $m$ +%% 32. Exponent of an Integer Modulo $m$ +%% 33. Another Proof of Fermat's General Theorem +%% 34. Definition of Primitive Roots +%% 35. Primitive Roots Modulo $p$ +%% 36. Primitive Roots Modulo $p^\alpha$, $p$ an Odd Prime +%% 37. Primitive Roots Modulo $2p^\alpha$, $p$ an Odd Prime +%% 38. Recapitulation +%% 39. Primitive $\lambda$-Roots +%% +%% CHAPTER VI. OTHER TOPICS +%% 40. Introduction +%% 41. Theory of Quadratic Residues +%% 42. Galois Imaginaries +%% 43. Arithmetic Forms +%% 44. Analytical Theory of Numbers +%% 45. Diophantine Equations +%% 46. Pythagorean Triangles +%% 47. The Equation $x^n+y^n = z^n$ + +\mainmatter + +\chapter{ELEMENTARY PROPERTIES OF INTEGERS} +\section{Fundamental Notions and Laws}\label{s1}% +\index{Fundamental notions} + +In the present chapter we are concerned primarily with certain +elementary properties of the positive integers 1, 2, 3, 4, \ldots It +will sometimes be convenient, when no confusion can arise, to employ +the word \emph{integer} or the word \emph{number} in the sense of +positive integer. + +We shall suppose that the integers are already defined, either by +the process of counting or otherwise. We assume further that the +meaning of the terms \emph{greater, less, equal, sum, difference, +product} is known. + +From the ideas and definitions thus assumed to be known follow +immediately the theorems: +\begin{table}[h] +\begin{tabular}{rl} + I.\ & The sum of any two integers is an integer. \\ + II.\ & The difference of any two integers is an integer. \\ + III.\ & The product of any two integers is an integer. +\end{tabular} +\end{table} + +Other fundamental theorems, which we take without proof, are +embodied in the following formulas: +\begin{table}[h] +\begin{tabular}{rrcl} + IV.\ & $a + b$ & = & $b + a$. \\ + V.\ & $a \times b$ & = & $b \times a$. \\ + VI.\ & $(a + b) + c$ & = & $a + (b + c)$. \\ + VII.\ & $(a \times b) \times c$ & = & $a \times (b \times c)$. \\ +VIII.\ & $a \times (b + c)$ & = & $a \times b + a \times c$. +\end{tabular} +\end{table} +Here $a$, $b$, $c$ denote any positive integers. + +\newpage +These formulas are equivalent in order to the following five +theorems: addition is commutative; multiplication is commutative; +addition is associative; multiplication is associative; +multiplication is distributive with respect to addition. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove the following relations: +\begin{align*} + 1 + 2 + 3 \ldots + n &= \frac{n(n+1)}{2} \\ + 1 + 3 + 5 + \ldots + (2n - 1) &= n^2, \\ +1^3 + 2^3 + 3^3 + \ldots + n^3 &= \left(\frac{n(n+1)}{2}\right)^2 + = (1+2+\ldots+n)^2. +\end{align*} + +\item[2.] Find the sum of each of the following series: +\begin{align*} +1^2 + 2^2 + 3^2 + &\ldots + n^2, \\ +1^2 + 3^2 + 5^2 + &\ldots + (2n - 1)^2, \\ +1^3 + 3^3 + 5^3 + &\ldots + (2n - 1)^3. +\end{align*} + +\item[3.] Discover and establish the law suggested by the equations +$1^2 = 0 + 1$, $2^2 = 1 + 3$, $3^2 = 3 + 6$, $4^2 = 6 + 10$, +$\ldots$; by the equations $1 = 1^3$, $3 + 5 = 2^3$, $7 + 9 + 11 = +3^3$, $13 + 15 + 17 + 19 = 4^3$, $\ldots$. +\end{enumerate} \normalsize + +\section{Definition of Divisibility. The Unit}\label{s2}% +\index{Divisibility}\index{Unit} + +\textsc{Definitions.} An integer $a$ is said to be divisible by an +integer $b$ if there exists an integer $c$ such that $a = bc$. It is +clear from this definition that $a$ is also divisible by $c$. The +integers $b$ and $c$ are said to be divisors or factors of $a$; and +$a$ is said to be a multiple of $b$ or of $c$. The process of +finding two integers $b$ and $c$ such that $bc$ is equal to a given +integer $a$ is called the process of resolving $a$ into factors or +of factoring $a$; and $a$ is said to be resolved into factors or to +be factored. + +We have the following fundamental theorems: + +\smallskip I.~\emph{If $b$ is a divisor of $a$ and $c$ is a divisor +of $b$, then $c$ is a divisor of $a$.} + +Since $b$ is a divisor of a there exists an integer $\beta$ such +that $a = b\beta$. Since $c$ is a divisor of $b$ there exists an +integer $\gamma$ such that $b = c\gamma$. Substituting this value of +$b$ in the equation $a = b\gamma$ we have $a = c\gamma\beta$. But +from theorem III of \S~\ref{s1} it follows that $\gamma\beta$ is an +integer; hence, $c$ is a divisor of $a$, as was to be proved. + +\smallskip II.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the sum of $a$ and $b$.} + +From the hypothesis of the theorem it follows that integers $\alpha$ +and $\beta$ exist such that +\begin{gather*} +a = c\alpha,\quad b = c\beta. \\ +\intertext{Adding, we have} +a + b = c\alpha + c\beta = c(\alpha + \beta) = c\delta, +\end{gather*} +where $\delta$ is an integer. Hence, $c$ is a divisor of $a+b$. + +\smallskip III.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the difference of $a$ and $b$.} + +The proof is analogous to that of the preceding theorem. + +\smallskip \textsc{Definitions.} If $a$ and $b$ are both divisible +by $c$, then $c$ is said to be a common divisor or a common factor +of $a$ and $b$. Every two integers have the common factor 1. The +greatest integer which divides both $a$ and $b$ is called the +greatest common divisor of $a$ and $b$. More generally, we define in +a similar way a common divisor and the greatest common divisor of +$n$ integers $a_1$, $a_2$, $\ldots$, $a_n$.\index{Common!divisors} + +\smallskip \textsc{Definitions.} If an integer $a$ is a multiple of +each of two or more integers it is called a common multiple of these +integers. The product of any set of integers is a common multiple of +the set. The least integer which is a multiple of each of two or +more integers is called their least common multiple.% +\index{Common!multiples} + +It is evident that the integer $1$ is a divisor of every integer and +that it is the only integer which has this property. It is called +the unit. + +\smallskip \textsc{Definition.} Two or more integers which have no +common factor except $1$ are said to be prime to each other or to be +relatively prime.\index{Relatively prime} + +\smallskip \textsc{Definition.} If a set of integers is such that no +two of them have a common divisor besides $1$ they are said to be +prime each to each.\index{Prime each to each} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove that $n^3 - n$ is divisible by $6$ for every +positive integer $n$. + +\item[2.] If the product of four consecutive integers is increased by +$1$ the result is a square number. + +\item[3.] Show that $2^{4n + 2} + 1$ has a factor different from itself +and $1$ when $n$ is a positive integer. +\end{enumerate} \normalsize + +\section{Prime Numbers. The Sieve of Eratosthenes}\label{s3}% +\index{Eratosthenes}\index{Sieve of Eratosthenes} + +\textsc{Definition.} If an integer $p$ is different from 1 and has +no divisor except itself and 1 it is said to be a prime number or to +be a prime. + +\smallskip \textsc{Definition.} An integer which has at least one +divisor other than itself and 1 is said to be a composite number or +to be composite. + +All integers are thus divided into three classes: +\begin{table}[h] +\begin{tabular}{rl} +1.\ & The unit; \\ +2.\ & Prime numbers; \\ +3.\ & Composite numbers. +\end{tabular} +\end{table}\index{Composite numbers}\index{Prime numbers} + +We have seen that the first class contains only a single number. The +third class evidently contains an infinitude of numbers; for, it +contains all the numbers $2^2, 2^3, 2^4, \ldots$ In the next section +we shall show that the second class also contains an infinitude of +numbers. We shall now show that every number of the third class +contains one of the second class as a factor, by proving the +following theorem: + +\smallskip I.~\emph{Every integer greater than 1 has a prime factor.} + +Let $m$ be any integer which is greater than 1. We have to show that +it has a prime factor. If $m$ is prime there is the prime factor $m$ +itself. If $m$ is not prime we have +\begin{equation*} +m = m_1 m_2 +\end{equation*} +where $m_1$ and $m_2$ are positive integers both of which are less +than $m$. If either $m_1$ or $m_2$ is prime we have thus obtained a +prime factor of $m$. If neither of these numbers is prime, then +write +\begin{equation*} +m_1 = m'_1 m'_2,\quad m'_1 > 1, m'_2 > 1. +\end{equation*} +Both $m'_1$ and $m'_2$ are factors of $m$ and each of them is less +than $m_1$. Either we have not found in $m'_1$ or $m'_2$ a prime +factor of $m$ or the process can be continued by separating one of +these numbers into factors. Since for any given $m$ there is +evidently only a finite number of such steps possible, it is clear +that we must finally arrive at a prime factor of $m$. From this +conclusion, the theorem follows immediately. + +Eratosthenes has given a useful means of finding the prime numbers +which are less than any given integer $m$. It may be described as +follows: + +Every prime except 2 is odd. Hence if we write down every odd number +from 3 up to $m$ we shall have it the list every prime less than $m$ +except 2. Now 3 is prime. Leave it in the list; but beginning to +count from 3 strike out every third number in the list. Thus every +number divisible by 3, except 3 itself, is cancelled. Then begin +from 5 and cancel every fifth number. Then begin from from the next +uncancelled number, namely 7, and strike out every seventh number. +Then begin from the next uncancelled number, namely 11, and strike +out every eleventh number. Proceed in this way up to $m$. The +uncancelled numbers remaining will be the odd primes not greater +than $m$. + +It is obvious that this process of cancellation need not be carried +altogether so far as indicated; for if $p$ is a prime greater than +$\sqrt{m}$, the cancellation of any $p^\text{th}$ number from $p$ +will be merely a repetition of cancellations effected by means of +another factor smaller than $p$, as one my see by the use of the +following theorem. + +\smallskip II.~\emph{An integer $m$ is prime if it has no prime +factor equal or less than $I$, where $I$ is the greatest integer +whose square is equal to or less than $m$.} + +Since $m$ has no prime factor less than $I$, it follows from theorem +I that is has no factor but unity less than $I$. Hence, if $m$ is +not prime it must be the product of two numbers each greater than +$I$; and hence it must be equal to or greater than $(I+1)^2$. This +contradicts the hypothesis on $I$; and hence we conclude that $m$ is +prime. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] By means of the method of Eratosthenes determine the primes +less than 200. +\end{enumerate} +\normalsize + +\section{The Number of Primes is Infinite}\label{s4}% +\index{Prime numbers} + +I.~\emph{The number of primes is infinite.} + +We shall prove this theorem by supposing that the number of primes +is not infinite and showing that this leads to a contradiction. If +the number of primes is not infinite there is a greatest prime +number, which we shall denote by $p$. Then form the number +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p + 1. +\end{equation*} +Now by theorem 1 of \S~\ref{s3} $N$ has a prime divisor $q$. But +every non-unit divisor of $N$ is obviously greater than $p$. Hence +$q$ is greater than $p$, in contradiction to the conclusion that $p$ +is the greatest prime. Thus the proof of the theorem is complete. + +In a similar way we may prove the following theorem: + +\smallskip II.~\emph{Among the integers of the arithmetic +progression $5$, $11$, $17$, $23$, $\ldots$, there is an infinite +number of primes.} + +If the number of primes in this sequence is not infinite there is a +greatest prime number in the sequence; supposing that this greatest +prime number exists we shall denote it by $p$. Then the number $N$, +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p-1, +\end{equation*} +is not divisible by any number less than or equal to $p$. This +number $N$, which is of the form $6n - 1$, has a prime factor. If +this factor is of the form $6k - 1$ we have already reached a +contradiction, and our theorem is proved. If the prime is of the +form $6k_1 + 1$ the complementary factor is of the form $6k_2 - 1$. +Every prime factor of $6k_2 - 1$ is greater than $p$. Hence we may +treat $6k_2 - 1$ as we did $6n - 1$, and with a like result. Hence +we must ultimately reach a prime factor of the form $6k_3 - 1$; for, +otherwise, we should have $6n - 1$ expressed as a product of prime +factors all of the form $6t + 1$---a result which is clearly +impossible. Hence we must in any case reach a contradiction of the +hypothesis. Thus the theorem is proved. + +The preceding results are special cases of the following more +general theorem: + +\smallskip III.~\emph{Among the integers of the arithmetic +progression $a$, $a + d$, $a + 2d$, $a + 3d$, $\ldots$, there is an +infinite number of +primes, provided that $a$ and $b$ are relatively prime.}% +\index{Arithmetic progression} + +For the special case given in theorem II we have an elementary +proof; but for the general theorem the proof is difficult. We shall +not give it here. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Prove that there is an infinite number of primes of the +form $4n - 1$. + +\item[2.] Show that an odd prime number can be represented as the +difference of two squares in one and in only one way. + +\item[3.] The expression $m^p - n^p$, in which $m$ and $n$ are integers +and $p$ is a prime, is either prime to $p$ or is divisible by $p^2$. + +\item[4.] Prove that any prime number except $2$ and $3$ is of one of +the forms $6n + 1$, $6n - 1$. +\end{enumerate}\normalsize + +\section{The Fundamental Theorem of Euclid}\label{s5}% +\index{Euclid, Theorem of} + +\emph{If $a$ and $b$ are any two positive integers there exist +integers $q$ and $r$, $q\stackrel{=}{>} 0, 0 \leqq r < b$, such +that} +\begin{equation*} +a = qb + r. +\end{equation*} + +If $a$ is a multiple of $b$ the theorem is at once verified, $r$ +being in this case $0$. If $a$ is not a multiple of $b$ it must lie +between two consecutive multiples of $b$; that is, there exists a +$q$ such that +\begin{equation*} +qb < a < (q + 1)b. +\end{equation*} +Hence there is an integer $r$, $0 < r < b$, such that $a = qb + r$. +In case $b$ is greater than $a$ it is evident that $q = 0$ and $r = +a$. Thus the proof of the theorem is complete. + +\section{Divisibility by a Prime Number}\label{s6}\index{Prime numbers} + +I.~\emph{If $p$ is a prime number and $m$ is any integer, then $m$ +either is divisible by $p$ or is prime to $p$.} + +This theorem follows at once from the fact that the only divisors of +$p$ are $1$ and $p$. + +\smallskip II.~\emph{The product of two integers each less than a +given prime number $p$ is not divisible by $p$.} + +Let $a$ be a number which is less than $p$ and suppose that $b$ is a +number less than $p$ such that $ab$ is divisible by $p$, and let $b$ +be the least number for which $ab$ is so divisible. Evidently there +exists an integer $m$ such that +\begin{equation*} +mb < p < (m + 1)b. +\end{equation*} +Then $p - mb < b$. Since $ab$ is divisible by $p$ it is clear that +$mab$ is divisible by $p$; so is $ap$ also; and hence their +difference $ap - mab$, $=a(p - mb)$, is divisible by $p$. That is, +the product of $a$ by an integer less than $b$ is divisible by $p$, +contrary to the assumption that $b$ is the least integer such that +$ab$ is divisible by $p$. The assumption that the theorem is not +true has thus led to a contradiction; and thus the theorem is +proved. + +\smallskip III.~\emph{If neither of two integers is divisible by a +given prime number $p$ their product is not divisible by $p$.} + +Let $a$ and $b$ be two integers neither of which is divisible by the +prime $p$. According to the fundamental theorem of Euclid there +exist integers $m$, $n$, $\alpha$, $\beta$ such that +\begin{align*} +a &= mp + \alpha,& 0 &< \alpha < p, \\ +b &= np + \beta, & 0 &< \beta < p. +\end{align*} +Then +\begin{equation*} +ab = (mp + \alpha)(np + \beta) + = (mnp + \alpha + \beta)p + \alpha\beta. +\end{equation*} +If now we suppose $ab$ to be divisible by $p$ we have $\alpha\beta$ +divisible by $p$. This contradicts II, since $\alpha$ and $\beta$ +are less than $p$. Hence $ab$ is not divisible by $p$. + +By an application of this theorem to the continued product of +several factors, the following result is readily obtained: + +\smallskip IV.~\emph{If no one of several integers is divisible by a +given prime $p$ their product is not divisible by $p$.} + +\section{The Unique Factorization Theorem}\label{s7}% +\index{Factorization theorem}\index{Factors} + +I.~\emph{Every integer greater than unity can be represented in one +and in only one way as a product of prime numbers.} + +In the first place we shall show that it is always possible to +resolve a given integer $m$ greater than unity into prime factors by +a finite number of operations. In the proof of theorem I, +\S~\ref{s3}, we showed how to find a prime factor $p_1$ of $m$ by a +finite number of operations. Let us write +\begin{equation*} +m = p_1 m_1. +\end{equation*} +If $m_1$ is not unity we may now find a prime factor $p_2$ of $m_1$. +Then we may write +\begin{equation*} +m = p_1 m_1 = p_1 p_2 m_2. +\end{equation*} +If $m_2$ is not unity we may apply to it the same process as that +applied to $m_1$ and thus obtain a third prime factor of $m$. Since +$m_1 > m_2 > m_3 > \ldots$ it is clear that after a finite number of +operations we shall arrive at a decomposition of $m$ into prime +factors. Thus we shall have +\begin{equation*} +m = p_1 p_2 \ldots p_r +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_r$ are prime numbers. We have thus +proved the first part of our theorem, which says that the +decomposition of an integer (greater than unity) into prime factors +is always possible. + +Let us now suppose that we have also a decomposition of $m$ into +prime factors as follows: +\begin{gather*} +m = q_1 q_2 \ldots q_s. \\ +\intertext{Then we have} +p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_s. +\end{gather*} +Now $p_1$ divides the first member of this equation. Hence it also +divides the second member of the equation. But $p_1$ is prime; and +therefore by theorem IV of the preceding section we see that $p_1$ +divides some one of the factors $q$; we suppose that $p_1$ is a +factor of $q_1$. It must then be equal to $q_1$. Hence we have +\begin{equation*} +p_2 p_3 \ldots p_r = q_2 q_3 \ldots q_s. +\end{equation*} +By the same argument we prove that $p_2$ is equal to some $q$, say +$q_2$. Then we have +\begin{equation*} +p_3 p_4 \ldots p_r = q_3 q_4 \ldots q_s. +\end{equation*} +Evidently the process may be continued until one side of the +equation is reduced to $1$. The other side must also be reduced to +$1$ at the same time. Hence it follows that the two decompositions +of $m$ are in fact identical. + +This completes the proof of the theorem. + +\smallskip The result which we have thus demonstrated is easily the +most important theorem in the theory of integers. It can also be +stated in a different form more convenient for some purposes: + +\smallskip II.~\emph{Every non-unit positive integer $m$ can be +represented in one and in only one way in the form +\begin{equation*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_n$ are different primes and +$\alpha_1$, $\alpha_2$, $\ldots$, $\alpha_n$ are positive integers.}% +\index{Factors} + +This comes immediately from the preceding representation of $m$ in +the form $m = p_1 p_2 \ldots p_r$ by combining into a power of $p_1$ +all the primes which are equal to $p_1$. + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ and $b$ are relatively +prime integers and $c$ is divisible by both $a$ and $b$, then $c$ is +divisible by $ab$.} + +\smallskip \textsc{Corollary 2.}~\emph{If $a$ and $b$ are each prime +to $c$ then $ab$ is prime to $c$.} + +\smallskip \textsc{Corollary 3.}~\emph{If $a$ is prime to $c$ and +$ab$ is divisible by $c$, then $b$ is divisible by $c$.} + +\section{The Divisors of an Integer}\label{s8}% +\index{Divisors of a number|(}\index{Factors} + +The following theorem is an immediate corollary of the results in +the preceding section: + +I.~\emph{All the divisors of $m$, +\begin{gather*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}, \\ +\intertext{are of the form} +p_1^{\beta_1} p_2^{\beta_2} \ldots p_n^{\beta_n},\ + 0 \leqq \beta_i \leqq \alpha_i; +\end{gather*} +and every such number is a divisor of $m$.} + +From this it is clear that every divisor of $m$ is included once and +only once among the terms of the product +\begin{multline*} +(1 + p_1 + p_1^2 + \ldots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \ldots + + p_2^{\alpha_2}) \ldots \\ +(1 + p_n + p_n^2 + \ldots + p_n^{\alpha_n}), +\end{multline*} +when this product is expanded by multiplication. It is obvious that +the number of terms in the expansion is $(\alpha_1 + 1)(\alpha_2 + +1) \ldots (\alpha_n+1)$. Hence we have the theorem: + +\smallskip II.~\emph{The number of divisors of $m$ is} +$(\alpha_1 + 1) (\alpha_2 + 1) \ldots (\alpha_n+1)$. + +Again we have +\begin{equation*} +\prod_i(1 + p_i + p_i^2 + \ldots + p_i^{\alpha_i}) = + \prod_i\frac{p_i^{\alpha_i+1} - 1}{p_i - 1}. +\end{equation*} +Hence, + +\smallskip III.~\emph{The sum of the divisors of $m$ is} +\begin{equation*} +\frac{p_1^{\alpha_1 + 1} - 1}{p_1 - 1} \cdot + \frac{p_2^{\alpha_2 + 1} - 1}{p_2 - 1} \cdot + \ldots \cdot + \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}. +\end{equation*} + +In a similar manner we may prove the following theorem: + +\smallskip IV.~\emph{The sum of the $h^{th}$ powers of the divisors +of $m$ is} +\begin{equation*} +\frac{p_1^{h(\alpha_1 + 1)} - 1}{p_1^h - 1} \cdot + \ldots \cdot + \frac{p_n^{h(\alpha_n + 1)} - 1}{p_n^h - 1}. +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find numbers $x$ such that the sum of the divisors of $x$ +is a perfect square. + +\item[2.] Show that the sum of the divisors of each of the following +integers is twice the integer itself: 6, 28, 496, 8128, 33550336. +Find other integers $x$ such that the sum of the divisors of $x$ is +a multiple of $x$. + +\item[3.] Prove that the sum of two odd squares cannot be a square. + +\item[4.] Prove that the cube of any integer is the difference of the +squares of two integers. + +\item[5.] In order that a number shall be the sum of consecutive +integers, it is necessary and sufficient that it shall not be a +power of 2. + +\item[6.] Show that there exist no integers $x$ and $y$ (zero excluded) +such that $y^2 = 2x^2$. Hence, show that there does not exist a +rational fraction whose square is 2. + +\item[7.] The number $m = p_1^{\alpha_1} p_2^{\alpha_2} \cdots +p_n^{\alpha_n}$, where the $p$'s are different primes and the +$\alpha$'s are positive integers, may be separated into relatively +prime factors in $2^{n-1}$ different ways. + +\item[8.] The product of the divisors of $m$ is $\sqrt{m^v}$ where $v$ +is the number of divisors of $m$. +\end{enumerate} \normalsize\index{Divisors of a number|)} + +\section{The Greatest Common Factor of Two or More +Integers}\label{s9}% +\index{Common!divisors|(}\index{Factors}% +\index{Greatest common factor|(} + +Let $m$ and $n$ be two positive integers such that $m$ is greater +than $n$. Then, according to the fundamental theorem of Euclid, we +can form the set of equations +\begin{align*} +m &= qn + n_1, & 0 &< n_1 < n, \\ +n &= q_1 n_1 + n_2, & 0 &< n_2 < n_1, \\ +n_1 &= q_2 n_2 + n_3, & 0 &< n_3 < n_2, \\ +&\vdots \qquad \vdots &&\vdots \qquad \vdots \\ +n_{k - 2} &= q_{k - 1} n_{k-1} + n_k, & 0 &< n_k < n_{k - 1}, \\ +n_{k - 1} &= q _k n_k. & & +\end{align*} +If $m$ is a multiple of $n$ we write $n = n_0$, $k = 0$, in the +above equations. + +\smallskip \textsc{Definition.} The process of reckoning involved in +determining the above set of equations is called the Euclidian +Algorithm.\index{Euclidian algorithm} + +\smallskip I.~\emph{The number $n_k$ to which the Euclidian +algorithm leads is the greatest common divisor of $m$ and $n$.} + +In order to prove this theorem we have to show two things: + +1)~That $n_k$ is a divisor of both $m$ and $n$; + +2)~That the greatest common divisor $d$ of $m$ and $n$ is a divisor +of $n_k$. + +To prove the first statement we examine the above set of equations, +working from the last to the first. From the last equation we see +that $n_k$ is a divisor of $n_{k-1}$. Using this result we see that +the second member of next to the last equation is divisible by $n_k$ +Hence its first member $n_{k-2}$ must be divisible by $n_k$. +Proceeding in this way step by step we show that $n_2$ and $n_1$, +and finally that $n$ and $m$, are divisible by $n_k$. + +For the second part of the proof we employ the same set of equations +and work from the first one to the last one. Let $d$ be any common +divisor of $m$ and $n$. From the first equation we see that $d$ is a +divisor of $n_1$. Then from the second equation it follows that $d$ +is a divisor of $n_2$. Proceeding in this way we show finally that +$d$ is a divisor of $n_k$. Hence any common divisor, and in +particular the greatest common divisor, of $m$ and $n$ is a factor +of $n_k$. + +This completes the proof of the theorem. + +\smallskip \textsc{Corollary.} \emph{Every common divisor of $m$ and +$n$ is a factor of their greatest common divisor.} + +\smallskip II.~\emph{Any number $n_i$ in the above set of equations +is the difference of multiples of $m$ and $n$.} + +From the first equation we have +\begin{equation*} +n_i = m - qn +\end{equation*} +so that the theorem is true for $i = 1$. We shall suppose that the +theorem is true for every subscript up to $i - 1$ and prove it true +for the subscript $i$. Thus by hypothesis we have\footnote{If $i = +2$ we must replace $n_{i-2}$ by $n$.} +\begin{align*} +n_{i-2} &= \pm(\alpha_{i-2}m - \beta_{i-2}n ), \\ +n_{i-1} &= \mp(\alpha_{i-1}m - \beta_{i-1}n). +\intertext{Substituting in the equation} +n_i &= -q_{i-1}n_{n-1} + n_{i-2} \\ +\intertext{we have a result of the form} +n_i &= \pm (\alpha_i m - \beta_i n). +\end{align*} +From this we conclude at once to the truth of the theorem. + +Since $n_k$ is the greatest common divisor of $m$ and $n$, we have +as a corollary the following important theorem: + +\smallskip III.~\emph{If $d$ is the greatest common divisor of the +positive integers $m$ and $n$, then there exist positive integers +$\alpha$ and $\beta$ such that} +\begin{equation*} +\alpha m - \beta n = \pm d. +\end{equation*} + +If we consider the particular case in which $m$ and $n$ are +relatively prime, so that $d = 1$, we see that there exist positive +integers $\alpha$ and $\beta$ such that $\alpha m - \beta n = \pm +1$. Obviously, if $m$ and $n$ have a common divisor $d$, greater +than $1$, there do not exist integers $\alpha$ and $\beta$ +satisfying this relation; for, if so, $d$ would be a divisor of the +first member of the equation and not of the second. Thus we have the +following theorem: + +\smallskip IV.~\emph{A necessary and sufficient condition that $m$ +and $n$ are relatively prime is that there exist integers $\alpha$ +and $\beta$ such that $\alpha m - \beta n = \pm 1$.} + +The theory of the greatest common divisor of three or more numbers +is based directly on that of the greatest common divisor of two +numbers; consequently it does not require to be developed in detail. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] If $d$ is the greatest common divisor of $m$ and $n$, +then $m / d$ and $n / d$ are relatively prime. + +\item[2.] If $d$ is the greatest common divisor of $m$ and $n$ and +$k$ is prime to $n$, then $d$ is the greatest common divisor of $km$ +and $n$. + +\item[3.] The number of multiplies of $6$ in the sequence $a, 2a, 3a, +\cdots, ba$ is equal to the greatest common divisor of $a$ and $b$. + +\item[4.] If the sum or the difference of two irreducible fractions is +an integer, the denominators of the fractions are equal. + +\item[5.] The algebraic sum of any number of irreducible fractions, +whose denominators are prime each to each, cannot be an integer. + +\item[6*.] The number of divisions to be effected in finding the +greatest common divisor of two numbers by the Euclidian algorithm +does not exceed five times the number of digits in the smaller +number (when this number is written in the usual scale of 10). +\end{enumerate}\normalsize% +\index{Common!divisors|)}\index{Greatest common factor|)} + +\section{The Least Common Multiple of Two or More +Integers}\label{s10}% +\index{Common!multiples|(}\index{Least common multiple|(} + +I.~\emph{The common multiples of two or more numbers are the +multiples of their least common multiple.} + +This may be readily proved by means of the unique factorization +theorem. The method is obvious. We shall, however, give a proof +independent of this theorem. + +Consider first the case of two numbers; denote them by $m$ and $n$ +and their greatest common divisor by $d$. Then we have +\begin{equation*} +m = d\mu, \quad n = d\nu, +\end{equation*} +where $\mu$ and $\nu$ are relatively prime +integers.\index{Common!divisors}\index{Greatest common factor} The +common multiples sought are multiples of $m$ and are all comprised +in the numbers $am=ad\mu$, where $a$ is any integer whatever. In +order that these numbers shall be multiples of $n$ it is necessary +and sufficient that $ad\mu$ shall be a multiple of $d\nu$; that is, +that $a\mu$ shall be a multiple of $\nu$; that is, that $a$ shall be +a multiple of $\nu$, since $\mu$ and $\nu$ are relatively prime. +Writing $a = \delta\nu$ we have as the multiples in question the set +$\delta d\mu\nu$ where $\delta$ is an arbitrary integer. This proves +the theorem for the case of two numbers; for $d\mu\nu$ is evidently +the least common multiple of $m$ and $n$. + +We shall now extend the proposition to any number of integers $m, n, +p, q,\ldots$. The multiples in question must be common multiples of +$m$ and $n$ and hence of their least common multiple $\mu$. Then the +multiples must be multiples of $\mu$ and $p$ and hence of their +least common multiple $\mu_1$. But $\mu_1$ is evidently the least +common multiple of $m, n, p$. Continuing in a similar manner we may +show that every multiple in question is a multiple of $\mu$, the +least common multiple of $m, n, p, q, \ldots$. And evidently every +such number is a multiple of each of the numbers $m, n, p, q, +\ldots$. + +Thus the proof of the theorem is complete. + +When the two integers $m$ and $n$ are relatively prime their +greatest common divisor is $1$ and their least common multiple is +their product. Again if $p$ is prime to both $m$ and $n$ it is prime +to their product $mn$; and hence the least common multiple of $m, n, +p$ is in this case $mnp$. Continuing in a similar manner we have the +theorem: + +\smallskip II.~\emph{The least common multiple of several integers, +prime each to each, is equal to their product.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] In order that a common multiple of $n$ numbers shall be +the least, it is necessary and sufficient that the quotients +obtained by dividing it successively by the numbers shall be +relatively prime. + +\item[2.] The product of $n$ numbers is equal to the product of +their least common multiple by the greatest common divisor of their +products $n - 1$ at a time. + +\item[3.] The least common multiple of $n$ numbers is equal to any +common multiple $M$ divided by the greatest common divisor of the +quotients obtained on dividing this common multiple by each of the +numbers. + +\item[4.] The product of $n$ numbers is equal to the product of their +greatest common divisor by the least common multiple of the products +of the numbers taken $n - 1$ at a time. +\end{enumerate} \normalsize% +\index{Common!multiples|)}\index{Least common multiple|)} + +\section{Scales of Notation}\label{s11}\index{Scales of notation|(} + +I.~\emph{If $m$ and $n$ are positive integers and $n > 1$, then $m$ +can be represented in terms of $n$ in one and in only one way in the +form} +\begin{gather*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1} n + a_h, \\ +\intertext{where} +a_0 \ne 0,\ 0 \leqq a_i < n, \quad i = 0, 1, 2, \ldots, h. +\end{gather*} + +That such a representation of $m$ exists is readily proved by means +of the fundamental theorem of Euclid. For we have +\begin{align*} +m &= n_0 n + a_h, & 0 &\leqq a_h < n, \\ +n_0 &= n_1n + a_{h-1}, & 0 &\leqq a_{h-1} < n, \\ +n_1 &= n_2 n + a_{h-2}, & 0 &\leqq a_{h-2} < n, \\ +\hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots & + \hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots \\ +n_{h-3} &= n_{h-2} n + a_2, & 0 &\leqq a_2 < n, \\ +n_{h-2} &= n_{h-1} n + a_1, & 0 &\leqq a_1 < n, \\ +n_{h-1} &= a_0, & 0 &< a_0 < n. +\end{align*} +If the value of $n_{h-1}$ given in the last of these equations is +substituted in the second last we have +\begin{equation*} +n_{h-2} = a_0n + a_1. +\end{equation*} +This with the preceding gives +\begin{equation*} +n_{h-3} = a_0 n^2 + a_1n + a_2. +\end{equation*} +Substituting from this in the preceding and continuing the process +we have finally +\begin{equation*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1}n + a_h, +\end{equation*} +a representation of $m$ in the form specified in the theorem. + +To prove that this representation is unique, we shall suppose that +$m$ has the representation +\begin{gather*} +m = b_0 n^k + b_1 n^{k-1} + \ldots + b_{k-1}n + b_k, \\ +\intertext{where} +b_0 \ne 0,\ 0 < b_i < n,\quad i=0, 1, 2, \ldots, k, \\ +\intertext{and show that the two representations are identical. We +have} +a_0 n^h + \ldots + a_{h-1} n + a_h = + b_0 n^k + \ldots + b_{k-1} n + b_k. +\intertext{Then} +a_0 n^h + \ldots + a_{h-1} n - + (b_0 n^k + \ldots + b_{k-1} n) = b_k - a_h. +\end{gather*} +The first member is divisible by $n$. Hence the second is also. But +the second member is less than $n$ in absolute value; and hence, in +order to be divisible by $n$, it must be zero. That is, $b_k = a_h$. +Dividing the equation through by $n$ and transposing we have +\begin{equation*} +a_0 n^{h-1} + \ldots + a_{h-2} n - (b_0 n^{k-1} + \ldots + + b_{k-2} n) + = b_{k-1} - a_{h-1}. +\end{equation*} +It may now be seen that $b_{k-1} = a_{h-1}$. It is evident that this +process may be continued until either the $a$'s are all eliminated +from the equation or the $b$'s are all eliminated. But it is obvious +that when one of these sets is eliminated the other is also. Hence, +$h = k$. Also, every $a$ equals the $b$ which multiplies the same +power of $n$ as the corresponding $a$. That is, the two +representations of $m$ are identical. Hence the representation in +the theorem is unique. + +From this theorem it follows as a special case that any positive +integer can be represented in one and in only one way in the scale +of 10; that is, in the familiar Hindoo notation. It can also be +represented in one and in only one way in any other scale. Thus +\begin{equation*} +120759 = 1 \cdot 7^6 + 0 \cdot 7^5 + 1 \cdot 7^4 + 2 \cdot 7^3 + + 0 \cdot 7^2 + 3 \cdot 7^1 + 2. +\end{equation*} +Or, using a subscript to denote the scale of notation, this may be +written +\begin{equation*} +(120759)_{10} = (1012032)_7. +\end{equation*} + +For the case in which $n$ (of theorem I) is equal to 2, the only +possible values for the $a$'s are 0 and 1. Hence we have at once the +following theorem: + +II.~\emph{Any positive integer can be represented in one and in only +one way as a sum of different powers of 2.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Any positive integer can be represented as an aggregate of +different powers of $3$, the terms in the aggregate being combined +by the signs $+$ and $-$ appropriately chosen. + +\item[2.] Let $m$ and $n$ be two positive integers of which $n$ is the +smaller and suppose that $2^k \leq n < 2^{k+1}$. By means of the +representation of $m$ and $n$ in the scale of 2 prove that the +number of divisions to be effected in finding the greatest common +divisor of $m$ and $n$ by the Euclidian algorithm does not exceed +$2k$. +\end{enumerate}\normalsize\index{Scales of notation|)} + +\section{Highest Power of a Prime $p$ Contained in $n!$.}\label{s12}% +\index{Highest power of \emph{p} in \emph{n}"!|(} + +Let $n$ be any positive integer and $p$ any prime number not greater +than $n$. We inquire as to what is the highest power $p^\nu$ of the +prime $p$ contained in $n!$. + +In solving this problem we shall find it convenient to employ the +notation +\begin{equation*} +\left [ \frac{r}{s} \right ] +\end{equation*} to denote the greatest integer $\alpha$ such that +$\alpha s \leq r$. With this notation it is evident that we have +\begin{gather} +\left [ + \frac{\left [ \frac{n}{p} \right ]} + {p} +\right ] = \left [ \frac{n}{p^2} \right ]; \tag{1} \\ +\intertext{and more generally} +\left [ + \frac{\left [ \frac{n}{p^i} \right ]} + {p^j} +\right ] = \left [ \frac{n}{p^{i+j}} \right ]. \notag +\end{gather} + +If now we use $H\{x\}$ to denote the index of the highest power of +$p$ contained in an integer $x$, it is clear that we have +\begin{gather*} +H\{n!\} = + H \left \{ p \cdot 2p \cdot 3p \ldots + \left [ \frac{n}{p} \right ] p \right \}, \\ +\intertext{since only multiples of $p$ contain the factor $p$. +Hence} +H\{n!\} = + \left [ \frac{n}{p} \right ] + + H \left \{ 1 \cdot 2 \ldots \left [ \frac{n}{p} \right ] + \right \}. +\end{gather*} +Applying the same process to the $H$-function in the second member +and remembering relation (1) it is easy to see that we have +\begin{align*} +H\{n!\} &= \left[ \frac{n}{p} \right] + + H\left\{ p \cdot 2p \cdot \ldots \cdot + \left[ \frac{n}{p^2} \right]p\right\} \\ + &= \left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + + H \left\{\cdot 1 \cdot 2 \cdot 3 + \ldots \left[ \frac{n}{p^2} \right] \right\}. \\ +\intertext{Continuing the process we have finally} +H\{n1\} &= \left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] + + \ldots, +\end{align*} +the series on the right containing evidently only a finite number of +terms different from zero. Thus we have the theorem: + +\smallskip I.~\emph{The index of the highest power of a prime $p$ +contained in $n!$ is} +\begin{gather*} +\left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + + \left[ \frac{n}{p^3} \right] + \ldots. +\end{gather*} + +The theorem just obtained may be written in a different form, more +convenient for certain of its applications. Let $n$ be expressed in +the scale of $p$ in the form +\begin{gather*} +n = a_0p^h + a_1p^{h-1} + \ldots + a_{h-1}p + a_h, \\ +\intertext{where} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h. +\end{gather*} +Then evidently +\begin{align*} +\left[ \frac{n}{p} \right] &= a_0p^{h-1} + a_1p^{h-2} + \ldots + + a_{h-2}p + a_{h-1}, \\ +\left[ \frac{n}{p^2} \right] &= a_0p^{h-2} + a_1p^{h-3} + \ldots + + a_{h-2}, \\ +.\ \ .\ \ .\ \ .\ \ &.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ +.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ . +\end{align*} +Adding these equations member by member and combining the second +members in columns as written, we have +\begin{align*} +\left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] &+ + \left[ \frac{n}{p^3} \right] + \ldots \\ +&= \sum_{i=0}^h \frac{a_i(p^{h-i} - 1)}{p - 1} \\ +&= \frac{a_0p^h + a_1p^{h-1} + \ldots + a_h - + (a_0 + a_1 + \ldots + a_h)}{p-1} \\ +&= \frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{align*} +Comparing this result with theorem I we have the following theorem: + +\smallskip II.~\emph{If $n$ is represented in the scale of $p$ in +the form +\begin{gather*} +n = a_0 p^h + a_1 p^{h-1} + \ldots + a_h, \\ +\intertext{where $p$ is prime and} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h, \\ +\intertext{then the index of the highest power of $p$ contained in +$n!$ is} +\frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{gather*}} + +Note the simple form of the theorem for the case $p = 2$; in this +case the denominator $p - 1$ is unity. + +We shall make a single application of these theorems by proving the +following theorem: + +\smallskip III.~\emph{If $n$, $\alpha$, $\beta$, $\ldots$, $\lambda$ +are any positive integers such that $n = \alpha + \beta + \ldots + +\lambda$, then +\begin{equation} +\frac{n!}{\alpha! \beta! \ldots \lambda!} \tag{A} +\end{equation} +is an integer.} + +Let $p$ be any prime factor of the denominator of the fraction (A). +To prove the theorem it is sufficient to show that the index of the +highest power of $p$ contained in the numerator is at least as great +as the index of the highest power of $p$ contained in the +denominator. This index for the denominator is the sum of the +expressions +\begin{equation} + \left . + \begin{gathered} + \left [ \frac{\alpha}{p} \right ] + + \left [ \frac{\alpha}{p^2} \right ] + + \left [ \frac{\alpha}{p^3} \right ] + + \ldots \\ + \left [ \frac{\beta}{p} \right ] + + \left [ \frac{\beta}{p^2} \right ] + + \left [ \frac{\beta}{p^3} \right ] + + \ldots \\ + \vdots \\ + \left [ \frac{\lambda}{p} \right ] + + \left [ \frac{\lambda}{p^2} \right ] + + \left [ \frac{\lambda}{p^3} \right ] + + \ldots + \end{gathered} + \right \} \tag{B} +\end{equation} + +The corresponding index for the numerator is +\begin{equation} +\left [ \frac{n}{p} \right ] + +\left [ \frac{n}{p^2} \right ] + +\left [ \frac{n}{p^3} \right ] + +\ldots \tag{C} +\end{equation} +But, since $n = \alpha + \beta + \ldots + \lambda$, it is evident +that +\begin{equation*} + \left [ \frac{n}{p^r} \right ] \stackrel{=}{>} + \left [ \frac{\alpha}{p^r} \right ] + + \left [ \frac{\beta}{p^r} \right ] + + \ldots + + \left [ \frac{\lambda}{p^r} \right ]. +\end{equation*} +From this and the expressions in (B) and (C) it follows that the +index of the highest power of any prime $p$ in the numerator of (A) +is equal to or greater than the index of the highest power of p +contained in its denominator. The theorem now follows at once. + +\smallskip \textsc{Corollary.}~\emph{The product of $n$ consecutive +integers is divisible by $n!$.} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the highest power of 2 contained in 1000! is +$2^{994}$; in 1900! is $2^{1893}$. Show that the highest power of 7 +contained in 10000! is $7^{1665}$. + +\item[2.] Find the highest power of 72 contained in 1000! + +\item[3.] Show that 1000! ends with 249 zeros. + +\item[4.] Show that there is no number $n$ such that $3^7$ is the +highest power of 3 contained in $n!$. + +\item[5.] Find the smallest number $n$ such that the highest power +of 5 contained in $n!$ is $5^{31}$. What other numbers have the same +property? + +\item[6.] If $n = rs$, $r$ and $s$ being positive integers, show that +$n!$ is divisible by $(r!)^s$ by $(s!)^r$; by the least common +multiple of $(r!)^s$ and $(s!)^r$. + +\item[7.] If $n = \alpha + \beta + pq + rs$, where $\alpha, \beta, p, +q, r, s$, are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha ! \beta ! (q!)^p (s!)^r. +\end{equation*} + +\item[8.] When $m$ and $n$ are two relatively prime positive integers +the quotient +\begin{equation*} +Q = \frac{(m + n + 1)!} {m! n!} +\end{equation*} +as an integer. + +\item[9*.] If $m$ and $n$ are positive integers, then each of the +quotients +\begin{equation*} +Q = \frac{(mn)!} {n! (m!)^n},\quad +Q = \frac{(2m)! (2n)!} {m! n! (m+n)!}, +\end{equation*} +is an integer. Generalize to $k$ integers $m, n, p, \ldots$. + +\item[10*.] If $n = \alpha + \beta + pq + rs$ where $\alpha, \beta, +p, q, r, s$ are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha! \beta! r! p! (q!)^p (s!)^r. +\end{equation*} + +\item[11*.] Show that +\begin{equation*} +\frac{(rst)!} {t! (s!)^t (r!)^{st}}, +\end{equation*} is an integer ($r, s, t$ being positive integers). +Generalize to the case of $n$ integers $r, s, t, u, \ldots$. +\end{enumerate}\normalsize% +\index{Highest power of \emph{p} in \emph{n}"!|)} + +\section{Remarks Concerning Prime Numbers}\label{s13}% +\index{Prime numbers|(} + +We have seen that the number of primes is infinite. But the integers +which have actually been identified as prime are finite in number. +Moreover, the question as to whether a large number, as for instance +$2^{257}-1$, is prime is in general very difficult to answer. Among +the large primes actually identified as such are the following: +\begin{equation*} +2^{61}-1, \quad 2^{75} \cdot 5+1, \quad 2^{89}-1, \quad 2^{127}-1. +\end{equation*} + +\emph{No analytical expression for the representation of prime +numbers has yet been discovered.} Fermat believed, though he +confessed that he was unable to prove, that he had found such an +analytical expression in +\begin{equation*} +2^{2^n} + 1. +\end{equation*} +Euler showed the error of this opinion by finding that $641$ is a +factor of this number for the case when $n = 5$.% +\index{Euler}\index{Fermat} + +The subject of prime numbers is in general one of exceeding +difficulty. In fact it is an easy matter to propose problems about +prime numbers which no one has been able to solve. Some of the +simplest of these are the following: + +\begin{enumerate} +\item Is there an infinite number of pairs of primes differing by +2? +\item Is every even number (other than 2) the sum of two primes or +the sum of a prime and the unit? +\item Is every even number the difference of two primes or the +difference of 1 and a prime number? +\item To find a prime number greater than a given prime. +\item To find the prime number which follows a given prime. +\item To find the number of primes not greater than a given number. +\item To compute directly the $n^\text{th}$ prime number, when $n$ +is given. +\end{enumerate}\index{Prime numbers|)} + +\chapter{ON THE INDICATOR OF AN INTEGER}% +\index{Indicator|(} + +\section{Definition. Indicator of a Prime Power}\label{s14}% +\index{Indicator!of a prime power} + +\emph{Definition.} If $m$ is any given positive integer the number +of positive integers not greater than $m$ and prime to it is called +the indicator of $m$. It is usually denoted by $\phi(m)$, and is +sometimes called Euler's $\phi$-function of $m$.% +\index{Euler's!$\phi$-function}\index{$\phi(m)$} More rarely, it has +been given the name of totient of $m$.\index{Totient} + +As examples we have +\begin{equation*} +\phi(1) = 1,\ \phi(2) = 1,\ \phi(3) = 2,\ \phi(4) = 2. +\end{equation*} + +If $p$ is a prime number it is obvious that +\begin{equation*} +\phi(p) = p - 1; +\end{equation*} +for each of the integers 1, 2, 3, $\ldots$, $p-1$ is prime to $p$. + +Instead of taking $m = p$ let us assume that $m = p^\alpha$, where +$\alpha$ is a positive integer, and seek the value of +$\phi(p^\alpha)$. Obviously, every number of the set 1, 2, 3, +$\ldots$, $p^\alpha$ either is divisible by $p$ or is prime to +$p^\alpha$. The number of integers in the set divisible by $p$ is +$p^{\alpha - 1}$. Hence $p^\alpha-p^{\alpha-1}$ of them are prime to +$p$. Hence $\phi(p^\alpha) = p^\alpha-p^{\alpha-1}$. Therefore + +\emph{If $p$ is any prime number and $\alpha$ is any positive +integer, then} +\begin{equation*} +\phi(p^\alpha) = p^\alpha \left ( 1 - \frac{1}{p} \right ). +\end{equation*} + +\section{The Indicator of a Product}\label{s15}% +\index{Indicator!of a product|(} + +I.~\emph{If $\mu$ and $\nu$ are any two relatively prime positive +integers, then} +\begin{equation*} +\phi(\mu\nu) = \phi(\mu) \phi(\nu). +\end{equation*} + +In order to prove this theorem let us write all the integers up to +$\mu\nu$ in a rectangular array as follows: +\footnotesize\begin{equation} + \left . + \begin{aligned} + 1 && 2 && 3 && + \ldots && h && \ldots && \mu \\ + \mu + 1 && \mu + 2 && \mu + 3 && + \ldots && \mu + h && \ldots && 2\mu \\ + 2 \mu + 1 && 2 \mu + 2 && 2 \mu + 3 && + \ldots && 2 \mu + h && \ldots && 3\mu \\ + \vdots && \vdots && \vdots && + && \vdots && && \vdots \\ + (\nu - 1)\mu + 1 && (\nu - 1)\mu + 2 && (\nu - 1)\mu + 3 && + \ldots && (\nu - 1)\mu + h && \ldots && \nu\mu \\ + \end{aligned} + \right \} \tag{A} +\end{equation}\normalsize + +If a number $h$ in the first line of this array has a factor in +common with $\mu$ then every number in the same column with $h$ has +a factor in common with $\mu$. On the other hand if $h$ is prime to +$\mu$, so is every number in the column with $h$ at the top. But the +number of integers in the first row prime to $\mu$ is $\phi(\mu)$. +Hence the number of columns containing integers prime to $\mu$ is +$\phi(\mu)$ and every integer in these columns is prime to $\mu$. + +Let us now consider what numbers in one of these columns are prime +to $\nu$; for instance, the column with $h$ at the top. We wish to +determine how many integers of the set +\begin{gather*} +h,\ \mu + h,\ 2\mu + h,\ \ldots,\ (\nu - 1)\mu + h \\ +\intertext{are prime to $\nu$. Write} +s\mu + h = q_s\nu + r_s +\end{gather*} where s ranges over the numbers $s = 0,\ 1,\ 2,\ +\ldots,\ \nu - 1$ and $0\leqq r_s < \nu$. Clearly $s\mu + h$ is or +is not prime to $\nu$ according as $r_s$ is or is not prime to +$\nu$. Our problem is then reduced to that of determining how many +of the quantities $r_s$ are prime to $\nu$. + +First let us notice that all the numbers $r_s$ are different; for, +if $r_s = r_t$ then from +\begin{equation*} +s\mu + h = q_s\nu + r_s,\quad t\mu + h = q_t\nu + r_t, +\end{equation*} +we have by subtraction that $(s-t)\mu$ is divisible by $\nu$. But +$\mu$ is prime to $\nu$ and $s$ and $t$ are each less than $\nu$. +Hence $(s-t)\mu$ can be a multiple of $\nu$ only by being zero; that +is, $s$ must equal $t$. Hence no two of the remainders $r_s$ can be +equal. + +Now the remainders $r_s$ are $\nu$ in number, are all zero or +positive, each is less than $\nu$, and they are all distinct. Hence +they are in some order the numbers 0, 1, 2, $\ldots$, $\nu-1$. The +number of integers in this set prime to $\nu$ is evidently +$\phi(\nu)$. + +Hence it follows that in any column of the array (A) in which the +numbers are prime to $\mu$ there are just $\phi(\nu)$ numbers which +are prime to $\nu$. That is, in this column there are just +$\phi(\nu)$ numbers which are prime to $\mu\nu$. But there are +$\phi(\mu)$ such columns. Hence the number of integers in the array +(A) prime to $\mu\nu$ is $\phi(\mu)\phi(\nu)$. + +But from the definition of the $\phi$-function it follows that the +number of integers in the array (A) prime to $\mu\nu$ is +$\phi(\mu\nu).$ Hence, +\begin{equation*} +\phi(\mu\nu) = \phi(\mu)\phi(\nu), +\end{equation*} which is the theorem to be proved. + +\smallskip \textsc{Corollary.}~\emph{In the series of $n$ +consecutive terms of an arithmetical progression the common +difference of which is prime to $n$, the number of terms prime to +$n$ is $\phi(n)$.} + +From theorem I we have readily the following more general result: + +\smallskip II.~\emph{If $m_1, m_2, \ldots, m_k$ are $k$ positive +integers which are prime each to each, then} +\begin{equation*} +\phi(m_1 m_2 \ldots m_k) = \phi(m_1) \phi(m_2) \ldots \phi(m_k). +\end{equation*}\index{Indicator!of a product|)} + +\section{The Indicator of any Positive Integer}\label{s16}% +\index{Indicator!of any integer|(} + +From the results of \S\S \ref{s14} and \ref{s15} we have an +immediate proof of the following fundamental theorem: + +\emph{If $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}$ +where $p_1, p_2, \ldots, p_n$ are different primes and $\alpha_1, +\alpha_2, \ldots, \alpha_n$ are positive integers, then} +\begin{equation*} +\phi(m) = m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{equation*} + +For, +\begin{align*} +\phi(m) &= \phi(p_1^{\alpha_1}) \phi(p_2^{\alpha_2}) \ldots + \phi(p_n^{\alpha_n}) \\ + &= p_1^{\alpha_1} \left ( 1-\frac{1}{p_1} \right ) + p_2^{\alpha_2} \left ( 1-\frac{1}{p_2} \right ) + \ldots + p_n^{\alpha_n} \left ( 1-\frac{1}{p_n} \right ) \\ + &= m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{align*} + +On account of the great importance of this theorem we shall give a +second demonstration of it. + +It is clear that the number of integers less than $m$ and divisible +by $p_1$ is +\begin{gather*} +\frac{m}{p_1}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_2$ is} +\frac{m}{p_2}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_1 p_2$ is} +\frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and divisible +by either $p_1$ or $p_2$ is} +\frac{m}{p_1} + \frac{m}{p_2} - \frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and prime to +$p_1 p_2$ is} +m - \frac{m}{p_1} - \frac{m}{p_2} + \frac{m}{p_1 p_2} = + m \left ( 1-\frac{1}{p_1} \right ) \left ( 1-\frac{1}{p_2} \right ). +\end{gather*} + +We shall now show that if the number of integers less than $m$ and +prime to $p_1 p_2 \ldots p_i$, where $i$ is less than $n$, is +\begin{gather*} +m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_i} \right ), \\ +\intertext{then the number of integers less than $m$ and prime to +$p_1 p_2 \ldots p_i p_{i+1}$ is} + m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_{i+1}} \right ). +\end{gather*} +From this our theorem will follow at once by induction. + +From our hypothesis it follows that the number of integers less than +$m$ and divisible by at least one of the primes $p_1$, $p_2$, +$\ldots$, $p_i$ is +\begin{gather} +m - + m \left (1 - \frac{1}{p_1}\right ) + \ldots + \left (1 - \frac{1}{p_i}\right ), \notag \\ +\intertext{or} +\sum \frac{m}{p_1} - \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - \ldots, \tag{A} +\end{gather} +where the summation in each case runs over all numbers of the type +indicated, the subscripts of the $p$'s being equal to or less than +$i$. + +Let us consider the integers less than $m$ and having the factor +$p_{i+1}$ but not having any of the factors $p_1$, $p_2$, $\ldots$, +$p_i$. Their number is +\begin{gather} +\frac{m}{p_{i+1}} - + \frac{1}{p_{i+1}} \left \{ + \sum \frac{m}{p_1} - + \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - + \ldots + \right \}, \tag{B} +\end{gather} +where the summation signs have the same significance as before. For +the number in question is evidently $\frac{m}{p_{i+1}}$ \emph{minus} +the number of integers not greater than $\frac{m}{p_{i+1}}$ and +divisible by at least one of the primes $p_1$, $p_2$, $\ldots$, +$p_i$. + +If we add (A) and (B) we have the number of integers less than $m$ +and divisible by one at least of the numbers $p_1$, $p_2$, $\ldots$, +$p_{i+1}$. Hence the number of integers less than $m$ and prime to +$p_1$, $p_2$, $\ldots$, $p_{i+1}$ is +\begin{gather*} +m - + \sum \frac{m}{p_1} + + \sum \frac{m}{p_1 p_2} - + \sum \frac{m}{p_1 p_2 p_3} + + \ldots, \\ +\intertext{where now in the summations the subscripts run from 1 to +$i+1$. This number is clearly equal to} +m + \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_{i+1}} \right ). +\end{gather*} +From this result, as we have seen above, our theorem follows at once +by induction.\index{Indicator!of any integer|)} + +\section{Sum of the Indicators of the Divisors of a Number}% +\label{s17} + +We shall first prove the following lemma: + +\smallskip \emph{Lemma. If $d$ is any divisor of $m$ and $m = nd$, +the number of integers not greater than $m$ which have with $m$ the +greatest common divisor $d$ is $\phi(n)$.} + +Every integer not greater than $m$ and having the divisor $d$ is +contained in the set $d$, $2d$, $3d$, $\ldots$, $nd$. The number of +these integers which have with $m$ the greatest common divisor $d$ +is evidently the same as the number of integers of the set 1, 2, +$\ldots$, $n$ which are prime to $\frac{m}{d}$, or $n$; for $\alpha +d$ and $n$ have or have not the greatest common divisor $d$ +according as $\alpha$ is or is not prime to $\frac{m}{d}=n$. Hence +the number in question is $\phi(n)$. + +From this lemma follows readily the proof of the following theorem: + +\smallskip \emph{If $d_1$, $d_2$, $\ldots$, $d_r$ are the different +divisors of $m$, then} +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = m. +\end{equation*} + +Let us define integers $m_1$, $m_2$, $\ldots$, $m_r$ by the +relations +\begin{equation*} +m = d_1 m_1 = d_2 m_2 = \ldots = d_r m_r. +\end{equation*} +Now consider the set of $m$ positive integers not greater than $m$, +and classify them as follows into $r$ classes. Place in the first +class those integers of the set which have with $m$ the greatest +common divisor $m_1$; their number is $\phi(d_1)$, as may be seen +from the lemma. Place in the second class those integers of the set +which have with $m$ the greatest common divisor $m_2$; their number +is $\phi(d_2)$. Proceeding in this way throughout, we place finally +in the last class those integers of the set which have with $m$ the +greatest common divisor $m_r$; their number is $\phi(d_r)$. It is +evident that every integer in the set falls into one and into just +one of these $r$ classes. Hence the total number $m$ of integers in +the set is $\phi(d_1) + \phi(d_r) + \ldots + \phi(d_r)$. From this +the theorem follows immediately. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the indicator of any integer greater than $2$ +is even. + +\item[2.] Prove that the number of irreducible fractions not greater +than $1$ and with denominator equal to $n$ is $\phi(n)$. + +\item[3.] Prove that the number of irreducible fractions not greater +than $1$ and with denominators not greater than $n$ is +\begin{equation*} +\phi(1) + \phi(2) + \phi(3) + \ldots + \phi(n). +\end{equation*} + +\item[4.] Show that the sum of the integers less than $n$ and prime to +$n$ is $\frac{1}{2} n \phi(n)$ if $n > 1$. + +\item[5.] Find ten values of $x$ such that $\phi(x) = 24$. + +\item[6.] Find seventeen values of $x$ such that $\phi(x) = 72$. + +\item[7.] Find three values of $n$ for which there is no $x$ satisfying +the equation $\phi(x) = 2n$. + +\item[8.] Show that if the equation +\begin{equation*} +\phi(x) = n +\end{equation*} +has one solution it always has a second solution, $n$ being given +and $x$ being the unknown. + +\item[9.] Prove that all the solutions of the equation +\begin{equation*} +\phi(x) = 4n - 2, n > 1, +\end{equation*} +are of the form $p^\alpha$ and $2p^\alpha$, where $p$ is a prime of +the form $4s-1$. + +\item[10.] How many integers prime to $n$ are there in the set +\begin{enumerate} +\item $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots, n(n+1)$? +\item $1 \cdot 2 \cdot 3, 2 \cdot 3 \cdot 4, + 3 \cdot 4 \cdot 5, \ldots, n(n+1)(n+2)$? +\item $\frac{1 \cdot 2}{2}, \frac{2 \cdot 3}{2}, + \frac{3 \cdot 4}{2}, \ldots, \frac{n(n+1)}{2}$? +\item $\frac{1 \cdot 2 \cdot 3}{6}, + \frac{2 \cdot 3 \cdot 4}{6}, + \frac{3 \cdot 4 \cdot 5}{6}, + \ldots, + \frac{n(n+1)(n+2)}{6}$? +\end{enumerate} + +\item[11*.] Find a method for determining all the solutions of the +equation +\begin{equation*} +\phi(x) = n, +\end{equation*} +where $n$ is given and $x$ is to be sought. + +\item[12*.] A number theory function $\phi(n)$ is defined for every +positive integer $n$; and for every such number $n$ it satisfies the +relation +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = n, +\end{equation*} +where $d_1, d_2, \ldots, d_r$ are the divisors of $n$. From this +property alone show that +\begin{equation*} +\phi(n) = n \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_k} \right ), +\end{equation*} +where $p_1, p_2, \ldots, p_k$ are the different prime factors of +$n$. \end{enumerate} \normalsize\index{Indicator|)} + +\chapter{ELEMENTARY PROPERTIES OF CONGRUENCES}% +\index{Congruences|(} + +\section{Congruences Modulo $m$}\label{s18} + +\textsc{Definitions.} If $a$ and $b$ are any two integers, positive +or zero or negative, whose difference is divisible by $m$, $a$ and +$b$ are said to be congruent modulo $m$, or congruent for the +modulus $m$, or congruent according to the modulus $m$. Each of +the numbers $a$ and $b$ is said to be a residue of the other.% +\index{Residue} + +\smallskip To express the relation thus defined we may write +\begin{equation*} +a = b + cm, +\end{equation*} +where $c$ is an integer (positive or zero or negative). It is more +convenient, however, to use a special notation due to Gauss, and to +write +\begin{equation*} +a \equiv b \mod m, +\end{equation*} +an expression which is read $a$ is congruent to $b$ modulo $m$, or +$a$ is congruent to $b$ for the modulus $m$, or $a$ is congruent to +$b$ according to the modulus $m$.\index{Gauss} This notation has the +advantage that it involves only the quantities which are essential +to the idea involved, whereas in the preceding expression we had the +irrelevant integer $c$. The Gaussian notation is of great value and +convenience in the study of the theory of divisibility. In the +present chapter we develop some of the fundamental elementary +properties of congruences. It will be seen that many theorems +concerning equations are likewise true of congruences with fixed +modulus; and it is this analogy with equations which gives +congruences (as such) one of their chief claims to attention. + +As immediate consequences of our definitions we have the following +fundamental theorems: + +\smallskip I.~\emph{If} $a\equiv c \mod m$, $b\equiv c\mod m$, +\emph{then} $a\equiv b\mod m$; \noindent \emph{that is, for a given +modulus, numbers congruent to the same number are congruent to each +other.} + +For, by hypothesis, $a - c = c_1 m$, $b - c = c_2 m$, where $c_1$ +and $c_2$ are integers. Then by subtraction we have $a - b = (c_1 - +c_2) m$; whence $a \equiv b \mod m$. + +\smallskip II.~\emph{If} $a \equiv b \mod m$, $\alpha \equiv +\beta \mod m$, \emph{then} $a \pm \alpha \equiv b \pm \beta \mod m$; +\emph{that is, congruences with the same modulus may be added or +subtracted member by member.} + +For, by hypothesis, $a - b = c_1 m$, $\alpha - \beta = c_2 m$; +whence $(a \pm \alpha) - (b \pm \beta) = (c_1 \pm c_2)m$. Hence $a +\pm \alpha = b \pm \beta \mod m$. + +\smallskip III.~\emph{If} $a = b \mod m$, \emph{then} +$ca = cb \mod m$, \emph{$c$ being any integer whatever.} + +The proof is obvious and need not be stated. + +\smallskip IV.~\emph{If} $a \equiv b \mod m$, +$\alpha \equiv \beta \mod m$, \emph{then} $a \alpha \equiv b \beta +\mod m$; \emph{that is, two congruences with the same modulus may be +multiplied member by member.} + +For, we have $a = b + c_1 m$, $\alpha = \beta + c_2 m$. Multiplying +these equations member by member we have $a \alpha = b \beta + m (b +c_2 + \beta c_1 + c_1 c_2 m)$. Hence $a \alpha \equiv b \beta \mod +m$. + +\smallskip A repeated use of this theorem gives the following +result: + +\smallskip V.~\emph{If} $a \equiv b \mod m$, \emph{then} +$a^n \equiv b^n \mod m$ \emph{where $n$ is any positive integer.} + +\smallskip As a corollary of theorems II, III and V we have the +following more general result: + +\smallskip VI.~\emph{If $f(x)$ denotes any polynomial in $x$ with +coefficients which are integers (positive or zero or negative) and +if further $a\equiv b \bmod m$, then} +\begin{equation*} +f(a) \equiv f(b) \bmod m. +\end{equation*} + +\section{Solutions of Congruences by Trial}\label{s19}% +\index{Congruences!Solution by trial|(} + +Let $f(x)$ be any polynomial in $x$ with coefficients which are +integers (positive or negative or zero). Then if $x$ and $c$ are any +two integers it follows from the last theorem of the preceding +section that +\begin{gather*} +f(x) \equiv f(x + cm) \bmod m. \tag{1} \\ +\intertext{Hence if $a$ is any value of $x$ for which the +congruence} +f(x)\equiv 0\bmod m. \tag{2} +\end{gather*} +is satisfied, then the congruence is also satisfied for $x = \alpha ++ cm$, where $c$ is any integer whatever. The numbers $\alpha + cm$ +are said to form a \emph{solution} (or to be a \emph{root}) of the +congruence, $c$ being a variable integer. Any one of the integers +$\alpha + cm$ may be taken as the representative of the solution. We +shall often speak of one of these numbers as the solution itself. + +Among the integers in a solution of the congruence (2) there is +evidently one which is positive and not greater than $m$. Hence all +solutions of a congruence of the type (2) may be found by trial, a +substitution of each of the numbers $1, 2, \ldots, m$ being made for +$x$. It is clear also that $m$ is the maximum number of solutions +which (2) can have whatever be the function $f(x)$. By means of an +example it is easy to show that this maximum number of solutions is +not always possessed by a congruence; in fact, it is not even +necessary that the congruence have a solution at all. + +This is illustrated by the example +\begin{equation*} +x^2 - 3 \equiv 0 \bmod 5. +\end{equation*} +In order to show that no solution is possible it is necessary to +make trial only of the values $1, 2, 3, 4, 5$ for $x$. A direct +substitution verifies the conclusion that none of them satisfies the +congruence; and hence that the congruence has no solution at all. + +On the other hand the congruence +\begin{equation*} +x^5 - x \equiv 0 \bmod 5 +\end{equation*} +has the solutions $x = 1, 2, 3, 4, 5$ as one readily verifies; that +is, this congruence has five solutions---the maximum number possible +in accordance with the results obtained above. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $(a + b)^p \equiv a^p + b^p \bmod p$ +where $a$ and $b$ are any integers and $p$ is any prime. + +\item[2.] From the preceding result prove that +$\alpha^p \equiv \alpha \bmod p$ for every integer $\alpha$. + +\item[3.] Find all the solutions of each of the congruences $x^{11} +\equiv x \bmod 11, x^{10} \equiv 1 \bmod 11, x^{5} \equiv 1 \bmod +11$. +\end{enumerate} \normalsize\index{Congruences!Solution by trial|)} + +\section{Properties of Congruences Relative to Division}\label{s20} + +The properties of congruences relative to addition, subtraction and +multiplication are entirely analogous to the properties of algebraic +equations. But the properties relative to division are essentially +different. These we shall now give. + +\smallskip I.~\emph{If two numbers are congruent modulo $m$ they are +congruent modulo $d$, where $d$ is any divisor of $m$.} + +For, from $a \equiv b \bmod m$, we have $a = b + cm = b + c'd$. +Hence $a\equiv b \bmod d$. + +\smallskip II.~\emph{If two numbers are congruent for different +moduli they are congruent for a modulus which is the least common +multiple of the given moduli.} + +The proof is obvious, since the difference of the given numbers is +divisible by each of the moduli. + +\smallskip III.~\emph{When the two members of a congruence are +multiples of an integer $c$ prime to the modulus, each member of the +congruence may be divided by $c$.} + +For, if $ca \equiv cb \bmod m$ then $ca - cb$ is divisible by $m$. +Since $c$ is prime to $m$ it follows that $a - b$ is divisible by +$m$. Hence $a\equiv b \bmod m$. + +\smallskip IV.~\emph{If the two members of a congruence are +divisible by an integer $c$, having with the modulus the greatest +common divisor $\delta$, one obtains a congruence equivalent to the +given congruence by dividing the two members by $c$ and the modulus +by $\delta$.} + +By hypothesis $ac \equiv bc \bmod m,\quad c = \delta c_1,\quad m = +\delta m_1$. Hence $c(a-b)$ is divisible by $m$. A necessary and +sufficient condition for this is evidently that $c_1(a-b)$ is +divisible by $m_1$. This leads at once to the desired result. + +\section{Congruences with a Prime Modulus}\label{s21}% +\index{Congruences!with prime modulus|(} + +\emph{The congruence\footnote{The sign $\not\equiv$ is read \emph{is +not congruent to}.}} +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod p, + \quad a_0 \not\equiv 0 \bmod p +\end{equation*} +\emph{where $p$ is a prime number and the $a$'s are any integers, +has not more than $n$ solutions.} + +Denote the first member of this congruence by $f(x)$ so that the +congruence may be written +\begin{gather} +f(x) \equiv 0 \bmod p \tag{1} \\ +\intertext{Suppose that $a$ is a root of the congruence, so that} +f(a) \equiv 0 \bmod p. \notag \\ +\intertext{Then we have} f(x) +\equiv f(x) - f(a) \bmod p. \notag \\ +\intertext{But, from algebra, $f(x) - f(a)$ is divisible by $x - a$. +Let $(x-a)^{\alpha}$ be the highest power of $x - a$ contained in +$f(x) - f(a)$. Then we may write} +f(x) - f(a) = (x - a)^{\alpha} f_1(x), \tag{2} \\ +\intertext{where $f_1(x)$ is evidently a polynomial with integral +coefficients. Hence we have} +f(x) \equiv (x - a)^{\alpha} f_1(x) \bmod p. \tag{3} +\end{gather} +We shall say that $a$ occurs $\alpha$ times as a solution of (1); or +that the congruence has $\alpha$ solutions each equal to $a$. + +Now suppose that congruence (1) has a root $b$ such that +$b\not\equiv a \bmod p$. Then from (3) we have +\begin{gather*} +f(b) \equiv (b-a)^{\alpha}f_1(b) \bmod p. \\ +\intertext{But} +f(b)\equiv 0 \bmod p,\quad (b-a)^{\alpha} \not\equiv 0 \bmod p. \\ +\intertext{Hence, since $p$ is a prime number, we must have} +f_1(b)\equiv 0 \bmod p. +\end{gather*} + +By an argument similar to that just used above, we may show that +$f_1(x) - f_1(b)$ may be written in the form +\begin{gather*} +f_1(x) - f_1(b) = (x-b)^{\beta}f_2(x), \\ +\intertext{where $\beta$ is some positive integer. Then we have} +f(x) \equiv (x-a)^{\alpha}(x-b)^{\beta}f_2(x) \bmod p. +\end{gather*} + +Now this process can be continued until either all the solutions of +(1) are exhausted or the second member is a product of linear +factors multiplied by the integer $a_0$. In the former case there +will be fewer than $n$ solutions of (1), so that our theorem is true +for this case. In the other case we have +\begin{equation*} +f(x) \equiv a_0(x-a)^{\alpha}(x-b)^{\beta} + \ldots (x-l)^{\lambda} \bmod p. +\end{equation*} +We have now $n$ solutions of (1): $a$ counted $\alpha$ times, $b$ +counted $\beta$ times, \ldots, $l$ counted $\lambda$ times; $\alpha ++ \beta + \ldots +\lambda = n$. + +Now let $\eta$ be any solution of (1). Then +\begin{equation*} +f(\eta) \equiv a_0(\eta-a)^{\alpha}(\eta-b)^{\beta} \ldots + (\eta-l)^{\lambda} \equiv 0 \bmod p. +\end{equation*} +Since $p$ is prime it follows now that some one of the factors +$\eta-a, \eta-b, \ldots, \eta-l$ is divisible by $p$. Hence $\eta$ +coincides with one of the solutions $a, b, c, \ldots, l$. That is, +(1) can have only the $n$ solutions already found. + +This completes the proof of the theorem. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Construct a congruence of the form +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod m, \quad + a_0 \not\equiv 0 \bmod m, +\end{equation*} +having more than $n$ solutions and thus show that the limitation to +a prime modulus in the theorem of this section is essential. + +\item[2.] Prove that +\begin{equation*} +x^6-1 \equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) \bmod 7 +\end{equation*} +for every integer $x$. + +\item[3.] How many solutions has the congruence $x^5 \equiv 1 \bmod +11$? the congruence $x^5\equiv 2 \bmod 11$? +\end{enumerate}\normalsize\index{Congruences!with prime modulus|)} + +\section{Linear Congruences}\label{s22}% +\index{Congruences!Linear|(} + +From the theorem of the preceding section it follows that the +congruence +\begin{equation*} +ax \equiv c \bmod p,\quad a \not\equiv 0 \bmod p, +\end{equation*} +where $p$ is a prime number, has not more than one solution. In this +section we shall prove that it always has a solution. More +generally, we shall consider the congruence +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +where $m$ is any integer. The discussion will be broken up into +parts for convenience in the proofs. + +\smallskip I.~\emph{The congruence} +\begin{equation} +ax \equiv 1 \bmod m, \tag{1} +\end{equation} +\emph{in which a and m are relatively prime, has one and only one +solution.} + +The question as to the existence and number of the solutions of (1) +is equivalent to the question as to the existence and number of +integer pairs $x, y$ satisfying the equation, +\begin{equation} +ax - my = 1, \tag{2} +\end{equation} +the integers $x$ being incongruent modulo $m$. Since $a$ and $m$ are +relatively prime it follows from theorem IV of \S~\ref{s9} that +there exists a solution of equation (2). Let $x = \alpha$ and $y = +\beta$ be a particular solution of (2) and let $x = \bar{\alpha}$ +and $y = \bar{\beta}$ be any solution of (2). Then we have +\begin{gather*} +a\alpha-m\beta = 1, \\ +a \bar{\alpha} - m\bar{\beta} = 1; \\ +\intertext{whence} +a(\alpha - \bar{\alpha}) - m(\beta - \bar{\beta}) = 0. +\end{gather*} +Hence $\alpha-\bar{\alpha}$ is divisible by $m$, since $a$ and $m$ +are relatively prime. That is, $a \equiv \bar{\alpha} \mod m$. Hence +$\alpha$ and $\bar{\alpha}$ are representatives of the same solution +of (1). Hence (1) has one and only one solution, as was to be +proved. + +\smallskip II.~\emph{The solution $x = \alpha$ of the congruence +$ax \equiv 1 \mod m$, in which $a$ and $m$ are relatively prime, is +prime to $m$.} + +For, if $a\alpha - 1$ is divisible by $m$, $\alpha$ is divisible by +no factor of $m$ except $1$. + +\smallskip III.~\emph{The congruence} +\begin{equation} +ax \equiv c \mod m \tag{3} +\end{equation} +\emph{in which $a$ and $m$ and also $c$ and $m$ are relatively +prime, has one and only one solution.} + +Let $x = \gamma$ be the unique solution of the congruence $cx = 1 +\mod m$. Then we have $a\gamma x \equiv c\gamma \equiv 1 \mod m$. +Now, by I we see that there is one and only one solution of the +congruence $a\gamma x \equiv 1 \mod m$; and from this the theorem +follows at once. + +Suppose now that $a$ is prime to $m$ but that $c$ and $m$ have the +greatest common divisor $\delta$ which is different from 1. Then it +is easy to see that any solution $x$ of the congruence $ax \equiv c +\mod m$ must be divisible by $\delta$. The question of the existence +of solutions of the congruence $ax \equiv c \bmod m$ is then +equivalent to the question of the existence of solutions of the +congruence +\begin{equation*} +a \frac{x}{\delta} \equiv \frac{c}{\delta} \bmod \frac{m}{\delta}, +\end{equation*} +where $\frac{x}{\delta}$ is the unknown integer. From III it follows +that this congruence has a unique solution $\frac{x}{\delta} = +\alpha$. Hence the congruence $ax \equiv c \bmod m$ has the unique +solution $x = \delta\alpha$. Thus we have the following theorem: + +\smallskip IV.~\emph{The congruence $ax \equiv c \bmod m$, in which +$a$ and $m$ are relatively prime, has one and only one solution.} + + +\smallskip\textsc{Corollary.}~\emph{The congruence +$ax \equiv c \bmod p$, $a \not\equiv 0 \bmod p$, where $p$ is a +prime number, has one and only one solution.} + +It remains to examine the case of the congruence $ax =c \bmod m$ in +which $a$ and $m$ have the greatest common divisor $d$. It is +evident that there is no solution unless $c$ also contains this +divisor $d$. Then let us suppose that $a = \alpha d$, $c = \gamma +d$, $m = \mu d$. Then for every $x$ such that $ax = c \bmod m$ we +have $\alpha x = \gamma \bmod \mu$; and conversely every $x$ +satisfying the latter congruence also satisfies the former. Now +$\alpha x = \gamma \bmod \mu$, has only one solution. Let $\beta$ be +a non-negative number less than $\mu$, which satisfies the +congruence $\alpha x = \gamma \bmod \mu$. All integers which satisfy +this congruence are then of the form $\beta + \mu\nu$, where $\nu$ +is an integer. Hence all integers satisfying the congruence $ax = c +\bmod m$ are of the form $\beta + \mu\nu$; and every such integer is +a representative of a solution of this congruence. It is clear that +the numbers +\begin{equation} +\beta,\ \beta + \mu,\ \beta + 2\mu,\ \ldots,\ \beta + (d-1)\mu +\tag{A} +\end{equation} +are incongruent modulo $m$ while every integer of the form $\beta + +\mu\nu$ is congruent modulo $m$ to a number of the set (A). Hence +the congruence $ax = c \bmod m$ has the $d$ solutions (A). + +This leads us to an important theorem which includes all the other +theorems of this section as special cases. It may be stated as +follows: + +\smallskip V.~\emph{Let} +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +\emph{be any linear congruence and let $a$ and $m$ have the greatest +common divisor $d (d \geq 1)$. Then a necessary and sufficient +condition for the existence of solutions of the congruence is that +$c$ be divisible by $d$. If this condition is satisfied the +congruence has just $d$ solutions, and all the solutions are +congruent modulo $m / d$.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find the remainder when $2^{40}$ is divided by $31$; when +$2^{43}$ is divided by $31$. + +\item[2.] Show that $2^{2^5}+1$ has the factor $641$. + +\item[3.] Prove that a number is a multiple of $9$ if and only if the +sum of its digits is a multiple of $9$. + +\item[4.] Prove that a number is a multiple of $11$ if and only if the +sum of the digits in the odd numbered places diminished by the sum +of the digits in the even numbered places is a multiple of $11$. +\end{enumerate} \normalsize% +\index{Congruences|)}\index{Congruences!Linear|)} + +\chapter{THE THEOREMS OF FERMAT AND WILSON} + +\section{Fermat's General Theorem}\label{s23}% +\index{Fermat's!General Theorem} + +Let $m$ be any positive integer and let +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)} \tag{A} +\end{equation} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$. Let $a$ be any integer prime to $m$ and form the set +of integers +\begin{equation} +aa_1,\ aa_2,\ \ldots,\ aa_{\phi(m)} \tag{B} +\end{equation} +No number $aa_i$ of the set (B) is congruent to a number $aa_j$, +unless $j = i;$ for, from +\begin{equation*} +aa_i \equiv aa_j \bmod m +\end{equation*} +we have $a_i \equiv a_j \bmod m$; whence $a_i = a_j$ since both +$a_i$ and $a_j$ are positive and not greater than $m$. Therefore $j += i$. Furthermore, every number of the set (B) is congruent to some +number of the set (A). Hence we have congruences of the form +\begin{align*} +aa_1 & \equiv a_{i_1} \bmod m, \\ +aa_2 & \equiv a_{i_2} \bmod m, \\ + & \vdots \\ +aa_{\phi(m)} & \equiv a_{i_{\phi(m)}} \bmod m. +\end{align*} +No two numbers in the second members are equal, since $aa_i +\not\equiv aa_j$ unless $i= j$. Hence the numbers $a_{i_1},\ +a_{i_2},\ \ldots,\ a_{i_{\phi(m)}}$ are the numbers $a_1,\ a_2,\ +\ldots,\ a_{\phi(m)}$ in some order. Therefore, if we multiply the +above system of congruences together member by member and divide +each member of the resulting congruence by $a_1\cdot a_2\ldots +a_{\phi(m)}$ (which is prime to $m$), we have +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} +This result is known as Fermat's general theorem.% +\index{Fermat's!general theorem} It may be stated as follows: + +\emph{If $m$ is any positive integer and $a$ is any integer prime to +$m$, then} +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ is any integer +not divisible by a prime number $p$, then} +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +\smallskip \textsc{Corollary 2.}~\emph{If $p$ is any prime number +and $a$ is any integer, then} +\begin{equation*} +a^p \equiv a \bmod p. +\end{equation*} + +\section{Euler's Proof of the Simple Fermat Theorem}\label{s24}% +\index{Euler}\index{Fermat}\index{Fermat's!Simple Theorem} + +The theorem of Cor.\ 1, \S~\ref{s23}, is often spoken of as the +simple Fermat theorem. It was first announced by Fermat in 1679, but +without proof. The first proof of it was given by Euler in 1736. +This proof may be stated as follows: + +From the Binomial Theorem it follows readily that +\begin{gather*} +(a+1)^p \equiv a^p + 1 \bmod p \\ +\intertext{since} +\frac{p!}{r!(p-r)!}, \quad 0 < r < p, \\ +\intertext{is obviously divisible by $p$. Subtracting $a + 1$ from +each side of the foregoing congruence, we have} +(a+1)^p - (a+1) \equiv a^p - a \bmod p. +\end{gather*} +Hence if $a^p - a$ is divisible by $p$, so is $(a + 1)^p - (a + 1)$. +But $1^p - 1$ is divisible by $p$. Hence $2^p - 2$ is divisible by +$p$; and then $3^p - 3$; and so on. Therefore, in general, we have +\begin{equation*} +a^p \equiv a \mod p. +\end{equation*} +If $a$ is prime to $p$ this gives $a^{p-1} \equiv 1 \mod p$, as was +to be proved. + +If instead of the Binomial Theorem one employs the Polynomial +Theorem, an even simpler proof is obtained. For, from the latter +theorem, we have readily +\begin{gather*} +(\alpha_1 + \alpha_2 + \ldots + \alpha_a)^p \equiv + \alpha_1^p + \alpha_2^p + \ldots + \alpha_a^p \mod p. \\ +\intertext{Putting $\alpha_1 = \alpha_2 = \ldots = \alpha_a = 1$ we +have} +a^p\equiv a \mod p, +\end{gather*} +from which the theorem follows as before. + +\section{Wilson's Theorem}\label{s25}\index{Wilson's theorem|(} + +From the simple Fermat theorem it follows that the congruence +\begin{gather*} +x^{p-1} \equiv 1\mod p \\ +\intertext{has the $p-1$ solutions $1$, $2$, $3$, $\ldots$, $p-1$. +Hence from the discussion in \S \ref{s21} it follows that} +x^{p-1} \equiv (x-1)(x-2)\ldots(x-\overline{p-1}) \mod p, \\ +\intertext{this relation being satisfied for every value of $x$. +Putting $x = 0$ we have} +(-1) = (-1)^{p - 1}\cdot 1\cdot 2\cdot 3 \ldots + \overline{p-1}\mod p. \\ +\intertext{If $p$ is an odd prime this leads to the congruence} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} + 1 = 0 \mod p. +\end{gather*} +Now for $p = 2$ this congruence is evidently satisfied. Hence +we have the Wilson theorem: + +\smallskip \emph{Every prime number $p$ satisfies the relation} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{p+1} + 1 \equiv 0 \mod p. +\end{equation*} + +An interesting proof of this theorem on wholly different principles +may be given. Let $p$ points be distributed at equal intervals on +the circumference of a circle. The whole number of $p$-gons which +can be formed by joining up these $p$ points in every possible order +is evidently +\begin{equation*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1; +\end{equation*} +for the first vertex can be chosen in $p$ ways, the second in $p - +1$ ways, $\ldots$, the $(p-1)^{\mathrm{th}}$ in two ways, and the +last in one way; and in counting up thus we have evidently counted +each polygon $2p$ times, once for each vertex and for each direction +from the vertex around the polygon. Of the total number of polygons +$\frac{1}{2}(p-1)$ are regular (convex or stellated) so that a +revolution through $\frac{360^\circ}{p}$ brings each of these into +coincidence with its former position. The number of remaining +$p$-gons must be divisible by $p$; for with each such $p$-gon we may +associate the $p-1$ $p$-gons which can be obtained from it by +rotating it through successive angles of $\frac{360^\circ}{p}$. That +is, +\begin{gather*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - + \frac 12 (p-1) \equiv 0 \bmod p. \\ +\intertext{Hence} +(p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - p + 1 \equiv 0 \bmod p; \\ +\intertext{and from this it follows that} +1 \cdot 2 \ldots \overline{p-1} + 1 \equiv 0 \bmod p, \\ +\end{gather*} +as was to be proved. + +\section{The Converse of Wilson's Theorem}\label{s26} + +Wilson's theorem is noteworthy in that its converse is also true. +The converse may be stated as follows: + +\smallskip \emph{Every integer $n$ such that the congruence} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n +\end{equation*} +\emph{is satisfied is a prime number.} + +For, if $n$ is not prime, there is some divisor $d$ of $n$ different +from $1$ and less than $n$. For such a $d$ we have $1 \cdot 2 \cdot +3 \ldots \overline{n-1} \equiv 0 \bmod d$; so that $1 \cdot 2 \ldots +\overline{n-1}+1 \not\equiv 0 \bmod d$; and hence $1 \cdot 2 \ldots +\overline{n-1}+1 \equiv 0 \bmod n$. Since this contradicts our +hypothesis the truth of the theorem follows. + +\smallskip Wilson's theorem and its converse may be combined into +the following elegant theorem: + +\smallskip \emph{A necessary and sufficient condition that an +integer $n$ is prime is that} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n. +\end{equation*}\index{Prime numbers} + +Theoretically this furnishes a complete and elegant test as to +whether a given number is prime. But, practically, the labor of +applying it is so great that it is useless for verifying large +primes. + +\section{Impossibility of $1 \cdot 2 \cdot 3 \cdots +\overline{n-1} + 1 = n^k$ for $n > 5$.}\label{s27} + +In this section we shall prove the following theorem: + +\emph{There exists no integer $k$ for which the equation} +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-1} + 1 = n^k +\end{equation*} +is true when $n$ is greater than $5$. + +If $n$ contains a divisor $d$ different from $1$ and $n$, the +equation is obviously false; for the second member is divisible by +$d$ while the first is not. Hence we need to prove the theorem only +for primes $n$. + +Transposing $1$ to the second member and dividing by $n - 1$ we have +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-2} = n^{k-1} + n^{k-2} + + \ldots + n+1. +\end{equation*} +If $n>5$ the product on the left contains both the factor $2$ and +the factor $\frac{1}{2} (n-1)$; that is, the first member contains +the factor $n - 1$. But the second member does not contain this +factor, since for $n = 1$ the expression $n^{k-1} + \ldots n + 1$ is +equal to $k \neq 0$. Hence the theorem follows at once. + +\section{Extension of Fermat's Theorem}\label{s28}% +\index{Fermat's!theorem extended|(} + +The object of this section is to extend Fermat's general theorem and +incidentally to give a new proof of it. We shall base this proof on +the simple Fermat theorem, of which we have already given a simple +independent proof. This theorem asserts that for every prime $p$ and +integer $a$ not divisible by $p$, we have the congruence +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +Then let us write +\begin{gather} +a^{p-1} = 1 + hp. \tag{1} \\ +\intertext{Raising each member of this equation to the +$p^{\text{th}}$ power we may write the result in the form} +a^{p(p-1)} = 1 + h_1p^2. \tag{2} \\ +\intertext{where $h_1$ is an integer. Hence} +a^{p(p-1)} \equiv 1 \bmod p^2. \notag \\ +\intertext{By raising each member of (2) to the $p^{\text{th}}$ +power we can readily show that} +a^{p^2(p-1)} \equiv 1 \bmod p^3. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{p^{\alpha - 1}(p-1)} \equiv 1 \bmod p^{\alpha}. \notag \\ +\intertext{where $\alpha$ is a positive integer; that is,} +a^{\phi(p^{\alpha})} \equiv 1 \bmod p^{\alpha}. \notag +\end{gather} + +For the special case when $p$ is 2 this result can be extended. For +this case (1) becomes +\begin{gather} +a = 1 + 2h. \notag \\ +\intertext{Squaring we have} +a^2 = 1 + 4h(h+1). \notag \\ +\intertext{Hence,} +a^2 = 1+8h_1, \tag{3} \\ +\intertext{where $h_1$ is an integer. Therefore} +a^2 \equiv 1 \bmod 2^3. \notag \\ +\intertext{Squaring (3) we have} +a^{2^2} = 1 + 2^4h_2; \notag \\ +\intertext{or} +a^{2^2} \equiv 1 \bmod 2^4. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{2^{\alpha-2}} \equiv 1 \bmod 2^{\alpha} \notag \\ +\intertext{if $\alpha > 2$. That is,} +a^{\frac{1}{2}\phi(2^{\alpha})} \equiv 1 \bmod 2^{\alpha} + \text{ if } a > 2. +\end{gather} + +Now in terms of the $\phi$-function let us define a new function +$\lambda(m)$ as follows: +\begin{align*} +\lambda(2^{\alpha}) &= \phi(2^{\alpha}) \text{ if $a = 0, 1, 2$;} \\ +\lambda(2^{\alpha}) &= \frac{1}{2}\phi(2^{\alpha}) + \text{ if $a > 2$;} \\ +\lambda(p^{\alpha}) &= \phi(p^{\alpha}) + \text{ if $p$ is an odd prime;} \\ +\lambda(2^{\alpha} p_1^{\alpha_1} p_2^{\alpha_2} \cdots + p_n^{\alpha_n}) &= M, +\end{align*} +where $M$ is the least common multiple of +\begin{equation*} + \lambda(2^{\alpha}), + \lambda(p_1^{\alpha_1}), + \lambda(p_2^{\alpha_2}), \ldots, \lambda(p_n^{\alpha_n}), +\end{equation*} +$2, p_1, p_2, \ldots, p_n$ being different primes.% +\index{$\lambda(m)$} + +Denote by $m$ the number +\begin{equation*} +m = 2^{\alpha}p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_n^{\alpha_n}. +\end{equation*} +Let $a$ be any number prime to $m$. From our preceding results we +have +\begin{align*} +a^{\lambda(2^{\alpha})} &\equiv 1 \bmod 2^{\alpha}, \\ +a^{\lambda(p_1^{\alpha_1})} &\equiv 1 \bmod p_1^{\alpha_1},\\ +a^{\lambda(p_2^{\alpha_2})} &\equiv 1 \bmod p_2^{\alpha_2}, \\ +\ldots \\ +a^{\lambda(p_n^{\alpha_n})} &\equiv 1 \bmod p_2^{\alpha_n}. \\ +\end{align*} + +Now any one of these congruences remains true if both of its members +are raised to the same positive integral power, whatever that power +may be. Then let us raise both members of the first congruence to +the power $\frac{\lambda(m)}{\lambda(2^\alpha)}$; both members of +the second congruence to the power +$\frac{\lambda(m)}{\lambda(p_1^{\alpha_1})}$; $\ldots$; both members +of the last congruence to the power +$\frac{\lambda(m)}{\lambda(p_n^{\alpha_n})}$. Then we have +\begin{align*} +a^{\lambda(m)} &\equiv 1 \mod 2^\alpha, \\ +a^{\lambda(m)} &\equiv 1 \mod p_1^{\alpha_1}, \\ +\ldots \ldots \\ +a^{\lambda(m)} &\equiv 1 \mod p_n^{\alpha_n}. \\ +\intertext{From these congruences we have immediately} +a^{\lambda(m)} &\equiv 1 \mod m. +\end{align*} + +We may state this result in full in the following theorem: + +\smallskip \emph{If $a$ and $m$ are any two relatively prime positive +integers, the congruence} +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +\emph{is satisfied.} + +As an excellent example to show the possible difference between the +exponent $\lambda(m)$ in this theorem and the exponent $\phi(m)$ in +Fermat's general theorem, let us take +\begin{gather*} +m = 2^6 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 19 + \cdot 37 \cdot 73. \\ +\intertext{Here} +\lambda(m) = 2^4 \cdot 3^2, \quad \phi(m) = 2^{31} \cdot 3^{10}. +\end{gather*} + +In a later chapter we shall show that there is no exponent $\nu$ +less than $\lambda(m)$ for which the congruence +\begin{equation*} +a^\nu = 1 \mod m +\end{equation*} +is verified for every integer $a$ prime to $m$. + +From our theorem, as stated above, Fermat's general theorem follows +as a corollary, since $\lambda(m)$ is obviously a factor of +$\phi(m)$, +\begin{equation*} +\phi(m) = \phi(2^\alpha) \phi(p_1^{\alpha_1}) \ldots + \phi(p_n^{\alpha_n}). +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $a^{16} \equiv 1 \bmod 16320$, for every $a$ +which is prime to $16320$. + +\item[2.] Show that $a^{12} \equiv 1 \bmod 65520$, for every $a$ which +is prime to $65520$. + +\item[3*.] Find one or more composite numbers $P$ such that +\begin{equation*} +a^{P-1} \equiv 1 \bmod P +\end{equation*} +for every a prime to $P$. (Compare this problem with the next +section.) \end{enumerate} \normalsize% +\index{Fermat's!theorem extended|)} + +\section{On the Converse of Fermat's Simple Theorem}\label{s29}% +\index{Fermat's!Simple Theorem} + +The fact that the converse of Wilson's theorem is a true proposition +leads one naturally to inquire whether the converse of Fermat's +simple theorem is true. Thus, we may ask the question: Does the +existence of the congruence $2^{n-1} \equiv 1 \bmod n$ require that +$n$ be a prime number? The Chinese answered this question in the +affirmative and the answer passed unchallenged among them for many +years. An example is sufficient to show that the theorem is not +true. We shall show that +\begin{equation*} +2^{340} \equiv 1 \bmod 341 +\end{equation*} +although $341 = 11 \cdot 31$, is not a prime number. Now $2^{10}-1 = +3 \cdot 11 \cdot 31$. Hence $2^{10} \equiv 1 \bmod 341$. Hence +$2^{340} \equiv 1 \bmod 341$. From this it follows that the direct +converse of Fermat's theorem is not true. The following theorem, +however, which is a converse with an extended hypothesis, is readily +proved. + +\smallskip \emph{If there exists an integer $a$ such that} +\begin{equation*} +a^{n-1} \equiv 1 \bmod n +\end{equation*} +\emph{and if further there does not exist an integer $\nu$ less than +$n - 1$ such that} +\begin{equation*} +a^{\nu} \equiv 1 \bmod n, +\end{equation*} +\emph{then the integer $n$ is a prime number.} + +For, if $n$ is not prime, $\phi(n) < n - 1$. Then for $\nu = +\phi(n)$ we have $a^{\nu} \equiv 1 \bmod n$, contrary to the +hypothesis of the theorem. + +\section{Application of Previous Results to Linear +Congruences}\label{s30}% +\index{Congruences!Linear} + +The theorems of the present chapter afford us a ready means of +writing down a solution of the congruence +\begin{equation} +ax \equiv c \bmod m. \tag{1} +\end{equation} +We shall consider only the case in which $a$ and $m$ are relatively +prime, since the general case is easily reducible to this one, as we +saw in the preceding chapter. + +Since $a$ and $m$ are relatively prime we have the congruences +\begin{gather*} +a^{\lambda(m)} \equiv 1,\quad a^{\phi(m)} \equiv 1 \bmod m. \\ +\intertext{Hence either of the numbers $x$,} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +is a representative of the solution of (1). Hence the following +theorem: + +\smallskip \emph{If} +\begin{gather*} +ax \equiv c \bmod m \\ +\intertext{\emph{is any linear congruence in which $a$ and $m$ are +relatively prime, then either of the numbers $x$,}} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +\emph{is a representative of the solution of the congruence.} + +The former representative of the solution is the more convenient of +the two, since the power of $a$ is in general much less in this case +than in the other. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] Find a solution of $7x \equiv 1 \bmod 2^6 \cdot 3 \cdot 5 \cdot +17.$ Note the greater facility in applying the first of the above +representatives of the solution rather than the second. +\end{enumerate} \normalsize + +\section{Application of the Preceding Results to the Theory +of Quadratic Residues}\label{s31}\index{Quadratic residues|(} + +In this section we shall apply the preceding results of this chapter +to the problem of finding the solutions of congruences of the form +\begin{equation*} +\alpha z^2 + \beta z + \gamma \equiv 0 \mod \mu +\end{equation*} +where $\alpha, \beta, \gamma, \mu$ are integers. These are called +quadratic congruences. + +The problem of the solution of the quadratic congruence (1) can be +reduced to that of the solution of a simpler form of congruence as +follows: Congruence (1) is evidently equivalent to the congruence +\begin{gather} +4\alpha^2 z^2 + 4\alpha\beta z + 4\alpha\gamma \equiv + 0 \mod 4\alpha\mu. \tag{1} \\ +\intertext{But this may be written in the form} +(2\alpha z + \beta)^2 \equiv \beta^2 - 4\alpha\gamma + \mod 4\alpha\mu. \notag \\ +\intertext{Now if we put} +2\alpha z + \beta\equiv x \mod 4\alpha\mu \tag{2} \\ +\intertext{and} +\beta^2 - 4\alpha\gamma = a,\quad 4\alpha\mu = m, \notag \\ +\intertext{we have} +x^2 \equiv a\mod m. \tag{3} +\end{gather} +We have thus reduced the problem of solving the general congruence +(1) to that of solving the binomial congruence (3) and the linear +congruence (2). The solution of the latter may be effected by means +of the results of \S \ref{s30}. We shall therefore confine ourselves +now to a study of congruence (3). We shall make a further limitation +by assuming that $a$ and $m$ are relatively prime, since it is +obvious that the more general case is readily reducible to this one. + +The example +\begin{equation*} +x^2 \equiv 3 \mod 5 +\end{equation*} +shows at once that the congruence (3) does not always have a +solution. First of all, then, it is necessary to find out in what +cases (3) has a solution. Before taking up the question it will be +convenient to introduce some definitions. + +\smallskip\textsc{Definitions.} An integer $a$ is said to be a +quadratic residue modulo $m$ or a quadratic non-residue modulo $m$ +according as the congruence +\begin{equation*} +x^2 = a \mod m +\end{equation*} +has or has not a solution. We shall confine our attention to the +case when $m > 2$.\index{Residue} + +We shall now prove the following theorem: + +\smallskip I.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\lambda(m)} \equiv 1 \mod m. +\end{equation*} + +Suppose that the congruence $x^2 = a \mod m$ has the solution $x = +\alpha$. Then $a^2 \equiv a \mod m$. Hence +\begin{equation*} +a^{\lambda(m)} \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +Since $a$ is prime to $m$ it is clear from $\alpha^2 \equiv a \mod +m$ that $a$ is prime to $m$. Hence $\alpha^{\lambda(m)} \equiv 1 +\mod m$. Therefore we have +\begin{equation*} +1 \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +That is, this is a necessary condition in order that $a$ shall be a +quadratic residue modulo $m$. + +In a similar way one may prove the following theorem: + +\smallskip II.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\phi(m)} \equiv 1 \mod m. +\end{equation*} + +When $m$ is a prime number $p$ each of the above results takes the +following form: If $a$ is prime to $p$ and is a quadratic residue +modulo $p$, then +\begin{equation*} +a^{\frac{1}{2}(p-1)} \equiv 1 \mod p. +\end{equation*} +We shall now prove the following more complete theorem, without the +use of I or II. + +\smallskip III.~\emph{If $p$ is an odd prime number and $a$ is an +integer not divisible by $p$, then $a$ is a quadratic residue or a +quadratic non-residue modulo $p$ according as} +\begin{equation*} +a^{\tfrac{1}{2}(p-1)} \equiv +1 \quad \text{or} \quad +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p. +\end{equation*} + +This is called Euler's criterion.\index{Euler's!criterion} + +Given a number $a$, not divisible by $p$, we have to determine +whether or not the congruence +\begin{gather} +x^2 \equiv a \bmod p \notag \\ +\intertext{has a solution. Let $r$ be any number of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 \tag{A} \\ +\intertext{and consider the congruence} +rx \equiv a \bmod p. +\end{gather} +This has always one and just one solution $x$ equal to a number $s$ +of the set (A). Two cases can arise: either for every $r$ of the set +(A) the corresponding $s$ is different from $r$ or for some $r$ of +the set (A) the corresponding $s$ is equal to $r$. The former is the +case when $a$ is a quadratic non-residue modulo $p$; the latter is +the case when $a$ is a quadratic residue modulo $p$. We consider the +two cases separately. + +In the first case the numbers of the set (A) go in pairs such that +the product of the numbers in the pair is congruent to a modulo $p$. +Hence, taking the product of all $\tfrac{1}{2}(p - 1)$ pairs, we +have +\begin{align*} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &\equiv + +a^{\tfrac{1}{2}(p-1)} \bmod p. \\ +\intertext{But} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &= -1 \bmod p. \\ +\intertext{Hence} +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p, +\end{align*} +whence the truth of one part of the theorem. + +In the other case, namely that in which some $r$ and corresponding +$s$ are equal, we have for this $r$ +\begin{gather*} +r^{2} \equiv a \bmod p \\ +\intertext{and} +(p - r)^{2} \equiv a \bmod p. +\end{gather*} +Since $x^{2} \equiv a \bmod p$ has at most two solutions it follows +that all the integers in the set (A) except $r$ and $p - r$ fall in +pairs such that the product of the numbers in each pair is congruent +to a modulo $p$. Hence, taking the product of all these pairs, which +are $\frac{1}{2}(p - 1) - 1$ in number, and multiplying by $r(p-r)$ +we have +\begin{align*} +1 \cdot 2 \cdot 3 \cdots \overline{p -1} + &\equiv (p - r) r a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -r^{2} a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a^{\frac{1}{2}(p -1)} \bmod p. \\ +\intertext{Since $1 \cdot 2 \cdot 3 \cdots \overline{p - 1} \equiv +\bmod p$ we have} +a^{\frac{1}{2}(p -1)} &\equiv + 1 \bmod p +\end{align*} +whence the truth of another part of the theorem. + +Thus the proof of the entire theorem is complete.% +\index{Quadratic residues|)}\index{Wilson's theorem|)} + +\chapter{PRIMITIVE ROOTS MODULO $m$.} + +\section{Exponent of an Integer Modulo $m$}\label{s32}% +\index{Exponent of an integer|(}\index{Primitive roots|(} + +Let +\begin{equation*} +a_{1},\ a_{2},\ \cdots,\ a_{\phi(m)} \tag{A} +\end{equation*} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$; and let $a$ denote any integer of the set (A). Now any +positive integral power of $a$ is prime to $m$ and hence is +congruent modulo $m$ to a number of the set (A). Hence, among all +the powers of a there must be two, say $a^{n}$ and $a^{\nu}$, $n > +\nu$, which, are congruent to the same integer of the set (A). These +two powers are then congruent to each other; that is, +\begin{equation*} +a^{n} \equiv a^{\nu} \bmod m +\end{equation*} +Since $a^{\nu}$ is prime to $m$ the members of this congruence may +be divided by $a^{\nu}$. Thus we have +\begin{equation*} +a^{n - \nu} \equiv 1 \bmod m. +\end{equation*} +That is, among the powers of $a$ there is one at least which is +congruent to $1$ modulo $m$. + +\smallskip Now, in the set of all powers of $a$ which are congruent +to $1$ modulo $m$ there is one in which the exponent is less than in +any other of the set. Let the exponent of this power be $d$, so that +$a^{d}$ is the lowest power of $a$ such that +\begin{equation} +a^{d} \equiv 1 \bmod m. \tag{1} +\end{equation} + +We shall now show that if $a^{\alpha} \equiv 1 \bmod m$, then +$\alpha$ is a multiple of $d$. Let us write +\begin{gather} +\alpha = d\delta + \beta, \quad 0 \leqq \beta < d. \notag \\ +\intertext{Then} +a^{\alpha} \equiv 1 \bmod m, \tag{2} \\ +a^{d\delta} \equiv 1 \bmod m, \tag{3} \\ +\intertext{the last congruence being obtained by raising (1) to the +power $\delta$. From (3) we have} +a^{d\delta + \beta} \equiv a^{\beta} \bmod m; \notag \\ +\intertext{or} +a^{\beta}\equiv 1 \bmod m. \notag +\end{gather} +Hence $\beta = 0$, for otherwise $d$ is not the exponent of the +lowest power of $a$ which is congruent to 1 modulo $m$. Hence $d$ is +a divisor of $\alpha$. + +\smallskip These results may be stated as follows: + +\smallskip I.~\emph{If $m$ is any integer and $a$ is any integer +prime to $m$, then there exists an integer $d$ such that} +\begin{gather*} +a^d\equiv 1 \bmod m \\ +\intertext{\emph{while there is no integer $\beta$ less than $d$ for +which}} +a^\beta\equiv 1 \bmod m. \\ +\intertext{\emph{Further, a necessary and sufficient condition +that}} +a^\nu \equiv 1 \bmod m +\end{gather*} +\emph{is that $\nu$ is a multiple of $d$.} + +\smallskip \textsc{Definition.} The integer $d$ which is thus +uniquely determined when the two relatively prime integers $a$ and +$m$ are given is called the exponent of $a$ modulo $m$. Also, $d$ is +said to be the exponent to which $a$ belongs modulo $m$. + +Now, in every case we have +\begin{equation*} +a^{\phi(m)} \equiv 1,\quad a^{\lambda(m)} \equiv 1 \bmod m, +\end{equation*} +if $a$ and $m$ are relatively prime. Hence from the preceding +theorem we have at once the following: + +\smallskip II.~\textit{The exponent $d$ to which $a$ belongs modulo +$m$ is a divisor of both $\phi(m)$ and $\lambda(m)$.}% +\index{Exponent of an integer|)} + +\section{Another Proof of Fermat's General Theorem}\label{s33} + +In this section we shall give an independent proof of the theorem +that the exponent $d$ of $a$ modulo $m$ is a divisor of $\phi(m)$; +from this result we have obviously a new proof of Fermat's theorem +itself. + +We retain the notation of the preceding section. We shall first +prove the following theorem: + +\smallskip I.~\textit{The numbers} +\begin{equation} +1,\ a,\ a^2,\ \ldots,\ a^{a-1} \tag{A} +\end{equation} +\textit{are incongruent each to each modulo $m$.} + +For, if $a^\alpha \equiv a^\beta \bmod m$, where $0 \leqq \alpha < +d$ and $0 \leqq \beta < d$, $\alpha > \beta$, we have +$a^{\alpha-\beta} \equiv 1 \bmod m$, so that $d$ is not the exponent +to which $a$ belongs modulo $m$, contrary to hypothesis. + +\smallskip Now any number of the set (A) is congruent to some number +of the set +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)}. \tag{B} +\end{equation} +Let us undertake to separate the numbers (B) into classes after the +following manner: Let the first class consist of the numbers +\begin{equation} +\alpha_1,\ \alpha_2,\ \ldots,\ \alpha_{a-1}, \tag{I} +\end{equation} +where $\alpha_i$ is the number of the set (B) to which $a^i$ is +congruent modulo $m$. + +If the class (I) does not contain all the numbers of the set (B), +let $a_i$ be any number of the set (B) not contained in (I) and form +the following set of numbers: +\begin{equation} +\alpha_0 a_i,\ \alpha_1 a_i,\ \alpha_2 a_i,\ \ldots,\ + \alpha_{d-1}a_i. \tag{II'} +\end{equation} +We shall now show that no number of this set is congruent to a +number of class (I). For, if so, we should have a congruence of the +form +\begin{gather*} +a_i a_j \equiv a_k \bmod m; \\ +\intertext{hence} +a_i a^j \equiv a^k \bmod m, \\ +\intertext{so that} +a_i a^d \equiv a^{k+d-j} \bmod m; \\ +\intertext{or} +a_i \equiv a^{k+d-j} \bmod m, +\end{gather*} +so that $a_i$ would belong to the set (I) contrary to hypothesis. + +Now the numbers of the set (II$'$) are all congruent to numbers of +the set (B); and no two are congruent to the same number of this +set. For, if so, we should have two numbers of (II') congruent; that +is, $\alpha_k a_i \equiv \alpha_j a_i \bmod m,$ or $\alpha_k \equiv +\alpha_j \bmod m;$ and this we have seen to be impossible. + +Now let the numbers of the set (B) to which the numbers of the set +(II') are congruent be in order the following: +\begin{equation} +\beta_0,\ \beta_1,\ \beta_2,\ \ldots,\ \beta_{d-1}. \tag{II} +\end{equation} +These numbers constitute our class (II). + +If classes (I) and (II) do not contain all the numbers of the set +(B), let $a_j$ be a number of the set ($B$) not contained in either +of the classes (I) and (II): and form the set of numbers +\begin{equation} +\alpha_0 a_j,\ \alpha_1 a_j,\ \alpha_2 a_j,\ \ldots,\ + \alpha_{d-1} a_j. \tag{III'} +\end{equation} +Just as in the preceding case it may be shown that no number of this +set is congruent to a number of class (I) and that the numbers of +(III') are incongruent each to each. We shall also show that no +number of (III') is congruent to a number of class (II). For, if so, +we should have $a_k a_j \equiv \beta_l \bmod m$. Hence $a^k a_j +\equiv a^l a_i \bmod m$; or $a_j \equiv a^{l+d-k} \bmod m$, from +which it follows that $a_j$ is of class (II), contrary to +hypothesis. + +Now let the numbers of the set (B) to which the numbers of the set +(III') are congruent be in order the following: +\begin{equation} +\gamma_0,\ \gamma_1,\ \gamma_2,\ \ldots,\ \gamma_{d-1}. \tag{III} +\end{equation} +These numbers form our class (III). + +It is now evident that the process may be continued until all the +numbers of the set (B) have been separated into classes, each class +containing $d$ integers, thus: +\begin{equation*} +\begin{matrix} +(\text{I}) & \alpha_0, & \alpha_1, & \alpha_2, + & \ldots, & \alpha_{d-1}, \\ +(\text{II}) & \beta_0, & \beta_1, & \beta_2, + & \ldots, & \beta_{d-1}, \\ +(\text{III}) & \gamma_0, & \gamma_1, & \gamma_2, + & \ldots, & \gamma_{d-1}, \\ +&\hdotsfor{5} \\ +(\quad ) & \lambda_0, & \lambda_1, & \lambda_2, + & \ldots, & \lambda_{d-1}. +\end{matrix} +\end{equation*} +The set (B), which consists of $\phi(m)$ integers, has thus been +separated into classes, each class containing $d$ integers. Hence we +conclude that $d$ is a divisor of $\phi(m)$. Thus we have a second +proof of the theorem: + +\smallskip II.~\emph{If $a$ and $m$ are any two relatively prime +integers and $d$ is the exponent to which $a$ belongs modulo $m$, +then $d$ is a divisor of $\phi(m)$.} + +In our classification of the numbers (B) into the rectangular array +above we have proved much more than theorem II; in fact, theorem II +is to be regarded as one only of the consequences of the more +general result contained in the array. + +If we raise each member of the congruence +\begin{equation*} +a^d \equiv 1 \bmod m +\end{equation*} +to the (integral) power $\phi(m)/d$, the preceding theorem leads +immediately to an independent proof of Fermat's general theorem. + +\section{Definition of Primitive Roots}\label{s34} + +\textsc{Definition.} Let $a$ and $m$ be two relatively prime +integers. If the exponent to which $a$ belongs modulo $m$ is +$\phi(m)$, $a$ is said to be a primitive root modulo $m$ (or a +primitive root of $m$). + +In a previous chapter we saw that the congruence +\begin{equation*} +a^{\lambda(m)} \equiv 1 \bmod m +\end{equation*} +is verified by every pair of relatively prime integers $a$ and $m$. +Hence, primitive roots can exist only for such a modulus $m$ as +satisfies the equation +\begin{equation*} +\phi(m) = \lambda(m). \tag{1} +\end{equation*} +We shall show later that this is also sufficient for the existence +of primitive roots. + +From the relation which exists in general between the +$\phi$-function and the $\lambda$-function in virtue of the +definition of the latter, it follows that (1) can be satisfied only +when $m$ is a prime power or is twice an odd prime power. + +Suppose first that $m$ is a power of $2$, say $m = 2^\alpha$. Then +(1) is satisfied only if $\alpha = 0,\ 1,\ 2$. For $\alpha = 0$ or +$1$, $1$ itself is a primitive root. For $\alpha = 2$, $3$ is a +primitive root. We have therefore left to examine only the cases +\begin{equation*} +m = p^\alpha,\quad m = 2p^\alpha +\end{equation*} +where $p$ is an odd prime number. The detailed study of these cases +follows in the next sections. + +\section{Primitive roots modulo $p$.}\label{s35} + +We have seen that if $p$ is a prime number and $d$ is the exponent +to which $a$ belongs modulo $p$, then $d$ is a divisor of $\phi(p) = +p - 1$. Now, let +\begin{gather*} +d_1,\ d_2,\ d_3,\ \ldots,\ d_r \\ +\intertext{be all the divisors of $p-1$ and let $\psi(d_i)$ denote +the number of integers of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 +\end{gather*} +which belong to the exponent $d_i$. If there is no integer of the +set belonging to this exponent, then $\psi(d_i) = 0$. + +Evidently every integer of the set belongs to some one and only one +of the exponents $d_1, d_2, \ldots, d_r$. Hence we have the relation +\begin{gather} +\psi(d_1) + \psi(d_2) + \ldots + \psi(d_r) = p-1. \tag{1} \\ +\intertext{But} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = p-1. \tag{2} \\ +\intertext{If then we can show that} +\psi(d_i) \leqq \phi(d_i) \tag{3} \\ +\intertext{for $i = 1, 2, \ldots, r$, it will follow from a +comparison of (1) and (2) that} +\psi(d_i) = \phi(d_i). \notag +\end{gather} +Accordingly, we shall examine into the truth of (3). + +Now the congruence +\begin{equation} +x^{d_i} \equiv 1 \mod p \tag{4} +\end{equation} +has not more than $d_i$ roots. If no root of this congruence belongs +to the exponent $d_i$, then if $\psi(d_i) = 0$ and therefore in this +case we have $\psi(d_i) < \phi(d_i)$. On the other hand if $a$ is a +root of (4) belonging to the exponent $d_i$, then +\begin{equation} +a, a^2, a^3, \ldots, a^{d_i} \tag{5} +\end{equation} +are a set of $d_i$ incongruent roots of (4); and hence they are the +complete set of roots of (4). + +But it is easy to see that $a^k$ does or does not belong to the +exponent $d_i$ according as $k$ is or is not prime to $d_i$; for, if +$a^k$ belongs to the exponent $t$, then $t$ is the least integer +such that $kt$ is a multiple of $d_i$. Consequently the number of +roots in the set (5) belonging to the exponent $d_i$ is $\phi(d_i)$. +That is, in this case $\psi(d_i) = \phi(d_i)$. Hence in general +$\psi(d_i) \leqq \phi(d_i)$ Therefore from (1) and (2) we conclude +that +\begin{equation*} +\psi(d_i) = \phi(d_i), \quad i = 1,\ 2,\ \ldots,\ r. +\end{equation*} +The result thus obtained may be stated in the form of the following +theorem: + +\smallskip I.~\emph{If $p$ is a prime number and $d$ is any divisor +of $p-1$, then the number of integers belonging to the exponent $d$ +modulo $p$ is $\phi(d)$.} + +In particular: + +\smallskip II.~\emph{There exist primitive roots modulo $p$ and their +number is $\psi(p-i)$.} + +\section{Primitive Roots Modulo $p^\alpha$, $p$ an Odd +Prime}\label{s36} + +In proving that there exist primitive roots modulo $p^\alpha$, where +$p$ is an odd prime and $\alpha > 1$, we shall need the following +theorem: + +I.~\emph{There always exists a primitive root $\gamma$ modulo $p$ +for which $\gamma^{p-1}$ is not divisible by $p^2$.} + +Let $g$ be any primitive root modulo $p$. If $g^{p-1}$ is not +divisible by $p^2$ our theorem is verified. Then suppose that +$g^{p-1}-1$ is divisible by $p^2$, so that we have +\begin{gather*} +g^{p-1}-1 = kp^2 \\ +\intertext{where $k$ is an integer. Then put} +\gamma = g + xp \\ +\intertext{where $x$ is an integer. Then $\gamma = g \mod p$, and +hence} +\gamma^h \equiv g^h \mod p; +\end{gather*} +whence we conclude that $\gamma$ is a primitive root modulo $p$. But +\begin{align*} +\gamma^{p-1}-1 &= + g^{p-1} - 1 + \frac{p-1}{1!}g^{p-2}xp + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p^2 + \ldots \\ + &= p\left(kp + \frac{p-1}{1!}g^{p-2}x + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p + \ldots\right). +\end{align*} +Hence +\begin{equation*} +\gamma^{p-1}-1 \equiv p(-g^{p-2}x) \mod p^2. +\end{equation*} +Therefore it is evident that $x$ can be so chosen that +$\gamma^{p-1}-1$ is not divisible by $p^2$. Hence there exists a +primitive root $\gamma$ modulo $p$ such that $\gamma^{p-1}-1$ is not +divisible by $p^2$. Q.~E.~D. + +\smallskip We shall now prove that this integer $\gamma$ is a +primitive root modulo $p^\alpha$, where $\alpha$ is any positive +integer. + +If +\begin{equation*} +\gamma^k \equiv 1\mod p. +\end{equation*} +then $k$ is a multiple of $p-1$, since $\gamma$ is a primitive root +modulo $p$. Hence, if +\begin{equation*} +\gamma^k \equiv 1 \mod p^\alpha, +\end{equation*} +then $k$ is a multiple of $p-1$. + +Now, write +\begin{equation*} +\gamma^{p-1} = 1 + hp. +\end{equation*} +Since $\gamma^{p-1}$ is not divisible by $p^2$, it follows that $h$ +is prime to $p$. If we raise each member of this equation to the +power $\beta p^{\alpha-2}$, $\alpha \stackrel{=}{>}2$, we have +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} = + 1 + \beta p^{\alpha-1}h + p^\alpha I, +\end{equation*} +where $I$ is an integer. Then if +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} \equiv 1 \mod p^\alpha, +\end{equation*} +$\beta$ must be divisible by $p$. Therefore the exponent of the +lowest power of $\gamma$ which is congruent to $1$ modulo $p^\alpha$ +is divisible by $p^{\alpha-1}$. But we have seen that this exponent +is also divisible by $p-1$. Hence the exponent of $\gamma$ modulo +$p^\alpha$ is $p^{\alpha-1}(p-1)$ since $\phi(p^\alpha) = +p^{\alpha-1}(p-1)$. That is, $\gamma$ is a primitive root modulo +$p^\alpha$. + +It is easy to see that no two numbers of the set +\begin{equation} +\gamma, \gamma^2, \gamma^3, \ldots, \gamma^{p^{\alpha-1}(p-1)} +\tag{A} +\end{equation} +are congruent modulo $p^\alpha$; for, if so, $\gamma$ would belong +modulo $p^\alpha$ to an exponent less than $p^{\alpha-1}(p-1)$ and +would therefore not be a primitive root modulo $p^\alpha$. Now every +number in the set (A) is prime to $p^\alpha$; their number is +$\phi(p^\alpha) = p^{\alpha -1}(p-1)$. Hence the numbers of the set +(A) are congruent in some order to the numbers of the set (B): +\begin{equation} +a_1,\ a_2,\ a_3,\ \ldots ,\ a_{p^{\alpha-1}(p-1)}, \tag{B} +\end{equation} +where the integers (B) are the positive integers less than +$p^\alpha$ and prime to $p^\alpha$. + +But any number of the set (B) is a solution of the congruence +\begin{equation} +x^{p^{\alpha-1} (p-1)} \equiv 1 \bmod p^\alpha. \tag{1} +\end{equation} +Further, every solution of this congruence is prime to $p^\alpha$. +Hence the integers (B) are a complete set of solutions of (1). +Therefore the integers (A) are a complete set of solutions of (1). +But it is easy to see that an integer $\gamma^k$ of the set (A) is +or is not a primitive root modulo $p^\alpha$ according as $k$ is or +is not prime to $p^{\alpha-1} (p-1)$. Hence the number of primitive +roots modulo $p^\alpha$ is $\phi \{p^{\alpha-1} (p-1) \}.$ + +The results thus obtained may be stated as follows: + +\smallskip II.~\emph{If $p$ is any odd prime number and $\alpha$ is +any positive integer, then there exist primitive roots modulo +$p^\alpha$ and their number is $\phi \{ \phi(p^\alpha) \}$}. + +\section{Primitive Roots Modulo $2p^\alpha$, $p$ an Odd +Prime}\label{s37} + +In this section we shall prove the following theorem: + +\emph{If $p$ is any odd prime number and $\alpha$ is any positive +integer, then there exist primitive roots modulo $2p^\alpha$ and +their number is $\phi \{\phi(2 p^{\alpha} )\}.$} + +Since $2 p^\alpha$ is even it follows that every primitive root +modulo $2 p^\alpha$ is an odd number. Any odd primitive root modulo +$p^\alpha$ is obviously a primitive root modulo $2p^\alpha$. Again, +if $\gamma$ is an even primitive root modulo $p^\alpha$ then $\gamma ++ p^\alpha$ is a primitive root modulo $2 p^\alpha$. It is evident +that these two classes contain (without repetition) all the +primitive roots modulo $2 p^\alpha$. Hence the theorem follows as +stated above. + +\section{Recapitulation}\label{s38} + +The results which we have obtained in \S\S \ref{s34}--\ref{s37} +inclusive may be gathered into the following theorem: + +\emph{In order that there shall exist primitive roots modulo $m$, it +is necessary and sufficient that $m$ shall have one of the values} +\begin{equation*} +m = 1, 2, 4, p^\alpha, 2p^\alpha +\end{equation*} +\emph{where $p$ is an odd prime and $\alpha$ is a positive integer.} + +\emph{If $m$ has one of these values then the number of primitive +roots modulo $m$ is $\phi\{\phi(m)\}$.} + +\section{Primitive $\lambda$-roots}\label{s39}% +\index{Primitive roots!$\lambda$-roots|(} + +In the preceding sections of this chapter we have developed the +theory of primitive roots in the way in which it is usually +presented. But if one approaches the subject from a more general +point of view the results which may be obtained are more general and +at the same time more elegant. It is our purpose in this section to +develop the more general theory. + +\smallskip We have seen that if $a$ and $m$ are any two relatively +prime positive integers, then +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +Consequently there is no integer belonging modulo $m$ to an exponent +greater than $\lambda(m)$. It is natural to enquire if there are any +integers $a$ which belong to the exponent $\lambda(m)$. It turns out +that the question is to be answered in the affirmative, as we shall +show. Accordingly, we introduce the following definition: + +\smallskip \textsc{Definition.} If $a^{\lambda(m)}$ is the lowest +power of $a$ which is congruent to $1$ modulo $m$, $a$ is said to be +a primitive $\lambda$-root modulo $m$. We shall also say that it is +a primitive $\lambda$-root of the congruence $x^{\lambda(m)} = 1 +\mod m$. To distinguish we may speak of the usual primitive root as +a primitive $\phi$-root modulo $m$.% +\index{Primitive roots!$\phi$-roots} + +From the theory of primitive $\phi$-roots already developed it +follows that primitive $\lambda$-roots always exist when $m$ is a +power of any odd prime, and also when $m = 1,\ 2,\ 4$; for, for such +values of $m$ we have $\lambda(m) = \phi(m)$. + +We shall next show that primitive $\lambda$-roots exist when $m = +2^{\alpha}$, $a > 2$, by showing that 5 is such a root. It is +necessary and sufficient to prove that $5$ belongs modulo +$2^{\alpha}$ to the exponent $2^{\alpha-2} = \lambda(2^{\alpha})$. +Let $d$ be the exponent to which $5$ belongs modulo $2^{\alpha}$. +Then from theorem II of \S \ref{s32} it follows that $d$ is a +divisor of $2^{\alpha-2} = \lambda(2^{\alpha})$. Hence if $d$ is +different from $2^{\alpha-2}$ it is $2^{\alpha-3}$ or is a divisor +of $2^{\alpha-3}$. Hence if we can show that $5^{2^{\alpha-3}}$ is +not congruent to $1$ modulo $2^{\alpha}$ we will have proved that +$5$ belongs to the exponent $2^{\alpha-2}$. But, clearly, +\begin{gather*} +5^{2^{\alpha-3}} = (1+2^2)^{2^{\alpha-3}} + = 1+2^{\alpha-1}+ I\cdot 2^{\alpha}, \\ +\intertext{where $I$ is an integer. Hence} +5^{2^{\alpha-3}} \not\equiv 1 \bmod 2^{\alpha}. +\end{gather*} +Hence 5 belongs modulo $2^{\alpha}$ to the exponent +$\lambda(2^{\alpha})$. + +By means of these special results we are now in position to prove +readily the following general theorem which includes them as special +cases: + +\smallskip I.~\emph{For every congruence of the form} +\begin{gather*} +x^{\lambda(m)} \equiv 1 \bmod m +\end{gather*} +\emph{a solution $g$ exists which is a primitive $\lambda$-root, and +for any such solution $g$ there are $\phi\{\lambda(m)\}$ primitive +roots congruent to powers of $g$.} + +If any primitive $\lambda$-root $g$ exists, $g^\nu$ is or is not a +primitive $\lambda$-root according as $\nu$ is or is not prime to +$\lambda(m)$; and therefore the number of primitive $\lambda$-roots +which are congruent to powers of any such root $g$ is +$\phi\{\lambda(m)\}$. + +The existence of a primitive $\lambda$-root in every case may easily +be shown by induction. In case $m$ is a power of a prime the theorem +has already been established. We will suppose that it is true when +$m$ is the product of powers of $r$ different primes and show that +it is true when $m$ is the product of powers of $r+1$ different +primes; from this will follow the theorem in general. + +Put $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_r^{\alpha_r} +p_{r+1}^{\alpha_{r+1}}, \quad n = p_1^{\alpha_1} p_2^{\alpha_2} +\ldots p_r^{\alpha_r}$, and let $h$ be a primitive $\lambda$-root of +\begin{gather} +x^{\lambda(n)} \equiv 1 \mod n. \tag{1} \\ +\intertext{Then} +h + ny \notag +\end{gather} +is a form of the same root if $y$ is an integer. + +Likewise, if $c$ is any primitive $\lambda$-root of +\begin{equation} +x^\lambda(p_{r+1}^{\alpha_{r+1}}) + \equiv 1 \mod p_{r+1}^{\alpha_{r+1}} \tag{2} +\end{equation} +a form of this root is +\begin{equation*} +c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +where $z$ is any integer. + +Now, if $y$ and $z$ can be chosen so that +\begin{equation*} +h+ny = c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +the number in either member of this equation will be a common +primitive $\lambda$-root of congruences (1) and (2); that is, a +common primitive $\lambda$-root of the two congruences may always be +obtained provided that the equation +\begin{equation*} +p_1^{\alpha_1} \ldots p_r^{\alpha_r}y - p_{r+1}^{\alpha_{r+1}}z = c-h +\end{equation*} +has always a solution in which $y$ and $z$ are integers. That this +equation has such a solution follows readily from theorem III of \S +\ref{s9}; for, if $c-h$ is replaced by $1$, the new equation has a +solution $\bar{y}$, $\bar{z}$; and therefore for $y$ and $z$ we may +take $y = \bar{y}(c-h)$, $z = \bar{z}(c-h)$. + +Now let $g$ be a common primitive $\lambda$-root of congruences (1) +and (2) and write +\begin{equation*} +g^\nu \equiv 1 \mod m, +\end{equation*} +where $\nu$ is to be the smallest exponent for which the congruence +is true. Since $g$ is a primitive $\lambda$-root of (1) $\nu$ is a +multiple of $\lambda(p_1^{\alpha_1} \ldots p_r^{\alpha_r})$. Since +$g$ is a primitive $\lambda$-root of (2) $\nu$ is a multiple of +$\lambda\left(p_{r+1}^{\alpha_{r+1}} \right)$. Hence it is a +multiple of $\lambda(m)$. But $g^{\lambda(m)} \equiv 1 \bmod m$; +therefore $\nu = \lambda(m)$. That is, $g$ is a primitive +$\lambda$-root modulo $m$. + +The theorem as stated now follows at once by induction. + +\smallskip There is nothing in the preceding argument to indicate +that the primitive $\lambda$-roots modulo $m$ are all in a single +set obtained by taking powers of some root $g$; in fact it is not in +general true when $m$ contains more than one prime factor. + +By taking powers of a primitive $\lambda$-root $g$ modulo $m$ one +obtains $\phi\{\lambda(m)\}$ different primitive $\lambda$-roots +modulo $m$. It is evident that if $\gamma$ is any one of these +primitive $\lambda$-roots, then the same set is obtained again by +taking the powers of $\gamma$. We may say then that the set thus +obtained is the set belonging to $g$. + +\smallskip II.~\emph{If $\lambda(m)>2$ the product of the +$\phi\{\lambda(m)\}$ primitive $\lambda$-roots in the set belonging +to any primitive $\lambda$-root $g$ is congruent to $1$ modulo $m$.} + +These primitive $\lambda$-roots are +\begin{gather*} +g,\ g^{c_1},\ g^{c_2},\ \ldots,\ g^{c_\mu} \\ +\intertext{where} +1,\ c_1,\ c_2,\ \ldots,\ c_\mu \\ +\end{gather*} +are the integers less than $\lambda(m)$ and prime to $\lambda(m)$. +If any one of these is $c$ another is $\lambda(m)-c$, since +$\lambda(m) > 2$. Hence +\begin{gather*} +1 + c_1 + c_2 + \ldots + c_\mu \equiv 0 \bmod \lambda(m). \\ +\intertext{Therefore} +g^{1 + c_1 + c_2 + \ldots + c_\mu} \equiv 1 \bmod m. +\end{gather*} +From this the theorem follows. + +\smallskip \textsc{Corollary.}\emph{The product of all the +primitive $\lambda$-roots modulo $m$ is congruent to $1$ modulo $m$ +when $\lambda(m) > 2$.}\index{Primitive roots!$\lambda$-roots|)} + +\begin{center} +EXERCISES +\end{center} + +\small\begin{enumerate} +\item[1.] If $x_1$ is the largest value of $x$ satisfying the equation +$\lambda(x) = a$, where $a$ is a given integer, then any solution +$x_2$ of the equation is a factor of $x_1$. + +\item[2*.] Obtain an effective rule for solving the equation +$\lambda(x) = a$. + +\item[3*.] Obtain an effective rule for solving the equation +$\phi(x) = a$. + +\item[4.] A necessary and sufficient condition that $a^{P-1} \equiv 1 +\mod P$ for every integer $a$ prime to $P$ is that $P \equiv 1 \mod +\lambda(P)$. + +\item[5.] If $a^{P-1} \equiv 1\mod P$ for every a prime to $P$, then +(1) $P$ does not contain a square factor other than $1$, (2) $P$ +either is prime or contains at least three different prime factors. + +\item[6.] Let $p$ be a prime number. If $a$ is a root of the congruence +$x^q \equiv 1 \mod p$ and $\alpha$ is a root of the congruence +$x^\delta\equiv 1 \mod p$, then $a\alpha$ is a root of the +congruence $x^{d\delta}\equiv 1 \mod p$. If $a$ is a primitive root +of the first congruence and $\alpha$ of the second and if $d$ and +$\delta$ are relatively prime, then $a\alpha$ is a primitive root of +the congruence $x^{d\delta} \equiv 1\mod p$. +\end{enumerate} \normalsize\index{Primitive roots|)} + +\chapter{OTHER TOPICS} + +\section{Introduction}\label{s40} + +The theory of numbers is a vast discipline and no single volume can +adequately treat of it in all of its phases. A short book can serve +only as an introduction; but where the field is so vast such an +introduction is much needed. That is the end which the present +volume is intended to serve; and it will best accomplish this end +if, in addition to the detailed theory already developed, some +account is given of the various directions in which the matter might +be carried further. + +To do even this properly it is necessary to limit the number of +subjects considered. Consequently we shall at once lay aside many +topics of interest which would find a place in an exhaustive +treatise. We shall say nothing, for instance, about the vast domain +of algebraic numbers, even though this is one of the most +fascinating subjects in the whole field of +mathematics.\index{Algebraic numbers} Consequently, we shall not +refer to any of the extensive theory connected with the division of +the circle into equal parts.\index{Circle, Division of} Again, we +shall leave unmentioned many topics connected with the theory of +positive integers; such, for instance, is the frequency of prime +numbers in the ordered system of integers---a subject which contains +in itself an extensive and elegant theory.\index{Prime numbers} + +In \S\S \ref{s41}--\ref{s44} we shall speak briefly of each of the +following topics: theory of quadratic residues, Galois imaginaries, +arithmetic forms, analytical theory of numbers. Each of these alone +would require a considerable volume for its proper development. All +that we can do is to indicate the nature of the problem in each case +and in some cases to give a few of the fundamental results. + +In the remaining three sections we shall give a brief introduction +to the theory of Diophantine equations, developing some of the more +elementary properties of certain special cases. We shall carry this +far enough to indicate the nature of the problem connected with the +now famous Last Theorem of Fermat. The earlier sections of this +chapter are not required as a preliminary to reading this latter +part. + +\section{Theory of Quadratic Residues}\label{s41}% +\index{Quadratic residues|(} + +Let $a$ and $m$ be any two relatively prime integers. In \S +\ref{s31} we agreed to say that $a$ is a quadratic residue modulo +$m$ or a quadratic non-residue modulo $m$ according as the +congruence +\begin{equation*} +x^2 \equiv a \bmod m +\end{equation*} +has or has not a solution. We saw that if $m$ is chosen equal to an +odd prime number $p$, then $a$ is a quadratic residue modulo $p$ or +a quadratic non-residue modulo $p$ according as +\begin{equation*} +a^{\frac{1}{2} (p-1)} \equiv 1\quad \mathrm{or}\quad + a^{\frac{1}{2} (p-1)} \equiv -1 \bmod p. +\end{equation*} +This is known as Euler's criterion.\index{Euler's!criterion} + +It is convenient to employ the Legendre symbol +\begin{equation*} +\left( \frac{a}{p} \right ) +\end{equation*} +to denote the quadratic character of $a$ with respect to $p$.% +\index{Legendre symbol} This symbol is to have the value $+1$ or the +value $-1$ according as $a$ is a quadratic residue modulo $p$ or a +quadratic non-residue modulo $p$. We shall now derive some of the +fundamental properties of this symbol, understanding always that the +numbers in the numerator and the denominator are relatively prime. + +From the definition of quadratic residues and non-residues it is +obvious that +\begin{equation} +\left ( \frac{a}{p} \right ) = \left ( \frac{b}{p} \right ) + \quad \text{if}\quad a \equiv b \bmod p. \tag{1} +\end{equation} + +It is easy to prove in general that +\begin{equation} +\left ( \frac{a}{p} \right ) \left ( \frac{b}{p} \right ) = + \left (\frac {ab}{p} \right ). \tag{2} +\end{equation} +This comes readily from Euler's criterion. We have to consider the +three cases +\begin{align*} +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=+1; & +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=-1; \\ +&& \left( \frac{a}{p} \right ) &=-1,& + \left( \frac{b}{p} \right ) &=-1. +\end{align*} +The method will be sufficiently illustrated by the treatment +of the last case. Here we have +\begin{gather*} +a^{\frac 12 (p-1)}\equiv -1 \bmod p,\quad + b^{\frac 12 (p-1)}\equiv -1 \bmod p. \\ +\intertext{Multiplying these two congruences together member by +member we have} +(ab)^{\frac 12 (p-1)} \equiv 1 \bmod p, \\ +\intertext{whence} +\left( \frac {ab}{p} \right ) = 1 = + \left( \frac ap \right ) \left( \frac bp \right ), +\end{gather*} +as was to be proved. + +If $m$ is any number prime to $p$ and we write $m$ as the product of +factors +\begin{equation*} +m = \epsilon \cdot 2^\alpha \cdot q' q'' q''' \cdots +\end{equation*} +where $q',\ q'',\ q''',\ \ldots$ are odd primes, $\alpha$ is zero or +a positive integer and $\epsilon$ is $+1$ or $-1$ according as $m$ +is positive or negative, we have +\begin{equation} +\left( \frac{m}{p} \right ) = +\left( \frac{\epsilon}{p} \right ) +\left( \frac{2}{p} \right ) ^\alpha +\left( \frac{q'}{p} \right ) +\left( \frac{q''}{p} \right ) +\left( \frac{q'''}{p} \right ) \ldots, \tag{3} +\end{equation} +as one shows easily by repeated application of relation (2). +Obviously, +\begin{equation*} +\left( \frac{1}{p} \right ) = 1. +\end{equation*} +Hence, it follows from (3) that we can readily determine the +quadratic character of $m$ with respect to the odd prime $p$, that +is, the value of +\begin{equation*} +\left( \frac{m}{p} \right ), +\end{equation*} +provided that we know the value of each of the expressions +\begin{equation} +\left( \frac{-1}{p} \right ),\quad + \left( \frac{2}{p} \right ),\quad + \left( \frac{q}{p} \right ),\tag{4} +\end{equation} +where $q$ is an odd prime. + +The first of these can be evaluated at once by means of Euler's +criterion; for, we have +\begin{gather*} +\left( \frac{-1}{p} \right ) \equiv + (-1)^{\frac{1}{2} (p-1)} \bmod p \\ +\intertext{and hence} +\left( \frac{-1}{p} \right ) = (-1)^{\frac{1}{2} (p-1)}. +\end{gather*} +Thus we have the following result: The number $-1$ is a quadratic +residue of every prime number of the form $4k + 1$ and a quadratic +non-residue of every prime number of the form $4k + 3$. + +The value of the second symbol in (4) is given by the formula +\begin{equation*} +\left( \frac{2}{p} \right ) = (-1)^{\frac{1}{8} (p^2 -1)}. +\end{equation*} +The theorem contained in this equation may be stated in the +following words: The number $2$ is a quadratic residue of every +prime number of either of the forms $8k + 1, 8k + 7$; it is a +quadratic non-residue of every prime number of either of the forms +$8k + 3, 8k + 5$. + +The proof of this result is not so immediate as that of the +preceding one. To evaluate the third expression in (4) is still more +difficult. We shall omit the demonstration in both of these cases. +For the latter we have the very elegant relation +\begin{equation*} +\left( \frac{p}{q} \right ) \left( \frac{q}{p} \right ) = + (-1)^{\frac{1}{4}(p-1)(q-1)}. +\end{equation*} +This equation states the law which connects the quadratic character +of $q$ with respect to $p$ with the quadratic character of $p$ with +respect to $q$. It is known as the Law of Quadratic Reciprocity. +About fifty proofs of it have been given. Its history has been a +very interesting one; see Bachmann's Niedere Zablentheorie, Teil I, +pp.\ 180--318, especially pp.\ 200--206.\index{Bachmann}% +\index{Law of quadratic reciprocity}\index{Quadratic reciprocity} + +For a further account of this beautiful and interesting subject we +refer the reader to Bachmann, loc.\ cit., and to the memoirs to +which this author gives reference.\index{Quadratic residues|)} + +\section{Galois Imaginaries}\label{s42}% +\index{Galois imaginaries}\index{Imaginaries of Galois} + +If one is working in the domain of real numbers the equation +\begin{equation*} +x^2 + 1 = 0 +\end{equation*} +has no solution; for there is no real number whose square is $-1$. +If, however, one enlarges the ``number system'' so as to include not +only all real numbers but all complex numbers as well, then it is +true that every algebraic equation has a root. It is on account of +the existence of this theorem for the enlarged domain that much of +the general theory of algebra takes the elegant form in which we +know it. + +The question naturally arises as to whether we can make a similar +extension in the case of congruences. The congruence +\begin{equation*} +x^2 = 3 \bmod 5 +\end{equation*} +has no solution, if we employ the term solution in the sense in +which we have so far used it. But we may if we choose introduce an +imaginary quantity, or mark, $j$ such that +\begin{equation*} +j^2 \equiv 3 \bmod 5, +\end{equation*} +just as in connection with the equation $x^2 + 1 = 0$ we would +introduce the symbol $i$ having the property expressed by the +equation +\begin{equation*} +i^2 = -1. +\end{equation*} + +It is found to be possible to introduce in this way a general set of +imaginaries satisfying congruences with prime moduli; and the new +quantities or marks have the property of combining according to the +laws of algebra. + +The quantities so introduced are called Galois imaginaries. + +We cannot go into a development of the important theory which is +introduced in this way. We shall be content with indicating two +directions in which it leads. + +In the first place there is the general Galois field theory which is +of fundamental importance in the study of certain finite groups. It +may be developed from the point of view indicated here. An excellent +exposition, along somewhat different lines, is to be found in +Dickson's \emph{Linear Groups with an Exposition of the Galois Field +Theory.}\index{Dickson} + +Again, the whole matter may be looked upon from the geometric point +of view. In this way we are led to the general theory of finite +geometries, that is, geometries in which there is only a finite +number of points. For a development of the ideas which arise here +see Veblen and Young's \emph{Projective Geometry} and the memoir by +Veblen and Bussey in the Transactions of the American Mathematical +Society, vol.\ 7, pp.\ 241--259.\index{Bussey}\index{Veblen}% +\index{Young} + +\section{Arithmetic Forms}\label{s43}% +\index{Arithmetic forms|(}\index{Forms|(} + +The simplest arithmetic form is $ax + b$ where $a$ and $b$ are fixed +integers different from zero and $x$ is a variable integer. By +varying $x$ in this case we have the terms of an arithmetic +progression. We have already referred to Dirichlet's celebrated +theorem which asserts that the form $ax + b$ has an infinite number +of prime values if only $a$ and $b$ are relatively +prime.\index{Dirichlet} This is an illustration of one type of +theorem connected with arithmetic forms in general, namely, those in +which it is asserted that numbers of a given form have in addition a +given property.\index{Prime numbers} + +Another type of theorem is illustrated by a result stated in \S +\ref{s41}, provided that we look at that result in the proper way. +We saw that the number $2$ is a quadratic residue of every prime of +either of the forms $8k + 1$ and $8k + 7$ and a quadratic +non-residue of every prime of either of the forms $8k + 3$ and $8k + +5$. We may state that result as follows: A given prime number of +either of the forms $8k + 1$ and $8k + 7$ is a divisor of some +number of the form $x^2 - 2$, where $x$ is an integer; no prime +number of either of the forms $8k + 3$ and $8k + 5$ is a divisor of +a number of the form $x^2 - 2$, where $x$ is an integer. + +The result just stated is a theorem in a discipline of vast extent, +namely, the theory of quadratic forms. Here a large number of +questions arise among which are the following: What numbers can be +represented in a given form? What is the character of the divisors +of a given form? As a special case of the first we have the question +as to what numbers can be represented as the sum of three squares. +To this category belong also the following two theorems: Every +positive integer is the sum of four squares of integers; every prime +number of the form $4n + 1$ may be represented (and in only one way) +as the sum of two squares.\index{Prime numbers} + +For an extended development of the theory of quadratic forms we +refer the reader to Bachmann's Arithmetik der Quadratischen Formen +of which the first part has appeared in a volume of nearly seven +hundred pages.\index{Bachmann} + +It is clear that one may further extend the theory of arithmetic +forms by investigating the properties of those of the third and +higher degrees. Naturally the development of this subject has not +been carried so far as that of quadratic forms; but there is a +considerable number of memoirs devoted to various parts of this +extensive field, and especially to the consideration of various +special forms. + +Probably the most interesting of these special forms are the +following: +\begin{equation*} +\alpha^n + \beta^n , \quad + \frac{\alpha^n - \beta^n}{\alpha - \beta} = + \alpha^{n-1} + \alpha^{n-2} \beta + \cdots + \beta^{n-1}, +\end{equation*} +where $\alpha$ and $\beta$ are relatively prime integers, or, more +generally, where $\alpha$ and $\beta$ are the roots of the quadratic +equation $x^2 - ux + v = 0$ where $u$ and $v$ are relatively prime +integers. A development of the theory of these forms has been given +by the present author in a memoir published in 1913 in the Annals of +Mathematics, vol.\ 13, pp.\ 30--70.% +\index{Arithmetic forms|)}\index{Carmichael}\index{Forms|)}% +\index{Quadratic forms} + +\section{Analytical theory of numbers}\label{s44}% +\index{Analytical theory of numbers|(} + +Let us consider the function +\begin{equation*} +P(x) = \frac{1}{\prod_{k=0}^\infty (1-x^{2^k} )} , \quad + |x|\leqq \rho < 1. +\end{equation*} +It is clear that we have +\begin{align*} +P(x) = \prod_{k=0}^\infty \frac{1}{(1-x^{2^k} )} &= + \prod_{k=0}^\infty + ( 1 + x^{2k} + x^{2\cdot 2^k} + x^{3\cdot 2^k} + \cdots ) \\ +&= \sum_{s=0}^\infty G(s) x^s, +\end{align*} +where $G(0) = 1$ and $G(s)$ (for $s$ greater than $0$) is the number +of ways in which the positive integer $s$ may be separated into like +or distinct summands each of which is a power of $2$. + +We have readily +\begin{equation*} +(1-x)\sum_{s=0}^\infty G(s) x^s = (1-x)P(x) = P(x^2) = + \sum_{s=0}^\infty x^{2^s}; +\end{equation*} +whence +\begin{equation} +G(2s + 1) = G(2s) = G(2s - 1) + G(s), \tag{A} +\end{equation} +as one readily verifies by equating coefficients of like powers of +$x$. From this we have in particular +\begin{gather*} +G(0) = 1, \quad G(1) = 1, \quad G(2) = 2, \quad G(3) = 2, \\ +G(4) = 4, \quad G(5) = 4, \quad G(6) = 6, \quad G(7) = 6. +\end{gather*} +Thus in (A) we have recurrence relations by means of which we may +readily reckon out the values of the number theoretic function +$G(s)$. Thus we may determine the number of ways in which a given +positive integer $s$ may be represented as a sum of powers of $2$. + +We have given this example as an elementary illustration of the +analytical theory of numbers, that is, of that part of the theory of +numbers in which one employs (as above) the theory of a continuous +variable or some analogous theory in order to derive properties of +sets of integers. This general subject has been developed in several +directions. For a systematic account of it the reader is referred to +Bachmann's Analytische Zahlentheorie.% +\index{Analytical theory of numbers|)}\index{Bachmann} + +\section{Diophantine equations}\label{s45}% +\index{Diophantine equations}\index{Equations!Diophantine} + +If $f(x, y, z, \ldots)$ is a polynomial in the variables $x, y, z, +\ldots$ with integral coefficients, then the equation +\begin{equation*} +f(x, y, z, \ldots) = 0 +\end{equation*} +is called a Diophantine equation when we look at it from the point +of view of determining the integers (or the positive integers) $x, +y, z, \ldots$ which satisfy it. Similarly, if we have several such +functions $f_i(x, y, z, \ldots)$, in number less than the number of +variables $x, y, z, \ldots$, then the set of equations +\begin{equation*} +f_i(x, y, z, \ldots) = 0,\quad i = i, 2, \ldots, +\end{equation*} +is said to be a Diophantine system of equations. Any set of integers +$x, y, z, \ldots$ which satisfies the equation [system] is said to +be a solution of the equation [system]. + +We may likewise define Diophantine inequalities by replacing the +sign of equality above by the sign of inequality. But little has +been done toward developing a theory of Diophantine inequalities. +Even for Diophantine equations the theory is in a rather fragmentary +state. + +In the next two sections we shall illustrate the nature of the ideas +and the methods of the theory of Diophantine equations by developing +some of the results for two important special cases. + +\section{Pythagorean triangles}\label{s46}% +\index{Pythagorean triangles|(} + +\textsc{Definitions.} If three positive integers $x, y, z$ satisfy +the relation +\begin{equation} +x^2 + y^2 = z^2 \tag{1} +\end{equation} +they are said to form a Pythagorean triangle or a numerical right +triangle; $z$ is called the hypotenuse of the triangle and $x$ and +$y$ are called its legs. The area of the triangle is said to be +$\frac{1}{2} xy$.\index{Triangles, Numerical} + +We shall determine the general form of the integers $x$, $y$, $z$, +such that equation (1) may be satisfied. Let us denote by $\nu$ the +greatest common divisor of $x$ and $y$ in a particular solution of +(1). Then $\nu$ is a divisor of $z$ and we may write +\begin{equation*} +x = \nu u, \quad y = \nu v,\quad z = \nu w. +\end{equation*} +Substituting these values in (1) and reducing we have +\begin{equation} +u^2 + v^2 = w^2, \tag{2} +\end{equation} +where $u, v, w$ are obviously prime each to each, since $u$ and $v$ +have the greatest common divisor $1$. + +Now an odd square is of the form $4k + 1$. Hence the sum of two odd +squares is divisible by $2$ but not by $4$; and therefore the sum of +two odd squares cannot be a square. Hence one of the numbers $u$, +$v$ is even. Suppose that $u$ is even and write equation (2) in the +form +\begin{equation} +u^2 = (w - v)(w + v). \tag{3} +\end{equation} +Every common divisor of $w - v$ and $w + v$ is a divisor of their +difference $2v$. Therefore, since $w$ and $v$ are relatively prime, +it follows that $2$ is the greatest common divisor of $w - v$ and $w ++ v$. Then from (3) we see that each of these numbers is twice a +square, so that we may write +\begin{equation*} +w - v = 2b^2,\quad w + v = 2a^2 +\end{equation*} +where $a$ and $b$ are relatively prime integers. From these two +equations and equation (3) we have +\begin{equation} +w = a^2 + b^2, \quad v = a^2 -b^2,\quad u = 2ab. \tag{4} +\end{equation} +Since $u$ and $v$ are relatively prime it is evident that one of the +numbers $a$, $b$ is even and the other odd. + +The forms of $u$, $v$, $w$ given in (4) are necessary in order that +(2) may be satisfied. A direct substitution in (2) shows that this +equation is indeed satisfied by these values. Hence we have in (4) +the general solution of (2) where $u$ is restricted to be even. A +similar solution would be obtained if $v$ were restricted to be +even. Therefore \emph{the general solution of (1) is +\begin{gather*} +x = 2\nu ab,\quad y = \nu (a^2 - b^2),\quad z = \nu (a^2 + b^2)\\ +\intertext{and} +x = 2\nu (a^2 - b^2 ),\quad y = 2\nu ab,\quad z = \nu (a^2 + b^2) +\end{gather*} +where $a$, $b$, $\nu$ are arbitrary integers except that $a$ and $b$ +are relatively prime and one of them is even and the other odd.} + +By means of this general solution of (1) we shall now prove the +following theorem: + +\smallskip I.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero, such that} +\begin{equation} +q^2 + n^2 = m^2 , \quad m^2 + n^2 = p^2. \tag{5} +\end{equation} + +It is obvious that an equivalent theorem is the following: + +\smallskip II.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero such that} +\begin{equation} +p^2 + q^2 = 2m^2, \quad p^2 - q^2 = 2n^2. \tag{6} +\end{equation} + +Obviously, we may without loss of generality take $m$, $n$, $p$, $q$ +to be positive; and this we do. + +The method of proof is to assume the existence of integers +satisfying equations (5) and (6) and to show that we are thus led to +a contradiction. The argument we give is an illustration of Fermat's +famous method of ``infinite descent.''% +\index{Descent, Infinite}\index{Fermat}\index{Infinite descent} + +If any two of the numbers $p$, $q$, $m$, $n$ have a common prime +factor $t$, it follows at once from (5) and (6) that all four of +them have this factor. For, consider an equation in (5) or in (6) in +which these two numbers occur; this equation contains a third +number, and it is readily seen that this third number is divisible +by $t$. Then from one of the equations containing the fourth number +it follows that this fourth number is divisible by $t$. Now let us +divide each equation of system (6) through by $t^2$; the resulting +system is of the same form as (6). If any two numbers in this +resulting system have a common prime factor $t_1$, we may divide +through by $t_1^2$; and so on. Hence if a pair of simultaneous +equations (6) exists then there exists a pair of equations of the +same form in which no two of the numbers $m$, $n$, $p$, $q$ have a +common factor other than unity. Let this system of equations be +\begin{equation} +p_1^2 + q_1^2 = 2m_1^2, \quad p_1^2 - q_1^2 = 2n_1^2. \tag{7} +\end{equation} + +From the first equation in (7) it follows that $p_1$ and $q_1$ are +both even or both odd; and, since they are relatively prime, it +follows that they are both odd. Evidently $p_1 > q_1$. Then we may +write +\begin{equation*} +p_1 = q_1 + 2\alpha, +\end{equation*} +where $\alpha$ is a positive integer. If we substitute this value of +$p_1$ in the first equation of (7), the result may readily be put in +the form +\begin{equation} +(q_1 + \alpha)^2 + a^2 = m_1^2. \tag{8} +\end{equation} +Since $q_1$ and $m_1$ have no common prime factor it is easy to see +from this equation that $\alpha$ is prime to both $q_1$ and $m_1$, +and hence that no two of the numbers $q_1 + \alpha, \alpha, m_1$ +have a common factor. + +Now we have seen that if $a$, $b$, $c$ are positive integers no two +of which have a common prime factor, while +\begin{equation*} +a^2 + b^2 = c^2, +\end{equation*} +then there exist relatively prime integers $r$ and $s$, $r > s$, +such that +\begin{gather} +c = r^2 + s^2,\quad a = 2rs,\quad b = r^2 - s^2 \notag \\ +\intertext{or} +c = r^2 + s^2,\quad a = r^2 - s^2,\quad b = 2rs. \notag \\ +\intertext{Hence from (8) we see that we may write} +q_1 + \alpha = 2rs,\quad \alpha = r^2 - s^2 \tag{9} \\ +\intertext{or} +q_1 + \alpha = r^2 - s^2, \alpha = 2rs. \tag{10} \\ +\intertext{In either case we have} +p_1^2 - q_1^2 = (p_1 - q_1)(p_1 + q_1) = + 2\alpha \cdot 2(q_1 + \alpha) = 8rs(r^2 - s^2). \notag \\ +\intertext{If we substitute in the second equation of (7) and divide +by 2 we have} 4rs(r^2 - s^2) = n_1^2. \notag +\end{gather} + +From this equation and the fact that $r$ and $s$ are relatively +prime it follows at once that $r$, $s$, $r^2 - s^2$ are all square +numbers; say, +\begin{gather} +r = u^2,\quad s = v^2,\quad r^2 - s^2 = w^2. \notag \\ +\intertext{Now $r - s$ and $r + s$ can have no common factor other +than 1 or 2; hence from} +w^2 = (r^2-s^2) = (r-s)(r+s) = (u^2-v^2)(u^2+v^2) \notag \\ +\intertext{we see that either} +u^2 + v^2 = 2w_1^2,\quad u^2 - v^2 = 2w_2^2 \tag{11} \\ +\intertext{or} +u^2 + v^2 = w_1^2,\quad u^2 - v^2 = w_2^2. \notag \\ +\intertext{And if it is the latter case which arises, then} +w_1^2 + w_2^2 = 2u^2,\quad w_1^2 - w_2^2 = 2v^2. \tag{12} +\end{gather} +Hence, assuming equations of the form (6) we are led either to +equations (11) or to equations (12); that is, we are led to new +equations of the form with which we started. Let us write the +equations thus: +\begin{equation} +p_2^2 + q_2^2 = 2m_2^2,\quad p_2^2 - q_2^2 = 2n_2^2; \tag{13} +\end{equation} +that is, system (13) is identical with that one of systems (11), +(12) which actually arises. + +Now from (9) and (10) and the relations $p_1 = q_1 + 2\alpha, r +> s$, we see that +\begin{gather*} +p_1 = 2rs + r^2 - s^2 > 2s^2 + r^2 - s^2 = + r^2 + s^2 = u^4 + v^4. \\ +\intertext{Hence $u < p_1$. Also,} +w_1^2 \leqq w^2 \leqq r+s < r^2 + s^2. +\end{gather*} +Hence $w_1 < p_1$. Since $u$ and $w_1$ are both less than $p_1$ it +follows that $p_2$ is less than $p_1$. Hence, obviously, $p_2 < p$. +Moreover, it is clear that all the numbers $p_2, q_2, m_2, n_2$ are +different from zero. + +From these results we have the following conclusion: If we assume a +system of the form (6) we are led to a new system (13) of the same +form; and in the new system $p_2$ is less than $p$. + +Now if we start with (13) and carry out a similar argument +we shall be led to a new system +\begin{gather*} +p_3^2 + q_3^2 = 2m_3^2,\quad p_3^2 - q_3^2 = 2n_3^2, +\end{gather*} +with the relation $p_3 < p_2$, starting from this last system we +shall be led to a new one of the same form, with a similar relation +of inequality; and so on \emph{ad infinitum.} But, since there is +only a finite number of positive integers less than the given +positive integer $p$ this is impossible. We are thus led to a +contradiction; whence we conclude at once to the truth of II and +likewise of I. + +By means of theorems I and II we may readily prove the following +theorem: + +\smallskip III.~\emph{The area of a numerical right triangle is +never a square number.} + +Let the sides and hypotenuse of a numerical right triangle be $u, v, +w$, respectively. The area of this triangle is $\frac{1}{2} uv$. If +we assume this to be a square number $t^2$ we shall have the +following simultaneous Diophantine equations +\begin{equation} +u^2 + v^2 = w^2,\quad uv = 2t^2. \tag{14} +\end{equation} +We shall prove our theorem by showing that the assumption of such a +system leads to a contradiction. + +If any two of the numbers $u, v, w$ have a common prime factor $p$ +then the remaining one also has this factor, as one sees readily +from the first equation in (14). From the second equation in (14) it +follows that $t$ also has the same factor. Then if we put $u = pu_1, +v = pv_1, w = pw_1, t = pt_1$, we have +\begin{equation*} +u_1^2 + v_1^2 = w_1^2,\quad u_1 v_1 = 2t_1^2, +\end{equation*} +a system of the same form as (14). It is clear that we may start +with this new system and proceed in the same manner as before, and +so on, until we arrive at a system +\begin{equation} +\bar{u}^2 + \bar{v}^2 = \bar{w}^2,\quad + \bar{u}\bar{v} = 2\bar{t}^2, \tag{15} +\end{equation} +where $\bar{u}$, $\bar{v}$, $\bar{w}$ are prime each to each. + +Now the general solution of the first equation (15) may be written +in one of the forms +\begin{gather*} +\bar{u} = 2ab,\quad \bar{v} = a^2 - b^2,\quad \bar{w} = a^2 + b^2 \\ +\bar{u} = a^2 b^2,\quad \bar{v} = 2ab, \quad \bar{w} = a^2 + b^2. \\ +\intertext{Then from the second equation in (15) we have} +\bar{t}^2 = ab(a^2 - b^2 ) = ab(a-b)(a+b). +\end{gather*} +It is easy to see that no two of the numbers $a$, $b$, $a - b$, $a + +b$ in the last member of this equation have a common factor; for, if +so, $\bar{u}$ and $\bar{v}$ would have a common factor, contrary to +hypothesis. Hence each of these four numbers is a square. That is, +we have equations of the form +\begin{gather*} +a = m^2,\quad b = n^2,\quad a + b = p^2,\quad a - b = q^2; \\ +\intertext{whence} +m^2 - n^2 = q^2,\quad m^2 + n^2 = p^2. +\end{gather*} +But, according to theorem I, no such system of equations can exist. +That is, the assumption of equations (14) leads to a contradiction. +Hence the theorem follows as stated above.% +\index{Pythagorean triangles|)} + +\section{The Equation $x^n + y^n = z^n$.}\label{s47}% +\index{Equation $x^n + y^n = z^n$|(}\index{Fermat's!last theorem} + +The following theorem, which is commonly known as Fermat's Last +Theorem, was stated without proof by Fermat in the seventeenth +century: + +\smallskip\emph{If n is an integer greater than 2 there do not exist +integers x, y, z, all different from zero, such that} +\begin{equation} +x^n + y^n = z^n. \tag{1} +\end{equation} + +No general proof of this theorem has yet been given. For various +special values of $n$ the proof has been found; in particular, for +every value of $n$ not greater than 100. + +In the study of equation (1) it is convenient to make some +preliminary reductions. If there exists any particular solution of +(1) there exists also a solution in which $x$, $y$, $z$ are prime +each to each, as one may show readily by the method employed in the +first part of \S \ref{s46}. Hence in proving the impossibility of +equation (1) it is sufficient to treat only the case in which $x$, +$y$, $z$ are prime each to each. + +Again, since $n$ is greater than 2 it must contain the factor +4 or an odd prime factor $p$. If $n$ contains the factor $p$ we write +$n = mp$, whence we have +\begin{gather*} +(x^m)^p + (y^m)^p = (z^m)^p). \\ +\intertext{If $n$ contains the factor 4 we write $n = 4m$, whence we +have} +(x^m)^4 + (y^m)^4 = (z^m)^4. +\end{gather*} +From this we see that in order to prove the impossibility of (1) in +general it is sufficient to prove it for the special cases when $n$ +is 4 and when $n$ is an odd prime $p$. For the latter case the proof +has not been found. For the former case we give a proof below. The +theorem may be stated as follows: + +\smallskip I.~\emph{There are no integers $x, y, z$, all different +from zero, such that} +\begin{equation*} +x^4 + y^4 = z^4. +\end{equation*} + +This is obviously a special case of the more general theorem: + +\smallskip II.~\emph{There are no integers $p$, $q$, $\alpha$, all +different from zero, such that} +\begin{equation} +p^4 - q^4 = \alpha^2. \tag{2} +\end{equation} + +The latter theorem is readily proved by means of theorem III of \S +\ref{s46}. For, if we assume an equation of the form (2), we have +\begin{gather} +(p^4 - q^4)p^2 q^2 = p^2 q^2 \alpha^2. \tag{3} \\ +\intertext{But, obviously,} +(2p^2 q^2)^2 + (p^4 - q^4)^2 = (p^4 + q^4)^2. \tag{4} +\end{gather} +Now, from (3) we see that the numerical right triangle determined by +(4) has its area $p^2 q^2(p^4 - q^4)$ equal to the square number +$p^2 q^2 \alpha^2$. But this is impossible. Hence no equation of the +form (2) exists. + +\begin{center} +EXERCISES +\end{center} + +\begin{enumerate} +\item[1.] Show that the equation $\alpha^4 + 4\beta^4 = \gamma^2$ is +impossible in integers $\alpha$, $\beta$, $\gamma$ all of which are +different from zero. + +\item[2.] Show that the system $p^2 - q^2 = km^2$, $p^2 + q^2 = kn^2$ +impossible in integers $p$, $q$, $k$, $m$, $n$, all of which are +different from zero. + +\item[3*.] Show that neither of the equations $m^4 - 4n^4 = \pm t^2$ +is possible in integers $m$, $n$, $t$, all of which are different +from zero. + +\item[4*.] Prove that the area of a numerical right triangle is not +twice a square number. + +\item[5*.] Prove that the equation $m^4 + n^4 = \alpha^2$ is not +possible in integers $m$, $n$, $\alpha$ all of which are different +from zero. + +\item[6*.] In the numerical right triangle $a^2 + b^2 = c^2$, +not more than one of the numbers $a$, $b$, $c$ is a square. + +\item[7.] Prove that the equation $x^{2k} + y^{2k} = z^{2k}$ implies +an equation of the form $m^k + n^k = 2^{k-2} t^k$. + +\item[8.] Find the general solution in integers of the equation +$x^2 + 2y^2 = t^2$. + +\item[9.] Find the general solution in integers of the equation +$x^2 + y^2 = z^4$. + +\item[10.] Obtain solutions of each of the following Diophantine +equations: +\begin{align*} +x^3 + y^3 + z^3 &= 2t^3, \\ +x^3 + 2y^3 + 3z^3 &= t^3, \\ +x^4 + y^4 + 4z^4 &= t^4, \\ +x^4 + y^4 + z^4 &= 2t^4. +\end{align*} +\end{enumerate}\index{Equation $x^n + y^n = z^n$|)} + +\addcontentsline{toc}{chapter}{Index} +\printindex + + +\newpage +\chapter{PROJECT GUTENBERG "SMALL PRINT"} +\small +\pagenumbering{gobble} +\begin{verbatim} + +End of Project Gutenberg's The Theory of Numbers, by Robert D. 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Hart was the originator of the Project Gutenberg-tm +concept of a library of electronic works that could be freely shared +with anyone. For forty years, he produced and distributed Project +Gutenberg-tm eBooks with only a loose network of volunteer support. + +Project Gutenberg-tm eBooks are often created from several printed +editions, all of which are confirmed as Public Domain in the U.S. +unless a copyright notice is included. Thus, we do not necessarily +keep eBooks in compliance with any particular paper edition. + +Most people start at our Web site which has the main PG search facility: + + www.gutenberg.org + +This Web site includes information about Project Gutenberg-tm, +including how to make donations to the Project Gutenberg Literary +Archive Foundation, how to help produce our new eBooks, and how to +subscribe to our email newsletter to hear about new eBooks. + +\end{verbatim} +\normalsize + +\iffalse +-----File: 092.png---\ggna\JohnHa0\---------------------------------------- +INDEX + +Algebraic numbers, 76. + +Analytical theory of numbers, 83--84. + +Arithmetic forms, 81--83. + +Arithmetic progression, 13. + + + +Bachmann, 80, 82, 84. + +Bussey, 81. + + + +Carmichael, 83. + +Circle, Division of, 76. + +Common divisors, 9, 18--20, 21. + multiples, 9, 20--21. + +Composite numbers, 10. + +Congruences, 37--46. + Linear, 43--46, 56. + Solution by trial, 39--40. + with prime modulus, 41--43. + + + +Descent, Infinite, 86. + +Dickson, 81. + +Diophantine equations, 84. + +Dirichlet, 81. + +Divisibility, 8. + +Divisors of a numbers, 16, 17. + + + +Equations, Diophantine, 84. + +Equation $x^n + y^n =z^n$, 91--92. + +Eratosthenes, 11. + +Euclid, Theorem of, 13. + +Euclidian algorithm, 18. + +Euler, 28, 48. + +Euler's criterion, 59, 77. + $\phi$-function, 30. + +Exponent of an integer, 61--63. + + + +Factorization theorem, 14. + +Factors, 14, 16, 17, 18. + +Fermat, 28, 48, 86. + +Fermat's general theorem, 47, 63. + last theorem, 91. + simple theorem, 48, 55. + theorem extended, 52--54. + +Forms, 81--83. + +Fundamental notions, 7. + + + +Galois imaginaries, 80. + +Gauss, 37. + +Greatest common factor, 18--20, 21. + + + +Highest power of $p$ in $n!$, 24--28. + + + +Imaginaries of Galois, 80. + +Indicator, 30--36. + of any integer, 32--34. + of a prime power, 30. + of a product, 30--32. + +Infinite descent, 86. + + + +$\lambda(m), 53. + +Law of quadratic reciprocity, 80. + +Least common multiple, 20--21. + +Legendre symbol, 77. + + + +$\phi(m), 30. + +Prime each to each, 9. + +Prime numbers, 10, 12, 13, 28--29, 51, 76, 81, 82. + +Primitive roots, 61--75. + $\lambda$-roots, 71--74. + $\psi$-roots, 71. + +Pythagorean triangles, 85--90. +-----File: 093.png---\ggna\JohnHa0\---------------------------------------- + +Quadratic forms, 82. + +Quadratic reciprocity, 80. + +Quadratic residues, 57--60, 77--80. + + + +Relatively prime, 10. + +Residue, 37, 58. + + + +Scales of notation, 22--24. + +Sieve of Eratosthenes, 10. + + + +Totient, 30. + +Triangles, Numerical, 85. + + + +Unit, 8. + + + +Veblen, 81. + + + +Wilson's theorem, 49--81. + + + +Young, 81. +\fi +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% End of Project Gutenberg's The Theory of Numbers, by Robert D. Carmichael +% % +% *** END OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** % +% % +% ***** This file should be named 13693-t.tex or 13693-t.zip ***** % +% This and all associated files of various formats will be found in: % +% http://www.gutenberg.org/1/3/6/9/13693/ % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\end{document} + +This is pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) (format=pdflatex 2012.9.24) 8 APR 2013 03:28 +entering extended mode + %&-line parsing enabled. +**13693-t.tex +(./13693-t.tex +LaTeX2e <2009/09/24> +Babel <v3.8l> and hyphenation patterns for english, usenglishmax, dumylang, noh +yphenation, farsi, arabic, croatian, bulgarian, ukrainian, russian, czech, slov +ak, danish, dutch, finnish, french, basque, ngerman, german, german-x-2009-06-1 +9, ngerman-x-2009-06-19, ibycus, monogreek, greek, ancientgreek, hungarian, san +skrit, italian, latin, latvian, lithuanian, mongolian2a, mongolian, bokmal, nyn 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Carmichael[] + [] + +[1 + +{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] [1 + +] +LaTeX Font Info: Try loading font information for U+msa on input line 186. +(/usr/share/texmf-texlive/tex/latex/amsfonts/umsa.fd +File: umsa.fd 2009/06/22 v3.00 AMS symbols A +) +LaTeX Font Info: Try loading font information for U+msb on input line 186. +(/usr/share/texmf-texlive/tex/latex/amsfonts/umsb.fd +File: umsb.fd 2009/06/22 v3.00 AMS symbols B +) [2] [3 + +] [4 + +] (./13693-t.toc [5 + +]) +\tf@toc=\write4 +\openout4 = `13693-t.toc'. + +[6] +Chapter 1. +[1 + + +] +LaTeX Font Info: Try loading font information for OMS+cmr on input line 401. + +(/usr/share/texmf-texlive/tex/latex/base/omscmr.fd +File: omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions +) +LaTeX Font Info: Font shape `OMS/cmr/m/n' in size <10> not available +(Font) Font shape `OMS/cmsy/m/n' tried instead on input line 401. +[2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [1 +9] +Chapter 2. +[20 + +] [21] +Overfull \hbox (15.17194pt too wide) in paragraph at lines 1607--1608 +[]\OT1/cmr/m/it/10 If $\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 p[ +]p[] [] p[]$ \OT1/cmr/m/it/10 where $\OML/cmm/m/it/10 p[]; p[]; [] ; p[]$ \OT1/ +cmr/m/it/10 are dif-fer-ent primes and $\OML/cmm/m/it/10 []; []; [] ; []$ + [] + +[22] [23] [24] [25] +Chapter 3. +[26 + +] [27] [28] [29] [30] [31] [32] [33] +Chapter 4. +[34 + +] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] +Chapter 5. +[46 + +] [47] [48] [49] [50] [51] [52] +Overfull \hbox (2.6285pt too wide) in paragraph at lines 3312--3312 +[]\OT1/cmr/bx/n/14.4 Primitive Roots Mod-ulo $\OT1/cmr/m/n/14.4 2\OML/cmm/m/it/ +14.4 p[]$\OT1/cmr/bx/n/14.4 , $\OML/cmm/m/it/14.4 p$ \OT1/cmr/bx/n/14.4 an Odd +Prime + [] + +[53] [54] [55] [56] [57] +Chapter 6. +[58 + +] [59] [60] [61] [62] [63] [64] [65] [66] [67] [68] [69] [70] (./13693-t.ind [7 +1] [72 + +] [73 + +]) +Chapter 7. +[1 + + +] [2] [3] [4] [5] [6] [7] [8] (./13693-t.aux) + + 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Carmichael +% % +% This eBook is for the use of anyone anywhere in the United States and % +% most other parts of the world at no cost and with almost no restrictions% +% whatsoever. You may copy it, give it away or re-use it under the terms % +% of the Project Gutenberg License included with this eBook or online at % +% www.gutenberg.org. If you are not located in the United States, you % +% will have to check the laws of the country where you are located before % +% using this eBook. % +% % +% % +% Title: The Theory of Numbers % +% % +% Author: Robert D. Carmichael % +% % +% Release Date: April 8, 2013 [EBook #13693] % +% % +% Language: English % +% % +% Character set encoding: TeX % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{13693} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: Additional mathematical symbols. Required. %% +%% %% +%% makeidx: Indexing. 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TWO times %% +%% makeindex 13693-t.idx %% +%% pdflatex 13693-t.tex %% +%% %% +%% pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) %% +%% %% +%% November 2021: okrick. %% +%% MiKTeX Console 4.3, Windows 10 Home %% +%% TeXworks 0.6.6 used to generate PDF output. %% +%% %% +%% %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\listfiles +\documentclass[oneside]{book} +\usepackage[latin1]{inputenc} +\usepackage[reqno]{amsmath} +\usepackage{amssymb} +\usepackage{makeidx} +\makeindex +\begin{document} + +\thispagestyle{empty} +\small +\begin{verbatim} +The Project Gutenberg EBook of The Theory of Numbers, by Robert D. Carmichael + +This eBook is for the use of anyone anywhere in the United States and +most other parts of the world at no cost and with almost no restrictions +whatsoever. You may copy it, give it away or re-use it under the terms +of the Project Gutenberg License included with this eBook or online at +www.gutenberg.org. If you are not located in the United States, you +will have to check the laws of the country where you are located before +using this eBook. + + +Title: The Theory of Numbers + +Author: Robert D. Carmichael + +Release Date: October 10, 2003 [eBook #13693] +Revised Date: November 30, 2021 + +Language: English + +Character set encoding: TeX + +*** START OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +Produced by David Starner, Joshua Hutchinson, John Hagerson. +Revised by Richard Tonsing +\end{verbatim} +\normalsize +\newpage + +\frontmatter + +\begin{center} +\noindent \Large MATHEMATICAL MONOGRAPHS \\ + +\bigskip \footnotesize{EDITED BY} \\ +\normalsize \textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD.} \\ + +\bigskip\bigskip \huge +No. 13. + +\bigskip\bigskip \huge THE THEORY \\ +\bigskip\small \textsc{of} \\ +\bigskip\huge NUMBERS \\ + +\bigskip\bigskip\footnotesize\textsc{by} \\ +\bigskip\large ROBERT D. CARMICHAEL, \\ +\footnotesize\textsc{Associate Professor of Mathematics in Indiana +University} + +\bigskip\bigskip\normalsize NEW YORK: \\ +\medskip JOHN WILEY \& SONS. \\ +\medskip \textsc{London: CHAPMAN \& HALL, Limited.} \\ +\medskip 1914. + +\bigskip\bigskip +\tiny \textsc{Copyright 1914} \\ +\textsc{by} \\ +ROBERT D. CARMICHAEL. \\ +\medskip \textsc{the scientific press} \\ +\textsc{robert drummond and company} \\ +\textsc{brooklyn, n.~y.} +\end{center} + +\bigskip\bigskip +\scriptsize \noindent \textsc{Transcriber's Note:} \emph{I did my +best to recreate the index.} \normalsize + +\newpage + +\fbox{\parbox{11cm}{ +\begin{center} +\textbf{MATHEMATICAL MONOGRAPHS.} \\ +\small\textsc{edited by}\normalsize \\ +\textbf{Mansfield Merriman and Robert S. Woodward.} \\ +\footnotesize \textbf{Octavo. Cloth. \$1.00 each.} \\ + +\bigskip \textbf{No. 1. History of Modern Mathematics.} \\ +By \textsc{David Eugene Smith.} + +\smallskip \textbf{No. 2. Synthetic Projective Geometry.} \\ +By \textsc{George Bruce Halsted.} + +\smallskip \textbf{No. 3. Determinants.} \\ +By \textsc{Laenas Gifford Weld.} + +\smallskip \textbf{No. 4. Hyperbolic Functions.} \\ +By \textsc{James McMahon.} + +\smallskip \textbf{No. 5. Harmonic Functions.} \\ +By \textsc{William E. Byerly.} + +\smallskip \textbf{No. 6. Grassmann's Space Analysis.} \\ +By \textsc{Edward W. Hyde.} + +\smallskip \textbf{No. 7. Probability and Theory of Errors.} \\ +By \textsc{Robert S. Woodward.} + +\smallskip \textbf{No. 8. Vector Analysis and Quaternions.} \\ +By \textsc{Alexander Macfarlane.} + +\smallskip \textbf{No. 9. Differential Equations.} \\ +By \textsc{William Woolsey Johnson.} + +\smallskip \textbf{No. 10. The Solution of Equations.} \\ +By \textsc{Mansfield Merriman.} + +\smallskip \textbf{No. 11. Functions of a Complex Variable.} \\ +By \textsc{Thomas S. Fiske.} + +\smallskip \textbf{No. 12. The Theory of Relativity.} \\ +By \textsc{Robert D. Carmichael.} + +\smallskip \textbf{No. 13. The Theory of Numbers.} \\ +By \textsc{Robert D. Carmichael.} \normalsize + +\bigskip \small PUBLISHED BY \\ +\smallskip \textbf{JOHN WILEY \& SONS, Inc., NEW YORK. \\ +CHAPMAN \& HALL, Limited, LONDON.} +\end{center}}} + +\chapter{Editors' Preface.} + +The volume called Higher Mathematics, the third edition of which was +published in 1900, contained eleven chapters by eleven authors, each +chapter being independent of the others, but all supposing the +reader to have at least a mathematical training equivalent to that +given in classical and engineering colleges. The publication of that +volume was discontinued in 1906, and the chapters have since been +issued in separate Monographs, they being generally enlarged by +additional articles or appendices which either amplify the former +presentation or record recent advances. This plan of publication was +arranged in order to meet the demand of teachers and the convenience +of classes, and it was also thought that it would prove advantageous +to readers in special lines of mathematical literature. + +It is the intention of the publishers and editors to add other +monographs to the series from time to time, if the demand seems to +warrant it. Among the topics which are under consideration are those +of elliptic functions, the theory of quantics, the group theory, the +calculus of variations, and non-Euclidean geometry; possibly also +monographs on branches of astronomy, mechanics, and mathematical +physics may be included. It is the hope of the editors that this +Series of Monographs may tend to promote mathematical study and +research over a wider field than that which the former volume has +occupied. + +\chapter{Preface} + +The purpose of this little book is to give the reader a convenient +introduction to the theory of numbers, one of the most extensive and +most elegant disciplines in the whole body of mathematics. The +arrangement of the material is as follows: The first five chapters +are devoted to the development of those elements which are essential +to any study of the subject. The sixth and last chapter is intended +to give the reader some indication of the direction of further study +with a brief account of the nature of the material in each of the +topics suggested. The treatment throughout is made as brief as is +possible consistent with clearness and is confined entirely to +fundamental matters. This is done because it is believed that in +this way the book may best be made to serve its purpose as an +introduction to the theory of numbers. + +Numerous problems are supplied throughout the text. These have been +selected with great care so as to serve as excellent exercises for +the student's introductory training in the methods of number theory +and to afford at the same time a further collection of useful +results. The exercises marked with a star are more difficult than +the others; they will doubtless appeal to the best students. + +Finally, I should add that this book is made up from the material +used by me in lectures in Indiana University during the past two +years; and the selection of matter, especially of exercises, has +been based on the experience gained in this way. + +\hfill \textsc{R.~D.\ Carmichael.} + +\tableofcontents + +%% CHAPTER I. ELEMENTARY PROPERTIES OF INTEGERS +%% 1. Fundamental Notions and Laws +%% 2. Definition of Divisibility. The Unit +%% 3. Prime Numbers. The Sieve of Eratosthenes +%% 4. The Number of Primes is Infinite +%% 5. The Fundamental Theorem of Euclid +%% 6. Divisibility by a Prime Number +%% 7. The Unique Factorization Theorem +%% 8. The Divisors of an Integer +%% 9. The Greatest Common Factor of Two or More Integers +%% 10. The Least Common Multiple of Two or More Integers +%% 11. Scales of Notation +%% 12. Highest Power of a Prime $p$ Contained in $n!$ +%% 13. Remarks Concerning Prime Numbers +%% +%% CHAPTER II. ON THE INDICATOR OF AN INTEGER +%% 14. Definition. Indicator of a Prime Power +%% 15. The Indicator of a Product +%% 16. The Indicator of Any Positive Integer +%% 17. Sum of the Indicators of the Divisors of a Number +%% +%% CHAPTER III. ELEMENTARY PROPERTIES OF CONGRUENCES +%% 18. Congruences Modulo $m$ +%% 19. Solutions of Congruences by Trial +%% 20. Properties of Congruences Relative to Division +%% 21. Congruences with a Prime Modulus +%% 22. Linear Congruences +%% +%% CHAPTER IV. THE THEOREMS OF FERMAT AND WILSON +%% 23. Fermat's General Theorem +%% 24. Euler's Proof of the Simple Fermat Theorem +%% 25. Wilson's Theorem +%% 26. The Converse of Wilson's Theorem +%% 27. Impossibility of $1\cdot 2\cdot 3\cdot \ldots \cdot +%% \overline{n-1}+1=n^k, n>5$ +%% 28. Extension of Fermat's Theorem +%% 29. On the Converse of Fermat's Simple Theorem +%% 30. Application of Previous Results to Linear Congruences +%% 31. Application of the Preceding Results to the Theory of +%% Quadratic Residues +%% +%% CHAPTER V. PRIMITIVE ROOTS MODULO $m$ +%% 32. Exponent of an Integer Modulo $m$ +%% 33. Another Proof of Fermat's General Theorem +%% 34. Definition of Primitive Roots +%% 35. Primitive Roots Modulo $p$ +%% 36. Primitive Roots Modulo $p^\alpha$, $p$ an Odd Prime +%% 37. Primitive Roots Modulo $2p^\alpha$, $p$ an Odd Prime +%% 38. Recapitulation +%% 39. Primitive $\lambda$-Roots +%% +%% CHAPTER VI. OTHER TOPICS +%% 40. Introduction +%% 41. Theory of Quadratic Residues +%% 42. Galois Imaginaries +%% 43. Arithmetic Forms +%% 44. Analytical Theory of Numbers +%% 45. Diophantine Equations +%% 46. Pythagorean Triangles +%% 47. The Equation $x^n+y^n = z^n$ + +\mainmatter + +\chapter{ELEMENTARY PROPERTIES OF INTEGERS} +\section{Fundamental Notions and Laws}\label{s1}% +\index{Fundamental notions} + +In the present chapter we are concerned primarily with certain +elementary properties of the positive integers 1, 2, 3, 4, \ldots It +will sometimes be convenient, when no confusion can arise, to employ +the word \emph{integer} or the word \emph{number} in the sense of +positive integer. + +We shall suppose that the integers are already defined, either by +the process of counting or otherwise. We assume further that the +meaning of the terms \emph{greater, less, equal, sum, difference, +product} is known. + +From the ideas and definitions thus assumed to be known follow +immediately the theorems: +\begin{table}[h] +\begin{tabular}{rl} + I.\ & The sum of any two integers is an integer. \\ + II.\ & The difference of any two integers is an integer. \\ + III.\ & The product of any two integers is an integer. +\end{tabular} +\end{table} + +Other fundamental theorems, which we take without proof, are +embodied in the following formulas: +\begin{table}[h] +\begin{tabular}{rrcl} + IV.\ & $a + b$ & = & $b + a$. \\ + V.\ & $a \times b$ & = & $b \times a$. \\ + VI.\ & $(a + b) + c$ & = & $a + (b + c)$. \\ + VII.\ & $(a \times b) \times c$ & = & $a \times (b \times c)$. \\ +VIII.\ & $a \times (b + c)$ & = & $a \times b + a \times c$. +\end{tabular} +\end{table} +Here $a$, $b$, $c$ denote any positive integers. + +\newpage +These formulas are equivalent in order to the following five +theorems: addition is commutative; multiplication is commutative; +addition is associative; multiplication is associative; +multiplication is distributive with respect to addition. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove the following relations: +\begin{align*} + 1 + 2 + 3 \ldots + n &= \frac{n(n+1)}{2} \\ + 1 + 3 + 5 + \ldots + (2n - 1) &= n^2, \\ +1^3 + 2^3 + 3^3 + \ldots + n^3 &= \left(\frac{n(n+1)}{2}\right)^2 + = (1+2+\ldots+n)^2. +\end{align*} + +\item[2.] Find the sum of each of the following series: +\begin{align*} +1^2 + 2^2 + 3^2 + &\ldots + n^2, \\ +1^2 + 3^2 + 5^2 + &\ldots + (2n - 1)^2, \\ +1^3 + 3^3 + 5^3 + &\ldots + (2n - 1)^3. +\end{align*} + +\item[3.] Discover and establish the law suggested by the equations +$1^2 = 0 + 1$, $2^2 = 1 + 3$, $3^2 = 3 + 6$, $4^2 = 6 + 10$, +$\ldots$; by the equations $1 = 1^3$, $3 + 5 = 2^3$, $7 + 9 + 11 = +3^3$, $13 + 15 + 17 + 19 = 4^3$, $\ldots$. +\end{enumerate} \normalsize + +\section{Definition of Divisibility. The Unit}\label{s2}% +\index{Divisibility}\index{Unit} + +\textsc{Definitions.} An integer $a$ is said to be divisible by an +integer $b$ if there exists an integer $c$ such that $a = bc$. It is +clear from this definition that $a$ is also divisible by $c$. The +integers $b$ and $c$ are said to be divisors or factors of $a$; and +$a$ is said to be a multiple of $b$ or of $c$. The process of +finding two integers $b$ and $c$ such that $bc$ is equal to a given +integer $a$ is called the process of resolving $a$ into factors or +of factoring $a$; and $a$ is said to be resolved into factors or to +be factored. + +We have the following fundamental theorems: + +\smallskip I.~\emph{If $b$ is a divisor of $a$ and $c$ is a divisor +of $b$, then $c$ is a divisor of $a$.} + +Since $b$ is a divisor of a there exists an integer $\beta$ such +that $a = b\beta$. Since $c$ is a divisor of $b$ there exists an +integer $\gamma$ such that $b = c\gamma$. Substituting this value of +$b$ in the equation $a = b\gamma$ we have $a = c\gamma\beta$. But +from theorem III of \S~\ref{s1} it follows that $\gamma\beta$ is an +integer; hence, $c$ is a divisor of $a$, as was to be proved. + +\smallskip II.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the sum of $a$ and $b$.} + +From the hypothesis of the theorem it follows that integers $\alpha$ +and $\beta$ exist such that +\begin{gather*} +a = c\alpha,\quad b = c\beta. \\ +\intertext{Adding, we have} +a + b = c\alpha + c\beta = c(\alpha + \beta) = c\delta, +\end{gather*} +where $\delta$ is an integer. Hence, $c$ is a divisor of $a+b$. + +\smallskip III.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the difference of $a$ and $b$.} + +The proof is analogous to that of the preceding theorem. + +\smallskip \textsc{Definitions.} If $a$ and $b$ are both divisible +by $c$, then $c$ is said to be a common divisor or a common factor +of $a$ and $b$. Every two integers have the common factor 1. The +greatest integer which divides both $a$ and $b$ is called the +greatest common divisor of $a$ and $b$. More generally, we define in +a similar way a common divisor and the greatest common divisor of +$n$ integers $a_1$, $a_2$, $\ldots$, $a_n$.\index{Common!divisors} + +\smallskip \textsc{Definitions.} If an integer $a$ is a multiple of +each of two or more integers it is called a common multiple of these +integers. The product of any set of integers is a common multiple of +the set. The least integer which is a multiple of each of two or +more integers is called their least common multiple.% +\index{Common!multiples} + +It is evident that the integer $1$ is a divisor of every integer and +that it is the only integer which has this property. It is called +the unit. + +\smallskip \textsc{Definition.} Two or more integers which have no +common factor except $1$ are said to be prime to each other or to be +relatively prime.\index{Relatively prime} + +\smallskip \textsc{Definition.} If a set of integers is such that no +two of them have a common divisor besides $1$ they are said to be +prime each to each.\index{Prime each to each} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove that $n^3 - n$ is divisible by $6$ for every +positive integer $n$. + +\item[2.] If the product of four consecutive integers is increased by +$1$ the result is a square number. + +\item[3.] Show that $2^{4n + 2} + 1$ has a factor different from itself +and $1$ when $n$ is a positive integer. +\end{enumerate} \normalsize + +\section{Prime Numbers. The Sieve of Eratosthenes}\label{s3}% +\index{Eratosthenes}\index{Sieve of Eratosthenes} + +\textsc{Definition.} If an integer $p$ is different from 1 and has +no divisor except itself and 1 it is said to be a prime number or to +be a prime. + +\smallskip \textsc{Definition.} An integer which has at least one +divisor other than itself and 1 is said to be a composite number or +to be composite. + +All integers are thus divided into three classes: +\begin{table}[h] +\begin{tabular}{rl} +1.\ & The unit; \\ +2.\ & Prime numbers; \\ +3.\ & Composite numbers. +\end{tabular} +\end{table}\index{Composite numbers}\index{Prime numbers} + +We have seen that the first class contains only a single number. The +third class evidently contains an infinitude of numbers; for, it +contains all the numbers $2^2, 2^3, 2^4, \ldots$ In the next section +we shall show that the second class also contains an infinitude of +numbers. We shall now show that every number of the third class +contains one of the second class as a factor, by proving the +following theorem: + +\smallskip I.~\emph{Every integer greater than 1 has a prime factor.} + +Let $m$ be any integer which is greater than 1. We have to show that +it has a prime factor. If $m$ is prime there is the prime factor $m$ +itself. If $m$ is not prime we have +\begin{equation*} +m = m_1 m_2 +\end{equation*} +where $m_1$ and $m_2$ are positive integers both of which are less +than $m$. If either $m_1$ or $m_2$ is prime we have thus obtained a +prime factor of $m$. If neither of these numbers is prime, then +write +\begin{equation*} +m_1 = m'_1 m'_2,\quad m'_1 > 1, m'_2 > 1. +\end{equation*} +Both $m'_1$ and $m'_2$ are factors of $m$ and each of them is less +than $m_1$. Either we have not found in $m'_1$ or $m'_2$ a prime +factor of $m$ or the process can be continued by separating one of +these numbers into factors. Since for any given $m$ there is +evidently only a finite number of such steps possible, it is clear +that we must finally arrive at a prime factor of $m$. From this +conclusion, the theorem follows immediately. + +Eratosthenes has given a useful means of finding the prime numbers +which are less than any given integer $m$. It may be described as +follows: + +Every prime except 2 is odd. Hence if we write down every odd number +from 3 up to $m$ we shall have it the list every prime less than $m$ +except 2. Now 3 is prime. Leave it in the list; but beginning to +count from 3 strike out every third number in the list. Thus every +number divisible by 3, except 3 itself, is cancelled. Then begin +from 5 and cancel every fifth number. Then begin from from the next +uncancelled number, namely 7, and strike out every seventh number. +Then begin from the next uncancelled number, namely 11, and strike +out every eleventh number. Proceed in this way up to $m$. The +uncancelled numbers remaining will be the odd primes not greater +than $m$. + +It is obvious that this process of cancellation need not be carried +altogether so far as indicated; for if $p$ is a prime greater than +$\sqrt{m}$, the cancellation of any $p^\text{th}$ number from $p$ +will be merely a repetition of cancellations effected by means of +another factor smaller than $p$, as one my see by the use of the +following theorem. + +\smallskip II.~\emph{An integer $m$ is prime if it has no prime +factor equal or less than $I$, where $I$ is the greatest integer +whose square is equal to or less than $m$.} + +Since $m$ has no prime factor less than $I$, it follows from theorem +I that is has no factor but unity less than $I$. Hence, if $m$ is +not prime it must be the product of two numbers each greater than +$I$; and hence it must be equal to or greater than $(I+1)^2$. This +contradicts the hypothesis on $I$; and hence we conclude that $m$ is +prime. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] By means of the method of Eratosthenes determine the primes +less than 200. +\end{enumerate} +\normalsize + +\section{The Number of Primes is Infinite}\label{s4}% +\index{Prime numbers} + +I.~\emph{The number of primes is infinite.} + +We shall prove this theorem by supposing that the number of primes +is not infinite and showing that this leads to a contradiction. If +the number of primes is not infinite there is a greatest prime +number, which we shall denote by $p$. Then form the number +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p + 1. +\end{equation*} +Now by theorem 1 of \S~\ref{s3} $N$ has a prime divisor $q$. But +every non-unit divisor of $N$ is obviously greater than $p$. Hence +$q$ is greater than $p$, in contradiction to the conclusion that $p$ +is the greatest prime. Thus the proof of the theorem is complete. + +In a similar way we may prove the following theorem: + +\smallskip II.~\emph{Among the integers of the arithmetic +progression $5$, $11$, $17$, $23$, $\ldots$, there is an infinite +number of primes.} + +If the number of primes in this sequence is not infinite there is a +greatest prime number in the sequence; supposing that this greatest +prime number exists we shall denote it by $p$. Then the number $N$, +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p-1, +\end{equation*} +is not divisible by any number less than or equal to $p$. This +number $N$, which is of the form $6n - 1$, has a prime factor. If +this factor is of the form $6k - 1$ we have already reached a +contradiction, and our theorem is proved. If the prime is of the +form $6k_1 + 1$ the complementary factor is of the form $6k_2 - 1$. +Every prime factor of $6k_2 - 1$ is greater than $p$. Hence we may +treat $6k_2 - 1$ as we did $6n - 1$, and with a like result. Hence +we must ultimately reach a prime factor of the form $6k_3 - 1$; for, +otherwise, we should have $6n - 1$ expressed as a product of prime +factors all of the form $6t + 1$---a result which is clearly +impossible. Hence we must in any case reach a contradiction of the +hypothesis. Thus the theorem is proved. + +The preceding results are special cases of the following more +general theorem: + +\smallskip III.~\emph{Among the integers of the arithmetic +progression $a$, $a + d$, $a + 2d$, $a + 3d$, $\ldots$, there is an +infinite number of +primes, provided that $a$ and $b$ are relatively prime.}% +\index{Arithmetic progression} + +For the special case given in theorem II we have an elementary +proof; but for the general theorem the proof is difficult. We shall +not give it here. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Prove that there is an infinite number of primes of the +form $4n - 1$. + +\item[2.] Show that an odd prime number can be represented as the +difference of two squares in one and in only one way. + +\item[3.] The expression $m^p - n^p$, in which $m$ and $n$ are integers +and $p$ is a prime, is either prime to $p$ or is divisible by $p^2$. + +\item[4.] Prove that any prime number except $2$ and $3$ is of one of +the forms $6n + 1$, $6n - 1$. +\end{enumerate}\normalsize + +\section{The Fundamental Theorem of Euclid}\label{s5}% +\index{Euclid, Theorem of} + +\emph{If $a$ and $b$ are any two positive integers there exist +integers $q$ and $r$, $q\stackrel{=}{>} 0, 0 \leqq r < b$, such +that} +\begin{equation*} +a = qb + r. +\end{equation*} + +If $a$ is a multiple of $b$ the theorem is at once verified, $r$ +being in this case $0$. If $a$ is not a multiple of $b$ it must lie +between two consecutive multiples of $b$; that is, there exists a +$q$ such that +\begin{equation*} +qb < a < (q + 1)b. +\end{equation*} +Hence there is an integer $r$, $0 < r < b$, such that $a = qb + r$. +In case $b$ is greater than $a$ it is evident that $q = 0$ and $r = +a$. Thus the proof of the theorem is complete. + +\section{Divisibility by a Prime Number}\label{s6}\index{Prime numbers} + +I.~\emph{If $p$ is a prime number and $m$ is any integer, then $m$ +either is divisible by $p$ or is prime to $p$.} + +This theorem follows at once from the fact that the only divisors of +$p$ are $1$ and $p$. + +\smallskip II.~\emph{The product of two integers each less than a +given prime number $p$ is not divisible by $p$.} + +Let $a$ be a number which is less than $p$ and suppose that $b$ is a +number less than $p$ such that $ab$ is divisible by $p$, and let $b$ +be the least number for which $ab$ is so divisible. Evidently there +exists an integer $m$ such that +\begin{equation*} +mb < p < (m + 1)b. +\end{equation*} +Then $p - mb < b$. Since $ab$ is divisible by $p$ it is clear that +$mab$ is divisible by $p$; so is $ap$ also; and hence their +difference $ap - mab$, $=a(p - mb)$, is divisible by $p$. That is, +the product of $a$ by an integer less than $b$ is divisible by $p$, +contrary to the assumption that $b$ is the least integer such that +$ab$ is divisible by $p$. The assumption that the theorem is not +true has thus led to a contradiction; and thus the theorem is +proved. + +\smallskip III.~\emph{If neither of two integers is divisible by a +given prime number $p$ their product is not divisible by $p$.} + +Let $a$ and $b$ be two integers neither of which is divisible by the +prime $p$. According to the fundamental theorem of Euclid there +exist integers $m$, $n$, $\alpha$, $\beta$ such that +\begin{align*} +a &= mp + \alpha,& 0 &< \alpha < p, \\ +b &= np + \beta, & 0 &< \beta < p. +\end{align*} +Then +\begin{equation*} +ab = (mp + \alpha)(np + \beta) + = (mnp + \alpha + \beta)p + \alpha\beta. +\end{equation*} +If now we suppose $ab$ to be divisible by $p$ we have $\alpha\beta$ +divisible by $p$. This contradicts II, since $\alpha$ and $\beta$ +are less than $p$. Hence $ab$ is not divisible by $p$. + +By an application of this theorem to the continued product of +several factors, the following result is readily obtained: + +\smallskip IV.~\emph{If no one of several integers is divisible by a +given prime $p$ their product is not divisible by $p$.} + +\section{The Unique Factorization Theorem}\label{s7}% +\index{Factorization theorem}\index{Factors} + +I.~\emph{Every integer greater than unity can be represented in one +and in only one way as a product of prime numbers.} + +In the first place we shall show that it is always possible to +resolve a given integer $m$ greater than unity into prime factors by +a finite number of operations. In the proof of theorem I, +\S~\ref{s3}, we showed how to find a prime factor $p_1$ of $m$ by a +finite number of operations. Let us write +\begin{equation*} +m = p_1 m_1. +\end{equation*} +If $m_1$ is not unity we may now find a prime factor $p_2$ of $m_1$. +Then we may write +\begin{equation*} +m = p_1 m_1 = p_1 p_2 m_2. +\end{equation*} +If $m_2$ is not unity we may apply to it the same process as that +applied to $m_1$ and thus obtain a third prime factor of $m$. Since +$m_1 > m_2 > m_3 > \ldots$ it is clear that after a finite number of +operations we shall arrive at a decomposition of $m$ into prime +factors. Thus we shall have +\begin{equation*} +m = p_1 p_2 \ldots p_r +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_r$ are prime numbers. We have thus +proved the first part of our theorem, which says that the +decomposition of an integer (greater than unity) into prime factors +is always possible. + +Let us now suppose that we have also a decomposition of $m$ into +prime factors as follows: +\begin{gather*} +m = q_1 q_2 \ldots q_s. \\ +\intertext{Then we have} +p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_s. +\end{gather*} +Now $p_1$ divides the first member of this equation. Hence it also +divides the second member of the equation. But $p_1$ is prime; and +therefore by theorem IV of the preceding section we see that $p_1$ +divides some one of the factors $q$; we suppose that $p_1$ is a +factor of $q_1$. It must then be equal to $q_1$. Hence we have +\begin{equation*} +p_2 p_3 \ldots p_r = q_2 q_3 \ldots q_s. +\end{equation*} +By the same argument we prove that $p_2$ is equal to some $q$, say +$q_2$. Then we have +\begin{equation*} +p_3 p_4 \ldots p_r = q_3 q_4 \ldots q_s. +\end{equation*} +Evidently the process may be continued until one side of the +equation is reduced to $1$. The other side must also be reduced to +$1$ at the same time. Hence it follows that the two decompositions +of $m$ are in fact identical. + +This completes the proof of the theorem. + +\smallskip The result which we have thus demonstrated is easily the +most important theorem in the theory of integers. It can also be +stated in a different form more convenient for some purposes: + +\smallskip II.~\emph{Every non-unit positive integer $m$ can be +represented in one and in only one way in the form +\begin{equation*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_n$ are different primes and +$\alpha_1$, $\alpha_2$, $\ldots$, $\alpha_n$ are positive integers.}% +\index{Factors} + +This comes immediately from the preceding representation of $m$ in +the form $m = p_1 p_2 \ldots p_r$ by combining into a power of $p_1$ +all the primes which are equal to $p_1$. + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ and $b$ are relatively +prime integers and $c$ is divisible by both $a$ and $b$, then $c$ is +divisible by $ab$.} + +\smallskip \textsc{Corollary 2.}~\emph{If $a$ and $b$ are each prime +to $c$ then $ab$ is prime to $c$.} + +\smallskip \textsc{Corollary 3.}~\emph{If $a$ is prime to $c$ and +$ab$ is divisible by $c$, then $b$ is divisible by $c$.} + +\section{The Divisors of an Integer}\label{s8}% +\index{Divisors of a number|(}\index{Factors} + +The following theorem is an immediate corollary of the results in +the preceding section: + +I.~\emph{All the divisors of $m$, +\begin{gather*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}, \\ +\intertext{are of the form} +p_1^{\beta_1} p_2^{\beta_2} \ldots p_n^{\beta_n},\ + 0 \leqq \beta_i \leqq \alpha_i; +\end{gather*} +and every such number is a divisor of $m$.} + +From this it is clear that every divisor of $m$ is included once and +only once among the terms of the product +\begin{multline*} +(1 + p_1 + p_1^2 + \ldots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \ldots + + p_2^{\alpha_2}) \ldots \\ +(1 + p_n + p_n^2 + \ldots + p_n^{\alpha_n}), +\end{multline*} +when this product is expanded by multiplication. It is obvious that +the number of terms in the expansion is $(\alpha_1 + 1)(\alpha_2 + +1) \ldots (\alpha_n+1)$. Hence we have the theorem: + +\smallskip II.~\emph{The number of divisors of $m$ is} +$(\alpha_1 + 1) (\alpha_2 + 1) \ldots (\alpha_n+1)$. + +Again we have +\begin{equation*} +\prod_i(1 + p_i + p_i^2 + \ldots + p_i^{\alpha_i}) = + \prod_i\frac{p_i^{\alpha_i+1} - 1}{p_i - 1}. +\end{equation*} +Hence, + +\smallskip III.~\emph{The sum of the divisors of $m$ is} +\begin{equation*} +\frac{p_1^{\alpha_1 + 1} - 1}{p_1 - 1} \cdot + \frac{p_2^{\alpha_2 + 1} - 1}{p_2 - 1} \cdot + \ldots \cdot + \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}. +\end{equation*} + +In a similar manner we may prove the following theorem: + +\smallskip IV.~\emph{The sum of the $h^{th}$ powers of the divisors +of $m$ is} +\begin{equation*} +\frac{p_1^{h(\alpha_1 + 1)} - 1}{p_1^h - 1} \cdot + \ldots \cdot + \frac{p_n^{h(\alpha_n + 1)} - 1}{p_n^h - 1}. +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find numbers $x$ such that the sum of the divisors of $x$ +is a perfect square. + +\item[2.] Show that the sum of the divisors of each of the following +integers is twice the integer itself: 6, 28, 496, 8128, 33550336. +Find other integers $x$ such that the sum of the divisors of $x$ is +a multiple of $x$. + +\item[3.] Prove that the sum of two odd squares cannot be a square. + +\item[4.] Prove that the cube of any integer is the difference of the +squares of two integers. + +\item[5.] In order that a number shall be the sum of consecutive +integers, it is necessary and sufficient that it shall not be a +power of 2. + +\item[6.] Show that there exist no integers $x$ and $y$ (zero excluded) +such that $y^2 = 2x^2$. Hence, show that there does not exist a +rational fraction whose square is 2. + +\item[7.] The number $m = p_1^{\alpha_1} p_2^{\alpha_2} \cdots +p_n^{\alpha_n}$, where the $p$'s are different primes and the +$\alpha$'s are positive integers, may be separated into relatively +prime factors in $2^{n-1}$ different ways. + +\item[8.] The product of the divisors of $m$ is $\sqrt{m^v}$ where $v$ +is the number of divisors of $m$. +\end{enumerate} \normalsize\index{Divisors of a number|)} + +\section{The Greatest Common Factor of Two or More +Integers}\label{s9}% +\index{Common!divisors|(}\index{Factors}% +\index{Greatest common factor|(} + +Let $m$ and $n$ be two positive integers such that $m$ is greater +than $n$. Then, according to the fundamental theorem of Euclid, we +can form the set of equations +\begin{align*} +m &= qn + n_1, & 0 &< n_1 < n, \\ +n &= q_1 n_1 + n_2, & 0 &< n_2 < n_1, \\ +n_1 &= q_2 n_2 + n_3, & 0 &< n_3 < n_2, \\ +&\vdots \qquad \vdots &&\vdots \qquad \vdots \\ +n_{k - 2} &= q_{k - 1} n_{k-1} + n_k, & 0 &< n_k < n_{k - 1}, \\ +n_{k - 1} &= q _k n_k. & & +\end{align*} +If $m$ is a multiple of $n$ we write $n = n_0$, $k = 0$, in the +above equations. + +\smallskip \textsc{Definition.} The process of reckoning involved in +determining the above set of equations is called the Euclidian +Algorithm.\index{Euclidian algorithm} + +\smallskip I.~\emph{The number $n_k$ to which the Euclidian +algorithm leads is the greatest common divisor of $m$ and $n$.} + +In order to prove this theorem we have to show two things: + +1)~That $n_k$ is a divisor of both $m$ and $n$; + +2)~That the greatest common divisor $d$ of $m$ and $n$ is a divisor +of $n_k$. + +To prove the first statement we examine the above set of equations, +working from the last to the first. From the last equation we see +that $n_k$ is a divisor of $n_{k-1}$. Using this result we see that +the second member of next to the last equation is divisible by $n_k$ +Hence its first member $n_{k-2}$ must be divisible by $n_k$. +Proceeding in this way step by step we show that $n_2$ and $n_1$, +and finally that $n$ and $m$, are divisible by $n_k$. + +For the second part of the proof we employ the same set of equations +and work from the first one to the last one. Let $d$ be any common +divisor of $m$ and $n$. From the first equation we see that $d$ is a +divisor of $n_1$. Then from the second equation it follows that $d$ +is a divisor of $n_2$. Proceeding in this way we show finally that +$d$ is a divisor of $n_k$. Hence any common divisor, and in +particular the greatest common divisor, of $m$ and $n$ is a factor +of $n_k$. + +This completes the proof of the theorem. + +\smallskip \textsc{Corollary.} \emph{Every common divisor of $m$ and +$n$ is a factor of their greatest common divisor.} + +\smallskip II.~\emph{Any number $n_i$ in the above set of equations +is the difference of multiples of $m$ and $n$.} + +From the first equation we have +\begin{equation*} +n_i = m - qn +\end{equation*} +so that the theorem is true for $i = 1$. We shall suppose that the +theorem is true for every subscript up to $i - 1$ and prove it true +for the subscript $i$. Thus by hypothesis we have\footnote{If $i = +2$ we must replace $n_{i-2}$ by $n$.} +\begin{align*} +n_{i-2} &= \pm(\alpha_{i-2}m - \beta_{i-2}n ), \\ +n_{i-1} &= \mp(\alpha_{i-1}m - \beta_{i-1}n). +\intertext{Substituting in the equation} +n_i &= -q_{i-1}n_{n-1} + n_{i-2} \\ +\intertext{we have a result of the form} +n_i &= \pm (\alpha_i m - \beta_i n). +\end{align*} +From this we conclude at once to the truth of the theorem. + +Since $n_k$ is the greatest common divisor of $m$ and $n$, we have +as a corollary the following important theorem: + +\smallskip III.~\emph{If $d$ is the greatest common divisor of the +positive integers $m$ and $n$, then there exist positive integers +$\alpha$ and $\beta$ such that} +\begin{equation*} +\alpha m - \beta n = \pm d. +\end{equation*} + +If we consider the particular case in which $m$ and $n$ are +relatively prime, so that $d = 1$, we see that there exist positive +integers $\alpha$ and $\beta$ such that $\alpha m - \beta n = \pm +1$. Obviously, if $m$ and $n$ have a common divisor $d$, greater +than $1$, there do not exist integers $\alpha$ and $\beta$ +satisfying this relation; for, if so, $d$ would be a divisor of the +first member of the equation and not of the second. Thus we have the +following theorem: + +\smallskip IV.~\emph{A necessary and sufficient condition that $m$ +and $n$ are relatively prime is that there exist integers $\alpha$ +and $\beta$ such that $\alpha m - \beta n = \pm 1$.} + +The theory of the greatest common divisor of three or more numbers +is based directly on that of the greatest common divisor of two +numbers; consequently it does not require to be developed in detail. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] If $d$ is the greatest common divisor of $m$ and $n$, +then $m / d$ and $n / d$ are relatively prime. + +\item[2.] If $d$ is the greatest common divisor of $m$ and $n$ and +$k$ is prime to $n$, then $d$ is the greatest common divisor of $km$ +and $n$. + +\item[3.] The number of multiplies of $6$ in the sequence $a, 2a, 3a, +\cdots, ba$ is equal to the greatest common divisor of $a$ and $b$. + +\item[4.] If the sum or the difference of two irreducible fractions is +an integer, the denominators of the fractions are equal. + +\item[5.] The algebraic sum of any number of irreducible fractions, +whose denominators are prime each to each, cannot be an integer. + +\item[6*.] The number of divisions to be effected in finding the +greatest common divisor of two numbers by the Euclidian algorithm +does not exceed five times the number of digits in the smaller +number (when this number is written in the usual scale of 10). +\end{enumerate}\normalsize% +\index{Common!divisors|)}\index{Greatest common factor|)} + +\section{The Least Common Multiple of Two or More +Integers}\label{s10}% +\index{Common!multiples|(}\index{Least common multiple|(} + +I.~\emph{The common multiples of two or more numbers are the +multiples of their least common multiple.} + +This may be readily proved by means of the unique factorization +theorem. The method is obvious. We shall, however, give a proof +independent of this theorem. + +Consider first the case of two numbers; denote them by $m$ and $n$ +and their greatest common divisor by $d$. Then we have +\begin{equation*} +m = d\mu, \quad n = d\nu, +\end{equation*} +where $\mu$ and $\nu$ are relatively prime +integers.\index{Common!divisors}\index{Greatest common factor} The +common multiples sought are multiples of $m$ and are all comprised +in the numbers $am=ad\mu$, where $a$ is any integer whatever. In +order that these numbers shall be multiples of $n$ it is necessary +and sufficient that $ad\mu$ shall be a multiple of $d\nu$; that is, +that $a\mu$ shall be a multiple of $\nu$; that is, that $a$ shall be +a multiple of $\nu$, since $\mu$ and $\nu$ are relatively prime. +Writing $a = \delta\nu$ we have as the multiples in question the set +$\delta d\mu\nu$ where $\delta$ is an arbitrary integer. This proves +the theorem for the case of two numbers; for $d\mu\nu$ is evidently +the least common multiple of $m$ and $n$. + +We shall now extend the proposition to any number of integers $m, n, +p, q,\ldots$. The multiples in question must be common multiples of +$m$ and $n$ and hence of their least common multiple $\mu$. Then the +multiples must be multiples of $\mu$ and $p$ and hence of their +least common multiple $\mu_1$. But $\mu_1$ is evidently the least +common multiple of $m, n, p$. Continuing in a similar manner we may +show that every multiple in question is a multiple of $\mu$, the +least common multiple of $m, n, p, q, \ldots$. And evidently every +such number is a multiple of each of the numbers $m, n, p, q, +\ldots$. + +Thus the proof of the theorem is complete. + +When the two integers $m$ and $n$ are relatively prime their +greatest common divisor is $1$ and their least common multiple is +their product. Again if $p$ is prime to both $m$ and $n$ it is prime +to their product $mn$; and hence the least common multiple of $m, n, +p$ is in this case $mnp$. Continuing in a similar manner we have the +theorem: + +\smallskip II.~\emph{The least common multiple of several integers, +prime each to each, is equal to their product.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] In order that a common multiple of $n$ numbers shall be +the least, it is necessary and sufficient that the quotients +obtained by dividing it successively by the numbers shall be +relatively prime. + +\item[2.] The product of $n$ numbers is equal to the product of +their least common multiple by the greatest common divisor of their +products $n - 1$ at a time. + +\item[3.] The least common multiple of $n$ numbers is equal to any +common multiple $M$ divided by the greatest common divisor of the +quotients obtained on dividing this common multiple by each of the +numbers. + +\item[4.] The product of $n$ numbers is equal to the product of their +greatest common divisor by the least common multiple of the products +of the numbers taken $n - 1$ at a time. +\end{enumerate} \normalsize% +\index{Common!multiples|)}\index{Least common multiple|)} + +\section{Scales of Notation}\label{s11}\index{Scales of notation|(} + +I.~\emph{If $m$ and $n$ are positive integers and $n > 1$, then $m$ +can be represented in terms of $n$ in one and in only one way in the +form} +\begin{gather*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1} n + a_h, \\ +\intertext{where} +a_0 \ne 0,\ 0 \leqq a_i < n, \quad i = 0, 1, 2, \ldots, h. +\end{gather*} + +That such a representation of $m$ exists is readily proved by means +of the fundamental theorem of Euclid. For we have +\begin{align*} +m &= n_0 n + a_h, & 0 &\leqq a_h < n, \\ +n_0 &= n_1n + a_{h-1}, & 0 &\leqq a_{h-1} < n, \\ +n_1 &= n_2 n + a_{h-2}, & 0 &\leqq a_{h-2} < n, \\ +\hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots & + \hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots \\ +n_{h-3} &= n_{h-2} n + a_2, & 0 &\leqq a_2 < n, \\ +n_{h-2} &= n_{h-1} n + a_1, & 0 &\leqq a_1 < n, \\ +n_{h-1} &= a_0, & 0 &< a_0 < n. +\end{align*} +If the value of $n_{h-1}$ given in the last of these equations is +substituted in the second last we have +\begin{equation*} +n_{h-2} = a_0n + a_1. +\end{equation*} +This with the preceding gives +\begin{equation*} +n_{h-3} = a_0 n^2 + a_1n + a_2. +\end{equation*} +Substituting from this in the preceding and continuing the process +we have finally +\begin{equation*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1}n + a_h, +\end{equation*} +a representation of $m$ in the form specified in the theorem. + +To prove that this representation is unique, we shall suppose that +$m$ has the representation +\begin{gather*} +m = b_0 n^k + b_1 n^{k-1} + \ldots + b_{k-1}n + b_k, \\ +\intertext{where} +b_0 \ne 0,\ 0 < b_i < n,\quad i=0, 1, 2, \ldots, k, \\ +\intertext{and show that the two representations are identical. We +have} +a_0 n^h + \ldots + a_{h-1} n + a_h = + b_0 n^k + \ldots + b_{k-1} n + b_k. +\intertext{Then} +a_0 n^h + \ldots + a_{h-1} n - + (b_0 n^k + \ldots + b_{k-1} n) = b_k - a_h. +\end{gather*} +The first member is divisible by $n$. Hence the second is also. But +the second member is less than $n$ in absolute value; and hence, in +order to be divisible by $n$, it must be zero. That is, $b_k = a_h$. +Dividing the equation through by $n$ and transposing we have +\begin{equation*} +a_0 n^{h-1} + \ldots + a_{h-2} n - (b_0 n^{k-1} + \ldots + + b_{k-2} n) + = b_{k-1} - a_{h-1}. +\end{equation*} +It may now be seen that $b_{k-1} = a_{h-1}$. It is evident that this +process may be continued until either the $a$'s are all eliminated +from the equation or the $b$'s are all eliminated. But it is obvious +that when one of these sets is eliminated the other is also. Hence, +$h = k$. Also, every $a$ equals the $b$ which multiplies the same +power of $n$ as the corresponding $a$. That is, the two +representations of $m$ are identical. Hence the representation in +the theorem is unique. + +From this theorem it follows as a special case that any positive +integer can be represented in one and in only one way in the scale +of 10; that is, in the familiar Hindoo notation. It can also be +represented in one and in only one way in any other scale. Thus +\begin{equation*} +120759 = 1 \cdot 7^6 + 0 \cdot 7^5 + 1 \cdot 7^4 + 2 \cdot 7^3 + + 0 \cdot 7^2 + 3 \cdot 7^1 + 2. +\end{equation*} +Or, using a subscript to denote the scale of notation, this may be +written +\begin{equation*} +(120759)_{10} = (1012032)_7. +\end{equation*} + +For the case in which $n$ (of theorem I) is equal to 2, the only +possible values for the $a$'s are 0 and 1. Hence we have at once the +following theorem: + +II.~\emph{Any positive integer can be represented in one and in only +one way as a sum of different powers of 2.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Any positive integer can be represented as an aggregate of +different powers of $3$, the terms in the aggregate being combined +by the signs $+$ and $-$ appropriately chosen. + +\item[2.] Let $m$ and $n$ be two positive integers of which $n$ is the +smaller and suppose that $2^k \leq n < 2^{k+1}$. By means of the +representation of $m$ and $n$ in the scale of 2 prove that the +number of divisions to be effected in finding the greatest common +divisor of $m$ and $n$ by the Euclidian algorithm does not exceed +$2k$. +\end{enumerate}\normalsize\index{Scales of notation|)} + +\section{Highest Power of a Prime $p$ Contained in $n!$.}\label{s12}% +\index{Highest power of \emph{p} in \emph{n}"!|(} + +Let $n$ be any positive integer and $p$ any prime number not greater +than $n$. We inquire as to what is the highest power $p^\nu$ of the +prime $p$ contained in $n!$. + +In solving this problem we shall find it convenient to employ the +notation +\begin{equation*} +\left [ \frac{r}{s} \right ] +\end{equation*} to denote the greatest integer $\alpha$ such that +$\alpha s \leq r$. With this notation it is evident that we have +\begin{gather} +\left [ + \frac{\left [ \frac{n}{p} \right ]} + {p} +\right ] = \left [ \frac{n}{p^2} \right ]; \tag{1} \\ +\intertext{and more generally} +\left [ + \frac{\left [ \frac{n}{p^i} \right ]} + {p^j} +\right ] = \left [ \frac{n}{p^{i+j}} \right ]. \notag +\end{gather} + +If now we use $H\{x\}$ to denote the index of the highest power of +$p$ contained in an integer $x$, it is clear that we have +\begin{gather*} +H\{n!\} = + H \left \{ p \cdot 2p \cdot 3p \ldots + \left [ \frac{n}{p} \right ] p \right \}, \\ +\intertext{since only multiples of $p$ contain the factor $p$. +Hence} +H\{n!\} = + \left [ \frac{n}{p} \right ] + + H \left \{ 1 \cdot 2 \ldots \left [ \frac{n}{p} \right ] + \right \}. +\end{gather*} +Applying the same process to the $H$-function in the second member +and remembering relation (1) it is easy to see that we have +\begin{align*} +H\{n!\} &= \left[ \frac{n}{p} \right] + + H\left\{ p \cdot 2p \cdot \ldots \cdot + \left[ \frac{n}{p^2} \right]p\right\} \\ + &= \left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + + H \left\{\cdot 1 \cdot 2 \cdot 3 + \ldots \left[ \frac{n}{p^2} \right] \right\}. \\ +\intertext{Continuing the process we have finally} +H\{n1\} &= \left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] + + \ldots, +\end{align*} +the series on the right containing evidently only a finite number of +terms different from zero. Thus we have the theorem: + +\smallskip I.~\emph{The index of the highest power of a prime $p$ +contained in $n!$ is} +\begin{gather*} +\left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + + \left[ \frac{n}{p^3} \right] + \ldots. +\end{gather*} + +The theorem just obtained may be written in a different form, more +convenient for certain of its applications. Let $n$ be expressed in +the scale of $p$ in the form +\begin{gather*} +n = a_0p^h + a_1p^{h-1} + \ldots + a_{h-1}p + a_h, \\ +\intertext{where} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h. +\end{gather*} +Then evidently +\begin{align*} +\left[ \frac{n}{p} \right] &= a_0p^{h-1} + a_1p^{h-2} + \ldots + + a_{h-2}p + a_{h-1}, \\ +\left[ \frac{n}{p^2} \right] &= a_0p^{h-2} + a_1p^{h-3} + \ldots + + a_{h-2}, \\ +.\ \ .\ \ .\ \ .\ \ &.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ +.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ . +\end{align*} +Adding these equations member by member and combining the second +members in columns as written, we have +\begin{align*} +\left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] &+ + \left[ \frac{n}{p^3} \right] + \ldots \\ +&= \sum_{i=0}^h \frac{a_i(p^{h-i} - 1)}{p - 1} \\ +&= \frac{a_0p^h + a_1p^{h-1} + \ldots + a_h - + (a_0 + a_1 + \ldots + a_h)}{p-1} \\ +&= \frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{align*} +Comparing this result with theorem I we have the following theorem: + +\smallskip II.~\emph{If $n$ is represented in the scale of $p$ in +the form +\begin{gather*} +n = a_0 p^h + a_1 p^{h-1} + \ldots + a_h, \\ +\intertext{where $p$ is prime and} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h, \\ +\intertext{then the index of the highest power of $p$ contained in +$n!$ is} +\frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{gather*}} + +Note the simple form of the theorem for the case $p = 2$; in this +case the denominator $p - 1$ is unity. + +We shall make a single application of these theorems by proving the +following theorem: + +\smallskip III.~\emph{If $n$, $\alpha$, $\beta$, $\ldots$, $\lambda$ +are any positive integers such that $n = \alpha + \beta + \ldots + +\lambda$, then +\begin{equation} +\frac{n!}{\alpha! \beta! \ldots \lambda!} \tag{A} +\end{equation} +is an integer.} + +Let $p$ be any prime factor of the denominator of the fraction (A). +To prove the theorem it is sufficient to show that the index of the +highest power of $p$ contained in the numerator is at least as great +as the index of the highest power of $p$ contained in the +denominator. This index for the denominator is the sum of the +expressions +\begin{equation} + \left . + \begin{gathered} + \left [ \frac{\alpha}{p} \right ] + + \left [ \frac{\alpha}{p^2} \right ] + + \left [ \frac{\alpha}{p^3} \right ] + + \ldots \\ + \left [ \frac{\beta}{p} \right ] + + \left [ \frac{\beta}{p^2} \right ] + + \left [ \frac{\beta}{p^3} \right ] + + \ldots \\ + \vdots \\ + \left [ \frac{\lambda}{p} \right ] + + \left [ \frac{\lambda}{p^2} \right ] + + \left [ \frac{\lambda}{p^3} \right ] + + \ldots + \end{gathered} + \right \} \tag{B} +\end{equation} + +The corresponding index for the numerator is +\begin{equation} +\left [ \frac{n}{p} \right ] + +\left [ \frac{n}{p^2} \right ] + +\left [ \frac{n}{p^3} \right ] + +\ldots \tag{C} +\end{equation} +But, since $n = \alpha + \beta + \ldots + \lambda$, it is evident +that +\begin{equation*} + \left [ \frac{n}{p^r} \right ] \stackrel{=}{>} + \left [ \frac{\alpha}{p^r} \right ] + + \left [ \frac{\beta}{p^r} \right ] + + \ldots + + \left [ \frac{\lambda}{p^r} \right ]. +\end{equation*} +From this and the expressions in (B) and (C) it follows that the +index of the highest power of any prime $p$ in the numerator of (A) +is equal to or greater than the index of the highest power of p +contained in its denominator. The theorem now follows at once. + +\smallskip \textsc{Corollary.}~\emph{The product of $n$ consecutive +integers is divisible by $n!$.} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the highest power of 2 contained in 1000! is +$2^{994}$; in 1900! is $2^{1893}$. Show that the highest power of 7 +contained in 10000! is $7^{1665}$. + +\item[2.] Find the highest power of 72 contained in 1000! + +\item[3.] Show that 1000! ends with 249 zeros. + +\item[4.] Show that there is no number $n$ such that $3^7$ is the +highest power of 3 contained in $n!$. + +\item[5.] Find the smallest number $n$ such that the highest power +of 5 contained in $n!$ is $5^{31}$. What other numbers have the same +property? + +\item[6.] If $n = rs$, $r$ and $s$ being positive integers, show that +$n!$ is divisible by $(r!)^s$ by $(s!)^r$; by the least common +multiple of $(r!)^s$ and $(s!)^r$. + +\item[7.] If $n = \alpha + \beta + pq + rs$, where $\alpha, \beta, p, +q, r, s$, are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha ! \beta ! (q!)^p (s!)^r. +\end{equation*} + +\item[8.] When $m$ and $n$ are two relatively prime positive integers +the quotient +\begin{equation*} +Q = \frac{(m + n + 1)!} {m! n!} +\end{equation*} +as an integer. + +\item[9*.] If $m$ and $n$ are positive integers, then each of the +quotients +\begin{equation*} +Q = \frac{(mn)!} {n! (m!)^n},\quad +Q = \frac{(2m)! (2n)!} {m! n! (m+n)!}, +\end{equation*} +is an integer. Generalize to $k$ integers $m, n, p, \ldots$. + +\item[10*.] If $n = \alpha + \beta + pq + rs$ where $\alpha, \beta, +p, q, r, s$ are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha! \beta! r! p! (q!)^p (s!)^r. +\end{equation*} + +\item[11*.] Show that +\begin{equation*} +\frac{(rst)!} {t! (s!)^t (r!)^{st}}, +\end{equation*} is an integer ($r, s, t$ being positive integers). +Generalize to the case of $n$ integers $r, s, t, u, \ldots$. +\end{enumerate}\normalsize% +\index{Highest power of \emph{p} in \emph{n}"!|)} + +\section{Remarks Concerning Prime Numbers}\label{s13}% +\index{Prime numbers|(} + +We have seen that the number of primes is infinite. But the integers +which have actually been identified as prime are finite in number. +Moreover, the question as to whether a large number, as for instance +$2^{257}-1$, is prime is in general very difficult to answer. Among +the large primes actually identified as such are the following: +\begin{equation*} +2^{61}-1, \quad 2^{75} \cdot 5+1, \quad 2^{89}-1, \quad 2^{127}-1. +\end{equation*} + +\emph{No analytical expression for the representation of prime +numbers has yet been discovered.} Fermat believed, though he +confessed that he was unable to prove, that he had found such an +analytical expression in +\begin{equation*} +2^{2^n} + 1. +\end{equation*} +Euler showed the error of this opinion by finding that $641$ is a +factor of this number for the case when $n = 5$.% +\index{Euler}\index{Fermat} + +The subject of prime numbers is in general one of exceeding +difficulty. In fact it is an easy matter to propose problems about +prime numbers which no one has been able to solve. Some of the +simplest of these are the following: + +\begin{enumerate} +\item Is there an infinite number of pairs of primes differing by +2? +\item Is every even number (other than 2) the sum of two primes or +the sum of a prime and the unit? +\item Is every even number the difference of two primes or the +difference of 1 and a prime number? +\item To find a prime number greater than a given prime. +\item To find the prime number which follows a given prime. +\item To find the number of primes not greater than a given number. +\item To compute directly the $n^\text{th}$ prime number, when $n$ +is given. +\end{enumerate}\index{Prime numbers|)} + +\chapter{ON THE INDICATOR OF AN INTEGER}% +\index{Indicator|(} + +\section{Definition. Indicator of a Prime Power}\label{s14}% +\index{Indicator!of a prime power} + +\emph{Definition.} If $m$ is any given positive integer the number +of positive integers not greater than $m$ and prime to it is called +the indicator of $m$. It is usually denoted by $\phi(m)$, and is +sometimes called Euler's $\phi$-function of $m$.% +\index{Euler's!$\phi$-function}\index{$\phi(m)$} More rarely, it has +been given the name of totient of $m$.\index{Totient} + +As examples we have +\begin{equation*} +\phi(1) = 1,\ \phi(2) = 1,\ \phi(3) = 2,\ \phi(4) = 2. +\end{equation*} + +If $p$ is a prime number it is obvious that +\begin{equation*} +\phi(p) = p - 1; +\end{equation*} +for each of the integers 1, 2, 3, $\ldots$, $p-1$ is prime to $p$. + +Instead of taking $m = p$ let us assume that $m = p^\alpha$, where +$\alpha$ is a positive integer, and seek the value of +$\phi(p^\alpha)$. Obviously, every number of the set 1, 2, 3, +$\ldots$, $p^\alpha$ either is divisible by $p$ or is prime to +$p^\alpha$. The number of integers in the set divisible by $p$ is +$p^{\alpha - 1}$. Hence $p^\alpha-p^{\alpha-1}$ of them are prime to +$p$. Hence $\phi(p^\alpha) = p^\alpha-p^{\alpha-1}$. Therefore + +\emph{If $p$ is any prime number and $\alpha$ is any positive +integer, then} +\begin{equation*} +\phi(p^\alpha) = p^\alpha \left ( 1 - \frac{1}{p} \right ). +\end{equation*} + +\section{The Indicator of a Product}\label{s15}% +\index{Indicator!of a product|(} + +I.~\emph{If $\mu$ and $\nu$ are any two relatively prime positive +integers, then} +\begin{equation*} +\phi(\mu\nu) = \phi(\mu) \phi(\nu). +\end{equation*} + +In order to prove this theorem let us write all the integers up to +$\mu\nu$ in a rectangular array as follows: +\footnotesize\begin{equation} + \left . + \begin{aligned} + 1 && 2 && 3 && + \ldots && h && \ldots && \mu \\ + \mu + 1 && \mu + 2 && \mu + 3 && + \ldots && \mu + h && \ldots && 2\mu \\ + 2 \mu + 1 && 2 \mu + 2 && 2 \mu + 3 && + \ldots && 2 \mu + h && \ldots && 3\mu \\ + \vdots && \vdots && \vdots && + && \vdots && && \vdots \\ + (\nu - 1)\mu + 1 && (\nu - 1)\mu + 2 && (\nu - 1)\mu + 3 && + \ldots && (\nu - 1)\mu + h && \ldots && \nu\mu \\ + \end{aligned} + \right \} \tag{A} +\end{equation}\normalsize + +If a number $h$ in the first line of this array has a factor in +common with $\mu$ then every number in the same column with $h$ has +a factor in common with $\mu$. On the other hand if $h$ is prime to +$\mu$, so is every number in the column with $h$ at the top. But the +number of integers in the first row prime to $\mu$ is $\phi(\mu)$. +Hence the number of columns containing integers prime to $\mu$ is +$\phi(\mu)$ and every integer in these columns is prime to $\mu$. + +Let us now consider what numbers in one of these columns are prime +to $\nu$; for instance, the column with $h$ at the top. We wish to +determine how many integers of the set +\begin{gather*} +h,\ \mu + h,\ 2\mu + h,\ \ldots,\ (\nu - 1)\mu + h \\ +\intertext{are prime to $\nu$. Write} +s\mu + h = q_s\nu + r_s +\end{gather*} where s ranges over the numbers $s = 0,\ 1,\ 2,\ +\ldots,\ \nu - 1$ and $0\leqq r_s < \nu$. Clearly $s\mu + h$ is or +is not prime to $\nu$ according as $r_s$ is or is not prime to +$\nu$. Our problem is then reduced to that of determining how many +of the quantities $r_s$ are prime to $\nu$. + +First let us notice that all the numbers $r_s$ are different; for, +if $r_s = r_t$ then from +\begin{equation*} +s\mu + h = q_s\nu + r_s,\quad t\mu + h = q_t\nu + r_t, +\end{equation*} +we have by subtraction that $(s-t)\mu$ is divisible by $\nu$. But +$\mu$ is prime to $\nu$ and $s$ and $t$ are each less than $\nu$. +Hence $(s-t)\mu$ can be a multiple of $\nu$ only by being zero; that +is, $s$ must equal $t$. Hence no two of the remainders $r_s$ can be +equal. + +Now the remainders $r_s$ are $\nu$ in number, are all zero or +positive, each is less than $\nu$, and they are all distinct. Hence +they are in some order the numbers 0, 1, 2, $\ldots$, $\nu-1$. The +number of integers in this set prime to $\nu$ is evidently +$\phi(\nu)$. + +Hence it follows that in any column of the array (A) in which the +numbers are prime to $\mu$ there are just $\phi(\nu)$ numbers which +are prime to $\nu$. That is, in this column there are just +$\phi(\nu)$ numbers which are prime to $\mu\nu$. But there are +$\phi(\mu)$ such columns. Hence the number of integers in the array +(A) prime to $\mu\nu$ is $\phi(\mu)\phi(\nu)$. + +But from the definition of the $\phi$-function it follows that the +number of integers in the array (A) prime to $\mu\nu$ is +$\phi(\mu\nu).$ Hence, +\begin{equation*} +\phi(\mu\nu) = \phi(\mu)\phi(\nu), +\end{equation*} which is the theorem to be proved. + +\smallskip \textsc{Corollary.}~\emph{In the series of $n$ +consecutive terms of an arithmetical progression the common +difference of which is prime to $n$, the number of terms prime to +$n$ is $\phi(n)$.} + +From theorem I we have readily the following more general result: + +\smallskip II.~\emph{If $m_1, m_2, \ldots, m_k$ are $k$ positive +integers which are prime each to each, then} +\begin{equation*} +\phi(m_1 m_2 \ldots m_k) = \phi(m_1) \phi(m_2) \ldots \phi(m_k). +\end{equation*}\index{Indicator!of a product|)} + +\section{The Indicator of any Positive Integer}\label{s16}% +\index{Indicator!of any integer|(} + +From the results of \S\S \ref{s14} and \ref{s15} we have an +immediate proof of the following fundamental theorem: + +\emph{If $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}$ +where $p_1, p_2, \ldots, p_n$ are different primes and $\alpha_1, +\alpha_2, \ldots, \alpha_n$ are positive integers, then} +\begin{equation*} +\phi(m) = m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{equation*} + +For, +\begin{align*} +\phi(m) &= \phi(p_1^{\alpha_1}) \phi(p_2^{\alpha_2}) \ldots + \phi(p_n^{\alpha_n}) \\ + &= p_1^{\alpha_1} \left ( 1-\frac{1}{p_1} \right ) + p_2^{\alpha_2} \left ( 1-\frac{1}{p_2} \right ) + \ldots + p_n^{\alpha_n} \left ( 1-\frac{1}{p_n} \right ) \\ + &= m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{align*} + +On account of the great importance of this theorem we shall give a +second demonstration of it. + +It is clear that the number of integers less than $m$ and divisible +by $p_1$ is +\begin{gather*} +\frac{m}{p_1}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_2$ is} +\frac{m}{p_2}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_1 p_2$ is} +\frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and divisible +by either $p_1$ or $p_2$ is} +\frac{m}{p_1} + \frac{m}{p_2} - \frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and prime to +$p_1 p_2$ is} +m - \frac{m}{p_1} - \frac{m}{p_2} + \frac{m}{p_1 p_2} = + m \left ( 1-\frac{1}{p_1} \right ) \left ( 1-\frac{1}{p_2} \right ). +\end{gather*} + +We shall now show that if the number of integers less than $m$ and +prime to $p_1 p_2 \ldots p_i$, where $i$ is less than $n$, is +\begin{gather*} +m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_i} \right ), \\ +\intertext{then the number of integers less than $m$ and prime to +$p_1 p_2 \ldots p_i p_{i+1}$ is} + m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_{i+1}} \right ). +\end{gather*} +From this our theorem will follow at once by induction. + +From our hypothesis it follows that the number of integers less than +$m$ and divisible by at least one of the primes $p_1$, $p_2$, +$\ldots$, $p_i$ is +\begin{gather} +m - + m \left (1 - \frac{1}{p_1}\right ) + \ldots + \left (1 - \frac{1}{p_i}\right ), \notag \\ +\intertext{or} +\sum \frac{m}{p_1} - \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - \ldots, \tag{A} +\end{gather} +where the summation in each case runs over all numbers of the type +indicated, the subscripts of the $p$'s being equal to or less than +$i$. + +Let us consider the integers less than $m$ and having the factor +$p_{i+1}$ but not having any of the factors $p_1$, $p_2$, $\ldots$, +$p_i$. Their number is +\begin{gather} +\frac{m}{p_{i+1}} - + \frac{1}{p_{i+1}} \left \{ + \sum \frac{m}{p_1} - + \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - + \ldots + \right \}, \tag{B} +\end{gather} +where the summation signs have the same significance as before. For +the number in question is evidently $\frac{m}{p_{i+1}}$ \emph{minus} +the number of integers not greater than $\frac{m}{p_{i+1}}$ and +divisible by at least one of the primes $p_1$, $p_2$, $\ldots$, +$p_i$. + +If we add (A) and (B) we have the number of integers less than $m$ +and divisible by one at least of the numbers $p_1$, $p_2$, $\ldots$, +$p_{i+1}$. Hence the number of integers less than $m$ and prime to +$p_1$, $p_2$, $\ldots$, $p_{i+1}$ is +\begin{gather*} +m - + \sum \frac{m}{p_1} + + \sum \frac{m}{p_1 p_2} - + \sum \frac{m}{p_1 p_2 p_3} + + \ldots, \\ +\intertext{where now in the summations the subscripts run from 1 to +$i+1$. This number is clearly equal to} +m + \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_{i+1}} \right ). +\end{gather*} +From this result, as we have seen above, our theorem follows at once +by induction.\index{Indicator!of any integer|)} + +\section{Sum of the Indicators of the Divisors of a Number}% +\label{s17} + +We shall first prove the following lemma: + +\smallskip \emph{Lemma. If $d$ is any divisor of $m$ and $m = nd$, +the number of integers not greater than $m$ which have with $m$ the +greatest common divisor $d$ is $\phi(n)$.} + +Every integer not greater than $m$ and having the divisor $d$ is +contained in the set $d$, $2d$, $3d$, $\ldots$, $nd$. The number of +these integers which have with $m$ the greatest common divisor $d$ +is evidently the same as the number of integers of the set 1, 2, +$\ldots$, $n$ which are prime to $\frac{m}{d}$, or $n$; for $\alpha +d$ and $n$ have or have not the greatest common divisor $d$ +according as $\alpha$ is or is not prime to $\frac{m}{d}=n$. Hence +the number in question is $\phi(n)$. + +From this lemma follows readily the proof of the following theorem: + +\smallskip \emph{If $d_1$, $d_2$, $\ldots$, $d_r$ are the different +divisors of $m$, then} +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = m. +\end{equation*} + +Let us define integers $m_1$, $m_2$, $\ldots$, $m_r$ by the +relations +\begin{equation*} +m = d_1 m_1 = d_2 m_2 = \ldots = d_r m_r. +\end{equation*} +Now consider the set of $m$ positive integers not greater than $m$, +and classify them as follows into $r$ classes. Place in the first +class those integers of the set which have with $m$ the greatest +common divisor $m_1$; their number is $\phi(d_1)$, as may be seen +from the lemma. Place in the second class those integers of the set +which have with $m$ the greatest common divisor $m_2$; their number +is $\phi(d_2)$. Proceeding in this way throughout, we place finally +in the last class those integers of the set which have with $m$ the +greatest common divisor $m_r$; their number is $\phi(d_r)$. It is +evident that every integer in the set falls into one and into just +one of these $r$ classes. Hence the total number $m$ of integers in +the set is $\phi(d_1) + \phi(d_r) + \ldots + \phi(d_r)$. From this +the theorem follows immediately. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the indicator of any integer greater than $2$ +is even. + +\item[2.] Prove that the number of irreducible fractions not greater +than $1$ and with denominator equal to $n$ is $\phi(n)$. + +\item[3.] Prove that the number of irreducible fractions not greater +than $1$ and with denominators not greater than $n$ is +\begin{equation*} +\phi(1) + \phi(2) + \phi(3) + \ldots + \phi(n). +\end{equation*} + +\item[4.] Show that the sum of the integers less than $n$ and prime to +$n$ is $\frac{1}{2} n \phi(n)$ if $n > 1$. + +\item[5.] Find ten values of $x$ such that $\phi(x) = 24$. + +\item[6.] Find seventeen values of $x$ such that $\phi(x) = 72$. + +\item[7.] Find three values of $n$ for which there is no $x$ satisfying +the equation $\phi(x) = 2n$. + +\item[8.] Show that if the equation +\begin{equation*} +\phi(x) = n +\end{equation*} +has one solution it always has a second solution, $n$ being given +and $x$ being the unknown. + +\item[9.] Prove that all the solutions of the equation +\begin{equation*} +\phi(x) = 4n - 2, n > 1, +\end{equation*} +are of the form $p^\alpha$ and $2p^\alpha$, where $p$ is a prime of +the form $4s-1$. + +\item[10.] How many integers prime to $n$ are there in the set +\begin{enumerate} +\item $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots, n(n+1)$? +\item $1 \cdot 2 \cdot 3, 2 \cdot 3 \cdot 4, + 3 \cdot 4 \cdot 5, \ldots, n(n+1)(n+2)$? +\item $\frac{1 \cdot 2}{2}, \frac{2 \cdot 3}{2}, + \frac{3 \cdot 4}{2}, \ldots, \frac{n(n+1)}{2}$? +\item $\frac{1 \cdot 2 \cdot 3}{6}, + \frac{2 \cdot 3 \cdot 4}{6}, + \frac{3 \cdot 4 \cdot 5}{6}, + \ldots, + \frac{n(n+1)(n+2)}{6}$? +\end{enumerate} + +\item[11*.] Find a method for determining all the solutions of the +equation +\begin{equation*} +\phi(x) = n, +\end{equation*} +where $n$ is given and $x$ is to be sought. + +\item[12*.] A number theory function $\phi(n)$ is defined for every +positive integer $n$; and for every such number $n$ it satisfies the +relation +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = n, +\end{equation*} +where $d_1, d_2, \ldots, d_r$ are the divisors of $n$. From this +property alone show that +\begin{equation*} +\phi(n) = n \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_k} \right ), +\end{equation*} +where $p_1, p_2, \ldots, p_k$ are the different prime factors of +$n$. \end{enumerate} \normalsize\index{Indicator|)} + +\chapter{ELEMENTARY PROPERTIES OF CONGRUENCES}% +\index{Congruences|(} + +\section{Congruences Modulo $m$}\label{s18} + +\textsc{Definitions.} If $a$ and $b$ are any two integers, positive +or zero or negative, whose difference is divisible by $m$, $a$ and +$b$ are said to be congruent modulo $m$, or congruent for the +modulus $m$, or congruent according to the modulus $m$. Each of +the numbers $a$ and $b$ is said to be a residue of the other.% +\index{Residue} + +\smallskip To express the relation thus defined we may write +\begin{equation*} +a = b + cm, +\end{equation*} +where $c$ is an integer (positive or zero or negative). It is more +convenient, however, to use a special notation due to Gauss, and to +write +\begin{equation*} +a \equiv b \mod m, +\end{equation*} +an expression which is read $a$ is congruent to $b$ modulo $m$, or +$a$ is congruent to $b$ for the modulus $m$, or $a$ is congruent to +$b$ according to the modulus $m$.\index{Gauss} This notation has the +advantage that it involves only the quantities which are essential +to the idea involved, whereas in the preceding expression we had the +irrelevant integer $c$. The Gaussian notation is of great value and +convenience in the study of the theory of divisibility. In the +present chapter we develop some of the fundamental elementary +properties of congruences. It will be seen that many theorems +concerning equations are likewise true of congruences with fixed +modulus; and it is this analogy with equations which gives +congruences (as such) one of their chief claims to attention. + +As immediate consequences of our definitions we have the following +fundamental theorems: + +\smallskip I.~\emph{If} $a\equiv c \mod m$, $b\equiv c\mod m$, +\emph{then} $a\equiv b\mod m$; \noindent \emph{that is, for a given +modulus, numbers congruent to the same number are congruent to each +other.} + +For, by hypothesis, $a - c = c_1 m$, $b - c = c_2 m$, where $c_1$ +and $c_2$ are integers. Then by subtraction we have $a - b = (c_1 - +c_2) m$; whence $a \equiv b \mod m$. + +\smallskip II.~\emph{If} $a \equiv b \mod m$, $\alpha \equiv +\beta \mod m$, \emph{then} $a \pm \alpha \equiv b \pm \beta \mod m$; +\emph{that is, congruences with the same modulus may be added or +subtracted member by member.} + +For, by hypothesis, $a - b = c_1 m$, $\alpha - \beta = c_2 m$; +whence $(a \pm \alpha) - (b \pm \beta) = (c_1 \pm c_2)m$. Hence $a +\pm \alpha = b \pm \beta \mod m$. + +\smallskip III.~\emph{If} $a = b \mod m$, \emph{then} +$ca = cb \mod m$, \emph{$c$ being any integer whatever.} + +The proof is obvious and need not be stated. + +\smallskip IV.~\emph{If} $a \equiv b \mod m$, +$\alpha \equiv \beta \mod m$, \emph{then} $a \alpha \equiv b \beta +\mod m$; \emph{that is, two congruences with the same modulus may be +multiplied member by member.} + +For, we have $a = b + c_1 m$, $\alpha = \beta + c_2 m$. Multiplying +these equations member by member we have $a \alpha = b \beta + m (b +c_2 + \beta c_1 + c_1 c_2 m)$. Hence $a \alpha \equiv b \beta \mod +m$. + +\smallskip A repeated use of this theorem gives the following +result: + +\smallskip V.~\emph{If} $a \equiv b \mod m$, \emph{then} +$a^n \equiv b^n \mod m$ \emph{where $n$ is any positive integer.} + +\smallskip As a corollary of theorems II, III and V we have the +following more general result: + +\smallskip VI.~\emph{If $f(x)$ denotes any polynomial in $x$ with +coefficients which are integers (positive or zero or negative) and +if further $a\equiv b \bmod m$, then} +\begin{equation*} +f(a) \equiv f(b) \bmod m. +\end{equation*} + +\section{Solutions of Congruences by Trial}\label{s19}% +\index{Congruences!Solution by trial|(} + +Let $f(x)$ be any polynomial in $x$ with coefficients which are +integers (positive or negative or zero). Then if $x$ and $c$ are any +two integers it follows from the last theorem of the preceding +section that +\begin{gather*} +f(x) \equiv f(x + cm) \bmod m. \tag{1} \\ +\intertext{Hence if $a$ is any value of $x$ for which the +congruence} +f(x)\equiv 0\bmod m. \tag{2} +\end{gather*} +is satisfied, then the congruence is also satisfied for $x = \alpha ++ cm$, where $c$ is any integer whatever. The numbers $\alpha + cm$ +are said to form a \emph{solution} (or to be a \emph{root}) of the +congruence, $c$ being a variable integer. Any one of the integers +$\alpha + cm$ may be taken as the representative of the solution. We +shall often speak of one of these numbers as the solution itself. + +Among the integers in a solution of the congruence (2) there is +evidently one which is positive and not greater than $m$. Hence all +solutions of a congruence of the type (2) may be found by trial, a +substitution of each of the numbers $1, 2, \ldots, m$ being made for +$x$. It is clear also that $m$ is the maximum number of solutions +which (2) can have whatever be the function $f(x)$. By means of an +example it is easy to show that this maximum number of solutions is +not always possessed by a congruence; in fact, it is not even +necessary that the congruence have a solution at all. + +This is illustrated by the example +\begin{equation*} +x^2 - 3 \equiv 0 \bmod 5. +\end{equation*} +In order to show that no solution is possible it is necessary to +make trial only of the values $1, 2, 3, 4, 5$ for $x$. A direct +substitution verifies the conclusion that none of them satisfies the +congruence; and hence that the congruence has no solution at all. + +On the other hand the congruence +\begin{equation*} +x^5 - x \equiv 0 \bmod 5 +\end{equation*} +has the solutions $x = 1, 2, 3, 4, 5$ as one readily verifies; that +is, this congruence has five solutions---the maximum number possible +in accordance with the results obtained above. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $(a + b)^p \equiv a^p + b^p \bmod p$ +where $a$ and $b$ are any integers and $p$ is any prime. + +\item[2.] From the preceding result prove that +$\alpha^p \equiv \alpha \bmod p$ for every integer $\alpha$. + +\item[3.] Find all the solutions of each of the congruences $x^{11} +\equiv x \bmod 11, x^{10} \equiv 1 \bmod 11, x^{5} \equiv 1 \bmod +11$. +\end{enumerate} \normalsize\index{Congruences!Solution by trial|)} + +\section{Properties of Congruences Relative to Division}\label{s20} + +The properties of congruences relative to addition, subtraction and +multiplication are entirely analogous to the properties of algebraic +equations. But the properties relative to division are essentially +different. These we shall now give. + +\smallskip I.~\emph{If two numbers are congruent modulo $m$ they are +congruent modulo $d$, where $d$ is any divisor of $m$.} + +For, from $a \equiv b \bmod m$, we have $a = b + cm = b + c'd$. +Hence $a\equiv b \bmod d$. + +\smallskip II.~\emph{If two numbers are congruent for different +moduli they are congruent for a modulus which is the least common +multiple of the given moduli.} + +The proof is obvious, since the difference of the given numbers is +divisible by each of the moduli. + +\smallskip III.~\emph{When the two members of a congruence are +multiples of an integer $c$ prime to the modulus, each member of the +congruence may be divided by $c$.} + +For, if $ca \equiv cb \bmod m$ then $ca - cb$ is divisible by $m$. +Since $c$ is prime to $m$ it follows that $a - b$ is divisible by +$m$. Hence $a\equiv b \bmod m$. + +\smallskip IV.~\emph{If the two members of a congruence are +divisible by an integer $c$, having with the modulus the greatest +common divisor $\delta$, one obtains a congruence equivalent to the +given congruence by dividing the two members by $c$ and the modulus +by $\delta$.} + +By hypothesis $ac \equiv bc \bmod m,\quad c = \delta c_1,\quad m = +\delta m_1$. Hence $c(a-b)$ is divisible by $m$. A necessary and +sufficient condition for this is evidently that $c_1(a-b)$ is +divisible by $m_1$. This leads at once to the desired result. + +\section{Congruences with a Prime Modulus}\label{s21}% +\index{Congruences!with prime modulus|(} + +\emph{The congruence\footnote{The sign $\not\equiv$ is read \emph{is +not congruent to}.}} +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod p, + \quad a_0 \not\equiv 0 \bmod p +\end{equation*} +\emph{where $p$ is a prime number and the $a$'s are any integers, +has not more than $n$ solutions.} + +Denote the first member of this congruence by $f(x)$ so that the +congruence may be written +\begin{gather} +f(x) \equiv 0 \bmod p \tag{1} \\ +\intertext{Suppose that $a$ is a root of the congruence, so that} +f(a) \equiv 0 \bmod p. \notag \\ +\intertext{Then we have} f(x) +\equiv f(x) - f(a) \bmod p. \notag \\ +\intertext{But, from algebra, $f(x) - f(a)$ is divisible by $x - a$. +Let $(x-a)^{\alpha}$ be the highest power of $x - a$ contained in +$f(x) - f(a)$. Then we may write} +f(x) - f(a) = (x - a)^{\alpha} f_1(x), \tag{2} \\ +\intertext{where $f_1(x)$ is evidently a polynomial with integral +coefficients. Hence we have} +f(x) \equiv (x - a)^{\alpha} f_1(x) \bmod p. \tag{3} +\end{gather} +We shall say that $a$ occurs $\alpha$ times as a solution of (1); or +that the congruence has $\alpha$ solutions each equal to $a$. + +Now suppose that congruence (1) has a root $b$ such that +$b\not\equiv a \bmod p$. Then from (3) we have +\begin{gather*} +f(b) \equiv (b-a)^{\alpha}f_1(b) \bmod p. \\ +\intertext{But} +f(b)\equiv 0 \bmod p,\quad (b-a)^{\alpha} \not\equiv 0 \bmod p. \\ +\intertext{Hence, since $p$ is a prime number, we must have} +f_1(b)\equiv 0 \bmod p. +\end{gather*} + +By an argument similar to that just used above, we may show that +$f_1(x) - f_1(b)$ may be written in the form +\begin{gather*} +f_1(x) - f_1(b) = (x-b)^{\beta}f_2(x), \\ +\intertext{where $\beta$ is some positive integer. Then we have} +f(x) \equiv (x-a)^{\alpha}(x-b)^{\beta}f_2(x) \bmod p. +\end{gather*} + +Now this process can be continued until either all the solutions of +(1) are exhausted or the second member is a product of linear +factors multiplied by the integer $a_0$. In the former case there +will be fewer than $n$ solutions of (1), so that our theorem is true +for this case. In the other case we have +\begin{equation*} +f(x) \equiv a_0(x-a)^{\alpha}(x-b)^{\beta} + \ldots (x-l)^{\lambda} \bmod p. +\end{equation*} +We have now $n$ solutions of (1): $a$ counted $\alpha$ times, $b$ +counted $\beta$ times, \ldots, $l$ counted $\lambda$ times; $\alpha ++ \beta + \ldots +\lambda = n$. + +Now let $\eta$ be any solution of (1). Then +\begin{equation*} +f(\eta) \equiv a_0(\eta-a)^{\alpha}(\eta-b)^{\beta} \ldots + (\eta-l)^{\lambda} \equiv 0 \bmod p. +\end{equation*} +Since $p$ is prime it follows now that some one of the factors +$\eta-a, \eta-b, \ldots, \eta-l$ is divisible by $p$. Hence $\eta$ +coincides with one of the solutions $a, b, c, \ldots, l$. That is, +(1) can have only the $n$ solutions already found. + +This completes the proof of the theorem. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Construct a congruence of the form +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod m, \quad + a_0 \not\equiv 0 \bmod m, +\end{equation*} +having more than $n$ solutions and thus show that the limitation to +a prime modulus in the theorem of this section is essential. + +\item[2.] Prove that +\begin{equation*} +x^6-1 \equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) \bmod 7 +\end{equation*} +for every integer $x$. + +\item[3.] How many solutions has the congruence $x^5 \equiv 1 \bmod +11$? the congruence $x^5\equiv 2 \bmod 11$? +\end{enumerate}\normalsize\index{Congruences!with prime modulus|)} + +\section{Linear Congruences}\label{s22}% +\index{Congruences!Linear|(} + +From the theorem of the preceding section it follows that the +congruence +\begin{equation*} +ax \equiv c \bmod p,\quad a \not\equiv 0 \bmod p, +\end{equation*} +where $p$ is a prime number, has not more than one solution. In this +section we shall prove that it always has a solution. More +generally, we shall consider the congruence +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +where $m$ is any integer. The discussion will be broken up into +parts for convenience in the proofs. + +\smallskip I.~\emph{The congruence} +\begin{equation} +ax \equiv 1 \bmod m, \tag{1} +\end{equation} +\emph{in which a and m are relatively prime, has one and only one +solution.} + +The question as to the existence and number of the solutions of (1) +is equivalent to the question as to the existence and number of +integer pairs $x, y$ satisfying the equation, +\begin{equation} +ax - my = 1, \tag{2} +\end{equation} +the integers $x$ being incongruent modulo $m$. Since $a$ and $m$ are +relatively prime it follows from theorem IV of \S~\ref{s9} that +there exists a solution of equation (2). Let $x = \alpha$ and $y = +\beta$ be a particular solution of (2) and let $x = \bar{\alpha}$ +and $y = \bar{\beta}$ be any solution of (2). Then we have +\begin{gather*} +a\alpha-m\beta = 1, \\ +a \bar{\alpha} - m\bar{\beta} = 1; \\ +\intertext{whence} +a(\alpha - \bar{\alpha}) - m(\beta - \bar{\beta}) = 0. +\end{gather*} +Hence $\alpha-\bar{\alpha}$ is divisible by $m$, since $a$ and $m$ +are relatively prime. That is, $a \equiv \bar{\alpha} \mod m$. Hence +$\alpha$ and $\bar{\alpha}$ are representatives of the same solution +of (1). Hence (1) has one and only one solution, as was to be +proved. + +\smallskip II.~\emph{The solution $x = \alpha$ of the congruence +$ax \equiv 1 \mod m$, in which $a$ and $m$ are relatively prime, is +prime to $m$.} + +For, if $a\alpha - 1$ is divisible by $m$, $\alpha$ is divisible by +no factor of $m$ except $1$. + +\smallskip III.~\emph{The congruence} +\begin{equation} +ax \equiv c \mod m \tag{3} +\end{equation} +\emph{in which $a$ and $m$ and also $c$ and $m$ are relatively +prime, has one and only one solution.} + +Let $x = \gamma$ be the unique solution of the congruence $cx = 1 +\mod m$. Then we have $a\gamma x \equiv c\gamma \equiv 1 \mod m$. +Now, by I we see that there is one and only one solution of the +congruence $a\gamma x \equiv 1 \mod m$; and from this the theorem +follows at once. + +Suppose now that $a$ is prime to $m$ but that $c$ and $m$ have the +greatest common divisor $\delta$ which is different from 1. Then it +is easy to see that any solution $x$ of the congruence $ax \equiv c +\mod m$ must be divisible by $\delta$. The question of the existence +of solutions of the congruence $ax \equiv c \bmod m$ is then +equivalent to the question of the existence of solutions of the +congruence +\begin{equation*} +a \frac{x}{\delta} \equiv \frac{c}{\delta} \bmod \frac{m}{\delta}, +\end{equation*} +where $\frac{x}{\delta}$ is the unknown integer. From III it follows +that this congruence has a unique solution $\frac{x}{\delta} = +\alpha$. Hence the congruence $ax \equiv c \bmod m$ has the unique +solution $x = \delta\alpha$. Thus we have the following theorem: + +\smallskip IV.~\emph{The congruence $ax \equiv c \bmod m$, in which +$a$ and $m$ are relatively prime, has one and only one solution.} + + +\smallskip\textsc{Corollary.}~\emph{The congruence +$ax \equiv c \bmod p$, $a \not\equiv 0 \bmod p$, where $p$ is a +prime number, has one and only one solution.} + +It remains to examine the case of the congruence $ax =c \bmod m$ in +which $a$ and $m$ have the greatest common divisor $d$. It is +evident that there is no solution unless $c$ also contains this +divisor $d$. Then let us suppose that $a = \alpha d$, $c = \gamma +d$, $m = \mu d$. Then for every $x$ such that $ax = c \bmod m$ we +have $\alpha x = \gamma \bmod \mu$; and conversely every $x$ +satisfying the latter congruence also satisfies the former. Now +$\alpha x = \gamma \bmod \mu$, has only one solution. Let $\beta$ be +a non-negative number less than $\mu$, which satisfies the +congruence $\alpha x = \gamma \bmod \mu$. All integers which satisfy +this congruence are then of the form $\beta + \mu\nu$, where $\nu$ +is an integer. Hence all integers satisfying the congruence $ax = c +\bmod m$ are of the form $\beta + \mu\nu$; and every such integer is +a representative of a solution of this congruence. It is clear that +the numbers +\begin{equation} +\beta,\ \beta + \mu,\ \beta + 2\mu,\ \ldots,\ \beta + (d-1)\mu +\tag{A} +\end{equation} +are incongruent modulo $m$ while every integer of the form $\beta + +\mu\nu$ is congruent modulo $m$ to a number of the set (A). Hence +the congruence $ax = c \bmod m$ has the $d$ solutions (A). + +This leads us to an important theorem which includes all the other +theorems of this section as special cases. It may be stated as +follows: + +\smallskip V.~\emph{Let} +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +\emph{be any linear congruence and let $a$ and $m$ have the greatest +common divisor $d (d \geq 1)$. Then a necessary and sufficient +condition for the existence of solutions of the congruence is that +$c$ be divisible by $d$. If this condition is satisfied the +congruence has just $d$ solutions, and all the solutions are +congruent modulo $m / d$.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find the remainder when $2^{40}$ is divided by $31$; when +$2^{43}$ is divided by $31$. + +\item[2.] Show that $2^{2^5}+1$ has the factor $641$. + +\item[3.] Prove that a number is a multiple of $9$ if and only if the +sum of its digits is a multiple of $9$. + +\item[4.] Prove that a number is a multiple of $11$ if and only if the +sum of the digits in the odd numbered places diminished by the sum +of the digits in the even numbered places is a multiple of $11$. +\end{enumerate} \normalsize% +\index{Congruences|)}\index{Congruences!Linear|)} + +\chapter{THE THEOREMS OF FERMAT AND WILSON} + +\section{Fermat's General Theorem}\label{s23}% +\index{Fermat's!General Theorem} + +Let $m$ be any positive integer and let +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)} \tag{A} +\end{equation} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$. Let $a$ be any integer prime to $m$ and form the set +of integers +\begin{equation} +aa_1,\ aa_2,\ \ldots,\ aa_{\phi(m)} \tag{B} +\end{equation} +No number $aa_i$ of the set (B) is congruent to a number $aa_j$, +unless $j = i;$ for, from +\begin{equation*} +aa_i \equiv aa_j \bmod m +\end{equation*} +we have $a_i \equiv a_j \bmod m$; whence $a_i = a_j$ since both +$a_i$ and $a_j$ are positive and not greater than $m$. Therefore $j += i$. Furthermore, every number of the set (B) is congruent to some +number of the set (A). Hence we have congruences of the form +\begin{align*} +aa_1 & \equiv a_{i_1} \bmod m, \\ +aa_2 & \equiv a_{i_2} \bmod m, \\ + & \vdots \\ +aa_{\phi(m)} & \equiv a_{i_{\phi(m)}} \bmod m. +\end{align*} +No two numbers in the second members are equal, since $aa_i +\not\equiv aa_j$ unless $i= j$. Hence the numbers $a_{i_1},\ +a_{i_2},\ \ldots,\ a_{i_{\phi(m)}}$ are the numbers $a_1,\ a_2,\ +\ldots,\ a_{\phi(m)}$ in some order. Therefore, if we multiply the +above system of congruences together member by member and divide +each member of the resulting congruence by $a_1\cdot a_2\ldots +a_{\phi(m)}$ (which is prime to $m$), we have +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} +This result is known as Fermat's general theorem.% +\index{Fermat's!general theorem} It may be stated as follows: + +\emph{If $m$ is any positive integer and $a$ is any integer prime to +$m$, then} +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ is any integer +not divisible by a prime number $p$, then} +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +\smallskip \textsc{Corollary 2.}~\emph{If $p$ is any prime number +and $a$ is any integer, then} +\begin{equation*} +a^p \equiv a \bmod p. +\end{equation*} + +\section{Euler's Proof of the Simple Fermat Theorem}\label{s24}% +\index{Euler}\index{Fermat}\index{Fermat's!Simple Theorem} + +The theorem of Cor.\ 1, \S~\ref{s23}, is often spoken of as the +simple Fermat theorem. It was first announced by Fermat in 1679, but +without proof. The first proof of it was given by Euler in 1736. +This proof may be stated as follows: + +From the Binomial Theorem it follows readily that +\begin{gather*} +(a+1)^p \equiv a^p + 1 \bmod p \\ +\intertext{since} +\frac{p!}{r!(p-r)!}, \quad 0 < r < p, \\ +\intertext{is obviously divisible by $p$. Subtracting $a + 1$ from +each side of the foregoing congruence, we have} +(a+1)^p - (a+1) \equiv a^p - a \bmod p. +\end{gather*} +Hence if $a^p - a$ is divisible by $p$, so is $(a + 1)^p - (a + 1)$. +But $1^p - 1$ is divisible by $p$. Hence $2^p - 2$ is divisible by +$p$; and then $3^p - 3$; and so on. Therefore, in general, we have +\begin{equation*} +a^p \equiv a \mod p. +\end{equation*} +If $a$ is prime to $p$ this gives $a^{p-1} \equiv 1 \mod p$, as was +to be proved. + +If instead of the Binomial Theorem one employs the Polynomial +Theorem, an even simpler proof is obtained. For, from the latter +theorem, we have readily +\begin{gather*} +(\alpha_1 + \alpha_2 + \ldots + \alpha_a)^p \equiv + \alpha_1^p + \alpha_2^p + \ldots + \alpha_a^p \mod p. \\ +\intertext{Putting $\alpha_1 = \alpha_2 = \ldots = \alpha_a = 1$ we +have} +a^p\equiv a \mod p, +\end{gather*} +from which the theorem follows as before. + +\section{Wilson's Theorem}\label{s25}\index{Wilson's theorem|(} + +From the simple Fermat theorem it follows that the congruence +\begin{gather*} +x^{p-1} \equiv 1\mod p \\ +\intertext{has the $p-1$ solutions $1$, $2$, $3$, $\ldots$, $p-1$. +Hence from the discussion in \S \ref{s21} it follows that} +x^{p-1}-1 \equiv (x-1)(x-2)\ldots(x-\overline{p-1}) \mod p, \\ +\intertext{this relation being satisfied for every value of $x$. +Putting $x = 0$ we have} +(-1) = (-1)^{p - 1}\cdot 1\cdot 2\cdot 3 \ldots + \overline{p-1}\mod p. \\ +\intertext{If $p$ is an odd prime this leads to the congruence} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} + 1 = 0 \mod p. +\end{gather*} +Now for $p = 2$ this congruence is evidently satisfied. Hence +we have the Wilson theorem: + +\smallskip \emph{Every prime number $p$ satisfies the relation} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{p+1} + 1 \equiv 0 \mod p. +\end{equation*} + +An interesting proof of this theorem on wholly different principles +may be given. Let $p$ points be distributed at equal intervals on +the circumference of a circle. The whole number of $p$-gons which +can be formed by joining up these $p$ points in every possible order +is evidently +\begin{equation*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1; +\end{equation*} +for the first vertex can be chosen in $p$ ways, the second in $p - +1$ ways, $\ldots$, the $(p-1)^{\mathrm{th}}$ in two ways, and the +last in one way; and in counting up thus we have evidently counted +each polygon $2p$ times, once for each vertex and for each direction +from the vertex around the polygon. Of the total number of polygons +$\frac{1}{2}(p-1)$ are regular (convex or stellated) so that a +revolution through $\frac{360^\circ}{p}$ brings each of these into +coincidence with its former position. The number of remaining +$p$-gons must be divisible by $p$; for with each such $p$-gon we may +associate the $p-1$ $p$-gons which can be obtained from it by +rotating it through successive angles of $\frac{360^\circ}{p}$. That +is, +\begin{gather*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - + \frac 12 (p-1) \equiv 0 \bmod p. \\ +\intertext{Hence} +(p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - p + 1 \equiv 0 \bmod p; \\ +\intertext{and from this it follows that} +1 \cdot 2 \ldots \overline{p-1} + 1 \equiv 0 \bmod p, \\ +\end{gather*} +as was to be proved. + +\section{The Converse of Wilson's Theorem}\label{s26} + +Wilson's theorem is noteworthy in that its converse is also true. +The converse may be stated as follows: + +\smallskip \emph{Every integer $n$ such that the congruence} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n +\end{equation*} +\emph{is satisfied is a prime number.} + +For, if $n$ is not prime, there is some divisor $d$ of $n$ different +from $1$ and less than $n$. For such a $d$ we have $1 \cdot 2 \cdot +3 \ldots \overline{n-1} \equiv 0 \bmod d$; so that $1 \cdot 2 \ldots +\overline{n-1}+1 \not\equiv 0 \bmod d$; and hence $1 \cdot 2 \ldots +\overline{n-1}+1 \equiv 0 \bmod n$. Since this contradicts our +hypothesis the truth of the theorem follows. + +\smallskip Wilson's theorem and its converse may be combined into +the following elegant theorem: + +\smallskip \emph{A necessary and sufficient condition that an +integer $n$ is prime is that} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n. +\end{equation*}\index{Prime numbers} + +Theoretically this furnishes a complete and elegant test as to +whether a given number is prime. But, practically, the labor of +applying it is so great that it is useless for verifying large +primes. + +\section{Impossibility of $1 \cdot 2 \cdot 3 \cdots +\overline{n-1} + 1 = n^k$ for $n > 5$.}\label{s27} + +In this section we shall prove the following theorem: + +\emph{There exists no integer $k$ for which the equation} +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-1} + 1 = n^k +\end{equation*} +is true when $n$ is greater than $5$. + +If $n$ contains a divisor $d$ different from $1$ and $n$, the +equation is obviously false; for the second member is divisible by +$d$ while the first is not. Hence we need to prove the theorem only +for primes $n$. + +Transposing $1$ to the second member and dividing by $n - 1$ we have +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-2} = n^{k-1} + n^{k-2} + + \ldots + n+1. +\end{equation*} +If $n>5$ the product on the left contains both the factor $2$ and +the factor $\frac{1}{2} (n-1)$; that is, the first member contains +the factor $n - 1$. But the second member does not contain this +factor, since for $n = 1$ the expression $n^{k-1} + \ldots n + 1$ is +equal to $k \neq 0$. Hence the theorem follows at once. + +\section{Extension of Fermat's Theorem}\label{s28}% +\index{Fermat's!theorem extended|(} + +The object of this section is to extend Fermat's general theorem and +incidentally to give a new proof of it. We shall base this proof on +the simple Fermat theorem, of which we have already given a simple +independent proof. This theorem asserts that for every prime $p$ and +integer $a$ not divisible by $p$, we have the congruence +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +Then let us write +\begin{gather} +a^{p-1} = 1 + hp. \tag{1} \\ +\intertext{Raising each member of this equation to the +$p^{\text{th}}$ power we may write the result in the form} +a^{p(p-1)} = 1 + h_1p^2. \tag{2} \\ +\intertext{where $h_1$ is an integer. Hence} +a^{p(p-1)} \equiv 1 \bmod p^2. \notag \\ +\intertext{By raising each member of (2) to the $p^{\text{th}}$ +power we can readily show that} +a^{p^2(p-1)} \equiv 1 \bmod p^3. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{p^{\alpha - 1}(p-1)} \equiv 1 \bmod p^{\alpha}. \notag \\ +\intertext{where $\alpha$ is a positive integer; that is,} +a^{\phi(p^{\alpha})} \equiv 1 \bmod p^{\alpha}. \notag +\end{gather} + +For the special case when $p$ is 2 this result can be extended. For +this case (1) becomes +\begin{gather} +a = 1 + 2h. \notag \\ +\intertext{Squaring we have} +a^2 = 1 + 4h(h+1). \notag \\ +\intertext{Hence,} +a^2 = 1+8h_1, \tag{3} \\ +\intertext{where $h_1$ is an integer. Therefore} +a^2 \equiv 1 \bmod 2^3. \notag \\ +\intertext{Squaring (3) we have} +a^{2^2} = 1 + 2^4h_2; \notag \\ +\intertext{or} +a^{2^2} \equiv 1 \bmod 2^4. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{2^{\alpha-2}} \equiv 1 \bmod 2^{\alpha} \notag \\ +\intertext{if $\alpha > 2$. That is,} +a^{\frac{1}{2}\phi(2^{\alpha})} \equiv 1 \bmod 2^{\alpha} + \text{ if } a > 2. +\end{gather} + +Now in terms of the $\phi$-function let us define a new function +$\lambda(m)$ as follows: +\begin{align*} +\lambda(2^{\alpha}) &= \phi(2^{\alpha}) \text{ if $a = 0, 1, 2$;} \\ +\lambda(2^{\alpha}) &= \frac{1}{2}\phi(2^{\alpha}) + \text{ if $a > 2$;} \\ +\lambda(p^{\alpha}) &= \phi(p^{\alpha}) + \text{ if $p$ is an odd prime;} \\ +\lambda(2^{\alpha} p_1^{\alpha_1} p_2^{\alpha_2} \cdots + p_n^{\alpha_n}) &= M, +\end{align*} +where $M$ is the least common multiple of +\begin{equation*} + \lambda(2^{\alpha}), + \lambda(p_1^{\alpha_1}), + \lambda(p_2^{\alpha_2}), \ldots, \lambda(p_n^{\alpha_n}), +\end{equation*} +$2, p_1, p_2, \ldots, p_n$ being different primes.% +\index{$\lambda(m)$} + +Denote by $m$ the number +\begin{equation*} +m = 2^{\alpha}p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_n^{\alpha_n}. +\end{equation*} +Let $a$ be any number prime to $m$. From our preceding results we +have +\begin{align*} +a^{\lambda(2^{\alpha})} &\equiv 1 \bmod 2^{\alpha}, \\ +a^{\lambda(p_1^{\alpha_1})} &\equiv 1 \bmod p_1^{\alpha_1},\\ +a^{\lambda(p_2^{\alpha_2})} &\equiv 1 \bmod p_2^{\alpha_2}, \\ +\ldots \\ +a^{\lambda(p_n^{\alpha_n})} &\equiv 1 \bmod p_2^{\alpha_n}. \\ +\end{align*} + +Now any one of these congruences remains true if both of its members +are raised to the same positive integral power, whatever that power +may be. Then let us raise both members of the first congruence to +the power $\frac{\lambda(m)}{\lambda(2^\alpha)}$; both members of +the second congruence to the power +$\frac{\lambda(m)}{\lambda(p_1^{\alpha_1})}$; $\ldots$; both members +of the last congruence to the power +$\frac{\lambda(m)}{\lambda(p_n^{\alpha_n})}$. Then we have +\begin{align*} +a^{\lambda(m)} &\equiv 1 \mod 2^\alpha, \\ +a^{\lambda(m)} &\equiv 1 \mod p_1^{\alpha_1}, \\ +\ldots \ldots \\ +a^{\lambda(m)} &\equiv 1 \mod p_n^{\alpha_n}. \\ +\intertext{From these congruences we have immediately} +a^{\lambda(m)} &\equiv 1 \mod m. +\end{align*} + +We may state this result in full in the following theorem: + +\smallskip \emph{If $a$ and $m$ are any two relatively prime positive +integers, the congruence} +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +\emph{is satisfied.} + +As an excellent example to show the possible difference between the +exponent $\lambda(m)$ in this theorem and the exponent $\phi(m)$ in +Fermat's general theorem, let us take +\begin{gather*} +m = 2^6 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 19 + \cdot 37 \cdot 73. \\ +\intertext{Here} +\lambda(m) = 2^4 \cdot 3^2, \quad \phi(m) = 2^{31} \cdot 3^{10}. +\end{gather*} + +In a later chapter we shall show that there is no exponent $\nu$ +less than $\lambda(m)$ for which the congruence +\begin{equation*} +a^\nu = 1 \mod m +\end{equation*} +is verified for every integer $a$ prime to $m$. + +From our theorem, as stated above, Fermat's general theorem follows +as a corollary, since $\lambda(m)$ is obviously a factor of +$\phi(m)$, +\begin{equation*} +\phi(m) = \phi(2^\alpha) \phi(p_1^{\alpha_1}) \ldots + \phi(p_n^{\alpha_n}). +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $a^{16} \equiv 1 \bmod 16320$, for every $a$ +which is prime to $16320$. + +\item[2.] Show that $a^{12} \equiv 1 \bmod 65520$, for every $a$ which +is prime to $65520$. + +\item[3*.] Find one or more composite numbers $P$ such that +\begin{equation*} +a^{P-1} \equiv 1 \bmod P +\end{equation*} +for every a prime to $P$. (Compare this problem with the next +section.) \end{enumerate} \normalsize% +\index{Fermat's!theorem extended|)} + +\section{On the Converse of Fermat's Simple Theorem}\label{s29}% +\index{Fermat's!Simple Theorem} + +The fact that the converse of Wilson's theorem is a true proposition +leads one naturally to inquire whether the converse of Fermat's +simple theorem is true. Thus, we may ask the question: Does the +existence of the congruence $2^{n-1} \equiv 1 \bmod n$ require that +$n$ be a prime number? The Chinese answered this question in the +affirmative and the answer passed unchallenged among them for many +years. An example is sufficient to show that the theorem is not +true. We shall show that +\begin{equation*} +2^{340} \equiv 1 \bmod 341 +\end{equation*} +although $341 = 11 \cdot 31$, is not a prime number. Now $2^{10}-1 = +3 \cdot 11 \cdot 31$. Hence $2^{10} \equiv 1 \bmod 341$. Hence +$2^{340} \equiv 1 \bmod 341$. From this it follows that the direct +converse of Fermat's theorem is not true. The following theorem, +however, which is a converse with an extended hypothesis, is readily +proved. + +\smallskip \emph{If there exists an integer $a$ such that} +\begin{equation*} +a^{n-1} \equiv 1 \bmod n +\end{equation*} +\emph{and if further there does not exist an integer $\nu$ less than +$n - 1$ such that} +\begin{equation*} +a^{\nu} \equiv 1 \bmod n, +\end{equation*} +\emph{then the integer $n$ is a prime number.} + +For, if $n$ is not prime, $\phi(n) < n - 1$. Then for $\nu = +\phi(n)$ we have $a^{\nu} \equiv 1 \bmod n$, contrary to the +hypothesis of the theorem. + +\section{Application of Previous Results to Linear +Congruences}\label{s30}% +\index{Congruences!Linear} + +The theorems of the present chapter afford us a ready means of +writing down a solution of the congruence +\begin{equation} +ax \equiv c \bmod m. \tag{1} +\end{equation} +We shall consider only the case in which $a$ and $m$ are relatively +prime, since the general case is easily reducible to this one, as we +saw in the preceding chapter. + +Since $a$ and $m$ are relatively prime we have the congruences +\begin{gather*} +a^{\lambda(m)} \equiv 1,\quad a^{\phi(m)} \equiv 1 \bmod m. \\ +\intertext{Hence either of the numbers $x$,} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +is a representative of the solution of (1). Hence the following +theorem: + +\smallskip \emph{If} +\begin{gather*} +ax \equiv c \bmod m \\ +\intertext{\emph{is any linear congruence in which $a$ and $m$ are +relatively prime, then either of the numbers $x$,}} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +\emph{is a representative of the solution of the congruence.} + +The former representative of the solution is the more convenient of +the two, since the power of $a$ is in general much less in this case +than in the other. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] Find a solution of $7x \equiv 1 \bmod 2^6 \cdot 3 \cdot 5 \cdot +17.$ Note the greater facility in applying the first of the above +representatives of the solution rather than the second. +\end{enumerate} \normalsize + +\section{Application of the Preceding Results to the Theory +of Quadratic Residues}\label{s31}\index{Quadratic residues|(} + +In this section we shall apply the preceding results of this chapter +to the problem of finding the solutions of congruences of the form +\begin{equation*} +\alpha z^2 + \beta z + \gamma \equiv 0 \mod \mu +\end{equation*} +where $\alpha, \beta, \gamma, \mu$ are integers. These are called +quadratic congruences. + +The problem of the solution of the quadratic congruence (1) can be +reduced to that of the solution of a simpler form of congruence as +follows: Congruence (1) is evidently equivalent to the congruence +\begin{gather} +4\alpha^2 z^2 + 4\alpha\beta z + 4\alpha\gamma \equiv + 0 \mod 4\alpha\mu. \tag{1} \\ +\intertext{But this may be written in the form} +(2\alpha z + \beta)^2 \equiv \beta^2 - 4\alpha\gamma + \mod 4\alpha\mu. \notag \\ +\intertext{Now if we put} +2\alpha z + \beta\equiv x \mod 4\alpha\mu \tag{2} \\ +\intertext{and} +\beta^2 - 4\alpha\gamma = a,\quad 4\alpha\mu = m, \notag \\ +\intertext{we have} +x^2 \equiv a\mod m. \tag{3} +\end{gather} +We have thus reduced the problem of solving the general congruence +(1) to that of solving the binomial congruence (3) and the linear +congruence (2). The solution of the latter may be effected by means +of the results of \S \ref{s30}. We shall therefore confine ourselves +now to a study of congruence (3). We shall make a further limitation +by assuming that $a$ and $m$ are relatively prime, since it is +obvious that the more general case is readily reducible to this one. + +The example +\begin{equation*} +x^2 \equiv 3 \mod 5 +\end{equation*} +shows at once that the congruence (3) does not always have a +solution. First of all, then, it is necessary to find out in what +cases (3) has a solution. Before taking up the question it will be +convenient to introduce some definitions. + +\smallskip\textsc{Definitions.} An integer $a$ is said to be a +quadratic residue modulo $m$ or a quadratic non-residue modulo $m$ +according as the congruence +\begin{equation*} +x^2 = a \mod m +\end{equation*} +has or has not a solution. We shall confine our attention to the +case when $m > 2$.\index{Residue} + +We shall now prove the following theorem: + +\smallskip I.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\lambda(m)} \equiv 1 \mod m. +\end{equation*} + +Suppose that the congruence $x^2 \equiv a \mod m$ has the solution $x = +\alpha$. Then $\alpha^2 \equiv a \mod m$. Hence +\begin{equation*} +\alpha^{\lambda(m)} \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +Since $a$ is prime to $m$ it is clear from $\alpha^2 \equiv a \mod +m$ that $\alpha$ is prime to $m$. Hence $\alpha^{\lambda(m)} \equiv 1 +\mod m$. Therefore we have +\begin{equation*} +1 \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +That is, this is a necessary condition in order that $a$ shall be a +quadratic residue modulo $m$. + +In a similar way one may prove the following theorem: + +\smallskip II.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\phi(m)} \equiv 1 \mod m. +\end{equation*} + +When $m$ is a prime number $p$ each of the above results takes the +following form: If $a$ is prime to $p$ and is a quadratic residue +modulo $p$, then +\begin{equation*} +a^{\frac{1}{2}(p-1)} \equiv 1 \mod p. +\end{equation*} +We shall now prove the following more complete theorem, without the +use of I or II. + +\smallskip III.~\emph{If $p$ is an odd prime number and $a$ is an +integer not divisible by $p$, then $a$ is a quadratic residue or a +quadratic non-residue modulo $p$ according as} +\begin{equation*} +a^{\tfrac{1}{2}(p-1)} \equiv +1 \quad \text{or} \quad +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p. +\end{equation*} + +This is called Euler's criterion.\index{Euler's!criterion} + +Given a number $a$, not divisible by $p$, we have to determine +whether or not the congruence +\begin{gather} +x^2 \equiv a \bmod p \notag \\ +\intertext{has a solution. Let $r$ be any number of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 \tag{A} \\ +\intertext{and consider the congruence} +rx \equiv a \bmod p. +\end{gather} +This has always one and just one solution $x$ equal to a number $s$ +of the set (A). Two cases can arise: either for every $r$ of the set +(A) the corresponding $s$ is different from $r$ or for some $r$ of +the set (A) the corresponding $s$ is equal to $r$. The former is the +case when $a$ is a quadratic non-residue modulo $p$; the latter is +the case when $a$ is a quadratic residue modulo $p$. We consider the +two cases separately. + +In the first case the numbers of the set (A) go in pairs such that +the product of the numbers in the pair is congruent to a modulo $p$. +Hence, taking the product of all $\tfrac{1}{2}(p - 1)$ pairs, we +have +\begin{align*} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &\equiv + +a^{\tfrac{1}{2}(p-1)} \bmod p. \\ +\intertext{But} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &= -1 \bmod p. \\ +\intertext{Hence} +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p, +\end{align*} +whence the truth of one part of the theorem. + +In the other case, namely that in which some $r$ and corresponding +$s$ are equal, we have for this $r$ +\begin{gather*} +r^{2} \equiv a \bmod p \\ +\intertext{and} +(p - r)^{2} \equiv a \bmod p. +\end{gather*} +Since $x^{2} \equiv a \bmod p$ has at most two solutions it follows +that all the integers in the set (A) except $r$ and $p - r$ fall in +pairs such that the product of the numbers in each pair is congruent +to a modulo $p$. Hence, taking the product of all these pairs, which +are $\frac{1}{2}(p - 1) - 1$ in number, and multiplying by $r(p-r)$ +we have +\begin{align*} +1 \cdot 2 \cdot 3 \cdots \overline{p -1} + &\equiv (p - r) r a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -r^{2} a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a^{\frac{1}{2}(p -1)} \bmod p. \\ +\intertext{Since $1 \cdot 2 \cdot 3 \cdots \overline{p - 1} \equiv -1 +\bmod p$ we have} +a^{\frac{1}{2}(p -1)} &\equiv + 1 \bmod p +\end{align*} +whence the truth of another part of the theorem. + +Thus the proof of the entire theorem is complete.% +\index{Quadratic residues|)}\index{Wilson's theorem|)} + +\chapter{PRIMITIVE ROOTS MODULO $m$.} + +\section{Exponent of an Integer Modulo $m$}\label{s32}% +\index{Exponent of an integer|(}\index{Primitive roots|(} + +Let +\begin{equation*} +a_{1},\ a_{2},\ \cdots,\ a_{\phi(m)} \tag{A} +\end{equation*} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$; and let $a$ denote any integer of the set (A). Now any +positive integral power of $a$ is prime to $m$ and hence is +congruent modulo $m$ to a number of the set (A). Hence, among all +the powers of a there must be two, say $a^{n}$ and $a^{\nu}$, $n > +\nu$, which, are congruent to the same integer of the set (A). These +two powers are then congruent to each other; that is, +\begin{equation*} +a^{n} \equiv a^{\nu} \bmod m +\end{equation*} +Since $a^{\nu}$ is prime to $m$ the members of this congruence may +be divided by $a^{\nu}$. Thus we have +\begin{equation*} +a^{n - \nu} \equiv 1 \bmod m. +\end{equation*} +That is, among the powers of $a$ there is one at least which is +congruent to $1$ modulo $m$. + +\smallskip Now, in the set of all powers of $a$ which are congruent +to $1$ modulo $m$ there is one in which the exponent is less than in +any other of the set. Let the exponent of this power be $d$, so that +$a^{d}$ is the lowest power of $a$ such that +\begin{equation} +a^{d} \equiv 1 \bmod m. \tag{1} +\end{equation} + +We shall now show that if $a^{\alpha} \equiv 1 \bmod m$, then +$\alpha$ is a multiple of $d$. Let us write +\begin{gather} +\alpha = d\delta + \beta, \quad 0 \leqq \beta < d. \notag \\ +\intertext{Then} +a^{\alpha} \equiv 1 \bmod m, \tag{2} \\ +a^{d\delta} \equiv 1 \bmod m, \tag{3} \\ +\intertext{the last congruence being obtained by raising (1) to the +power $\delta$. From (3) we have} +a^{d\delta + \beta} \equiv a^{\beta} \bmod m; \notag \\ +\intertext{or} +a^{\beta}\equiv 1 \bmod m. \notag +\end{gather} +Hence $\beta = 0$, for otherwise $d$ is not the exponent of the +lowest power of $a$ which is congruent to 1 modulo $m$. Hence $d$ is +a divisor of $\alpha$. + +\smallskip These results may be stated as follows: + +\smallskip I.~\emph{If $m$ is any integer and $a$ is any integer +prime to $m$, then there exists an integer $d$ such that} +\begin{gather*} +a^d\equiv 1 \bmod m \\ +\intertext{\emph{while there is no integer $\beta$ less than $d$ for +which}} +a^\beta\equiv 1 \bmod m. \\ +\intertext{\emph{Further, a necessary and sufficient condition +that}} +a^\nu \equiv 1 \bmod m +\end{gather*} +\emph{is that $\nu$ is a multiple of $d$.} + +\smallskip \textsc{Definition.} The integer $d$ which is thus +uniquely determined when the two relatively prime integers $a$ and +$m$ are given is called the exponent of $a$ modulo $m$. Also, $d$ is +said to be the exponent to which $a$ belongs modulo $m$. + +Now, in every case we have +\begin{equation*} +a^{\phi(m)} \equiv 1,\quad a^{\lambda(m)} \equiv 1 \bmod m, +\end{equation*} +if $a$ and $m$ are relatively prime. Hence from the preceding +theorem we have at once the following: + +\smallskip II.~\textit{The exponent $d$ to which $a$ belongs modulo +$m$ is a divisor of both $\phi(m)$ and $\lambda(m)$.}% +\index{Exponent of an integer|)} + +\section{Another Proof of Fermat's General Theorem}\label{s33} + +In this section we shall give an independent proof of the theorem +that the exponent $d$ of $a$ modulo $m$ is a divisor of $\phi(m)$; +from this result we have obviously a new proof of Fermat's theorem +itself. + +We retain the notation of the preceding section. We shall first +prove the following theorem: + +\smallskip I.~\textit{The numbers} +\begin{equation} +1,\ a,\ a^2,\ \ldots,\ a^{d-1} \tag{A} +\end{equation} +\textit{are incongruent each to each modulo $m$.} + +For, if $a^\alpha \equiv a^\beta \bmod m$, where $0 \leqq \alpha < +d$ and $0 \leqq \beta < d$, $\alpha > \beta$, we have +$a^{\alpha-\beta} \equiv 1 \bmod m$, so that $d$ is not the exponent +to which $a$ belongs modulo $m$, contrary to hypothesis. + +\smallskip Now any number of the set (A) is congruent to some number +of the set +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)}. \tag{B} +\end{equation} +Let us undertake to separate the numbers (B) into classes after the +following manner: Let the first class consist of the numbers +\begin{equation} +\alpha_1,\ \alpha_2,\ \ldots,\ \alpha_{d-1}, \tag{I} +\end{equation} +where $\alpha_i$ is the number of the set (B) to which $a^i$ is +congruent modulo $m$. + +If the class (I) does not contain all the numbers of the set (B), +let $a_i$ be any number of the set (B) not contained in (I) and form +the following set of numbers: +\begin{equation} +\alpha_0 a_i,\ \alpha_1 a_i,\ \alpha_2 a_i,\ \ldots,\ + \alpha_{d-1}a_i. \tag{II'} +\end{equation} +We shall now show that no number of this set is congruent to a +number of class (I). For, if so, we should have a congruence of the +form +\begin{gather*} +a_i \alpha_j \equiv \alpha_k \bmod m; \\ +\intertext{hence} +a_i a^j \equiv a^k \bmod m, \\ +\intertext{so that} +a_i a^d \equiv a^{k+d-j} \bmod m; \\ +\intertext{or} +a_i \equiv a^{k+d-j} \bmod m, +\end{gather*} +so that $a_i$ would belong to the set (I) contrary to hypothesis. + +Now the numbers of the set (II$'$) are all congruent to numbers of +the set (B); and no two are congruent to the same number of this +set. For, if so, we should have two numbers of (II') congruent; that +is, $\alpha_k a_i \equiv \alpha_j a_i \bmod m,$ or $\alpha_k \equiv +\alpha_j \bmod m;$ and this we have seen to be impossible. + +Now let the numbers of the set (B) to which the numbers of the set +(II') are congruent be in order the following: +\begin{equation} +\beta_0,\ \beta_1,\ \beta_2,\ \ldots,\ \beta_{d-1}. \tag{II} +\end{equation} +These numbers constitute our class (II). + +If classes (I) and (II) do not contain all the numbers of the set +(B), let $a_j$ be a number of the set ($B$) not contained in either +of the classes (I) and (II): and form the set of numbers +\begin{equation} +\alpha_0 a_j,\ \alpha_1 a_j,\ \alpha_2 a_j,\ \ldots,\ + \alpha_{d-1} a_j. \tag{III'} +\end{equation} +Just as in the preceding case it may be shown that no number of this +set is congruent to a number of class (I) and that the numbers of +(III') are incongruent each to each. We shall also show that no +number of (III') is congruent to a number of class (II). For, if so, +we should have $a_k a_j \equiv \beta_l \bmod m$. Hence $a^k a_j +\equiv a^l a_i \bmod m$; or $a_j \equiv a^{l+d-k} a_i \bmod m$, from +which it follows that $a_j$ is of class (II), contrary to +hypothesis. + +Now let the numbers of the set (B) to which the numbers of the set +(III') are congruent be in order the following: +\begin{equation} +\gamma_0,\ \gamma_1,\ \gamma_2,\ \ldots,\ \gamma_{d-1}. \tag{III} +\end{equation} +These numbers form our class (III). + +It is now evident that the process may be continued until all the +numbers of the set (B) have been separated into classes, each class +containing $d$ integers, thus: +\begin{equation*} +\begin{matrix} +(\text{I}) & \alpha_0, & \alpha_1, & \alpha_2, + & \ldots, & \alpha_{d-1}, \\ +(\text{II}) & \beta_0, & \beta_1, & \beta_2, + & \ldots, & \beta_{d-1}, \\ +(\text{III}) & \gamma_0, & \gamma_1, & \gamma_2, + & \ldots, & \gamma_{d-1}, \\ +&\hdotsfor{5} \\ +(\quad ) & \lambda_0, & \lambda_1, & \lambda_2, + & \ldots, & \lambda_{d-1}. +\end{matrix} +\end{equation*} +The set (B), which consists of $\phi(m)$ integers, has thus been +separated into classes, each class containing $d$ integers. Hence we +conclude that $d$ is a divisor of $\phi(m)$. Thus we have a second +proof of the theorem: + +\smallskip II.~\emph{If $a$ and $m$ are any two relatively prime +integers and $d$ is the exponent to which $a$ belongs modulo $m$, +then $d$ is a divisor of $\phi(m)$.} + +In our classification of the numbers (B) into the rectangular array +above we have proved much more than theorem II; in fact, theorem II +is to be regarded as one only of the consequences of the more +general result contained in the array. + +If we raise each member of the congruence +\begin{equation*} +a^d \equiv 1 \bmod m +\end{equation*} +to the (integral) power $\phi(m)/d$, the preceding theorem leads +immediately to an independent proof of Fermat's general theorem. + +\section{Definition of Primitive Roots}\label{s34} + +\textsc{Definition.} Let $a$ and $m$ be two relatively prime +integers. If the exponent to which $a$ belongs modulo $m$ is +$\phi(m)$, $a$ is said to be a primitive root modulo $m$ (or a +primitive root of $m$). + +In a previous chapter we saw that the congruence +\begin{equation*} +a^{\lambda(m)} \equiv 1 \bmod m +\end{equation*} +is verified by every pair of relatively prime integers $a$ and $m$. +Hence, primitive roots can exist only for such a modulus $m$ as +satisfies the equation +\begin{equation*} +\phi(m) = \lambda(m). \tag{1} +\end{equation*} +We shall show later that this is also sufficient for the existence +of primitive roots. + +From the relation which exists in general between the +$\phi$-function and the $\lambda$-function in virtue of the +definition of the latter, it follows that (1) can be satisfied only +when $m$ is a prime power or is twice an odd prime power. + +Suppose first that $m$ is a power of $2$, say $m = 2^\alpha$. Then +(1) is satisfied only if $\alpha = 0,\ 1,\ 2$. For $\alpha = 0$ or +$1$, $1$ itself is a primitive root. For $\alpha = 2$, $3$ is a +primitive root. We have therefore left to examine only the cases +\begin{equation*} +m = p^\alpha,\quad m = 2p^\alpha +\end{equation*} +where $p$ is an odd prime number. The detailed study of these cases +follows in the next sections. + +\section{Primitive roots modulo $p$.}\label{s35} + +We have seen that if $p$ is a prime number and $d$ is the exponent +to which $a$ belongs modulo $p$, then $d$ is a divisor of $\phi(p) = +p - 1$. Now, let +\begin{gather*} +d_1,\ d_2,\ d_3,\ \ldots,\ d_r \\ +\intertext{be all the divisors of $p-1$ and let $\psi(d_i)$ denote +the number of integers of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 +\end{gather*} +which belong to the exponent $d_i$. If there is no integer of the +set belonging to this exponent, then $\psi(d_i) = 0$. + +Evidently every integer of the set belongs to some one and only one +of the exponents $d_1, d_2, \ldots, d_r$. Hence we have the relation +\begin{gather} +\psi(d_1) + \psi(d_2) + \ldots + \psi(d_r) = p-1. \tag{1} \\ +\intertext{But} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = p-1. \tag{2} \\ +\intertext{If then we can show that} +\psi(d_i) \leqq \phi(d_i) \tag{3} \\ +\intertext{for $i = 1, 2, \ldots, r$, it will follow from a +comparison of (1) and (2) that} +\psi(d_i) = \phi(d_i). \notag +\end{gather} +Accordingly, we shall examine into the truth of (3). + +Now the congruence +\begin{equation} +x^{d_i} \equiv 1 \mod p \tag{4} +\end{equation} +has not more than $d_i$ roots. If no root of this congruence belongs +to the exponent $d_i$, then $\psi(d_i) = 0$ and therefore in this +case we have $\psi(d_i) < \phi(d_i)$. On the other hand if $a$ is a +root of (4) belonging to the exponent $d_i$, then +\begin{equation} +a, a^2, a^3, \ldots, a^{d_i} \tag{5} +\end{equation} +are a set of $d_i$ incongruent roots of (4); and hence they are the +complete set of roots of (4). + +But it is easy to see that $a^k$ does or does not belong to the +exponent $d_i$ according as $k$ is or is not prime to $d_i$; for, if +$a^k$ belongs to the exponent $t$, then $t$ is the least integer +such that $kt$ is a multiple of $d_i$. Consequently the number of +roots in the set (5) belonging to the exponent $d_i$ is $\phi(d_i)$. +That is, in this case $\psi(d_i) = \phi(d_i)$. Hence in general +$\psi(d_i) \leqq \phi(d_i)$. Therefore from (1) and (2) we conclude +that +\begin{equation*} +\psi(d_i) = \phi(d_i), \quad i = 1,\ 2,\ \ldots,\ r. +\end{equation*} +The result thus obtained may be stated in the form of the following +theorem: + +\smallskip I.~\emph{If $p$ is a prime number and $d$ is any divisor +of $p-1$, then the number of integers belonging to the exponent $d$ +modulo $p$ is $\phi(d)$.} + +In particular: + +\smallskip II.~\emph{There exist primitive roots modulo $p$ and their +number is $\phi(p-1)$.} + +\section{Primitive Roots Modulo $p^\alpha$, $p$ an Odd +Prime}\label{s36} + +In proving that there exist primitive roots modulo $p^\alpha$, where +$p$ is an odd prime and $\alpha > 1$, we shall need the following +theorem: + +I.~\emph{There always exists a primitive root $\gamma$ modulo $p$ +for which $\gamma^{p-1}$ is not divisible by $p^2$.} + +Let $g$ be any primitive root modulo $p$. If $g^{p-1}-1$ is not +divisible by $p^2$ our theorem is verified. Then suppose that +$g^{p-1}-1$ is divisible by $p^2$, so that we have +\begin{gather*} +g^{p-1}-1 = kp^2 \\ +\intertext{where $k$ is an integer. Then put} +\gamma \equiv g + xp \\ +\intertext{where $x$ is an integer. Then $\gamma \equiv g \mod p$, and +hence} +\gamma^h \equiv g^h \mod p; +\end{gather*} +whence we conclude that $\gamma$ is a primitive root modulo $p$. But +\begin{align*} +\gamma^{p-1}-1 &= + g^{p-1} - 1 + \frac{p-1}{1!}g^{p-2}xp + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p^2 + \ldots \\ + &= p\left(kp + \frac{p-1}{1!}g^{p-2}x + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p + \ldots\right). +\end{align*} +Hence +\begin{equation*} +\gamma^{p-1}-1 \equiv p(-g^{p-2}x) \mod p^2. +\end{equation*} +Therefore it is evident that $x$ can be so chosen that +$\gamma^{p-1}-1$ is not divisible by $p^2$. Hence there exists a +primitive root $\gamma$ modulo $p$ such that $\gamma^{p-1}-1$ is not +divisible by $p^2$. Q.~E.~D. + +\smallskip We shall now prove that this integer $\gamma$ is a +primitive root modulo $p^\alpha$, where $\alpha$ is any positive +integer. + +If +\begin{equation*} +\gamma^k \equiv 1\mod p, +\end{equation*} +then $k$ is a multiple of $p-1$, since $\gamma$ is a primitive root +modulo $p$. Hence, if +\begin{equation*} +\gamma^k \equiv 1 \mod p^\alpha, +\end{equation*} +then $k$ is a multiple of $p-1$. + +Now, write +\begin{equation*} +\gamma^{p-1} = 1 + hp. +\end{equation*} +Since $\gamma^{p-1}-1$ is not divisible by $p^2$, it follows that $h$ +is prime to $p$. If we raise each member of this equation to the +power $\beta p^{\alpha-2}$, $\alpha \stackrel{=}{>}2$, we have +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} = + 1 + \beta p^{\alpha-1}h + p^\alpha I, +\end{equation*} +where $I$ is an integer. Then if +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} \equiv 1 \mod p^\alpha, +\end{equation*} +$\beta$ must be divisible by $p$. Therefore the exponent of the +lowest power of $\gamma$ which is congruent to $1$ modulo $p^\alpha$ +is divisible by $p^{\alpha-1}$. But we have seen that this exponent +is also divisible by $p-1$. Hence the exponent of $\gamma$ modulo +$p^\alpha$ is $p^{\alpha-1}(p-1)$ since $\phi(p^\alpha) = +p^{\alpha-1}(p-1)$. That is, $\gamma$ is a primitive root modulo +$p^\alpha$. + +It is easy to see that no two numbers of the set +\begin{equation} +\gamma, \gamma^2, \gamma^3, \ldots, \gamma^{p^{\alpha-1}(p-1)} +\tag{A} +\end{equation} +are congruent modulo $p^\alpha$; for, if so, $\gamma$ would belong +modulo $p^\alpha$ to an exponent less than $p^{\alpha-1}(p-1)$ and +would therefore not be a primitive root modulo $p^\alpha$. Now every +number in the set (A) is prime to $p^\alpha$; their number is +$\phi(p^\alpha) = p^{\alpha -1}(p-1)$. Hence the numbers of the set +(A) are congruent in some order to the numbers of the set (B): +\begin{equation} +a_1,\ a_2,\ a_3,\ \ldots ,\ a_{p^{\alpha-1}(p-1)}, \tag{B} +\end{equation} +where the integers (B) are the positive integers less than +$p^\alpha$ and prime to $p^\alpha$. + +But any number of the set (B) is a solution of the congruence +\begin{equation} +x^{p^{\alpha-1} (p-1)} \equiv 1 \bmod p^\alpha. \tag{1} +\end{equation} +Further, every solution of this congruence is prime to $p^\alpha$. +Hence the integers (B) are a complete set of solutions of (1). +Therefore the integers (A) are a complete set of solutions of (1). +But it is easy to see that an integer $\gamma^k$ of the set (A) is +or is not a primitive root modulo $p^\alpha$ according as $k$ is or +is not prime to $p^{\alpha-1} (p-1)$. Hence the number of primitive +roots modulo $p^\alpha$ is $\phi \{p^{\alpha-1} (p-1) \}.$ + +The results thus obtained may be stated as follows: + +\smallskip II.~\emph{If $p$ is any odd prime number and $\alpha$ is +any positive integer, then there exist primitive roots modulo +$p^\alpha$ and their number is $\phi \{ \phi(p^\alpha) \}$}. + +\section{Primitive Roots Modulo $2p^\alpha$, $p$ an Odd +Prime}\label{s37} + +In this section we shall prove the following theorem: + +\emph{If $p$ is any odd prime number and $\alpha$ is any positive +integer, then there exist primitive roots modulo $2p^\alpha$ and +their number is $\phi \{\phi(2 p^{\alpha} )\}.$} + +Since $2 p^\alpha$ is even it follows that every primitive root +modulo $2 p^\alpha$ is an odd number. Any odd primitive root modulo +$p^\alpha$ is obviously a primitive root modulo $2p^\alpha$. Again, +if $\gamma$ is an even primitive root modulo $p^\alpha$ then $\gamma ++ p^\alpha$ is a primitive root modulo $2 p^\alpha$. It is evident +that these two classes contain (without repetition) all the +primitive roots modulo $2 p^\alpha$. Hence the theorem follows as +stated above. + +\section{Recapitulation}\label{s38} + +The results which we have obtained in \S\S \ref{s34}--\ref{s37} +inclusive may be gathered into the following theorem: + +\emph{In order that there shall exist primitive roots modulo $m$, it +is necessary and sufficient that $m$ shall have one of the values} +\begin{equation*} +m = 1, 2, 4, p^\alpha, 2p^\alpha +\end{equation*} +\emph{where $p$ is an odd prime and $\alpha$ is a positive integer.} + +\emph{If $m$ has one of these values then the number of primitive +roots modulo $m$ is $\phi\{\phi(m)\}$.} + +\section{Primitive $\lambda$-roots}\label{s39}% +\index{Primitive roots!$\lambda$-roots|(} + +In the preceding sections of this chapter we have developed the +theory of primitive roots in the way in which it is usually +presented. But if one approaches the subject from a more general +point of view the results which may be obtained are more general and +at the same time more elegant. It is our purpose in this section to +develop the more general theory. + +\smallskip We have seen that if $a$ and $m$ are any two relatively +prime positive integers, then +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +Consequently there is no integer belonging modulo $m$ to an exponent +greater than $\lambda(m)$. It is natural to enquire if there are any +integers $a$ which belong to the exponent $\lambda(m)$. It turns out +that the question is to be answered in the affirmative, as we shall +show. Accordingly, we introduce the following definition: + +\smallskip \textsc{Definition.} If $a^{\lambda(m)}$ is the lowest +power of $a$ which is congruent to $1$ modulo $m$, $a$ is said to be +a primitive $\lambda$-root modulo $m$. We shall also say that it is +a primitive $\lambda$-root of the congruence $x^{\lambda(m)} = 1 +\mod m$. To distinguish we may speak of the usual primitive root as +a primitive $\phi$-root modulo $m$.% +\index{Primitive roots!$\phi$-roots} + +From the theory of primitive $\phi$-roots already developed it +follows that primitive $\lambda$-roots always exist when $m$ is a +power of any odd prime, and also when $m = 1,\ 2,\ 4$; for, for such +values of $m$ we have $\lambda(m) = \phi(m)$. + +We shall next show that primitive $\lambda$-roots exist when $m = +2^{\alpha}$, $a > 2$, by showing that 5 is such a root. It is +necessary and sufficient to prove that $5$ belongs modulo +$2^{\alpha}$ to the exponent $2^{\alpha-2} = \lambda(2^{\alpha})$. +Let $d$ be the exponent to which $5$ belongs modulo $2^{\alpha}$. +Then from theorem II of \S \ref{s32} it follows that $d$ is a +divisor of $2^{\alpha-2} = \lambda(2^{\alpha})$. Hence if $d$ is +different from $2^{\alpha-2}$ it is $2^{\alpha-3}$ or is a divisor +of $2^{\alpha-3}$. Hence if we can show that $5^{2^{\alpha-3}}$ is +not congruent to $1$ modulo $2^{\alpha}$ we will have proved that +$5$ belongs to the exponent $2^{\alpha-2}$. But, clearly, +\begin{gather*} +5^{2^{\alpha-3}} = (1+2^2)^{2^{\alpha-3}} + = 1+2^{\alpha-1}+ I\cdot 2^{\alpha}, \\ +\intertext{where $I$ is an integer. Hence} +5^{2^{\alpha-3}} \not\equiv 1 \bmod 2^{\alpha}. +\end{gather*} +Hence 5 belongs modulo $2^{\alpha}$ to the exponent +$\lambda(2^{\alpha})$. + +By means of these special results we are now in position to prove +readily the following general theorem which includes them as special +cases: + +\smallskip I.~\emph{For every congruence of the form} +\begin{gather*} +x^{\lambda(m)} \equiv 1 \bmod m +\end{gather*} +\emph{a solution $g$ exists which is a primitive $\lambda$-root, and +for any such solution $g$ there are $\phi\{\lambda(m)\}$ primitive +roots congruent to powers of $g$.} + +If any primitive $\lambda$-root $g$ exists, $g^\nu$ is or is not a +primitive $\lambda$-root according as $\nu$ is or is not prime to +$\lambda(m)$; and therefore the number of primitive $\lambda$-roots +which are congruent to powers of any such root $g$ is +$\phi\{\lambda(m)\}$. + +The existence of a primitive $\lambda$-root in every case may easily +be shown by induction. In case $m$ is a power of a prime the theorem +has already been established. We will suppose that it is true when +$m$ is the product of powers of $r$ different primes and show that +it is true when $m$ is the product of powers of $r+1$ different +primes; from this will follow the theorem in general. + +Put $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_r^{\alpha_r} +p_{r+1}^{\alpha_{r+1}}, \quad n = p_1^{\alpha_1} p_2^{\alpha_2} +\ldots p_r^{\alpha_r}$, and let $h$ be a primitive $\lambda$-root of +\begin{gather} +x^{\lambda(n)} \equiv 1 \mod n. \tag{1} \\ +\intertext{Then} +h + ny \notag +\end{gather} +is a form of the same root if $y$ is an integer. + +Likewise, if $c$ is any primitive $\lambda$-root of +\begin{equation} +x^\lambda(p_{r+1}^{\alpha_{r+1}}) + \equiv 1 \mod p_{r+1}^{\alpha_{r+1}} \tag{2} +\end{equation} +a form of this root is +\begin{equation*} +c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +where $z$ is any integer. + +Now, if $y$ and $z$ can be chosen so that +\begin{equation*} +h+ny = c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +the number in either member of this equation will be a common +primitive $\lambda$-root of congruences (1) and (2); that is, a +common primitive $\lambda$-root of the two congruences may always be +obtained provided that the equation +\begin{equation*} +p_1^{\alpha_1} \ldots p_r^{\alpha_r}y - p_{r+1}^{\alpha_{r+1}}z = c-h +\end{equation*} +has always a solution in which $y$ and $z$ are integers. That this +equation has such a solution follows readily from theorem III of \S +\ref{s9}; for, if $c-h$ is replaced by $1$, the new equation has a +solution $\bar{y}$, $\bar{z}$; and therefore for $y$ and $z$ we may +take $y = \bar{y}(c-h)$, $z = \bar{z}(c-h)$. + +Now let $g$ be a common primitive $\lambda$-root of congruences (1) +and (2) and write +\begin{equation*} +g^\nu \equiv 1 \mod m, +\end{equation*} +where $\nu$ is to be the smallest exponent for which the congruence +is true. Since $g$ is a primitive $\lambda$-root of (1) $\nu$ is a +multiple of $\lambda(p_1^{\alpha_1} \ldots p_r^{\alpha_r})$. Since +$g$ is a primitive $\lambda$-root of (2) $\nu$ is a multiple of +$\lambda\left(p_{r+1}^{\alpha_{r+1}} \right)$. Hence it is a +multiple of $\lambda(m)$. But $g^{\lambda(m)} \equiv 1 \bmod m$; +therefore $\nu = \lambda(m)$. That is, $g$ is a primitive +$\lambda$-root modulo $m$. + +The theorem as stated now follows at once by induction. + +\smallskip There is nothing in the preceding argument to indicate +that the primitive $\lambda$-roots modulo $m$ are all in a single +set obtained by taking powers of some root $g$; in fact it is not in +general true when $m$ contains more than one prime factor. + +By taking powers of a primitive $\lambda$-root $g$ modulo $m$ one +obtains $\phi\{\lambda(m)\}$ different primitive $\lambda$-roots +modulo $m$. It is evident that if $\gamma$ is any one of these +primitive $\lambda$-roots, then the same set is obtained again by +taking the powers of $\gamma$. We may say then that the set thus +obtained is the set belonging to $g$. + +\smallskip II.~\emph{If $\lambda(m)>2$ the product of the +$\phi\{\lambda(m)\}$ primitive $\lambda$-roots in the set belonging +to any primitive $\lambda$-root $g$ is congruent to $1$ modulo $m$.} + +These primitive $\lambda$-roots are +\begin{gather*} +g,\ g^{c_1},\ g^{c_2},\ \ldots,\ g^{c_\mu} \\ +\intertext{where} +1,\ c_1,\ c_2,\ \ldots,\ c_\mu \\ +\end{gather*} +are the integers less than $\lambda(m)$ and prime to $\lambda(m)$. +If any one of these is $c$ another is $\lambda(m)-c$, since +$\lambda(m) > 2$. Hence +\begin{gather*} +1 + c_1 + c_2 + \ldots + c_\mu \equiv 0 \bmod \lambda(m). \\ +\intertext{Therefore} +g^{1 + c_1 + c_2 + \ldots + c_\mu} \equiv 1 \bmod m. +\end{gather*} +From this the theorem follows. + +\smallskip \textsc{Corollary.}\emph{The product of all the +primitive $\lambda$-roots modulo $m$ is congruent to $1$ modulo $m$ +when $\lambda(m) > 2$.}\index{Primitive roots!$\lambda$-roots|)} + +\begin{center} +EXERCISES +\end{center} + +\small\begin{enumerate} +\item[1.] If $x_1$ is the largest value of $x$ satisfying the equation +$\lambda(x) = a$, where $a$ is a given integer, then any solution +$x_2$ of the equation is a factor of $x_1$. + +\item[2*.] Obtain an effective rule for solving the equation +$\lambda(x) = a$. + +\item[3*.] Obtain an effective rule for solving the equation +$\phi(x) = a$. + +\item[4.] A necessary and sufficient condition that $a^{P-1} \equiv 1 +\mod P$ for every integer $a$ prime to $P$ is that $P \equiv 1 \mod +\lambda(P)$. + +\item[5.] If $a^{P-1} \equiv 1\mod P$ for every a prime to $P$, then +(1) $P$ does not contain a square factor other than $1$, (2) $P$ +either is prime or contains at least three different prime factors. + +\item[6.] Let $p$ be a prime number. If $a$ is a root of the congruence +$x^q \equiv 1 \mod p$ and $\alpha$ is a root of the congruence +$x^\delta\equiv 1 \mod p$, then $a\alpha$ is a root of the +congruence $x^{d\delta}\equiv 1 \mod p$. If $a$ is a primitive root +of the first congruence and $\alpha$ of the second and if $d$ and +$\delta$ are relatively prime, then $a\alpha$ is a primitive root of +the congruence $x^{d\delta} \equiv 1\mod p$. +\end{enumerate} \normalsize\index{Primitive roots|)} + +\chapter{OTHER TOPICS} + +\section{Introduction}\label{s40} + +The theory of numbers is a vast discipline and no single volume can +adequately treat of it in all of its phases. A short book can serve +only as an introduction; but where the field is so vast such an +introduction is much needed. That is the end which the present +volume is intended to serve; and it will best accomplish this end +if, in addition to the detailed theory already developed, some +account is given of the various directions in which the matter might +be carried further. + +To do even this properly it is necessary to limit the number of +subjects considered. Consequently we shall at once lay aside many +topics of interest which would find a place in an exhaustive +treatise. We shall say nothing, for instance, about the vast domain +of algebraic numbers, even though this is one of the most +fascinating subjects in the whole field of +mathematics.\index{Algebraic numbers} Consequently, we shall not +refer to any of the extensive theory connected with the division of +the circle into equal parts.\index{Circle, Division of} Again, we +shall leave unmentioned many topics connected with the theory of +positive integers; such, for instance, is the frequency of prime +numbers in the ordered system of integers---a subject which contains +in itself an extensive and elegant theory.\index{Prime numbers} + +In \S\S \ref{s41}--\ref{s44} we shall speak briefly of each of the +following topics: theory of quadratic residues, Galois imaginaries, +arithmetic forms, analytical theory of numbers. Each of these alone +would require a considerable volume for its proper development. All +that we can do is to indicate the nature of the problem in each case +and in some cases to give a few of the fundamental results. + +In the remaining three sections we shall give a brief introduction +to the theory of Diophantine equations, developing some of the more +elementary properties of certain special cases. We shall carry this +far enough to indicate the nature of the problem connected with the +now famous Last Theorem of Fermat. The earlier sections of this +chapter are not required as a preliminary to reading this latter +part. + +\section{Theory of Quadratic Residues}\label{s41}% +\index{Quadratic residues|(} + +Let $a$ and $m$ be any two relatively prime integers. In \S +\ref{s31} we agreed to say that $a$ is a quadratic residue modulo +$m$ or a quadratic non-residue modulo $m$ according as the +congruence +\begin{equation*} +x^2 \equiv a \bmod m +\end{equation*} +has or has not a solution. We saw that if $m$ is chosen equal to an +odd prime number $p$, then $a$ is a quadratic residue modulo $p$ or +a quadratic non-residue modulo $p$ according as +\begin{equation*} +a^{\frac{1}{2} (p-1)} \equiv 1\quad \mathrm{or}\quad + a^{\frac{1}{2} (p-1)} \equiv -1 \bmod p. +\end{equation*} +This is known as Euler's criterion.\index{Euler's!criterion} + +It is convenient to employ the Legendre symbol +\begin{equation*} +\left( \frac{a}{p} \right ) +\end{equation*} +to denote the quadratic character of $a$ with respect to $p$.% +\index{Legendre symbol} This symbol is to have the value $+1$ or the +value $-1$ according as $a$ is a quadratic residue modulo $p$ or a +quadratic non-residue modulo $p$. We shall now derive some of the +fundamental properties of this symbol, understanding always that the +numbers in the numerator and the denominator are relatively prime. + +From the definition of quadratic residues and non-residues it is +obvious that +\begin{equation} +\left ( \frac{a}{p} \right ) = \left ( \frac{b}{p} \right ) + \quad \text{if}\quad a \equiv b \bmod p. \tag{1} +\end{equation} + +It is easy to prove in general that +\begin{equation} +\left ( \frac{a}{p} \right ) \left ( \frac{b}{p} \right ) = + \left (\frac {ab}{p} \right ). \tag{2} +\end{equation} +This comes readily from Euler's criterion. We have to consider the +three cases +\begin{align*} +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=+1; & +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=-1; \\ +&& \left( \frac{a}{p} \right ) &=-1,& + \left( \frac{b}{p} \right ) &=-1. +\end{align*} +The method will be sufficiently illustrated by the treatment +of the last case. Here we have +\begin{gather*} +a^{\frac 12 (p-1)}\equiv -1 \bmod p,\quad + b^{\frac 12 (p-1)}\equiv -1 \bmod p. \\ +\intertext{Multiplying these two congruences together member by +member we have} +(ab)^{\frac 12 (p-1)} \equiv 1 \bmod p, \\ +\intertext{whence} +\left( \frac {ab}{p} \right ) = 1 = + \left( \frac ap \right ) \left( \frac bp \right ), +\end{gather*} +as was to be proved. + +If $m$ is any number prime to $p$ and we write $m$ as the product of +factors +\begin{equation*} +m = \epsilon \cdot 2^\alpha \cdot q' q'' q''' \cdots +\end{equation*} +where $q',\ q'',\ q''',\ \ldots$ are odd primes, $\alpha$ is zero or +a positive integer and $\epsilon$ is $+1$ or $-1$ according as $m$ +is positive or negative, we have +\begin{equation} +\left( \frac{m}{p} \right ) = +\left( \frac{\epsilon}{p} \right ) +\left( \frac{2}{p} \right ) ^\alpha +\left( \frac{q'}{p} \right ) +\left( \frac{q''}{p} \right ) +\left( \frac{q'''}{p} \right ) \ldots, \tag{3} +\end{equation} +as one shows easily by repeated application of relation (2). +Obviously, +\begin{equation*} +\left( \frac{1}{p} \right ) = 1. +\end{equation*} +Hence, it follows from (3) that we can readily determine the +quadratic character of $m$ with respect to the odd prime $p$, that +is, the value of +\begin{equation*} +\left( \frac{m}{p} \right ), +\end{equation*} +provided that we know the value of each of the expressions +\begin{equation} +\left( \frac{-1}{p} \right ),\quad + \left( \frac{2}{p} \right ),\quad + \left( \frac{q}{p} \right ),\tag{4} +\end{equation} +where $q$ is an odd prime. + +The first of these can be evaluated at once by means of Euler's +criterion; for, we have +\begin{gather*} +\left( \frac{-1}{p} \right ) \equiv + (-1)^{\frac{1}{2} (p-1)} \bmod p \\ +\intertext{and hence} +\left( \frac{-1}{p} \right ) = (-1)^{\frac{1}{2} (p-1)}. +\end{gather*} +Thus we have the following result: The number $-1$ is a quadratic +residue of every prime number of the form $4k + 1$ and a quadratic +non-residue of every prime number of the form $4k + 3$. + +The value of the second symbol in (4) is given by the formula +\begin{equation*} +\left( \frac{2}{p} \right ) = (-1)^{\frac{1}{8} (p^2 -1)}. +\end{equation*} +The theorem contained in this equation may be stated in the +following words: The number $2$ is a quadratic residue of every +prime number of either of the forms $8k + 1, 8k + 7$; it is a +quadratic non-residue of every prime number of either of the forms +$8k + 3, 8k + 5$. + +The proof of this result is not so immediate as that of the +preceding one. To evaluate the third expression in (4) is still more +difficult. We shall omit the demonstration in both of these cases. +For the latter we have the very elegant relation +\begin{equation*} +\left( \frac{p}{q} \right ) \left( \frac{q}{p} \right ) = + (-1)^{\frac{1}{4}(p-1)(q-1)}. +\end{equation*} +This equation states the law which connects the quadratic character +of $q$ with respect to $p$ with the quadratic character of $p$ with +respect to $q$. It is known as the Law of Quadratic Reciprocity. +About fifty proofs of it have been given. Its history has been a +very interesting one; see Bachmann's Niedere Zablentheorie, Teil I, +pp.\ 180--318, especially pp.\ 200--206.\index{Bachmann}% +\index{Law of quadratic reciprocity}\index{Quadratic reciprocity} + +For a further account of this beautiful and interesting subject we +refer the reader to Bachmann, loc.\ cit., and to the memoirs to +which this author gives reference.\index{Quadratic residues|)} + +\section{Galois Imaginaries}\label{s42}% +\index{Galois imaginaries}\index{Imaginaries of Galois} + +If one is working in the domain of real numbers the equation +\begin{equation*} +x^2 + 1 = 0 +\end{equation*} +has no solution; for there is no real number whose square is $-1$. +If, however, one enlarges the ``number system'' so as to include not +only all real numbers but all complex numbers as well, then it is +true that every algebraic equation has a root. It is on account of +the existence of this theorem for the enlarged domain that much of +the general theory of algebra takes the elegant form in which we +know it. + +The question naturally arises as to whether we can make a similar +extension in the case of congruences. The congruence +\begin{equation*} +x^2 = 3 \bmod 5 +\end{equation*} +has no solution, if we employ the term solution in the sense in +which we have so far used it. But we may if we choose introduce an +imaginary quantity, or mark, $j$ such that +\begin{equation*} +j^2 \equiv 3 \bmod 5, +\end{equation*} +just as in connection with the equation $x^2 + 1 = 0$ we would +introduce the symbol $i$ having the property expressed by the +equation +\begin{equation*} +i^2 = -1. +\end{equation*} + +It is found to be possible to introduce in this way a general set of +imaginaries satisfying congruences with prime moduli; and the new +quantities or marks have the property of combining according to the +laws of algebra. + +The quantities so introduced are called Galois imaginaries. + +We cannot go into a development of the important theory which is +introduced in this way. We shall be content with indicating two +directions in which it leads. + +In the first place there is the general Galois field theory which is +of fundamental importance in the study of certain finite groups. It +may be developed from the point of view indicated here. An excellent +exposition, along somewhat different lines, is to be found in +Dickson's \emph{Linear Groups with an Exposition of the Galois Field +Theory.}\index{Dickson} + +Again, the whole matter may be looked upon from the geometric point +of view. In this way we are led to the general theory of finite +geometries, that is, geometries in which there is only a finite +number of points. For a development of the ideas which arise here +see Veblen and Young's \emph{Projective Geometry} and the memoir by +Veblen and Bussey in the Transactions of the American Mathematical +Society, vol.\ 7, pp.\ 241--259.\index{Bussey}\index{Veblen}% +\index{Young} + +\section{Arithmetic Forms}\label{s43}% +\index{Arithmetic forms|(}\index{Forms|(} + +The simplest arithmetic form is $ax + b$ where $a$ and $b$ are fixed +integers different from zero and $x$ is a variable integer. By +varying $x$ in this case we have the terms of an arithmetic +progression. We have already referred to Dirichlet's celebrated +theorem which asserts that the form $ax + b$ has an infinite number +of prime values if only $a$ and $b$ are relatively +prime.\index{Dirichlet} This is an illustration of one type of +theorem connected with arithmetic forms in general, namely, those in +which it is asserted that numbers of a given form have in addition a +given property.\index{Prime numbers} + +Another type of theorem is illustrated by a result stated in \S +\ref{s41}, provided that we look at that result in the proper way. +We saw that the number $2$ is a quadratic residue of every prime of +either of the forms $8k + 1$ and $8k + 7$ and a quadratic +non-residue of every prime of either of the forms $8k + 3$ and $8k + +5$. We may state that result as follows: A given prime number of +either of the forms $8k + 1$ and $8k + 7$ is a divisor of some +number of the form $x^2 - 2$, where $x$ is an integer; no prime +number of either of the forms $8k + 3$ and $8k + 5$ is a divisor of +a number of the form $x^2 - 2$, where $x$ is an integer. + +The result just stated is a theorem in a discipline of vast extent, +namely, the theory of quadratic forms. Here a large number of +questions arise among which are the following: What numbers can be +represented in a given form? What is the character of the divisors +of a given form? As a special case of the first we have the question +as to what numbers can be represented as the sum of three squares. +To this category belong also the following two theorems: Every +positive integer is the sum of four squares of integers; every prime +number of the form $4n + 1$ may be represented (and in only one way) +as the sum of two squares.\index{Prime numbers} + +For an extended development of the theory of quadratic forms we +refer the reader to Bachmann's Arithmetik der Quadratischen Formen +of which the first part has appeared in a volume of nearly seven +hundred pages.\index{Bachmann} + +It is clear that one may further extend the theory of arithmetic +forms by investigating the properties of those of the third and +higher degrees. Naturally the development of this subject has not +been carried so far as that of quadratic forms; but there is a +considerable number of memoirs devoted to various parts of this +extensive field, and especially to the consideration of various +special forms. + +Probably the most interesting of these special forms are the +following: +\begin{equation*} +\alpha^n + \beta^n , \quad + \frac{\alpha^n - \beta^n}{\alpha - \beta} = + \alpha^{n-1} + \alpha^{n-2} \beta + \cdots + \beta^{n-1}, +\end{equation*} +where $\alpha$ and $\beta$ are relatively prime integers, or, more +generally, where $\alpha$ and $\beta$ are the roots of the quadratic +equation $x^2 - ux + v = 0$ where $u$ and $v$ are relatively prime +integers. A development of the theory of these forms has been given +by the present author in a memoir published in 1913 in the Annals of +Mathematics, vol.\ 13, pp.\ 30--70.% +\index{Arithmetic forms|)}\index{Carmichael}\index{Forms|)}% +\index{Quadratic forms} + +\section{Analytical theory of numbers}\label{s44}% +\index{Analytical theory of numbers|(} + +Let us consider the function +\begin{equation*} +P(x) = \frac{1}{\prod_{k=0}^\infty (1-x^{2^k} )} , \quad + |x|\leqq \rho < 1. +\end{equation*} +It is clear that we have +\begin{align*} +P(x) = \prod_{k=0}^\infty \frac{1}{(1-x^{2^k} )} &= + \prod_{k=0}^\infty + ( 1 + x^{2k} + x^{2\cdot 2^k} + x^{3\cdot 2^k} + \cdots ) \\ +&= \sum_{s=0}^\infty G(s) x^s, +\end{align*} +where $G(0) = 1$ and $G(s)$ (for $s$ greater than $0$) is the number +of ways in which the positive integer $s$ may be separated into like +or distinct summands each of which is a power of $2$. + +We have readily +\begin{equation*} +(1-x)\sum_{s=0}^\infty G(s) x^s = (1-x)P(x) = P(x^2) = + \sum_{s=0}^\infty x^{2^s}; +\end{equation*} +whence +\begin{equation} +G(2s + 1) = G(2s) = G(2s - 1) + G(s), \tag{A} +\end{equation} +as one readily verifies by equating coefficients of like powers of +$x$. From this we have in particular +\begin{gather*} +G(0) = 1, \quad G(1) = 1, \quad G(2) = 2, \quad G(3) = 2, \\ +G(4) = 4, \quad G(5) = 4, \quad G(6) = 6, \quad G(7) = 6. +\end{gather*} +Thus in (A) we have recurrence relations by means of which we may +readily reckon out the values of the number theoretic function +$G(s)$. Thus we may determine the number of ways in which a given +positive integer $s$ may be represented as a sum of powers of $2$. + +We have given this example as an elementary illustration of the +analytical theory of numbers, that is, of that part of the theory of +numbers in which one employs (as above) the theory of a continuous +variable or some analogous theory in order to derive properties of +sets of integers. This general subject has been developed in several +directions. For a systematic account of it the reader is referred to +Bachmann's Analytische Zahlentheorie.% +\index{Analytical theory of numbers|)}\index{Bachmann} + +\section{Diophantine equations}\label{s45}% +\index{Diophantine equations}\index{Equations!Diophantine} + +If $f(x, y, z, \ldots)$ is a polynomial in the variables $x, y, z, +\ldots$ with integral coefficients, then the equation +\begin{equation*} +f(x, y, z, \ldots) = 0 +\end{equation*} +is called a Diophantine equation when we look at it from the point +of view of determining the integers (or the positive integers) $x, +y, z, \ldots$ which satisfy it. Similarly, if we have several such +functions $f_i(x, y, z, \ldots)$, in number less than the number of +variables $x, y, z, \ldots$, then the set of equations +\begin{equation*} +f_i(x, y, z, \ldots) = 0,\quad i = i, 2, \ldots, +\end{equation*} +is said to be a Diophantine system of equations. Any set of integers +$x, y, z, \ldots$ which satisfies the equation [system] is said to +be a solution of the equation [system]. + +We may likewise define Diophantine inequalities by replacing the +sign of equality above by the sign of inequality. But little has +been done toward developing a theory of Diophantine inequalities. +Even for Diophantine equations the theory is in a rather fragmentary +state. + +In the next two sections we shall illustrate the nature of the ideas +and the methods of the theory of Diophantine equations by developing +some of the results for two important special cases. + +\section{Pythagorean triangles}\label{s46}% +\index{Pythagorean triangles|(} + +\textsc{Definitions.} If three positive integers $x, y, z$ satisfy +the relation +\begin{equation} +x^2 + y^2 = z^2 \tag{1} +\end{equation} +they are said to form a Pythagorean triangle or a numerical right +triangle; $z$ is called the hypotenuse of the triangle and $x$ and +$y$ are called its legs. The area of the triangle is said to be +$\frac{1}{2} xy$.\index{Triangles, Numerical} + +We shall determine the general form of the integers $x$, $y$, $z$, +such that equation (1) may be satisfied. Let us denote by $\nu$ the +greatest common divisor of $x$ and $y$ in a particular solution of +(1). Then $\nu$ is a divisor of $z$ and we may write +\begin{equation*} +x = \nu u, \quad y = \nu v,\quad z = \nu w. +\end{equation*} +Substituting these values in (1) and reducing we have +\begin{equation} +u^2 + v^2 = w^2, \tag{2} +\end{equation} +where $u, v, w$ are obviously prime each to each, since $u$ and $v$ +have the greatest common divisor $1$. + +Now an odd square is of the form $4k + 1$. Hence the sum of two odd +squares is divisible by $2$ but not by $4$; and therefore the sum of +two odd squares cannot be a square. Hence one of the numbers $u$, +$v$ is even. Suppose that $u$ is even and write equation (2) in the +form +\begin{equation} +u^2 = (w - v)(w + v). \tag{3} +\end{equation} +Every common divisor of $w - v$ and $w + v$ is a divisor of their +difference $2v$. Therefore, since $w$ and $v$ are relatively prime, +it follows that $2$ is the greatest common divisor of $w - v$ and $w ++ v$. Then from (3) we see that each of these numbers is twice a +square, so that we may write +\begin{equation*} +w - v = 2b^2,\quad w + v = 2a^2 +\end{equation*} +where $a$ and $b$ are relatively prime integers. From these two +equations and equation (3) we have +\begin{equation} +w = a^2 + b^2, \quad v = a^2 -b^2,\quad u = 2ab. \tag{4} +\end{equation} +Since $u$ and $v$ are relatively prime it is evident that one of the +numbers $a$, $b$ is even and the other odd. + +The forms of $u$, $v$, $w$ given in (4) are necessary in order that +(2) may be satisfied. A direct substitution in (2) shows that this +equation is indeed satisfied by these values. Hence we have in (4) +the general solution of (2) where $u$ is restricted to be even. A +similar solution would be obtained if $v$ were restricted to be +even. Therefore \emph{the general solution of (1) is +\begin{gather*} +x = 2\nu ab,\quad y = \nu (a^2 - b^2),\quad z = \nu (a^2 + b^2)\\ +\intertext{and} +x = 2\nu (a^2 - b^2 ),\quad y = 2\nu ab,\quad z = \nu (a^2 + b^2) +\end{gather*} +where $a$, $b$, $\nu$ are arbitrary integers except that $a$ and $b$ +are relatively prime and one of them is even and the other odd.} + +By means of this general solution of (1) we shall now prove the +following theorem: + +\smallskip I.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero, such that} +\begin{equation} +q^2 + n^2 = m^2 , \quad m^2 + n^2 = p^2. \tag{5} +\end{equation} + +It is obvious that an equivalent theorem is the following: + +\smallskip II.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero such that} +\begin{equation} +p^2 + q^2 = 2m^2, \quad p^2 - q^2 = 2n^2. \tag{6} +\end{equation} + +Obviously, we may without loss of generality take $m$, $n$, $p$, $q$ +to be positive; and this we do. + +The method of proof is to assume the existence of integers +satisfying equations (5) and (6) and to show that we are thus led to +a contradiction. The argument we give is an illustration of Fermat's +famous method of ``infinite descent.''% +\index{Descent, Infinite}\index{Fermat}\index{Infinite descent} + +If any two of the numbers $p$, $q$, $m$, $n$ have a common prime +factor $t$, it follows at once from (5) and (6) that all four of +them have this factor. For, consider an equation in (5) or in (6) in +which these two numbers occur; this equation contains a third +number, and it is readily seen that this third number is divisible +by $t$. Then from one of the equations containing the fourth number +it follows that this fourth number is divisible by $t$. Now let us +divide each equation of system (6) through by $t^2$; the resulting +system is of the same form as (6). If any two numbers in this +resulting system have a common prime factor $t_1$, we may divide +through by $t_1^2$; and so on. Hence if a pair of simultaneous +equations (6) exists then there exists a pair of equations of the +same form in which no two of the numbers $m$, $n$, $p$, $q$ have a +common factor other than unity. Let this system of equations be +\begin{equation} +p_1^2 + q_1^2 = 2m_1^2, \quad p_1^2 - q_1^2 = 2n_1^2. \tag{7} +\end{equation} + +From the first equation in (7) it follows that $p_1$ and $q_1$ are +both even or both odd; and, since they are relatively prime, it +follows that they are both odd. Evidently $p_1 > q_1$. Then we may +write +\begin{equation*} +p_1 = q_1 + 2\alpha, +\end{equation*} +where $\alpha$ is a positive integer. If we substitute this value of +$p_1$ in the first equation of (7), the result may readily be put in +the form +\begin{equation} +(q_1 + \alpha)^2 + a^2 = m_1^2. \tag{8} +\end{equation} +Since $q_1$ and $m_1$ have no common prime factor it is easy to see +from this equation that $\alpha$ is prime to both $q_1$ and $m_1$, +and hence that no two of the numbers $q_1 + \alpha, \alpha, m_1$ +have a common factor. + +Now we have seen that if $a$, $b$, $c$ are positive integers no two +of which have a common prime factor, while +\begin{equation*} +a^2 + b^2 = c^2, +\end{equation*} +then there exist relatively prime integers $r$ and $s$, $r > s$, +such that +\begin{gather} +c = r^2 + s^2,\quad a = 2rs,\quad b = r^2 - s^2 \notag \\ +\intertext{or} +c = r^2 + s^2,\quad a = r^2 - s^2,\quad b = 2rs. \notag \\ +\intertext{Hence from (8) we see that we may write} +q_1 + \alpha = 2rs,\quad \alpha = r^2 - s^2 \tag{9} \\ +\intertext{or} +q_1 + \alpha = r^2 - s^2, \alpha = 2rs. \tag{10} \\ +\intertext{In either case we have} +p_1^2 - q_1^2 = (p_1 - q_1)(p_1 + q_1) = + 2\alpha \cdot 2(q_1 + \alpha) = 8rs(r^2 - s^2). \notag \\ +\intertext{If we substitute in the second equation of (7) and divide +by 2 we have} 4rs(r^2 - s^2) = n_1^2. \notag +\end{gather} + +From this equation and the fact that $r$ and $s$ are relatively +prime it follows at once that $r$, $s$, $r^2 - s^2$ are all square +numbers; say, +\begin{gather} +r = u^2,\quad s = v^2,\quad r^2 - s^2 = w^2. \notag \\ +\intertext{Now $r - s$ and $r + s$ can have no common factor other +than 1 or 2; hence from} +w^2 = (r^2-s^2) = (r-s)(r+s) = (u^2-v^2)(u^2+v^2) \notag \\ +\intertext{we see that either} +u^2 + v^2 = 2w_1^2,\quad u^2 - v^2 = 2w_2^2 \tag{11} \\ +\intertext{or} +u^2 + v^2 = w_1^2,\quad u^2 - v^2 = w_2^2. \notag \\ +\intertext{And if it is the latter case which arises, then} +w_1^2 + w_2^2 = 2u^2,\quad w_1^2 - w_2^2 = 2v^2. \tag{12} +\end{gather} +Hence, assuming equations of the form (6) we are led either to +equations (11) or to equations (12); that is, we are led to new +equations of the form with which we started. Let us write the +equations thus: +\begin{equation} +p_2^2 + q_2^2 = 2m_2^2,\quad p_2^2 - q_2^2 = 2n_2^2; \tag{13} +\end{equation} +that is, system (13) is identical with that one of systems (11), +(12) which actually arises. + +Now from (9) and (10) and the relations $p_1 = q_1 + 2\alpha, r +> s$, we see that +\begin{gather*} +p_1 = 2rs + r^2 - s^2 > 2s^2 + r^2 - s^2 = + r^2 + s^2 = u^4 + v^4. \\ +\intertext{Hence $u < p_1$. Also,} +w_1^2 \leqq w^2 \leqq r+s < r^2 + s^2. +\end{gather*} +Hence $w_1 < p_1$. Since $u$ and $w_1$ are both less than $p_1$ it +follows that $p_2$ is less than $p_1$. Hence, obviously, $p_2 < p$. +Moreover, it is clear that all the numbers $p_2, q_2, m_2, n_2$ are +different from zero. + +From these results we have the following conclusion: If we assume a +system of the form (6) we are led to a new system (13) of the same +form; and in the new system $p_2$ is less than $p$. + +Now if we start with (13) and carry out a similar argument +we shall be led to a new system +\begin{gather*} +p_3^2 + q_3^2 = 2m_3^2,\quad p_3^2 - q_3^2 = 2n_3^2, +\end{gather*} +with the relation $p_3 < p_2$, starting from this last system we +shall be led to a new one of the same form, with a similar relation +of inequality; and so on \emph{ad infinitum.} But, since there is +only a finite number of positive integers less than the given +positive integer $p$ this is impossible. We are thus led to a +contradiction; whence we conclude at once to the truth of II and +likewise of I. + +By means of theorems I and II we may readily prove the following +theorem: + +\smallskip III.~\emph{The area of a numerical right triangle is +never a square number.} + +Let the sides and hypotenuse of a numerical right triangle be $u, v, +w$, respectively. The area of this triangle is $\frac{1}{2} uv$. If +we assume this to be a square number $t^2$ we shall have the +following simultaneous Diophantine equations +\begin{equation} +u^2 + v^2 = w^2,\quad uv = 2t^2. \tag{14} +\end{equation} +We shall prove our theorem by showing that the assumption of such a +system leads to a contradiction. + +If any two of the numbers $u, v, w$ have a common prime factor $p$ +then the remaining one also has this factor, as one sees readily +from the first equation in (14). From the second equation in (14) it +follows that $t$ also has the same factor. Then if we put $u = pu_1, +v = pv_1, w = pw_1, t = pt_1$, we have +\begin{equation*} +u_1^2 + v_1^2 = w_1^2,\quad u_1 v_1 = 2t_1^2, +\end{equation*} +a system of the same form as (14). It is clear that we may start +with this new system and proceed in the same manner as before, and +so on, until we arrive at a system +\begin{equation} +\bar{u}^2 + \bar{v}^2 = \bar{w}^2,\quad + \bar{u}\bar{v} = 2\bar{t}^2, \tag{15} +\end{equation} +where $\bar{u}$, $\bar{v}$, $\bar{w}$ are prime each to each. + +Now the general solution of the first equation (15) may be written +in one of the forms +\begin{gather*} +\bar{u} = 2ab,\quad \bar{v} = a^2 - b^2,\quad \bar{w} = a^2 + b^2 \\ +\bar{u} = a^2 - b^2,\quad \bar{v} = 2ab, \quad \bar{w} = a^2 + b^2. \\ +\intertext{Then from the second equation in (15) we have} +\bar{t}^2 = ab(a^2 - b^2 ) = ab(a-b)(a+b). +\end{gather*} +It is easy to see that no two of the numbers $a$, $b$, $a - b$, $a + +b$ in the last member of this equation have a common factor; for, if +so, $\bar{u}$ and $\bar{v}$ would have a common factor, contrary to +hypothesis. Hence each of these four numbers is a square. That is, +we have equations of the form +\begin{gather*} +a = m^2,\quad b = n^2,\quad a + b = p^2,\quad a - b = q^2; \\ +\intertext{whence} +m^2 - n^2 = q^2,\quad m^2 + n^2 = p^2. +\end{gather*} +But, according to theorem I, no such system of equations can exist. +That is, the assumption of equations (14) leads to a contradiction. +Hence the theorem follows as stated above.% +\index{Pythagorean triangles|)} + +\section{The Equation $x^n + y^n = z^n$.}\label{s47}% +\index{Equation $x^n + y^n = z^n$|(}\index{Fermat's!last theorem} + +The following theorem, which is commonly known as Fermat's Last +Theorem, was stated without proof by Fermat in the seventeenth +century: + +\smallskip\emph{If n is an integer greater than 2 there do not exist +integers x, y, z, all different from zero, such that} +\begin{equation} +x^n + y^n = z^n. \tag{1} +\end{equation} + +No general proof of this theorem has yet been given. For various +special values of $n$ the proof has been found; in particular, for +every value of $n$ not greater than 100. + +In the study of equation (1) it is convenient to make some +preliminary reductions. If there exists any particular solution of +(1) there exists also a solution in which $x$, $y$, $z$ are prime +each to each, as one may show readily by the method employed in the +first part of \S \ref{s46}. Hence in proving the impossibility of +equation (1) it is sufficient to treat only the case in which $x$, +$y$, $z$ are prime each to each. + +Again, since $n$ is greater than 2 it must contain the factor +4 or an odd prime factor $p$. If $n$ contains the factor $p$ we write +$n = mp$, whence we have +\begin{gather*} +(x^m)^p + (y^m)^p = (z^m)^p). \\ +\intertext{If $n$ contains the factor 4 we write $n = 4m$, whence we +have} +(x^m)^4 + (y^m)^4 = (z^m)^4. +\end{gather*} +From this we see that in order to prove the impossibility of (1) in +general it is sufficient to prove it for the special cases when $n$ +is 4 and when $n$ is an odd prime $p$. For the latter case the proof +has not been found. For the former case we give a proof below. The +theorem may be stated as follows: + +\smallskip I.~\emph{There are no integers $x, y, z$, all different +from zero, such that} +\begin{equation*} +x^4 + y^4 = z^4. +\end{equation*} + +This is obviously a special case of the more general theorem: + +\smallskip II.~\emph{There are no integers $p$, $q$, $\alpha$, all +different from zero, such that} +\begin{equation} +p^4 - q^4 = \alpha^2. \tag{2} +\end{equation} + +The latter theorem is readily proved by means of theorem III of \S +\ref{s46}. For, if we assume an equation of the form (2), we have +\begin{gather} +(p^4 - q^4)p^2 q^2 = p^2 q^2 \alpha^2. \tag{3} \\ +\intertext{But, obviously,} +(2p^2 q^2)^2 + (p^4 - q^4)^2 = (p^4 + q^4)^2. \tag{4} +\end{gather} +Now, from (3) we see that the numerical right triangle determined by +(4) has its area $p^2 q^2(p^4 - q^4)$ equal to the square number +$p^2 q^2 \alpha^2$. But this is impossible. Hence no equation of the +form (2) exists. + +\begin{center} +EXERCISES +\end{center} + +\begin{enumerate} +\item[1.] Show that the equation $\alpha^4 + 4\beta^4 = \gamma^2$ is +impossible in integers $\alpha$, $\beta$, $\gamma$ all of which are +different from zero. + +\item[2.] Show that the system $p^2 - q^2 = km^2$, $p^2 + q^2 = kn^2$ +impossible in integers $p$, $q$, $k$, $m$, $n$, all of which are +different from zero. + +\item[3*.] Show that neither of the equations $m^4 - 4n^4 = \pm t^2$ +is possible in integers $m$, $n$, $t$, all of which are different +from zero. + +\item[4*.] Prove that the area of a numerical right triangle is not +twice a square number. + +\item[5*.] Prove that the equation $m^4 + n^4 = \alpha^2$ is not +possible in integers $m$, $n$, $\alpha$ all of which are different +from zero. + +\item[6*.] In the numerical right triangle $a^2 + b^2 = c^2$, +not more than one of the numbers $a$, $b$, $c$ is a square. + +\item[7.] Prove that the equation $x^{2k} + y^{2k} = z^{2k}$ implies +an equation of the form $m^k + n^k = 2^{k-2} t^k$. + +\item[8.] Find the general solution in integers of the equation +$x^2 + 2y^2 = t^2$. + +\item[9.] Find the general solution in integers of the equation +$x^2 + y^2 = z^4$. + +\item[10.] Obtain solutions of each of the following Diophantine +equations: +\begin{align*} +x^3 + y^3 + z^3 &= 2t^3, \\ +x^3 + 2y^3 + 3z^3 &= t^3, \\ +x^4 + y^4 + 4z^4 &= t^4, \\ +x^4 + y^4 + z^4 &= 2t^4. +\end{align*} +\end{enumerate}\index{Equation $x^n + y^n = z^n$|)} + +\addcontentsline{toc}{chapter}{Index} +\printindex + + +\newpage +\chapter{PROJECT GUTENBERG "SMALL PRINT"} +\small +\pagenumbering{gobble} + +*** END OF THE PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +***** This file should be named 13693-pdf.pdf or 13693-pdf.zip ***** +This and all associated files of various formats will be found in: + https://www.gutenberg.org/1/3/6/9/13693/ + +Updated editions will replace the previous one--the old editions will +be renamed. + +Creating the works from print editions not protected by U.S. copyright +law means that no one owns a United States copyright in these works, +so the Foundation (and you!) can copy and distribute it in the +United States without permission and without paying copyright +royalties. 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Carmichael +% % +% This eBook is for the use of anyone anywhere in the United States and % +% most other parts of the world at no cost and with almost no restrictions% +% whatsoever. You may copy it, give it away or re-use it under the terms % +% of the Project Gutenberg License included with this eBook or online at % +% www.gutenberg.org. If you are not located in the United States, you % +% will have to check the laws of the country where you are located before % +% using this eBook. % +% % +% % +% Title: The Theory of Numbers % +% % +% Author: Robert D. 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If you are not located in the United States, you +will have to check the laws of the country where you are located before +using this eBook. + + +Title: The Theory of Numbers + +Author: Robert D. Carmichael + +Release Date: October 10, 2003 [eBook #13693] +Revised Date: December 6, 2021 + +Language: English + +Character set encoding: TeX + +*** START OF THIS PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +Produced by David Starner, Joshua Hutchinson, John Hagerson. +Revised by Richard Tonsing +\end{verbatim} +\normalsize +\newpage + +\frontmatter + +\begin{center} +\noindent \Large MATHEMATICAL MONOGRAPHS \\ + +\bigskip \footnotesize{EDITED BY} \\ +\normalsize \textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD.} \\ + +\bigskip\bigskip \huge +No. 13. + +\bigskip\bigskip \huge THE THEORY \\ +\bigskip\small \textsc{of} \\ +\bigskip\huge NUMBERS \\ + +\bigskip\bigskip\footnotesize\textsc{by} \\ +\bigskip\large ROBERT D. CARMICHAEL, \\ +\footnotesize\textsc{Associate Professor of Mathematics in Indiana +University} + +\bigskip\bigskip\normalsize NEW YORK: \\ +\medskip JOHN WILEY \& SONS. \\ +\medskip \textsc{London: CHAPMAN \& HALL, Limited.} \\ +\medskip 1914. + +\bigskip\bigskip +\tiny \textsc{Copyright 1914} \\ +\textsc{by} \\ +ROBERT D. CARMICHAEL. \\ +\medskip \textsc{the scientific press} \\ +\textsc{robert drummond and company} \\ +\textsc{brooklyn, n.~y.} +\end{center} + +\bigskip\bigskip +\scriptsize \noindent \textsc{Transcriber's Note:} \emph{I did my +best to recreate the index.} \normalsize + +\newpage + +\fbox{\parbox{11cm}{ +\begin{center} +\textbf{MATHEMATICAL MONOGRAPHS.} \\ +\small\textsc{edited by}\normalsize \\ +\textbf{Mansfield Merriman and Robert S. Woodward.} \\ +\footnotesize \textbf{Octavo. Cloth. \$1.00 each.} \\ + +\bigskip \textbf{No. 1. History of Modern Mathematics.} \\ +By \textsc{David Eugene Smith.} + +\smallskip \textbf{No. 2. Synthetic Projective Geometry.} \\ +By \textsc{George Bruce Halsted.} + +\smallskip \textbf{No. 3. Determinants.} \\ +By \textsc{Laenas Gifford Weld.} + +\smallskip \textbf{No. 4. Hyperbolic Functions.} \\ +By \textsc{James McMahon.} + +\smallskip \textbf{No. 5. Harmonic Functions.} \\ +By \textsc{William E. Byerly.} + +\smallskip \textbf{No. 6. Grassmann's Space Analysis.} \\ +By \textsc{Edward W. Hyde.} + +\smallskip \textbf{No. 7. Probability and Theory of Errors.} \\ +By \textsc{Robert S. Woodward.} + +\smallskip \textbf{No. 8. Vector Analysis and Quaternions.} \\ +By \textsc{Alexander Macfarlane.} + +\smallskip \textbf{No. 9. Differential Equations.} \\ +By \textsc{William Woolsey Johnson.} + +\smallskip \textbf{No. 10. The Solution of Equations.} \\ +By \textsc{Mansfield Merriman.} + +\smallskip \textbf{No. 11. Functions of a Complex Variable.} \\ +By \textsc{Thomas S. Fiske.} + +\smallskip \textbf{No. 12. The Theory of Relativity.} \\ +By \textsc{Robert D. Carmichael.} + +\smallskip \textbf{No. 13. The Theory of Numbers.} \\ +By \textsc{Robert D. Carmichael.} \normalsize + +\bigskip \small PUBLISHED BY \\ +\smallskip \textbf{JOHN WILEY \& SONS, Inc., NEW YORK. \\ +CHAPMAN \& HALL, Limited, LONDON.} +\end{center}}} + +\chapter{Editors' Preface.} + +The volume called Higher Mathematics, the third edition of which was +published in 1900, contained eleven chapters by eleven authors, each +chapter being independent of the others, but all supposing the +reader to have at least a mathematical training equivalent to that +given in classical and engineering colleges. The publication of that +volume was discontinued in 1906, and the chapters have since been +issued in separate Monographs, they being generally enlarged by +additional articles or appendices which either amplify the former +presentation or record recent advances. This plan of publication was +arranged in order to meet the demand of teachers and the convenience +of classes, and it was also thought that it would prove advantageous +to readers in special lines of mathematical literature. + +It is the intention of the publishers and editors to add other +monographs to the series from time to time, if the demand seems to +warrant it. Among the topics which are under consideration are those +of elliptic functions, the theory of quantics, the group theory, the +calculus of variations, and non-Euclidean geometry; possibly also +monographs on branches of astronomy, mechanics, and mathematical +physics may be included. It is the hope of the editors that this +Series of Monographs may tend to promote mathematical study and +research over a wider field than that which the former volume has +occupied. + +\chapter{Preface} + +The purpose of this little book is to give the reader a convenient +introduction to the theory of numbers, one of the most extensive and +most elegant disciplines in the whole body of mathematics. The +arrangement of the material is as follows: The first five chapters +are devoted to the development of those elements which are essential +to any study of the subject. The sixth and last chapter is intended +to give the reader some indication of the direction of further study +with a brief account of the nature of the material in each of the +topics suggested. The treatment throughout is made as brief as is +possible consistent with clearness and is confined entirely to +fundamental matters. This is done because it is believed that in +this way the book may best be made to serve its purpose as an +introduction to the theory of numbers. + +Numerous problems are supplied throughout the text. These have been +selected with great care so as to serve as excellent exercises for +the student's introductory training in the methods of number theory +and to afford at the same time a further collection of useful +results. The exercises marked with a star are more difficult than +the others; they will doubtless appeal to the best students. + +Finally, I should add that this book is made up from the material +used by me in lectures in Indiana University during the past two +years; and the selection of matter, especially of exercises, has +been based on the experience gained in this way. + +\hfill \textsc{R.~D.\ Carmichael.} + +\tableofcontents + +%% CHAPTER I. ELEMENTARY PROPERTIES OF INTEGERS +%% 1. Fundamental Notions and Laws +%% 2. Definition of Divisibility. The Unit +%% 3. Prime Numbers. The Sieve of Eratosthenes +%% 4. The Number of Primes is Infinite +%% 5. The Fundamental Theorem of Euclid +%% 6. Divisibility by a Prime Number +%% 7. The Unique Factorization Theorem +%% 8. The Divisors of an Integer +%% 9. The Greatest Common Factor of Two or More Integers +%% 10. The Least Common Multiple of Two or More Integers +%% 11. Scales of Notation +%% 12. Highest Power of a Prime $p$ Contained in $n!$ +%% 13. Remarks Concerning Prime Numbers +%% +%% CHAPTER II. ON THE INDICATOR OF AN INTEGER +%% 14. Definition. Indicator of a Prime Power +%% 15. The Indicator of a Product +%% 16. The Indicator of Any Positive Integer +%% 17. Sum of the Indicators of the Divisors of a Number +%% +%% CHAPTER III. ELEMENTARY PROPERTIES OF CONGRUENCES +%% 18. Congruences Modulo $m$ +%% 19. Solutions of Congruences by Trial +%% 20. Properties of Congruences Relative to Division +%% 21. Congruences with a Prime Modulus +%% 22. Linear Congruences +%% +%% CHAPTER IV. THE THEOREMS OF FERMAT AND WILSON +%% 23. Fermat's General Theorem +%% 24. Euler's Proof of the Simple Fermat Theorem +%% 25. Wilson's Theorem +%% 26. The Converse of Wilson's Theorem +%% 27. Impossibility of $1\cdot 2\cdot 3\cdot \ldots \cdot +%% \overline{n-1}+1=n^k, n>5$ +%% 28. Extension of Fermat's Theorem +%% 29. On the Converse of Fermat's Simple Theorem +%% 30. Application of Previous Results to Linear Congruences +%% 31. Application of the Preceding Results to the Theory of +%% Quadratic Residues +%% +%% CHAPTER V. PRIMITIVE ROOTS MODULO $m$ +%% 32. Exponent of an Integer Modulo $m$ +%% 33. Another Proof of Fermat's General Theorem +%% 34. Definition of Primitive Roots +%% 35. Primitive Roots Modulo $p$ +%% 36. Primitive Roots Modulo $p^\alpha$, $p$ an Odd Prime +%% 37. Primitive Roots Modulo $2p^\alpha$, $p$ an Odd Prime +%% 38. Recapitulation +%% 39. Primitive $\lambda$-Roots +%% +%% CHAPTER VI. OTHER TOPICS +%% 40. Introduction +%% 41. Theory of Quadratic Residues +%% 42. Galois Imaginaries +%% 43. Arithmetic Forms +%% 44. Analytical Theory of Numbers +%% 45. Diophantine Equations +%% 46. Pythagorean Triangles +%% 47. The Equation $x^n+y^n = z^n$ + +\mainmatter + +\chapter{ELEMENTARY PROPERTIES OF INTEGERS} +\section{Fundamental Notions and Laws}\label{s1}% +\index{Fundamental notions} + +In the present chapter we are concerned primarily with certain +elementary properties of the positive integers 1, 2, 3, 4, \ldots It +will sometimes be convenient, when no confusion can arise, to employ +the word \emph{integer} or the word \emph{number} in the sense of +positive integer. + +We shall suppose that the integers are already defined, either by +the process of counting or otherwise. We assume further that the +meaning of the terms \emph{greater, less, equal, sum, difference, +product} is known. + +From the ideas and definitions thus assumed to be known follow +immediately the theorems: +\begin{table}[h] +\begin{tabular}{rl} + I.\ & The sum of any two integers is an integer. \\ + II.\ & The difference of any two integers is an integer. \\ + III.\ & The product of any two integers is an integer. +\end{tabular} +\end{table} + +Other fundamental theorems, which we take without proof, are +embodied in the following formulas: +\begin{table}[h] +\begin{tabular}{rrcl} + IV.\ & $a + b$ & = & $b + a$. \\ + V.\ & $a \times b$ & = & $b \times a$. \\ + VI.\ & $(a + b) + c$ & = & $a + (b + c)$. \\ + VII.\ & $(a \times b) \times c$ & = & $a \times (b \times c)$. \\ +VIII.\ & $a \times (b + c)$ & = & $a \times b + a \times c$. +\end{tabular} +\end{table} +Here $a$, $b$, $c$ denote any positive integers. + +\newpage +These formulas are equivalent in order to the following five +theorems: addition is commutative; multiplication is commutative; +addition is associative; multiplication is associative; +multiplication is distributive with respect to addition. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove the following relations: +\begin{align*} + 1 + 2 + 3 \ldots + n &= \frac{n(n+1)}{2} \\ + 1 + 3 + 5 + \ldots + (2n - 1) &= n^2, \\ +1^3 + 2^3 + 3^3 + \ldots + n^3 &= \left(\frac{n(n+1)}{2}\right)^2 + = (1+2+\ldots+n)^2. +\end{align*} + +\item[2.] Find the sum of each of the following series: +\begin{align*} +1^2 + 2^2 + 3^2 + &\ldots + n^2, \\ +1^2 + 3^2 + 5^2 + &\ldots + (2n - 1)^2, \\ +1^3 + 3^3 + 5^3 + &\ldots + (2n - 1)^3. +\end{align*} + +\item[3.] Discover and establish the law suggested by the equations +$1^2 = 0 + 1$, $2^2 = 1 + 3$, $3^2 = 3 + 6$, $4^2 = 6 + 10$, +$\ldots$; by the equations $1 = 1^3$, $3 + 5 = 2^3$, $7 + 9 + 11 = +3^3$, $13 + 15 + 17 + 19 = 4^3$, $\ldots$. +\end{enumerate} \normalsize + +\section{Definition of Divisibility. The Unit}\label{s2}% +\index{Divisibility}\index{Unit} + +\textsc{Definitions.} An integer $a$ is said to be divisible by an +integer $b$ if there exists an integer $c$ such that $a = bc$. It is +clear from this definition that $a$ is also divisible by $c$. The +integers $b$ and $c$ are said to be divisors or factors of $a$; and +$a$ is said to be a multiple of $b$ or of $c$. The process of +finding two integers $b$ and $c$ such that $bc$ is equal to a given +integer $a$ is called the process of resolving $a$ into factors or +of factoring $a$; and $a$ is said to be resolved into factors or to +be factored. + +We have the following fundamental theorems: + +\smallskip I.~\emph{If $b$ is a divisor of $a$ and $c$ is a divisor +of $b$, then $c$ is a divisor of $a$.} + +Since $b$ is a divisor of a there exists an integer $\beta$ such +that $a = b\beta$. Since $c$ is a divisor of $b$ there exists an +integer $\gamma$ such that $b = c\gamma$. Substituting this value of +$b$ in the equation $a = b\gamma$ we have $a = c\gamma\beta$. But +from theorem III of \S~\ref{s1} it follows that $\gamma\beta$ is an +integer; hence, $c$ is a divisor of $a$, as was to be proved. + +\smallskip II.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the sum of $a$ and $b$.} + +From the hypothesis of the theorem it follows that integers $\alpha$ +and $\beta$ exist such that +\begin{gather*} +a = c\alpha,\quad b = c\beta. \\ +\intertext{Adding, we have} +a + b = c\alpha + c\beta = c(\alpha + \beta) = c\delta, +\end{gather*} +where $\delta$ is an integer. Hence, $c$ is a divisor of $a+b$. + +\smallskip III.~\emph{If $c$ is a divisor of both $a$ and $b$, then +$c$ is a divisor of the difference of $a$ and $b$.} + +The proof is analogous to that of the preceding theorem. + +\smallskip \textsc{Definitions.} If $a$ and $b$ are both divisible +by $c$, then $c$ is said to be a common divisor or a common factor +of $a$ and $b$. Every two integers have the common factor 1. The +greatest integer which divides both $a$ and $b$ is called the +greatest common divisor of $a$ and $b$. More generally, we define in +a similar way a common divisor and the greatest common divisor of +$n$ integers $a_1$, $a_2$, $\ldots$, $a_n$.\index{Common!divisors} + +\smallskip \textsc{Definitions.} If an integer $a$ is a multiple of +each of two or more integers it is called a common multiple of these +integers. The product of any set of integers is a common multiple of +the set. The least integer which is a multiple of each of two or +more integers is called their least common multiple.% +\index{Common!multiples} + +It is evident that the integer $1$ is a divisor of every integer and +that it is the only integer which has this property. It is called +the unit. + +\smallskip \textsc{Definition.} Two or more integers which have no +common factor except $1$ are said to be prime to each other or to be +relatively prime.\index{Relatively prime} + +\smallskip \textsc{Definition.} If a set of integers is such that no +two of them have a common divisor besides $1$ they are said to be +prime each to each.\index{Prime each to each} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Prove that $n^3 - n$ is divisible by $6$ for every +positive integer $n$. + +\item[2.] If the product of four consecutive integers is increased by +$1$ the result is a square number. + +\item[3.] Show that $2^{4n + 2} + 1$ has a factor different from itself +and $1$ when $n$ is a positive integer. +\end{enumerate} \normalsize + +\section{Prime Numbers. The Sieve of Eratosthenes}\label{s3}% +\index{Eratosthenes}\index{Sieve of Eratosthenes} + +\textsc{Definition.} If an integer $p$ is different from 1 and has +no divisor except itself and 1 it is said to be a prime number or to +be a prime. + +\smallskip \textsc{Definition.} An integer which has at least one +divisor other than itself and 1 is said to be a composite number or +to be composite. + +All integers are thus divided into three classes: +\begin{table}[h] +\begin{tabular}{rl} +1.\ & The unit; \\ +2.\ & Prime numbers; \\ +3.\ & Composite numbers. +\end{tabular} +\end{table}\index{Composite numbers}\index{Prime numbers} + +We have seen that the first class contains only a single number. The +third class evidently contains an infinitude of numbers; for, it +contains all the numbers $2^2, 2^3, 2^4, \ldots$ In the next section +we shall show that the second class also contains an infinitude of +numbers. We shall now show that every number of the third class +contains one of the second class as a factor, by proving the +following theorem: + +\smallskip I.~\emph{Every integer greater than 1 has a prime factor.} + +Let $m$ be any integer which is greater than 1. We have to show that +it has a prime factor. If $m$ is prime there is the prime factor $m$ +itself. If $m$ is not prime we have +\begin{equation*} +m = m_1 m_2 +\end{equation*} +where $m_1$ and $m_2$ are positive integers both of which are less +than $m$. If either $m_1$ or $m_2$ is prime we have thus obtained a +prime factor of $m$. If neither of these numbers is prime, then +write +\begin{equation*} +m_1 = m'_1 m'_2,\quad m'_1 > 1, m'_2 > 1. +\end{equation*} +Both $m'_1$ and $m'_2$ are factors of $m$ and each of them is less +than $m_1$. Either we have not found in $m'_1$ or $m'_2$ a prime +factor of $m$ or the process can be continued by separating one of +these numbers into factors. Since for any given $m$ there is +evidently only a finite number of such steps possible, it is clear +that we must finally arrive at a prime factor of $m$. From this +conclusion, the theorem follows immediately. + +Eratosthenes has given a useful means of finding the prime numbers +which are less than any given integer $m$. It may be described as +follows: + +Every prime except 2 is odd. Hence if we write down every odd number +from 3 up to $m$ we shall have in the list every prime less than $m$ +except 2. Now 3 is prime. Leave it in the list; but beginning to +count from 3 strike out every third number in the list. Thus every +number divisible by 3, except 3 itself, is cancelled. Then begin +from 5 and cancel every fifth number. Then begin from the next +uncancelled number, namely 7, and strike out every seventh number. +Then begin from the next uncancelled number, namely 11, and strike +out every eleventh number. Proceed in this way up to $m$. The +uncancelled numbers remaining will be the odd primes not greater +than $m$. + +It is obvious that this process of cancellation need not be carried +altogether so far as indicated; for if $p$ is a prime greater than +$\sqrt{m}$, the cancellation of any $p^\text{th}$ number from $p$ +will be merely a repetition of cancellations effected by means of +another factor smaller than $p$, as one may see by the use of the +following theorem. + +\smallskip II.~\emph{An integer $m$ is prime if it has no prime +factor equal or less than $I$, where $I$ is the greatest integer +whose square is equal to or less than $m$.} + +Since $m$ has no prime factor less than $I$, it follows from theorem +I that it has no factor but unity less than $I$. Hence, if $m$ is +not prime it must be the product of two numbers each greater than +$I$; and hence it must be equal to or greater than $(I+1)^2$. This +contradicts the hypothesis on $I$; and hence we conclude that $m$ is +prime. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] By means of the method of Eratosthenes determine the primes +less than 200. +\end{enumerate} +\normalsize + +\section{The Number of Primes is Infinite}\label{s4}% +\index{Prime numbers} + +I.~\emph{The number of primes is infinite.} + +We shall prove this theorem by supposing that the number of primes +is not infinite and showing that this leads to a contradiction. If +the number of primes is not infinite there is a greatest prime +number, which we shall denote by $p$. Then form the number +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p + 1. +\end{equation*} +Now by theorem 1 of \S~\ref{s3} $N$ has a prime divisor $q$. But +every non-unit divisor of $N$ is obviously greater than $p$. Hence +$q$ is greater than $p$, in contradiction to the conclusion that $p$ +is the greatest prime. Thus the proof of the theorem is complete. + +In a similar way we may prove the following theorem: + +\smallskip II.~\emph{Among the integers of the arithmetic +progression $5$, $11$, $17$, $23$, $\ldots$, there is an infinite +number of primes.} + +If the number of primes in this sequence is not infinite there is a +greatest prime number in the sequence; supposing that this greatest +prime number exists we shall denote it by $p$. Then the number $N$, +\begin{equation*} +N = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot p-1, +\end{equation*} +is not divisible by any number less than or equal to $p$. This +number $N$, which is of the form $6n - 1$, has a prime factor. If +this factor is of the form $6k - 1$ we have already reached a +contradiction, and our theorem is proved. If the prime is of the +form $6k_1 + 1$ the complementary factor is of the form $6k_2 - 1$. +Every prime factor of $6k_2 - 1$ is greater than $p$. Hence we may +treat $6k_2 - 1$ as we did $6n - 1$, and with a like result. Hence +we must ultimately reach a prime factor of the form $6k_3 - 1$; for, +otherwise, we should have $6n - 1$ expressed as a product of prime +factors all of the form $6t + 1$---a result which is clearly +impossible. Hence we must in any case reach a contradiction of the +hypothesis. Thus the theorem is proved. + +The preceding results are special cases of the following more +general theorem: + +\smallskip III.~\emph{Among the integers of the arithmetic +progression $a$, $a + d$, $a + 2d$, $a + 3d$, $\ldots$, there is an +infinite number of +primes, provided that $a$ and $d$ are relatively prime.}% +\index{Arithmetic progression} + +For the special case given in theorem II we have an elementary +proof; but for the general theorem the proof is difficult. We shall +not give it here. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Prove that there is an infinite number of primes of the +form $4n - 1$. + +\item[2.] Show that an odd prime number can be represented as the +difference of two squares in one and in only one way. + +\item[3.] The expression $m^p - n^p$, in which $m$ and $n$ are integers +and $p$ is a prime, is either prime to $p$ or is divisible by $p^2$. + +\item[4.] Prove that any prime number except $2$ and $3$ is of one of +the forms $6n + 1$, $6n - 1$. +\end{enumerate}\normalsize + +\section{The Fundamental Theorem of Euclid}\label{s5}% +\index{Euclid, Theorem of} + +\emph{If $a$ and $b$ are any two positive integers there exist +integers $q$ and $r$, $q\stackrel{=}{>} 0, 0 \leqq r < b$, such +that} +\begin{equation*} +a = qb + r. +\end{equation*} + +If $a$ is a multiple of $b$ the theorem is at once verified, $r$ +being in this case $0$. If $a$ is not a multiple of $b$ it must lie +between two consecutive multiples of $b$; that is, there exists a +$q$ such that +\begin{equation*} +qb < a < (q + 1)b. +\end{equation*} +Hence there is an integer $r$, $0 < r < b$, such that $a = qb + r$. +In case $b$ is greater than $a$ it is evident that $q = 0$ and $r = +a$. Thus the proof of the theorem is complete. + +\section{Divisibility by a Prime Number}\label{s6}\index{Prime numbers} + +I.~\emph{If $p$ is a prime number and $m$ is any integer, then $m$ +either is divisible by $p$ or is prime to $p$.} + +This theorem follows at once from the fact that the only divisors of +$p$ are $1$ and $p$. + +\smallskip II.~\emph{The product of two integers each less than a +given prime number $p$ is not divisible by $p$.} + +Let $a$ be a number which is less than $p$ and suppose that $b$ is a +number less than $p$ such that $ab$ is divisible by $p$, and let $b$ +be the least number for which $ab$ is so divisible. Evidently there +exists an integer $m$ such that +\begin{equation*} +mb < p < (m + 1)b. +\end{equation*} +Then $p - mb < b$. Since $ab$ is divisible by $p$ it is clear that +$mab$ is divisible by $p$; so is $ap$ also; and hence their +difference $ap - mab$, $=a(p - mb)$, is divisible by $p$. That is, +the product of $a$ by an integer less than $b$ is divisible by $p$, +contrary to the assumption that $b$ is the least integer such that +$ab$ is divisible by $p$. The assumption that the theorem is not +true has thus led to a contradiction; and thus the theorem is +proved. + +\smallskip III.~\emph{If neither of two integers is divisible by a +given prime number $p$ their product is not divisible by $p$.} + +Let $a$ and $b$ be two integers neither of which is divisible by the +prime $p$. According to the fundamental theorem of Euclid there +exist integers $m$, $n$, $\alpha$, $\beta$ such that +\begin{align*} +a &= mp + \alpha,& 0 &< \alpha < p, \\ +b &= np + \beta, & 0 &< \beta < p. +\end{align*} +Then +\begin{equation*} +ab = (mp + \alpha)(np + \beta) + = (mnp + \alpha + \beta)p + \alpha\beta. +\end{equation*} +If now we suppose $ab$ to be divisible by $p$ we have $\alpha\beta$ +divisible by $p$. This contradicts II, since $\alpha$ and $\beta$ +are less than $p$. Hence $ab$ is not divisible by $p$. + +By an application of this theorem to the continued product of +several factors, the following result is readily obtained: + +\smallskip IV.~\emph{If no one of several integers is divisible by a +given prime $p$ their product is not divisible by $p$.} + +\section{The Unique Factorization Theorem}\label{s7}% +\index{Factorization theorem}\index{Factors} + +I.~\emph{Every integer greater than unity can be represented in one +and in only one way as a product of prime numbers.} + +In the first place we shall show that it is always possible to +resolve a given integer $m$ greater than unity into prime factors by +a finite number of operations. In the proof of theorem I, +\S~\ref{s3}, we showed how to find a prime factor $p_1$ of $m$ by a +finite number of operations. Let us write +\begin{equation*} +m = p_1 m_1. +\end{equation*} +If $m_1$ is not unity we may now find a prime factor $p_2$ of $m_1$. +Then we may write +\begin{equation*} +m = p_1 m_1 = p_1 p_2 m_2. +\end{equation*} +If $m_2$ is not unity we may apply to it the same process as that +applied to $m_1$ and thus obtain a third prime factor of $m$. Since +$m_1 > m_2 > m_3 > \ldots$ it is clear that after a finite number of +operations we shall arrive at a decomposition of $m$ into prime +factors. Thus we shall have +\begin{equation*} +m = p_1 p_2 \ldots p_r +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_r$ are prime numbers. We have thus +proved the first part of our theorem, which says that the +decomposition of an integer (greater than unity) into prime factors +is always possible. + +Let us now suppose that we have also a decomposition of $m$ into +prime factors as follows: +\begin{gather*} +m = q_1 q_2 \ldots q_s. \\ +\intertext{Then we have} +p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_s. +\end{gather*} +Now $p_1$ divides the first member of this equation. Hence it also +divides the second member of the equation. But $p_1$ is prime; and +therefore by theorem IV of the preceding section we see that $p_1$ +divides some one of the factors $q$; we suppose that $p_1$ is a +factor of $q_1$. It must then be equal to $q_1$. Hence we have +\begin{equation*} +p_2 p_3 \ldots p_r = q_2 q_3 \ldots q_s. +\end{equation*} +By the same argument we prove that $p_2$ is equal to some $q$, say +$q_2$. Then we have +\begin{equation*} +p_3 p_4 \ldots p_r = q_3 q_4 \ldots q_s. +\end{equation*} +Evidently the process may be continued until one side of the +equation is reduced to $1$. The other side must also be reduced to +$1$ at the same time. Hence it follows that the two decompositions +of $m$ are in fact identical. + +This completes the proof of the theorem. + +\smallskip The result which we have thus demonstrated is easily the +most important theorem in the theory of integers. It can also be +stated in a different form more convenient for some purposes: + +\smallskip II.~\emph{Every non-unit positive integer $m$ can be +represented in one and in only one way in the form +\begin{equation*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} +\end{equation*} +where $p_1$, $p_2$, $\ldots$, $p_n$ are different primes and +$\alpha_1$, $\alpha_2$, $\ldots$, $\alpha_n$ are positive integers.}% +\index{Factors} + +This comes immediately from the preceding representation of $m$ in +the form $m = p_1 p_2 \ldots p_r$ by combining into a power of $p_1$ +all the primes which are equal to $p_1$. + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ and $b$ are relatively +prime integers and $c$ is divisible by both $a$ and $b$, then $c$ is +divisible by $ab$.} + +\smallskip \textsc{Corollary 2.}~\emph{If $a$ and $b$ are each prime +to $c$ then $ab$ is prime to $c$.} + +\smallskip \textsc{Corollary 3.}~\emph{If $a$ is prime to $c$ and +$ab$ is divisible by $c$, then $b$ is divisible by $c$.} + +\section{The Divisors of an Integer}\label{s8}% +\index{Divisors of a number|(}\index{Factors} + +The following theorem is an immediate corollary of the results in +the preceding section: + +I.~\emph{All the divisors of $m$, +\begin{gather*} +m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}, \\ +\intertext{are of the form} +p_1^{\beta_1} p_2^{\beta_2} \ldots p_n^{\beta_n},\ + 0 \leqq \beta_i \leqq \alpha_i; +\end{gather*} +and every such number is a divisor of $m$.} + +From this it is clear that every divisor of $m$ is included once and +only once among the terms of the product +\begin{multline*} +(1 + p_1 + p_1^2 + \ldots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \ldots + + p_2^{\alpha_2}) \ldots \\ +(1 + p_n + p_n^2 + \ldots + p_n^{\alpha_n}), +\end{multline*} +when this product is expanded by multiplication. It is obvious that +the number of terms in the expansion is $(\alpha_1 + 1)(\alpha_2 + +1) \ldots (\alpha_n+1)$. Hence we have the theorem: + +\smallskip II.~\emph{The number of divisors of $m$ is} +$(\alpha_1 + 1) (\alpha_2 + 1) \ldots (\alpha_n+1)$. + +Again we have +\begin{equation*} +\prod_i(1 + p_i + p_i^2 + \ldots + p_i^{\alpha_i}) = + \prod_i\frac{p_i^{\alpha_i+1} - 1}{p_i - 1}. +\end{equation*} +Hence, + +\smallskip III.~\emph{The sum of the divisors of $m$ is} +\begin{equation*} +\frac{p_1^{\alpha_1 + 1} - 1}{p_1 - 1} \cdot + \frac{p_2^{\alpha_2 + 1} - 1}{p_2 - 1} \cdot + \ldots \cdot + \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}. +\end{equation*} + +In a similar manner we may prove the following theorem: + +\smallskip IV.~\emph{The sum of the $h^{th}$ powers of the divisors +of $m$ is} +\begin{equation*} +\frac{p_1^{h(\alpha_1 + 1)} - 1}{p_1^h - 1} \cdot + \ldots \cdot + \frac{p_n^{h(\alpha_n + 1)} - 1}{p_n^h - 1}. +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find numbers $x$ such that the sum of the divisors of $x$ +is a perfect square. + +\item[2.] Show that the sum of the divisors of each of the following +integers is twice the integer itself: 6, 28, 496, 8128, 33550336. +Find other integers $x$ such that the sum of the divisors of $x$ is +a multiple of $x$. + +\item[3.] Prove that the sum of two odd squares cannot be a square. + +\item[4.] Prove that the cube of any integer is the difference of the +squares of two integers. + +\item[5.] In order that a number shall be the sum of consecutive +integers, it is necessary and sufficient that it shall not be a +power of 2. + +\item[6.] Show that there exist no integers $x$ and $y$ (zero excluded) +such that $y^2 = 2x^2$. Hence, show that there does not exist a +rational fraction whose square is 2. + +\item[7.] The number $m = p_1^{\alpha_1} p_2^{\alpha_2} \cdots +p_n^{\alpha_n}$, where the $p$'s are different primes and the +$\alpha$'s are positive integers, may be separated into relatively +prime factors in $2^{n-1}$ different ways. + +\item[8.] The product of the divisors of $m$ is $\sqrt{m^v}$ where $v$ +is the number of divisors of $m$. +\end{enumerate} \normalsize\index{Divisors of a number|)} + +\section{The Greatest Common Factor of Two or More +Integers}\label{s9}% +\index{Common!divisors|(}\index{Factors}% +\index{Greatest common factor|(} + +Let $m$ and $n$ be two positive integers such that $m$ is greater +than $n$. Then, according to the fundamental theorem of Euclid, we +can form the set of equations +\begin{align*} +m &= qn + n_1, & 0 &< n_1 < n, \\ +n &= q_1 n_1 + n_2, & 0 &< n_2 < n_1, \\ +n_1 &= q_2 n_2 + n_3, & 0 &< n_3 < n_2, \\ +&\vdots \qquad \vdots &&\vdots \qquad \vdots \\ +n_{k - 2} &= q_{k - 1} n_{k-1} + n_k, & 0 &< n_k < n_{k - 1}, \\ +n_{k - 1} &= q _k n_k. & & +\end{align*} +If $m$ is a multiple of $n$ we write $n = n_0$, $k = 0$, in the +above equations. + +\smallskip \textsc{Definition.} The process of reckoning involved in +determining the above set of equations is called the Euclidian +Algorithm.\index{Euclidian algorithm} + +\smallskip I.~\emph{The number $n_k$ to which the Euclidian +algorithm leads is the greatest common divisor of $m$ and $n$.} + +In order to prove this theorem we have to show two things: + +1)~That $n_k$ is a divisor of both $m$ and $n$; + +2)~That the greatest common divisor $d$ of $m$ and $n$ is a divisor +of $n_k$. + +To prove the first statement we examine the above set of equations, +working from the last to the first. From the last equation we see +that $n_k$ is a divisor of $n_{k-1}$. Using this result we see that +the second member of next to the last equation is divisible by $n_k$ +Hence its first member $n_{k-2}$ must be divisible by $n_k$. +Proceeding in this way step by step we show that $n_2$ and $n_1$, +and finally that $n$ and $m$, are divisible by $n_k$. + +For the second part of the proof we employ the same set of equations +and work from the first one to the last one. Let $d$ be any common +divisor of $m$ and $n$. From the first equation we see that $d$ is a +divisor of $n_1$. Then from the second equation it follows that $d$ +is a divisor of $n_2$. Proceeding in this way we show finally that +$d$ is a divisor of $n_k$. Hence any common divisor, and in +particular the greatest common divisor, of $m$ and $n$ is a factor +of $n_k$. + +This completes the proof of the theorem. + +\smallskip \textsc{Corollary.} \emph{Every common divisor of $m$ and +$n$ is a factor of their greatest common divisor.} + +\smallskip II.~\emph{Any number $n_i$ in the above set of equations +is the difference of multiples of $m$ and $n$.} + +From the first equation we have +\begin{equation*} +n_i = m - qn +\end{equation*} +so that the theorem is true for $i = 1$. We shall suppose that the +theorem is true for every subscript up to $i - 1$ and prove it true +for the subscript $i$. Thus by hypothesis we have\footnote{If $i = +2$ we must replace $n_{i-2}$ by $n$.} +\begin{align*} +n_{i-2} &= \pm(\alpha_{i-2}m - \beta_{i-2}n ), \\ +n_{i-1} &= \mp(\alpha_{i-1}m - \beta_{i-1}n). +\intertext{Substituting in the equation} +n_i &= -q_{i-1}n_{n-1} + n_{i-2} \\ +\intertext{we have a result of the form} +n_i &= \pm (\alpha_i m - \beta_i n). +\end{align*} +From this we conclude at once to the truth of the theorem. + +Since $n_k$ is the greatest common divisor of $m$ and $n$, we have +as a corollary the following important theorem: + +\smallskip III.~\emph{If $d$ is the greatest common divisor of the +positive integers $m$ and $n$, then there exist positive integers +$\alpha$ and $\beta$ such that} +\begin{equation*} +\alpha m - \beta n = \pm d. +\end{equation*} + +If we consider the particular case in which $m$ and $n$ are +relatively prime, so that $d = 1$, we see that there exist positive +integers $\alpha$ and $\beta$ such that $\alpha m - \beta n = \pm +1$. Obviously, if $m$ and $n$ have a common divisor $d$, greater +than $1$, there do not exist integers $\alpha$ and $\beta$ +satisfying this relation; for, if so, $d$ would be a divisor of the +first member of the equation and not of the second. Thus we have the +following theorem: + +\smallskip IV.~\emph{A necessary and sufficient condition that $m$ +and $n$ are relatively prime is that there exist integers $\alpha$ +and $\beta$ such that $\alpha m - \beta n = \pm 1$.} + +The theory of the greatest common divisor of three or more numbers +is based directly on that of the greatest common divisor of two +numbers; consequently it does not require to be developed in detail. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] If $d$ is the greatest common divisor of $m$ and $n$, +then $m / d$ and $n / d$ are relatively prime. + +\item[2.] If $d$ is the greatest common divisor of $m$ and $n$ and +$k$ is prime to $n$, then $d$ is the greatest common divisor of $km$ +and $n$. + +\item[3.] The number of multiple of $b$ in the sequence $a, 2a, 3a, +\cdots, ba$ is equal to the greatest common divisor of $a$ and $b$. + +\item[4.] If the sum or the difference of two irreducible fractions is +an integer, the denominators of the fractions are equal. + +\item[5.] The algebraic sum of any number of irreducible fractions, +whose denominators are prime each to each, cannot be an integer. + +\item[6*.] The number of divisions to be effected in finding the +greatest common divisor of two numbers by the Euclidian algorithm +does not exceed five times the number of digits in the smaller +number (when this number is written in the usual scale of 10). +\end{enumerate}\normalsize% +\index{Common!divisors|)}\index{Greatest common factor|)} + +\section{The Least Common Multiple of Two or More +Integers}\label{s10}% +\index{Common!multiples|(}\index{Least common multiple|(} + +I.~\emph{The common multiples of two or more numbers are the +multiples of their least common multiple.} + +This may be readily proved by means of the unique factorization +theorem. The method is obvious. We shall, however, give a proof +independent of this theorem. + +Consider first the case of two numbers; denote them by $m$ and $n$ +and their greatest common divisor by $d$. Then we have +\begin{equation*} +m = d\mu, \quad n = d\nu, +\end{equation*} +where $\mu$ and $\nu$ are relatively prime +integers.\index{Common!divisors}\index{Greatest common factor} The +common multiples sought are multiples of $m$ and are all comprised +in the numbers $am=ad\mu$, where $a$ is any integer whatever. In +order that these numbers shall be multiples of $n$ it is necessary +and sufficient that $ad\mu$ shall be a multiple of $d\nu$; that is, +that $a\mu$ shall be a multiple of $\nu$; that is, that $a$ shall be +a multiple of $\nu$, since $\mu$ and $\nu$ are relatively prime. +Writing $a = \delta\nu$ we have as the multiples in question the set +$\delta d\mu\nu$ where $\delta$ is an arbitrary integer. This proves +the theorem for the case of two numbers; for $d\mu\nu$ is evidently +the least common multiple of $m$ and $n$. + +We shall now extend the proposition to any number of integers $m, n, +p, q,\ldots$. The multiples in question must be common multiples of +$m$ and $n$ and hence of their least common multiple $\mu$. Then the +multiples must be multiples of $\mu$ and $p$ and hence of their +least common multiple $\mu_1$. But $\mu_1$ is evidently the least +common multiple of $m, n, p$. Continuing in a similar manner we may +show that every multiple in question is a multiple of $\mu$, the +least common multiple of $m, n, p, q, \ldots$. And evidently every +such number is a multiple of each of the numbers $m, n, p, q, +\ldots$. + +Thus the proof of the theorem is complete. + +When the two integers $m$ and $n$ are relatively prime their +greatest common divisor is $1$ and their least common multiple is +their product. Again if $p$ is prime to both $m$ and $n$ it is prime +to their product $mn$; and hence the least common multiple of $m, n, +p$ is in this case $mnp$. Continuing in a similar manner we have the +theorem: + +\smallskip II.~\emph{The least common multiple of several integers, +prime each to each, is equal to their product.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] In order that a common multiple of $n$ numbers shall be +the least, it is necessary and sufficient that the quotients +obtained by dividing it successively by the numbers shall be +relatively prime. + +\item[2.] The product of $n$ numbers is equal to the product of +their least common multiple by the greatest common divisor of their +products $n - 1$ at a time. + +\item[3.] The least common multiple of $n$ numbers is equal to any +common multiple $M$ divided by the greatest common divisor of the +quotients obtained on dividing this common multiple by each of the +numbers. + +\item[4.] The product of $n$ numbers is equal to the product of their +greatest common divisor by the least common multiple of the products +of the numbers taken $n - 1$ at a time. +\end{enumerate} \normalsize% +\index{Common!multiples|)}\index{Least common multiple|)} + +\section{Scales of Notation}\label{s11}\index{Scales of notation|(} + +I.~\emph{If $m$ and $n$ are positive integers and $n > 1$, then $m$ +can be represented in terms of $n$ in one and in only one way in the +form} +\begin{gather*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1} n + a_h, \\ +\intertext{where} +a_0 \ne 0,\ 0 \leqq a_i < n, \quad i = 0, 1, 2, \ldots, h. +\end{gather*} + +That such a representation of $m$ exists is readily proved by means +of the fundamental theorem of Euclid. For we have +\begin{align*} +m &= n_0 n + a_h, & 0 &\leqq a_h < n, \\ +n_0 &= n_1n + a_{h-1}, & 0 &\leqq a_{h-1} < n, \\ +n_1 &= n_2 n + a_{h-2}, & 0 &\leqq a_{h-2} < n, \\ +\hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots & + \hdots\hdots & \hdots\hdots\hdots\hdots\hdots\hdots \\ +n_{h-3} &= n_{h-2} n + a_2, & 0 &\leqq a_2 < n, \\ +n_{h-2} &= n_{h-1} n + a_1, & 0 &\leqq a_1 < n, \\ +n_{h-1} &= a_0, & 0 &< a_0 < n. +\end{align*} +If the value of $n_{h-1}$ given in the last of these equations is +substituted in the second last we have +\begin{equation*} +n_{h-2} = a_0n + a_1. +\end{equation*} +This with the preceding gives +\begin{equation*} +n_{h-3} = a_0 n^2 + a_1n + a_2. +\end{equation*} +Substituting from this in the preceding and continuing the process +we have finally +\begin{equation*} +m = a_0 n^h + a_1 n^{h-1} + \ldots + a_{h-1}n + a_h, +\end{equation*} +a representation of $m$ in the form specified in the theorem. + +To prove that this representation is unique, we shall suppose that +$m$ has the representation +\begin{gather*} +m = b_0 n^k + b_1 n^{k-1} + \ldots + b_{k-1}n + b_k, \\ +\intertext{where} +b_0 \ne 0,\ 0 < b_i < n,\quad i=0, 1, 2, \ldots, k, \\ +\intertext{and show that the two representations are identical. We +have} +a_0 n^h + \ldots + a_{h-1} n + a_h = + b_0 n^k + \ldots + b_{k-1} n + b_k. +\intertext{Then} +a_0 n^h + \ldots + a_{h-1} n - + (b_0 n^k + \ldots + b_{k-1} n) = b_k - a_h. +\end{gather*} +The first member is divisible by $n$. Hence the second is also. But +the second member is less than $n$ in absolute value; and hence, in +order to be divisible by $n$, it must be zero. That is, $b_k = a_h$. +Dividing the equation through by $n$ and transposing we have +\begin{equation*} +a_0 n^{h-1} + \ldots + a_{h-2} n - (b_0 n^{k-1} + \ldots + + b_{k-2} n) + = b_{k-1} - a_{h-1}. +\end{equation*} +It may now be seen that $b_{k-1} = a_{h-1}$. It is evident that this +process may be continued until either the $a$'s are all eliminated +from the equation or the $b$'s are all eliminated. But it is obvious +that when one of these sets is eliminated the other is also. Hence, +$h = k$. Also, every $a$ equals the $b$ which multiplies the same +power of $n$ as the corresponding $a$. That is, the two +representations of $m$ are identical. Hence the representation in +the theorem is unique. + +From this theorem it follows as a special case that any positive +integer can be represented in one and in only one way in the scale +of 10; that is, in the familiar Hindoo notation. It can also be +represented in one and in only one way in any other scale. Thus +\begin{equation*} +120759 = 1 \cdot 7^6 + 0 \cdot 7^5 + 1 \cdot 7^4 + 2 \cdot 7^3 + + 0 \cdot 7^2 + 3 \cdot 7^1 + 2. +\end{equation*} +Or, using a subscript to denote the scale of notation, this may be +written +\begin{equation*} +(120759)_{10} = (1012032)_7. +\end{equation*} + +For the case in which $n$ (of theorem I) is equal to 2, the only +possible values for the $a$'s are 0 and 1. Hence we have at once the +following theorem: + +II.~\emph{Any positive integer can be represented in one and in only +one way as a sum of different powers of 2.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} + +\item[1.] Any positive integer can be represented as an aggregate of +different powers of $3$, the terms in the aggregate being combined +by the signs $+$ and $-$ appropriately chosen. + +\item[2.] Let $m$ and $n$ be two positive integers of which $n$ is the +smaller and suppose that $2^k \leq n < 2^{k+1}$. By means of the +representation of $m$ and $n$ in the scale of 2 prove that the +number of divisions to be effected in finding the greatest common +divisor of $m$ and $n$ by the Euclidian algorithm does not exceed +$2k$. +\end{enumerate}\normalsize\index{Scales of notation|)} + +\section{Highest Power of a Prime $p$ Contained in $n!$.}\label{s12}% +\index{Highest power of \emph{p} in \emph{n}"!|(} + +Let $n$ be any positive integer and $p$ any prime number not greater +than $n$. We inquire as to what is the highest power $p^\nu$ of the +prime $p$ contained in $n!$. + +In solving this problem we shall find it convenient to employ the +notation +\begin{equation*} +\left [ \frac{r}{s} \right ] +\end{equation*} to denote the greatest integer $\alpha$ such that +$\alpha s \leq r$. With this notation it is evident that we have +\begin{gather} +\left [ + \frac{\left [ \frac{n}{p} \right ]} + {p} +\right ] = \left [ \frac{n}{p^2} \right ]; \tag{1} \\ +\intertext{and more generally} +\left [ + \frac{\left [ \frac{n}{p^i} \right ]} + {p^j} +\right ] = \left [ \frac{n}{p^{i+j}} \right ]. \notag +\end{gather} + +If now we use $H\{x\}$ to denote the index of the highest power of +$p$ contained in an integer $x$, it is clear that we have +\begin{gather*} +H\{n!\} = + H \left \{ p \cdot 2p \cdot 3p \ldots + \left [ \frac{n}{p} \right ] p \right \}, \\ +\intertext{since only multiples of $p$ contain the factor $p$. +Hence} +H\{n!\} = + \left [ \frac{n}{p} \right ] + + H \left \{ 1 \cdot 2 \ldots \left [ \frac{n}{p} \right ] + \right \}. +\end{gather*} +Applying the same process to the $H$-function in the second member +and remembering relation (1) it is easy to see that we have +\begin{align*} +H\{n!\} &= \left[ \frac{n}{p} \right] + + H\left\{ p \cdot 2p \cdot \ldots \cdot + \left[ \frac{n}{p^2} \right]p\right\} \\ + &= \left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + + H \left\{\cdot 1 \cdot 2 \cdot 3 + \ldots \left[ \frac{n}{p^2} \right] \right\}. \\ +\intertext{Continuing the process we have finally} +H\{n1\} &= \left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] + + \ldots, +\end{align*} +the series on the right containing evidently only a finite number of +terms different from zero. Thus we have the theorem: + +\smallskip I.~\emph{The index of the highest power of a prime $p$ +contained in $n!$ is} +\begin{gather*} +\left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + + \left[ \frac{n}{p^3} \right] + \ldots. +\end{gather*} + +The theorem just obtained may be written in a different form, more +convenient for certain of its applications. Let $n$ be expressed in +the scale of $p$ in the form +\begin{gather*} +n = a_0p^h + a_1p^{h-1} + \ldots + a_{h-1}p + a_h, \\ +\intertext{where} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h. +\end{gather*} +Then evidently +\begin{align*} +\left[ \frac{n}{p} \right] &= a_0p^{h-1} + a_1p^{h-2} + \ldots + + a_{h-2}p + a_{h-1}, \\ +\left[ \frac{n}{p^2} \right] &= a_0p^{h-2} + a_1p^{h-3} + \ldots + + a_{h-2}, \\ +.\ \ .\ \ .\ \ .\ \ &.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ +.\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ . +\end{align*} +Adding these equations member by member and combining the second +members in columns as written, we have +\begin{align*} +\left[ \frac{n}{p} \right] + + \left[ \frac{n}{p^2} \right] &+ + \left[ \frac{n}{p^3} \right] + \ldots \\ +&= \sum_{i=0}^h \frac{a_i(p^{h-i} - 1)}{p - 1} \\ +&= \frac{a_0p^h + a_1p^{h-1} + \ldots + a_h - + (a_0 + a_1 + \ldots + a_h)}{p-1} \\ +&= \frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{align*} +Comparing this result with theorem I we have the following theorem: + +\smallskip II.~\emph{If $n$ is represented in the scale of $p$ in +the form +\begin{gather*} +n = a_0 p^h + a_1 p^{h-1} + \ldots + a_h, \\ +\intertext{where $p$ is prime and} +a_0 \neq 0,\ 0 \leqq a_i < p,\ i = 0, 1, 2, \ldots, h, \\ +\intertext{then the index of the highest power of $p$ contained in +$n!$ is} +\frac{n - (a_0 + a_1 + \ldots + a_h)}{p - 1}. +\end{gather*}} + +Note the simple form of the theorem for the case $p = 2$; in this +case the denominator $p - 1$ is unity. + +We shall make a single application of these theorems by proving the +following theorem: + +\smallskip III.~\emph{If $n$, $\alpha$, $\beta$, $\ldots$, $\lambda$ +are any positive integers such that $n = \alpha + \beta + \ldots + +\lambda$, then +\begin{equation} +\frac{n!}{\alpha! \beta! \ldots \lambda!} \tag{A} +\end{equation} +is an integer.} + +Let $p$ be any prime factor of the denominator of the fraction (A). +To prove the theorem it is sufficient to show that the index of the +highest power of $p$ contained in the numerator is at least as great +as the index of the highest power of $p$ contained in the +denominator. This index for the denominator is the sum of the +expressions +\begin{equation} + \left . + \begin{gathered} + \left [ \frac{\alpha}{p} \right ] + + \left [ \frac{\alpha}{p^2} \right ] + + \left [ \frac{\alpha}{p^3} \right ] + + \ldots \\ + \left [ \frac{\beta}{p} \right ] + + \left [ \frac{\beta}{p^2} \right ] + + \left [ \frac{\beta}{p^3} \right ] + + \ldots \\ + \vdots \\ + \left [ \frac{\lambda}{p} \right ] + + \left [ \frac{\lambda}{p^2} \right ] + + \left [ \frac{\lambda}{p^3} \right ] + + \ldots + \end{gathered} + \right \} \tag{B} +\end{equation} + +The corresponding index for the numerator is +\begin{equation} +\left [ \frac{n}{p} \right ] + +\left [ \frac{n}{p^2} \right ] + +\left [ \frac{n}{p^3} \right ] + +\ldots \tag{C} +\end{equation} +But, since $n = \alpha + \beta + \ldots + \lambda$, it is evident +that +\begin{equation*} + \left [ \frac{n}{p^r} \right ] \stackrel{=}{>} + \left [ \frac{\alpha}{p^r} \right ] + + \left [ \frac{\beta}{p^r} \right ] + + \ldots + + \left [ \frac{\lambda}{p^r} \right ]. +\end{equation*} +From this and the expressions in (B) and (C) it follows that the +index of the highest power of any prime $p$ in the numerator of (A) +is equal to or greater than the index of the highest power of p +contained in its denominator. The theorem now follows at once. + +\smallskip \textsc{Corollary.}~\emph{The product of $n$ consecutive +integers is divisible by $n!$.} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the highest power of 2 contained in 1000! is +$2^{994}$; in 1900! is $2^{1893}$. Show that the highest power of 7 +contained in 10000! is $7^{1665}$. + +\item[2.] Find the highest power of 72 contained in 1000! + +\item[3.] Show that 1000! ends with 249 zeros. + +\item[4.] Show that there is no number $n$ such that $3^7$ is the +highest power of 3 contained in $n!$. + +\item[5.] Find the smallest number $n$ such that the highest power +of 5 contained in $n!$ is $5^{31}$. What other numbers have the same +property? + +\item[6.] If $n = rs$, $r$ and $s$ being positive integers, show that +$n!$ is divisible by $(r!)^s$; by $(s!)^r$; by the least common +multiple of $(r!)^s$ and $(s!)^r$. + +\item[7.] If $n = \alpha + \beta + pq + rs$, where $\alpha, \beta, p, +q, r, s$, are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha ! \beta ! (q!)^p (s!)^r. +\end{equation*} + +\item[8.] When $m$ and $n$ are two relatively prime positive integers +the quotient +\begin{equation*} +Q = \frac{(m + n + 1)!} {m! n!} +\end{equation*} +as an integer. + +\item[9*.] If $m$ and $n$ are positive integers, then each of the +quotients +\begin{equation*} +Q = \frac{(mn)!} {n! (m!)^n},\quad +Q = \frac{(2m)! (2n)!} {m! n! (m+n)!}, +\end{equation*} +is an integer. Generalize to $k$ integers $m, n, p, \ldots$. + +\item[10*.] If $n = \alpha + \beta + pq + rs$ where $\alpha, \beta, +p, q, r, s$ are positive integers, then $n!$ is divisible by +\begin{equation*} +\alpha! \beta! r! p! (q!)^p (s!)^r. +\end{equation*} + +\item[11*.] Show that +\begin{equation*} +\frac{(rst)!} {t! (s!)^t (r!)^{st}}, +\end{equation*} is an integer ($r, s, t$ being positive integers). +Generalize to the case of $n$ integers $r, s, t, u, \ldots$. +\end{enumerate}\normalsize% +\index{Highest power of \emph{p} in \emph{n}"!|)} + +\section{Remarks Concerning Prime Numbers}\label{s13}% +\index{Prime numbers|(} + +We have seen that the number of primes is infinite. But the integers +which have actually been identified as prime are finite in number. +Moreover, the question as to whether a large number, as for instance +$2^{257}-1$, is prime is in general very difficult to answer. Among +the large primes actually identified as such are the following: +\begin{equation*} +2^{61}-1, \quad 2^{75} \cdot 5+1, \quad 2^{89}-1, \quad 2^{127}-1. +\end{equation*} + +\emph{No analytical expression for the representation of prime +numbers has yet been discovered.} Fermat believed, though he +confessed that he was unable to prove, that he had found such an +analytical expression in +\begin{equation*} +2^{2^n} + 1. +\end{equation*} +Euler showed the error of this opinion by finding that $641$ is a +factor of this number for the case when $n = 5$.% +\index{Euler}\index{Fermat} + +The subject of prime numbers is in general one of exceeding +difficulty. In fact it is an easy matter to propose problems about +prime numbers which no one has been able to solve. Some of the +simplest of these are the following: + +\begin{enumerate} +\item Is there an infinite number of pairs of primes differing by +2? +\item Is every even number (other than 2) the sum of two primes or +the sum of a prime and the unit? +\item Is every even number the difference of two primes or the +difference of 1 and a prime number? +\item To find a prime number greater than a given prime. +\item To find the prime number which follows a given prime. +\item To find the number of primes not greater than a given number. +\item To compute directly the $n^\text{th}$ prime number, when $n$ +is given. +\end{enumerate}\index{Prime numbers|)} + +\chapter{ON THE INDICATOR OF AN INTEGER}% +\index{Indicator|(} + +\section{Definition. Indicator of a Prime Power}\label{s14}% +\index{Indicator!of a prime power} + +\emph{Definition.} If $m$ is any given positive integer the number +of positive integers not greater than $m$ and prime to it is called +the indicator of $m$. It is usually denoted by $\phi(m)$, and is +sometimes called Euler's $\phi$-function of $m$.% +\index{Euler's!$\phi$-function}\index{$\phi(m)$} More rarely, it has +been given the name of totient of $m$.\index{Totient} + +As examples we have +\begin{equation*} +\phi(1) = 1,\ \phi(2) = 1,\ \phi(3) = 2,\ \phi(4) = 2. +\end{equation*} + +If $p$ is a prime number it is obvious that +\begin{equation*} +\phi(p) = p - 1; +\end{equation*} +for each of the integers 1, 2, 3, $\ldots$, $p-1$ is prime to $p$. + +Instead of taking $m = p$ let us assume that $m = p^\alpha$, where +$\alpha$ is a positive integer, and seek the value of +$\phi(p^\alpha)$. Obviously, every number of the set 1, 2, 3, +$\ldots$, $p^\alpha$ either is divisible by $p$ or is prime to +$p^\alpha$. The number of integers in the set divisible by $p$ is +$p^{\alpha - 1}$. Hence $p^\alpha-p^{\alpha-1}$ of them are prime to +$p$. Hence $\phi(p^\alpha) = p^\alpha-p^{\alpha-1}$. Therefore + +\emph{If $p$ is any prime number and $\alpha$ is any positive +integer, then} +\begin{equation*} +\phi(p^\alpha) = p^\alpha \left ( 1 - \frac{1}{p} \right ). +\end{equation*} + +\section{The Indicator of a Product}\label{s15}% +\index{Indicator!of a product|(} + +I.~\emph{If $\mu$ and $\nu$ are any two relatively prime positive +integers, then} +\begin{equation*} +\phi(\mu\nu) = \phi(\mu) \phi(\nu). +\end{equation*} + +In order to prove this theorem let us write all the integers up to +$\mu\nu$ in a rectangular array as follows: +\footnotesize\begin{equation} + \left . + \begin{aligned} + 1 && 2 && 3 && + \ldots && h && \ldots && \mu \\ + \mu + 1 && \mu + 2 && \mu + 3 && + \ldots && \mu + h && \ldots && 2\mu \\ + 2 \mu + 1 && 2 \mu + 2 && 2 \mu + 3 && + \ldots && 2 \mu + h && \ldots && 3\mu \\ + \vdots && \vdots && \vdots && + && \vdots && && \vdots \\ + (\nu - 1)\mu + 1 && (\nu - 1)\mu + 2 && (\nu - 1)\mu + 3 && + \ldots && (\nu - 1)\mu + h && \ldots && \nu\mu \\ + \end{aligned} + \right \} \tag{A} +\end{equation}\normalsize + +If a number $h$ in the first line of this array has a factor in +common with $\mu$ then every number in the same column with $h$ has +a factor in common with $\mu$. On the other hand if $h$ is prime to +$\mu$, so is every number in the column with $h$ at the top. But the +number of integers in the first row prime to $\mu$ is $\phi(\mu)$. +Hence the number of columns containing integers prime to $\mu$ is +$\phi(\mu)$ and every integer in these columns is prime to $\mu$. + +Let us now consider what numbers in one of these columns are prime +to $\nu$; for instance, the column with $h$ at the top. We wish to +determine how many integers of the set +\begin{gather*} +h,\ \mu + h,\ 2\mu + h,\ \ldots,\ (\nu - 1)\mu + h \\ +\intertext{are prime to $\nu$. Write} +s\mu + h = q_s\nu + r_s +\end{gather*} where s ranges over the numbers $s = 0,\ 1,\ 2,\ +\ldots,\ \nu - 1$ and $0\leqq r_s < \nu$. Clearly $s\mu + h$ is or +is not prime to $\nu$ according as $r_s$ is or is not prime to +$\nu$. Our problem is then reduced to that of determining how many +of the quantities $r_s$ are prime to $\nu$. + +First let us notice that all the numbers $r_s$ are different; for, +if $r_s = r_t$ then from +\begin{equation*} +s\mu + h = q_s\nu + r_s,\quad t\mu + h = q_t\nu + r_t, +\end{equation*} +we have by subtraction that $(s-t)\mu$ is divisible by $\nu$. But +$\mu$ is prime to $\nu$ and $s$ and $t$ are each less than $\nu$. +Hence $(s-t)\mu$ can be a multiple of $\nu$ only by being zero; that +is, $s$ must equal $t$. Hence no two of the remainders $r_s$ can be +equal. + +Now the remainders $r_s$ are $\nu$ in number, are all zero or +positive, each is less than $\nu$, and they are all distinct. Hence +they are in some order the numbers 0, 1, 2, $\ldots$, $\nu-1$. The +number of integers in this set prime to $\nu$ is evidently +$\phi(\nu)$. + +Hence it follows that in any column of the array (A) in which the +numbers are prime to $\mu$ there are just $\phi(\nu)$ numbers which +are prime to $\nu$. That is, in this column there are just +$\phi(\nu)$ numbers which are prime to $\mu\nu$. But there are +$\phi(\mu)$ such columns. Hence the number of integers in the array +(A) prime to $\mu\nu$ is $\phi(\mu)\phi(\nu)$. + +But from the definition of the $\phi$-function it follows that the +number of integers in the array (A) prime to $\mu\nu$ is +$\phi(\mu\nu).$ Hence, +\begin{equation*} +\phi(\mu\nu) = \phi(\mu)\phi(\nu), +\end{equation*} which is the theorem to be proved. + +\smallskip \textsc{Corollary.}~\emph{In the series of $n$ +consecutive terms of an arithmetical progression the common +difference of which is prime to $n$, the number of terms prime to +$n$ is $\phi(n)$.} + +From theorem I we have readily the following more general result: + +\smallskip II.~\emph{If $m_1, m_2, \ldots, m_k$ are $k$ positive +integers which are prime each to each, then} +\begin{equation*} +\phi(m_1 m_2 \ldots m_k) = \phi(m_1) \phi(m_2) \ldots \phi(m_k). +\end{equation*}\index{Indicator!of a product|)} + +\section{The Indicator of any Positive Integer}\label{s16}% +\index{Indicator!of any integer|(} + +From the results of \S\S \ref{s14} and \ref{s15} we have an +immediate proof of the following fundamental theorem: + +\emph{If $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n}$ +where $p_1, p_2, \ldots, p_n$ are different primes and $\alpha_1, +\alpha_2, \ldots, \alpha_n$ are positive integers, then} +\begin{equation*} +\phi(m) = m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{equation*} + +For, +\begin{align*} +\phi(m) &= \phi(p_1^{\alpha_1}) \phi(p_2^{\alpha_2}) \ldots + \phi(p_n^{\alpha_n}) \\ + &= p_1^{\alpha_1} \left ( 1-\frac{1}{p_1} \right ) + p_2^{\alpha_2} \left ( 1-\frac{1}{p_2} \right ) + \ldots + p_n^{\alpha_n} \left ( 1-\frac{1}{p_n} \right ) \\ + &= m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_n} \right ). +\end{align*} + +On account of the great importance of this theorem we shall give a +second demonstration of it. + +It is clear that the number of integers less than $m$ and divisible +by $p_1$ is +\begin{gather*} +\frac{m}{p_1}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_2$ is} +\frac{m}{p_2}. \\ +\intertext{The number of integers less than $m$ and divisible by +$p_1 p_2$ is} +\frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and divisible +by either $p_1$ or $p_2$ is} +\frac{m}{p_1} + \frac{m}{p_2} - \frac{m}{p_1 p_2}. \\ +\intertext{Hence the number of integers less than $m$ and prime to +$p_1 p_2$ is} +m - \frac{m}{p_1} - \frac{m}{p_2} + \frac{m}{p_1 p_2} = + m \left ( 1-\frac{1}{p_1} \right ) \left ( 1-\frac{1}{p_2} \right ). +\end{gather*} + +We shall now show that if the number of integers less than $m$ and +prime to $p_1 p_2 \ldots p_i$, where $i$ is less than $n$, is +\begin{gather*} +m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_i} \right ), \\ +\intertext{then the number of integers less than $m$ and prime to +$p_1 p_2 \ldots p_i p_{i+1}$ is} + m \left ( 1-\frac{1}{p_1} \right ) + \left ( 1-\frac{1}{p_2} \right ) + \ldots + \left ( 1-\frac{1}{p_{i+1}} \right ). +\end{gather*} +From this our theorem will follow at once by induction. + +From our hypothesis it follows that the number of integers less than +$m$ and divisible by at least one of the primes $p_1$, $p_2$, +$\ldots$, $p_i$ is +\begin{gather} +m - + m \left (1 - \frac{1}{p_1}\right ) + \ldots + \left (1 - \frac{1}{p_i}\right ), \notag \\ +\intertext{or} +\sum \frac{m}{p_1} - \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - \ldots, \tag{A} +\end{gather} +where the summation in each case runs over all numbers of the type +indicated, the subscripts of the $p$'s being equal to or less than +$i$. + +Let us consider the integers less than $m$ and having the factor +$p_{i+1}$ but not having any of the factors $p_1$, $p_2$, $\ldots$, +$p_i$. Their number is +\begin{gather} +\frac{m}{p_{i+1}} - + \frac{1}{p_{i+1}} \left \{ + \sum \frac{m}{p_1} - + \sum \frac{m}{p_1p_2} + + \sum \frac{m}{p_1p_2p_3} - + \ldots + \right \}, \tag{B} +\end{gather} +where the summation signs have the same significance as before. For +the number in question is evidently $\frac{m}{p_{i+1}}$ \emph{minus} +the number of integers not greater than $\frac{m}{p_{i+1}}$ and +divisible by at least one of the primes $p_1$, $p_2$, $\ldots$, +$p_i$. + +If we add (A) and (B) we have the number of integers less than $m$ +and divisible by one at least of the numbers $p_1$, $p_2$, $\ldots$, +$p_{i+1}$. Hence the number of integers less than $m$ and prime to +$p_1$, $p_2$, $\ldots$, $p_{i+1}$ is +\begin{gather*} +m - + \sum \frac{m}{p_1} + + \sum \frac{m}{p_1 p_2} - + \sum \frac{m}{p_1 p_2 p_3} + + \ldots, \\ +\intertext{where now in the summations the subscripts run from 1 to +$i+1$. This number is clearly equal to} +m + \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_{i+1}} \right ). +\end{gather*} +From this result, as we have seen above, our theorem follows at once +by induction.\index{Indicator!of any integer|)} + +\section{Sum of the Indicators of the Divisors of a Number}% +\label{s17} + +We shall first prove the following lemma: + +\smallskip \emph{Lemma. If $d$ is any divisor of $m$ and $m = nd$, +the number of integers not greater than $m$ which have with $m$ the +greatest common divisor $d$ is $\phi(n)$.} + +Every integer not greater than $m$ and having the divisor $d$ is +contained in the set $d$, $2d$, $3d$, $\ldots$, $nd$. The number of +these integers which have with $m$ the greatest common divisor $d$ +is evidently the same as the number of integers of the set 1, 2, +$\ldots$, $n$ which are prime to $\frac{m}{d}$, or $n$; for $\alpha +d$ and $n$ have or have not the greatest common divisor $d$ +according as $\alpha$ is or is not prime to $\frac{m}{d}=n$. Hence +the number in question is $\phi(n)$. + +From this lemma follows readily the proof of the following theorem: + +\smallskip \emph{If $d_1$, $d_2$, $\ldots$, $d_r$ are the different +divisors of $m$, then} +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = m. +\end{equation*} + +Let us define integers $m_1$, $m_2$, $\ldots$, $m_r$ by the +relations +\begin{equation*} +m = d_1 m_1 = d_2 m_2 = \ldots = d_r m_r. +\end{equation*} +Now consider the set of $m$ positive integers not greater than $m$, +and classify them as follows into $r$ classes. Place in the first +class those integers of the set which have with $m$ the greatest +common divisor $m_1$; their number is $\phi(d_1)$, as may be seen +from the lemma. Place in the second class those integers of the set +which have with $m$ the greatest common divisor $m_2$; their number +is $\phi(d_2)$. Proceeding in this way throughout, we place finally +in the last class those integers of the set which have with $m$ the +greatest common divisor $m_r$; their number is $\phi(d_r)$. It is +evident that every integer in the set falls into one and into just +one of these $r$ classes. Hence the total number $m$ of integers in +the set is $\phi(d_1) + \phi(d_r) + \ldots + \phi(d_r)$. From this +the theorem follows immediately. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that the indicator of any integer greater than $2$ +is even. + +\item[2.] Prove that the number of irreducible fractions not greater +than $1$ and with denominator equal to $n$ is $\phi(n)$. + +\item[3.] Prove that the number of irreducible fractions not greater +than $1$ and with denominators not greater than $n$ is +\begin{equation*} +\phi(1) + \phi(2) + \phi(3) + \ldots + \phi(n). +\end{equation*} + +\item[4.] Show that the sum of the integers less than $n$ and prime to +$n$ is $\frac{1}{2} n \phi(n)$ if $n > 1$. + +\item[5.] Find ten values of $x$ such that $\phi(x) = 24$. + +\item[6.] Find seventeen values of $x$ such that $\phi(x) = 72$. + +\item[7.] Find three values of $n$ for which there is no $x$ satisfying +the equation $\phi(x) = 2n$. + +\item[8.] Show that if the equation +\begin{equation*} +\phi(x) = n +\end{equation*} +has one solution it always has a second solution, $n$ being given +and $x$ being the unknown. + +\item[9.] Prove that all the solutions of the equation +\begin{equation*} +\phi(x) = 4n - 2, n > 1, +\end{equation*} +are of the form $p^\alpha$ and $2p^\alpha$, where $p$ is a prime of +the form $4s-1$. + +\item[10.] How many integers prime to $n$ are there in the set +\begin{enumerate} +\item $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots, n(n+1)$? +\item $1 \cdot 2 \cdot 3, 2 \cdot 3 \cdot 4, + 3 \cdot 4 \cdot 5, \ldots, n(n+1)(n+2)$? +\item $\frac{1 \cdot 2}{2}, \frac{2 \cdot 3}{2}, + \frac{3 \cdot 4}{2}, \ldots, \frac{n(n+1)}{2}$? +\item $\frac{1 \cdot 2 \cdot 3}{6}, + \frac{2 \cdot 3 \cdot 4}{6}, + \frac{3 \cdot 4 \cdot 5}{6}, + \ldots, + \frac{n(n+1)(n+2)}{6}$? +\end{enumerate} + +\item[11*.] Find a method for determining all the solutions of the +equation +\begin{equation*} +\phi(x) = n, +\end{equation*} +where $n$ is given and $x$ is to be sought. + +\item[12*.] A number theory function $\phi(n)$ is defined for every +positive integer $n$; and for every such number $n$ it satisfies the +relation +\begin{equation*} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = n, +\end{equation*} +where $d_1, d_2, \ldots, d_r$ are the divisors of $n$. From this +property alone show that +\begin{equation*} +\phi(n) = n \left ( 1 - \frac{1}{p_1} \right ) + \left ( 1 - \frac{1}{p_2} \right ) + \ldots + \left ( 1 - \frac{1}{p_k} \right ), +\end{equation*} +where $p_1, p_2, \ldots, p_k$ are the different prime factors of +$n$. \end{enumerate} \normalsize\index{Indicator|)} + +\chapter{ELEMENTARY PROPERTIES OF CONGRUENCES}% +\index{Congruences|(} + +\section{Congruences Modulo $m$}\label{s18} + +\textsc{Definitions.} If $a$ and $b$ are any two integers, positive +or zero or negative, whose difference is divisible by $m$, $a$ and +$b$ are said to be congruent modulo $m$, or congruent for the +modulus $m$, or congruent according to the modulus $m$. Each of +the numbers $a$ and $b$ is said to be a residue of the other.% +\index{Residue} + +\smallskip To express the relation thus defined we may write +\begin{equation*} +a = b + cm, +\end{equation*} +where $c$ is an integer (positive or zero or negative). It is more +convenient, however, to use a special notation due to Gauss, and to +write +\begin{equation*} +a \equiv b \mod m, +\end{equation*} +an expression which is read $a$ is congruent to $b$ modulo $m$, or +$a$ is congruent to $b$ for the modulus $m$, or $a$ is congruent to +$b$ according to the modulus $m$.\index{Gauss} This notation has the +advantage that it involves only the quantities which are essential +to the idea involved, whereas in the preceding expression we had the +irrelevant integer $c$. The Gaussian notation is of great value and +convenience in the study of the theory of divisibility. In the +present chapter we develop some of the fundamental elementary +properties of congruences. It will be seen that many theorems +concerning equations are likewise true of congruences with fixed +modulus; and it is this analogy with equations which gives +congruences (as such) one of their chief claims to attention. + +As immediate consequences of our definitions we have the following +fundamental theorems: + +\smallskip I.~\emph{If} $a\equiv c \mod m$, $b\equiv c\mod m$, +\emph{then} $a\equiv b\mod m$; \noindent \emph{that is, for a given +modulus, numbers congruent to the same number are congruent to each +other.} + +For, by hypothesis, $a - c = c_1 m$, $b - c = c_2 m$, where $c_1$ +and $c_2$ are integers. Then by subtraction we have $a - b = (c_1 - +c_2) m$; whence $a \equiv b \mod m$. + +\smallskip II.~\emph{If} $a \equiv b \mod m$, $\alpha \equiv +\beta \mod m$, \emph{then} $a \pm \alpha \equiv b \pm \beta \mod m$; +\emph{that is, congruences with the same modulus may be added or +subtracted member by member.} + +For, by hypothesis, $a - b = c_1 m$, $\alpha - \beta = c_2 m$; +whence $(a \pm \alpha) - (b \pm \beta) = (c_1 \pm c_2)m$. Hence $a +\pm \alpha = b \pm \beta \mod m$. + +\smallskip III.~\emph{If} $a = b \mod m$, \emph{then} +$ca = cb \mod m$, \emph{$c$ being any integer whatever.} + +The proof is obvious and need not be stated. + +\smallskip IV.~\emph{If} $a \equiv b \mod m$, +$\alpha \equiv \beta \mod m$, \emph{then} $a \alpha \equiv b \beta +\mod m$; \emph{that is, two congruences with the same modulus may be +multiplied member by member.} + +For, we have $a = b + c_1 m$, $\alpha = \beta + c_2 m$. Multiplying +these equations member by member we have $a \alpha = b \beta + m (b +c_2 + \beta c_1 + c_1 c_2 m)$. Hence $a \alpha \equiv b \beta \mod +m$. + +\smallskip A repeated use of this theorem gives the following +result: + +\smallskip V.~\emph{If} $a \equiv b \mod m$, \emph{then} +$a^n \equiv b^n \mod m$ \emph{where $n$ is any positive integer.} + +\smallskip As a corollary of theorems II, III and V we have the +following more general result: + +\smallskip VI.~\emph{If $f(x)$ denotes any polynomial in $x$ with +coefficients which are integers (positive or zero or negative) and +if further $a\equiv b \bmod m$, then} +\begin{equation*} +f(a) \equiv f(b) \bmod m. +\end{equation*} + +\section{Solutions of Congruences by Trial}\label{s19}% +\index{Congruences!Solution by trial|(} + +Let $f(x)$ be any polynomial in $x$ with coefficients which are +integers (positive or negative or zero). Then if $x$ and $c$ are any +two integers it follows from the last theorem of the preceding +section that +\begin{gather*} +f(x) \equiv f(x + cm) \bmod m. \tag{1} \\ +\intertext{Hence if $a$ is any value of $x$ for which the +congruence} +f(x)\equiv 0\bmod m. \tag{2} +\end{gather*} +is satisfied, then the congruence is also satisfied for $x = \alpha ++ cm$, where $c$ is any integer whatever. The numbers $\alpha + cm$ +are said to form a \emph{solution} (or to be a \emph{root}) of the +congruence, $c$ being a variable integer. Any one of the integers +$\alpha + cm$ may be taken as the representative of the solution. We +shall often speak of one of these numbers as the solution itself. + +Among the integers in a solution of the congruence (2) there is +evidently one which is positive and not greater than $m$. Hence all +solutions of a congruence of the type (2) may be found by trial, a +substitution of each of the numbers $1, 2, \ldots, m$ being made for +$x$. It is clear also that $m$ is the maximum number of solutions +which (2) can have whatever be the function $f(x)$. By means of an +example it is easy to show that this maximum number of solutions is +not always possessed by a congruence; in fact, it is not even +necessary that the congruence have a solution at all. + +This is illustrated by the example +\begin{equation*} +x^2 - 3 \equiv 0 \bmod 5. +\end{equation*} +In order to show that no solution is possible it is necessary to +make trial only of the values $1, 2, 3, 4, 5$ for $x$. A direct +substitution verifies the conclusion that none of them satisfies the +congruence; and hence that the congruence has no solution at all. + +On the other hand the congruence +\begin{equation*} +x^5 - x \equiv 0 \bmod 5 +\end{equation*} +has the solutions $x = 1, 2, 3, 4, 5$ as one readily verifies; that +is, this congruence has five solutions---the maximum number possible +in accordance with the results obtained above. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $(a + b)^p \equiv a^p + b^p \bmod p$ +where $a$ and $b$ are any integers and $p$ is any prime. + +\item[2.] From the preceding result prove that +$\alpha^p \equiv \alpha \bmod p$ for every integer $\alpha$. + +\item[3.] Find all the solutions of each of the congruences $x^{11} +\equiv x \bmod 11, x^{10} \equiv 1 \bmod 11, x^{5} \equiv 1 \bmod +11$. +\end{enumerate} \normalsize\index{Congruences!Solution by trial|)} + +\section{Properties of Congruences Relative to Division}\label{s20} + +The properties of congruences relative to addition, subtraction and +multiplication are entirely analogous to the properties of algebraic +equations. But the properties relative to division are essentially +different. These we shall now give. + +\smallskip I.~\emph{If two numbers are congruent modulo $m$ they are +congruent modulo $d$, where $d$ is any divisor of $m$.} + +For, from $a \equiv b \bmod m$, we have $a = b + cm = b + c'd$. +Hence $a\equiv b \bmod d$. + +\smallskip II.~\emph{If two numbers are congruent for different +moduli they are congruent for a modulus which is the least common +multiple of the given moduli.} + +The proof is obvious, since the difference of the given numbers is +divisible by each of the moduli. + +\smallskip III.~\emph{When the two members of a congruence are +multiples of an integer $c$ prime to the modulus, each member of the +congruence may be divided by $c$.} + +For, if $ca \equiv cb \bmod m$ then $ca - cb$ is divisible by $m$. +Since $c$ is prime to $m$ it follows that $a - b$ is divisible by +$m$. Hence $a\equiv b \bmod m$. + +\smallskip IV.~\emph{If the two members of a congruence are +divisible by an integer $c$, having with the modulus the greatest +common divisor $\delta$, one obtains a congruence equivalent to the +given congruence by dividing the two members by $c$ and the modulus +by $\delta$.} + +By hypothesis $ac \equiv bc \bmod m,\quad c = \delta c_1,\quad m = +\delta m_1$. Hence $c(a-b)$ is divisible by $m$. A necessary and +sufficient condition for this is evidently that $c_1(a-b)$ is +divisible by $m_1$. This leads at once to the desired result. + +\section{Congruences with a Prime Modulus}\label{s21}% +\index{Congruences!with prime modulus|(} + +\emph{The congruence\footnote{The sign $\not\equiv$ is read \emph{is +not congruent to}.}} +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod p, + \quad a_0 \not\equiv 0 \bmod p +\end{equation*} +\emph{where $p$ is a prime number and the $a$'s are any integers, +has not more than $n$ solutions.} + +Denote the first member of this congruence by $f(x)$ so that the +congruence may be written +\begin{gather} +f(x) \equiv 0 \bmod p \tag{1} \\ +\intertext{Suppose that $a$ is a root of the congruence, so that} +f(a) \equiv 0 \bmod p. \notag \\ +\intertext{Then we have} f(x) +\equiv f(x) - f(a) \bmod p. \notag \\ +\intertext{But, from algebra, $f(x) - f(a)$ is divisible by $x - a$. +Let $(x-a)^{\alpha}$ be the highest power of $x - a$ contained in +$f(x) - f(a)$. Then we may write} +f(x) - f(a) = (x - a)^{\alpha} f_1(x), \tag{2} \\ +\intertext{where $f_1(x)$ is evidently a polynomial with integral +coefficients. Hence we have} +f(x) \equiv (x - a)^{\alpha} f_1(x) \bmod p. \tag{3} +\end{gather} +We shall say that $a$ occurs $\alpha$ times as a solution of (1); or +that the congruence has $\alpha$ solutions each equal to $a$. + +Now suppose that congruence (1) has a root $b$ such that +$b\not\equiv a \bmod p$. Then from (3) we have +\begin{gather*} +f(b) \equiv (b-a)^{\alpha}f_1(b) \bmod p. \\ +\intertext{But} +f(b)\equiv 0 \bmod p,\quad (b-a)^{\alpha} \not\equiv 0 \bmod p. \\ +\intertext{Hence, since $p$ is a prime number, we must have} +f_1(b)\equiv 0 \bmod p. +\end{gather*} + +By an argument similar to that just used above, we may show that +$f_1(x) - f_1(b)$ may be written in the form +\begin{gather*} +f_1(x) - f_1(b) = (x-b)^{\beta}f_2(x), \\ +\intertext{where $\beta$ is some positive integer. Then we have} +f(x) \equiv (x-a)^{\alpha}(x-b)^{\beta}f_2(x) \bmod p. +\end{gather*} + +Now this process can be continued until either all the solutions of +(1) are exhausted or the second member is a product of linear +factors multiplied by the integer $a_0$. In the former case there +will be fewer than $n$ solutions of (1), so that our theorem is true +for this case. In the other case we have +\begin{equation*} +f(x) \equiv a_0(x-a)^{\alpha}(x-b)^{\beta} + \ldots (x-l)^{\lambda} \bmod p. +\end{equation*} +We have now $n$ solutions of (1): $a$ counted $\alpha$ times, $b$ +counted $\beta$ times, \ldots, $l$ counted $\lambda$ times; $\alpha ++ \beta + \ldots +\lambda = n$. + +Now let $\eta$ be any solution of (1). Then +\begin{equation*} +f(\eta) \equiv a_0(\eta-a)^{\alpha}(\eta-b)^{\beta} \ldots + (\eta-l)^{\lambda} \equiv 0 \bmod p. +\end{equation*} +Since $p$ is prime it follows now that some one of the factors +$\eta-a, \eta-b, \ldots, \eta-l$ is divisible by $p$. Hence $\eta$ +coincides with one of the solutions $a, b, c, \ldots, l$. That is, +(1) can have only the $n$ solutions already found. + +This completes the proof of the theorem. + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Construct a congruence of the form +\begin{equation*} +a_0 x^n + a_1 x^{n-1} + \ldots + a_n \equiv 0 \bmod m, \quad + a_0 \not\equiv 0 \bmod m, +\end{equation*} +having more than $n$ solutions and thus show that the limitation to +a prime modulus in the theorem of this section is essential. + +\item[2.] Prove that +\begin{equation*} +x^6-1 \equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) \bmod 7 +\end{equation*} +for every integer $x$. + +\item[3.] How many solutions has the congruence $x^5 \equiv 1 \bmod +11$? the congruence $x^5\equiv 2 \bmod 11$? +\end{enumerate}\normalsize\index{Congruences!with prime modulus|)} + +\section{Linear Congruences}\label{s22}% +\index{Congruences!Linear|(} + +From the theorem of the preceding section it follows that the +congruence +\begin{equation*} +ax \equiv c \bmod p,\quad a \not\equiv 0 \bmod p, +\end{equation*} +where $p$ is a prime number, has not more than one solution. In this +section we shall prove that it always has a solution. More +generally, we shall consider the congruence +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +where $m$ is any integer. The discussion will be broken up into +parts for convenience in the proofs. + +\smallskip I.~\emph{The congruence} +\begin{equation} +ax \equiv 1 \bmod m, \tag{1} +\end{equation} +\emph{in which a and m are relatively prime, has one and only one +solution.} + +The question as to the existence and number of the solutions of (1) +is equivalent to the question as to the existence and number of +integer pairs $x, y$ satisfying the equation, +\begin{equation} +ax - my = 1, \tag{2} +\end{equation} +the integers $x$ being incongruent modulo $m$. Since $a$ and $m$ are +relatively prime it follows from theorem IV of \S~\ref{s9} that +there exists a solution of equation (2). Let $x = \alpha$ and $y = +\beta$ be a particular solution of (2) and let $x = \bar{\alpha}$ +and $y = \bar{\beta}$ be any solution of (2). Then we have +\begin{gather*} +a\alpha-m\beta = 1, \\ +a \bar{\alpha} - m\bar{\beta} = 1; \\ +\intertext{whence} +a(\alpha - \bar{\alpha}) - m(\beta - \bar{\beta}) = 0. +\end{gather*} +Hence $\alpha-\bar{\alpha}$ is divisible by $m$, since $a$ and $m$ +are relatively prime. That is, $a \equiv \bar{\alpha} \mod m$. Hence +$\alpha$ and $\bar{\alpha}$ are representatives of the same solution +of (1). Hence (1) has one and only one solution, as was to be +proved. + +\smallskip II.~\emph{The solution $x = \alpha$ of the congruence +$ax \equiv 1 \mod m$, in which $a$ and $m$ are relatively prime, is +prime to $m$.} + +For, if $a\alpha - 1$ is divisible by $m$, $\alpha$ is divisible by +no factor of $m$ except $1$. + +\smallskip III.~\emph{The congruence} +\begin{equation} +ax \equiv c \mod m \tag{3} +\end{equation} +\emph{in which $a$ and $m$ and also $c$ and $m$ are relatively +prime, has one and only one solution.} + +Let $x = \gamma$ be the unique solution of the congruence $cx = 1 +\mod m$. Then we have $a\gamma x \equiv c\gamma \equiv 1 \mod m$. +Now, by I we see that there is one and only one solution of the +congruence $a\gamma x \equiv 1 \mod m$; and from this the theorem +follows at once. + +Suppose now that $a$ is prime to $m$ but that $c$ and $m$ have the +greatest common divisor $\delta$ which is different from 1. Then it +is easy to see that any solution $x$ of the congruence $ax \equiv c +\mod m$ must be divisible by $\delta$. The question of the existence +of solutions of the congruence $ax \equiv c \bmod m$ is then +equivalent to the question of the existence of solutions of the +congruence +\begin{equation*} +a \frac{x}{\delta} \equiv \frac{c}{\delta} \bmod \frac{m}{\delta}, +\end{equation*} +where $\frac{x}{\delta}$ is the unknown integer. From III it follows +that this congruence has a unique solution $\frac{x}{\delta} = +\alpha$. Hence the congruence $ax \equiv c \bmod m$ has the unique +solution $x = \delta\alpha$. Thus we have the following theorem: + +\smallskip IV.~\emph{The congruence $ax \equiv c \bmod m$, in which +$a$ and $m$ are relatively prime, has one and only one solution.} + + +\smallskip\textsc{Corollary.}~\emph{The congruence +$ax \equiv c \bmod p$, $a \not\equiv 0 \bmod p$, where $p$ is a +prime number, has one and only one solution.} + +It remains to examine the case of the congruence $ax =c \bmod m$ in +which $a$ and $m$ have the greatest common divisor $d$. It is +evident that there is no solution unless $c$ also contains this +divisor $d$. Then let us suppose that $a = \alpha d$, $c = \gamma +d$, $m = \mu d$. Then for every $x$ such that $ax = c \bmod m$ we +have $\alpha x = \gamma \bmod \mu$; and conversely every $x$ +satisfying the latter congruence also satisfies the former. Now +$\alpha x = \gamma \bmod \mu$, has only one solution. Let $\beta$ be +a non-negative number less than $\mu$, which satisfies the +congruence $\alpha x = \gamma \bmod \mu$. All integers which satisfy +this congruence are then of the form $\beta + \mu\nu$, where $\nu$ +is an integer. Hence all integers satisfying the congruence $ax = c +\bmod m$ are of the form $\beta + \mu\nu$; and every such integer is +a representative of a solution of this congruence. It is clear that +the numbers +\begin{equation} +\beta,\ \beta + \mu,\ \beta + 2\mu,\ \ldots,\ \beta + (d-1)\mu +\tag{A} +\end{equation} +are incongruent modulo $m$ while every integer of the form $\beta + +\mu\nu$ is congruent modulo $m$ to a number of the set (A). Hence +the congruence $ax = c \bmod m$ has the $d$ solutions (A). + +This leads us to an important theorem which includes all the other +theorems of this section as special cases. It may be stated as +follows: + +\smallskip V.~\emph{Let} +\begin{equation*} +ax \equiv c \bmod m +\end{equation*} +\emph{be any linear congruence and let $a$ and $m$ have the greatest +common divisor $d (d \geq 1)$. Then a necessary and sufficient +condition for the existence of solutions of the congruence is that +$c$ be divisible by $d$. If this condition is satisfied the +congruence has just $d$ solutions, and all the solutions are +congruent modulo $m / d$.} + +\newpage +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Find the remainder when $2^{40}$ is divided by $31$; when +$2^{43}$ is divided by $31$. + +\item[2.] Show that $2^{2^5}+1$ has the factor $641$. + +\item[3.] Prove that a number is a multiple of $9$ if and only if the +sum of its digits is a multiple of $9$. + +\item[4.] Prove that a number is a multiple of $11$ if and only if the +sum of the digits in the odd numbered places diminished by the sum +of the digits in the even numbered places is a multiple of $11$. +\end{enumerate} \normalsize% +\index{Congruences|)}\index{Congruences!Linear|)} + +\chapter{THE THEOREMS OF FERMAT AND WILSON} + +\section{Fermat's General Theorem}\label{s23}% +\index{Fermat's!General Theorem} + +Let $m$ be any positive integer and let +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)} \tag{A} +\end{equation} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$. Let $a$ be any integer prime to $m$ and form the set +of integers +\begin{equation} +aa_1,\ aa_2,\ \ldots,\ aa_{\phi(m)} \tag{B} +\end{equation} +No number $aa_i$ of the set (B) is congruent to a number $aa_j$, +unless $j = i;$ for, from +\begin{equation*} +aa_i \equiv aa_j \bmod m +\end{equation*} +we have $a_i \equiv a_j \bmod m$; whence $a_i = a_j$ since both +$a_i$ and $a_j$ are positive and not greater than $m$. Therefore $j += i$. Furthermore, every number of the set (B) is congruent to some +number of the set (A). Hence we have congruences of the form +\begin{align*} +aa_1 & \equiv a_{i_1} \bmod m, \\ +aa_2 & \equiv a_{i_2} \bmod m, \\ + & \vdots \\ +aa_{\phi(m)} & \equiv a_{i_{\phi(m)}} \bmod m. +\end{align*} +No two numbers in the second members are equal, since $aa_i +\not\equiv aa_j$ unless $i= j$. Hence the numbers $a_{i_1},\ +a_{i_2},\ \ldots,\ a_{i_{\phi(m)}}$ are the numbers $a_1,\ a_2,\ +\ldots,\ a_{\phi(m)}$ in some order. Therefore, if we multiply the +above system of congruences together member by member and divide +each member of the resulting congruence by $a_1\cdot a_2\ldots +a_{\phi(m)}$ (which is prime to $m$), we have +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} +This result is known as Fermat's general theorem.% +\index{Fermat's!general theorem} It may be stated as follows: + +\emph{If $m$ is any positive integer and $a$ is any integer prime to +$m$, then} +\begin{equation*} +a^{\phi(m)} \equiv 1 \bmod m. +\end{equation*} + +\smallskip \textsc{Corollary 1.}~\emph{If $a$ is any integer +not divisible by a prime number $p$, then} +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +\smallskip \textsc{Corollary 2.}~\emph{If $p$ is any prime number +and $a$ is any integer, then} +\begin{equation*} +a^p \equiv a \bmod p. +\end{equation*} + +\section{Euler's Proof of the Simple Fermat Theorem}\label{s24}% +\index{Euler}\index{Fermat}\index{Fermat's!Simple Theorem} + +The theorem of Cor.\ 1, \S~\ref{s23}, is often spoken of as the +simple Fermat theorem. It was first announced by Fermat in 1679, but +without proof. The first proof of it was given by Euler in 1736. +This proof may be stated as follows: + +From the Binomial Theorem it follows readily that +\begin{gather*} +(a+1)^p \equiv a^p + 1 \bmod p \\ +\intertext{since} +\frac{p!}{r!(p-r)!}, \quad 0 < r < p, \\ +\intertext{is obviously divisible by $p$. Subtracting $a + 1$ from +each side of the foregoing congruence, we have} +(a+1)^p - (a+1) \equiv a^p - a \bmod p. +\end{gather*} +Hence if $a^p - a$ is divisible by $p$, so is $(a + 1)^p - (a + 1)$. +But $1^p - 1$ is divisible by $p$. Hence $2^p - 2$ is divisible by +$p$; and then $3^p - 3$; and so on. Therefore, in general, we have +\begin{equation*} +a^p \equiv a \mod p. +\end{equation*} +If $a$ is prime to $p$ this gives $a^{p-1} \equiv 1 \mod p$, as was +to be proved. + +If instead of the Binomial Theorem one employs the Polynomial +Theorem, an even simpler proof is obtained. For, from the latter +theorem, we have readily +\begin{gather*} +(\alpha_1 + \alpha_2 + \ldots + \alpha_a)^p \equiv + \alpha_1^p + \alpha_2^p + \ldots + \alpha_a^p \mod p. \\ +\intertext{Putting $\alpha_1 = \alpha_2 = \ldots = \alpha_a = 1$ we +have} +a^p\equiv a \mod p, +\end{gather*} +from which the theorem follows as before. + +\section{Wilson's Theorem}\label{s25}\index{Wilson's theorem|(} + +From the simple Fermat theorem it follows that the congruence +\begin{gather*} +x^{p-1} \equiv 1\mod p \\ +\intertext{has the $p-1$ solutions $1$, $2$, $3$, $\ldots$, $p-1$. +Hence from the discussion in \S \ref{s21} it follows that} +x^{p-1}-1 \equiv (x-1)(x-2)\ldots(x-\overline{p-1}) \mod p, \\ +\intertext{this relation being satisfied for every value of $x$. +Putting $x = 0$ we have} +(-1) = (-1)^{p - 1}\cdot 1\cdot 2\cdot 3 \ldots + \overline{p-1}\mod p. \\ +\intertext{If $p$ is an odd prime this leads to the congruence} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} + 1 = 0 \mod p. +\end{gather*} +Now for $p = 2$ this congruence is evidently satisfied. Hence +we have the Wilson theorem: + +\smallskip \emph{Every prime number $p$ satisfies the relation} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{p+1} + 1 \equiv 0 \mod p. +\end{equation*} + +An interesting proof of this theorem on wholly different principles +may be given. Let $p$ points be distributed at equal intervals on +the circumference of a circle. The whole number of $p$-gons which +can be formed by joining up these $p$ points in every possible order +is evidently +\begin{equation*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1; +\end{equation*} +for the first vertex can be chosen in $p$ ways, the second in $p - +1$ ways, $\ldots$, the $(p-1)^{\mathrm{th}}$ in two ways, and the +last in one way; and in counting up thus we have evidently counted +each polygon $2p$ times, once for each vertex and for each direction +from the vertex around the polygon. Of the total number of polygons +$\frac{1}{2}(p-1)$ are regular (convex or stellated) so that a +revolution through $\frac{360^\circ}{p}$ brings each of these into +coincidence with its former position. The number of remaining +$p$-gons must be divisible by $p$; for with each such $p$-gon we may +associate the $p-1$ $p$-gons which can be obtained from it by +rotating it through successive angles of $\frac{360^\circ}{p}$. That +is, +\begin{gather*} +\frac{1}{2p} p (p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - + \frac 12 (p-1) \equiv 0 \bmod p. \\ +\intertext{Hence} +(p-1) (p-2) \ldots 3 \cdot 2 \cdot 1 - p + 1 \equiv 0 \bmod p; \\ +\intertext{and from this it follows that} +1 \cdot 2 \ldots \overline{p-1} + 1 \equiv 0 \bmod p, \\ +\end{gather*} +as was to be proved. + +\section{The Converse of Wilson's Theorem}\label{s26} + +Wilson's theorem is noteworthy in that its converse is also true. +The converse may be stated as follows: + +\smallskip \emph{Every integer $n$ such that the congruence} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n +\end{equation*} +\emph{is satisfied is a prime number.} + +For, if $n$ is not prime, there is some divisor $d$ of $n$ different +from $1$ and less than $n$. For such a $d$ we have $1 \cdot 2 \cdot +3 \ldots \overline{n-1} \equiv 0 \bmod d$; so that $1 \cdot 2 \ldots +\overline{n-1}+1 \not\equiv 0 \bmod d$; and hence $1 \cdot 2 \ldots +\overline{n-1}+1 \equiv 0 \bmod n$. Since this contradicts our +hypothesis the truth of the theorem follows. + +\smallskip Wilson's theorem and its converse may be combined into +the following elegant theorem: + +\smallskip \emph{A necessary and sufficient condition that an +integer $n$ is prime is that} +\begin{equation*} +1 \cdot 2 \cdot 3 \ldots \overline{n-1} + 1 \equiv 0 \bmod n. +\end{equation*}\index{Prime numbers} + +Theoretically this furnishes a complete and elegant test as to +whether a given number is prime. But, practically, the labor of +applying it is so great that it is useless for verifying large +primes. + +\section{Impossibility of $1 \cdot 2 \cdot 3 \cdots +\overline{n-1} + 1 = n^k$ for $n > 5$.}\label{s27} + +In this section we shall prove the following theorem: + +\emph{There exists no integer $k$ for which the equation} +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-1} + 1 = n^k +\end{equation*} +is true when $n$ is greater than $5$. + +If $n$ contains a divisor $d$ different from $1$ and $n$, the +equation is obviously false; for the second member is divisible by +$d$ while the first is not. Hence we need to prove the theorem only +for primes $n$. + +Transposing $1$ to the second member and dividing by $n - 1$ we have +\begin{equation*} +1 \cdot 2 \cdot 3 \cdots \overline{n-2} = n^{k-1} + n^{k-2} + + \ldots + n+1. +\end{equation*} +If $n>5$ the product on the left contains both the factor $2$ and +the factor $\frac{1}{2} (n-1)$; that is, the first member contains +the factor $n - 1$. But the second member does not contain this +factor, since for $n = 1$ the expression $n^{k-1} + \ldots n + 1$ is +equal to $k \neq 0$. Hence the theorem follows at once. + +\section{Extension of Fermat's Theorem}\label{s28}% +\index{Fermat's!theorem extended|(} + +The object of this section is to extend Fermat's general theorem and +incidentally to give a new proof of it. We shall base this proof on +the simple Fermat theorem, of which we have already given a simple +independent proof. This theorem asserts that for every prime $p$ and +integer $a$ not divisible by $p$, we have the congruence +\begin{equation*} +a^{p-1} \equiv 1 \bmod p. +\end{equation*} + +Then let us write +\begin{gather} +a^{p-1} = 1 + hp. \tag{1} \\ +\intertext{Raising each member of this equation to the +$p^{\text{th}}$ power we may write the result in the form} +a^{p(p-1)} = 1 + h_1p^2. \tag{2} \\ +\intertext{where $h_1$ is an integer. Hence} +a^{p(p-1)} \equiv 1 \bmod p^2. \notag \\ +\intertext{By raising each member of (2) to the $p^{\text{th}}$ +power we can readily show that} +a^{p^2(p-1)} \equiv 1 \bmod p^3. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{p^{\alpha - 1}(p-1)} \equiv 1 \bmod p^{\alpha}. \notag \\ +\intertext{where $\alpha$ is a positive integer; that is,} +a^{\phi(p^{\alpha})} \equiv 1 \bmod p^{\alpha}. \notag +\end{gather} + +For the special case when $p$ is 2 this result can be extended. For +this case (1) becomes +\begin{gather} +a = 1 + 2h. \notag \\ +\intertext{Squaring we have} +a^2 = 1 + 4h(h+1). \notag \\ +\intertext{Hence,} +a^2 = 1+8h_1, \tag{3} \\ +\intertext{where $h_1$ is an integer. Therefore} +a^2 \equiv 1 \bmod 2^3. \notag \\ +\intertext{Squaring (3) we have} +a^{2^2} = 1 + 2^4h_2; \notag \\ +\intertext{or} +a^{2^2} \equiv 1 \bmod 2^4. \notag \\ +\intertext{It is now easy to see that we shall have in general} +a^{2^{\alpha-2}} \equiv 1 \bmod 2^{\alpha} \notag \\ +\intertext{if $\alpha > 2$. That is,} +a^{\frac{1}{2}\phi(2^{\alpha})} \equiv 1 \bmod 2^{\alpha} + \text{ if } a > 2. +\end{gather} + +Now in terms of the $\phi$-function let us define a new function +$\lambda(m)$ as follows: +\begin{align*} +\lambda(2^{\alpha}) &= \phi(2^{\alpha}) \text{ if $a = 0, 1, 2$;} \\ +\lambda(2^{\alpha}) &= \frac{1}{2}\phi(2^{\alpha}) + \text{ if $a > 2$;} \\ +\lambda(p^{\alpha}) &= \phi(p^{\alpha}) + \text{ if $p$ is an odd prime;} \\ +\lambda(2^{\alpha} p_1^{\alpha_1} p_2^{\alpha_2} \cdots + p_n^{\alpha_n}) &= M, +\end{align*} +where $M$ is the least common multiple of +\begin{equation*} + \lambda(2^{\alpha}), + \lambda(p_1^{\alpha_1}), + \lambda(p_2^{\alpha_2}), \ldots, \lambda(p_n^{\alpha_n}), +\end{equation*} +$2, p_1, p_2, \ldots, p_n$ being different primes.% +\index{$\lambda(m)$} + +Denote by $m$ the number +\begin{equation*} +m = 2^{\alpha}p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_n^{\alpha_n}. +\end{equation*} +Let $a$ be any number prime to $m$. From our preceding results we +have +\begin{align*} +a^{\lambda(2^{\alpha})} &\equiv 1 \bmod 2^{\alpha}, \\ +a^{\lambda(p_1^{\alpha_1})} &\equiv 1 \bmod p_1^{\alpha_1},\\ +a^{\lambda(p_2^{\alpha_2})} &\equiv 1 \bmod p_2^{\alpha_2}, \\ +\ldots \\ +a^{\lambda(p_n^{\alpha_n})} &\equiv 1 \bmod p_2^{\alpha_n}. \\ +\end{align*} + +Now any one of these congruences remains true if both of its members +are raised to the same positive integral power, whatever that power +may be. Then let us raise both members of the first congruence to +the power $\frac{\lambda(m)}{\lambda(2^\alpha)}$; both members of +the second congruence to the power +$\frac{\lambda(m)}{\lambda(p_1^{\alpha_1})}$; $\ldots$; both members +of the last congruence to the power +$\frac{\lambda(m)}{\lambda(p_n^{\alpha_n})}$. Then we have +\begin{align*} +a^{\lambda(m)} &\equiv 1 \mod 2^\alpha, \\ +a^{\lambda(m)} &\equiv 1 \mod p_1^{\alpha_1}, \\ +\ldots \ldots \\ +a^{\lambda(m)} &\equiv 1 \mod p_n^{\alpha_n}. \\ +\intertext{From these congruences we have immediately} +a^{\lambda(m)} &\equiv 1 \mod m. +\end{align*} + +We may state this result in full in the following theorem: + +\smallskip \emph{If $a$ and $m$ are any two relatively prime positive +integers, the congruence} +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +\emph{is satisfied.} + +As an excellent example to show the possible difference between the +exponent $\lambda(m)$ in this theorem and the exponent $\phi(m)$ in +Fermat's general theorem, let us take +\begin{gather*} +m = 2^6 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 19 + \cdot 37 \cdot 73. \\ +\intertext{Here} +\lambda(m) = 2^4 \cdot 3^2, \quad \phi(m) = 2^{31} \cdot 3^{10}. +\end{gather*} + +In a later chapter we shall show that there is no exponent $\nu$ +less than $\lambda(m)$ for which the congruence +\begin{equation*} +a^\nu = 1 \mod m +\end{equation*} +is verified for every integer $a$ prime to $m$. + +From our theorem, as stated above, Fermat's general theorem follows +as a corollary, since $\lambda(m)$ is obviously a factor of +$\phi(m)$, +\begin{equation*} +\phi(m) = \phi(2^\alpha) \phi(p_1^{\alpha_1}) \ldots + \phi(p_n^{\alpha_n}). +\end{equation*} + +\begin{center} +EXERCISES +\end{center} + +\small \begin{enumerate} +\item[1.] Show that $a^{16} \equiv 1 \bmod 16320$, for every $a$ +which is prime to $16320$. + +\item[2.] Show that $a^{12} \equiv 1 \bmod 65520$, for every $a$ which +is prime to $65520$. + +\item[3*.] Find one or more composite numbers $P$ such that +\begin{equation*} +a^{P-1} \equiv 1 \bmod P +\end{equation*} +for every a prime to $P$. (Compare this problem with the next +section.) \end{enumerate} \normalsize% +\index{Fermat's!theorem extended|)} + +\section{On the Converse of Fermat's Simple Theorem}\label{s29}% +\index{Fermat's!Simple Theorem} + +The fact that the converse of Wilson's theorem is a true proposition +leads one naturally to inquire whether the converse of Fermat's +simple theorem is true. Thus, we may ask the question: Does the +existence of the congruence $2^{n-1} \equiv 1 \bmod n$ require that +$n$ be a prime number? The Chinese answered this question in the +affirmative and the answer passed unchallenged among them for many +years. An example is sufficient to show that the theorem is not +true. We shall show that +\begin{equation*} +2^{340} \equiv 1 \bmod 341 +\end{equation*} +although $341 = 11 \cdot 31$, is not a prime number. Now $2^{10}-1 = +3 \cdot 11 \cdot 31$. Hence $2^{10} \equiv 1 \bmod 341$. Hence +$2^{340} \equiv 1 \bmod 341$. From this it follows that the direct +converse of Fermat's theorem is not true. The following theorem, +however, which is a converse with an extended hypothesis, is readily +proved. + +\smallskip \emph{If there exists an integer $a$ such that} +\begin{equation*} +a^{n-1} \equiv 1 \bmod n +\end{equation*} +\emph{and if further there does not exist an integer $\nu$ less than +$n - 1$ such that} +\begin{equation*} +a^{\nu} \equiv 1 \bmod n, +\end{equation*} +\emph{then the integer $n$ is a prime number.} + +For, if $n$ is not prime, $\phi(n) < n - 1$. Then for $\nu = +\phi(n)$ we have $a^{\nu} \equiv 1 \bmod n$, contrary to the +hypothesis of the theorem. + +\section{Application of Previous Results to Linear +Congruences}\label{s30}% +\index{Congruences!Linear} + +The theorems of the present chapter afford us a ready means of +writing down a solution of the congruence +\begin{equation} +ax \equiv c \bmod m. \tag{1} +\end{equation} +We shall consider only the case in which $a$ and $m$ are relatively +prime, since the general case is easily reducible to this one, as we +saw in the preceding chapter. + +Since $a$ and $m$ are relatively prime we have the congruences +\begin{gather*} +a^{\lambda(m)} \equiv 1,\quad a^{\phi(m)} \equiv 1 \bmod m. \\ +\intertext{Hence either of the numbers $x$,} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +is a representative of the solution of (1). Hence the following +theorem: + +\smallskip \emph{If} +\begin{gather*} +ax \equiv c \bmod m \\ +\intertext{\emph{is any linear congruence in which $a$ and $m$ are +relatively prime, then either of the numbers $x$,}} +x = ca^{\lambda(m)-1},\quad x = ca^{\phi(m)-1}, +\end{gather*} +\emph{is a representative of the solution of the congruence.} + +The former representative of the solution is the more convenient of +the two, since the power of $a$ is in general much less in this case +than in the other. + +\begin{center} +EXERCISE +\end{center} + +\small \begin{enumerate} +\item[ ] Find a solution of $7x \equiv 1 \bmod 2^6 \cdot 3 \cdot 5 \cdot +17.$ Note the greater facility in applying the first of the above +representatives of the solution rather than the second. +\end{enumerate} \normalsize + +\section{Application of the Preceding Results to the Theory +of Quadratic Residues}\label{s31}\index{Quadratic residues|(} + +In this section we shall apply the preceding results of this chapter +to the problem of finding the solutions of congruences of the form +\begin{equation*} +\alpha z^2 + \beta z + \gamma \equiv 0 \mod \mu +\end{equation*} +where $\alpha, \beta, \gamma, \mu$ are integers. These are called +quadratic congruences. + +The problem of the solution of the quadratic congruence (1) can be +reduced to that of the solution of a simpler form of congruence as +follows: Congruence (1) is evidently equivalent to the congruence +\begin{gather} +4\alpha^2 z^2 + 4\alpha\beta z + 4\alpha\gamma \equiv + 0 \mod 4\alpha\mu. \tag{1} \\ +\intertext{But this may be written in the form} +(2\alpha z + \beta)^2 \equiv \beta^2 - 4\alpha\gamma + \mod 4\alpha\mu. \notag \\ +\intertext{Now if we put} +2\alpha z + \beta\equiv x \mod 4\alpha\mu \tag{2} \\ +\intertext{and} +\beta^2 - 4\alpha\gamma = a,\quad 4\alpha\mu = m, \notag \\ +\intertext{we have} +x^2 \equiv a\mod m. \tag{3} +\end{gather} +We have thus reduced the problem of solving the general congruence +(1) to that of solving the binomial congruence (3) and the linear +congruence (2). The solution of the latter may be effected by means +of the results of \S \ref{s30}. We shall therefore confine ourselves +now to a study of congruence (3). We shall make a further limitation +by assuming that $a$ and $m$ are relatively prime, since it is +obvious that the more general case is readily reducible to this one. + +The example +\begin{equation*} +x^2 \equiv 3 \mod 5 +\end{equation*} +shows at once that the congruence (3) does not always have a +solution. First of all, then, it is necessary to find out in what +cases (3) has a solution. Before taking up the question it will be +convenient to introduce some definitions. + +\smallskip\textsc{Definitions.} An integer $a$ is said to be a +quadratic residue modulo $m$ or a quadratic non-residue modulo $m$ +according as the congruence +\begin{equation*} +x^2 = a \mod m +\end{equation*} +has or has not a solution. We shall confine our attention to the +case when $m > 2$.\index{Residue} + +We shall now prove the following theorem: + +\smallskip I.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\lambda(m)} \equiv 1 \mod m. +\end{equation*} + +Suppose that the congruence $x^2 \equiv a \mod m$ has the solution $x = +\alpha$. Then $\alpha^2 \equiv a \mod m$. Hence +\begin{equation*} +\alpha^{\lambda(m)} \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +Since $a$ is prime to $m$ it is clear from $\alpha^2 \equiv a \mod +m$ that $\alpha$ is prime to $m$. Hence $\alpha^{\lambda(m)} \equiv 1 +\mod m$. Therefore we have +\begin{equation*} +1 \equiv a^{\frac{1}{2}\lambda(m)} \mod m. +\end{equation*} +That is, this is a necessary condition in order that $a$ shall be a +quadratic residue modulo $m$. + +In a similar way one may prove the following theorem: + +\smallskip II.~\emph{If $a$ and $m$ are relatively prime integers, a +necessary condition that $a$ is a quadratic residue modulo $m$ is +that} +\begin{equation*} +a^{\frac{1}{2}\phi(m)} \equiv 1 \mod m. +\end{equation*} + +When $m$ is a prime number $p$ each of the above results takes the +following form: If $a$ is prime to $p$ and is a quadratic residue +modulo $p$, then +\begin{equation*} +a^{\frac{1}{2}(p-1)} \equiv 1 \mod p. +\end{equation*} +We shall now prove the following more complete theorem, without the +use of I or II. + +\smallskip III.~\emph{If $p$ is an odd prime number and $a$ is an +integer not divisible by $p$, then $a$ is a quadratic residue or a +quadratic non-residue modulo $p$ according as} +\begin{equation*} +a^{\tfrac{1}{2}(p-1)} \equiv +1 \quad \text{or} \quad +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p. +\end{equation*} + +This is called Euler's criterion.\index{Euler's!criterion} + +Given a number $a$, not divisible by $p$, we have to determine +whether or not the congruence +\begin{gather} +x^2 \equiv a \bmod p \notag \\ +\intertext{has a solution. Let $r$ be any number of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 \tag{A} \\ +\intertext{and consider the congruence} +rx \equiv a \bmod p. +\end{gather} +This has always one and just one solution $x$ equal to a number $s$ +of the set (A). Two cases can arise: either for every $r$ of the set +(A) the corresponding $s$ is different from $r$ or for some $r$ of +the set (A) the corresponding $s$ is equal to $r$. The former is the +case when $a$ is a quadratic non-residue modulo $p$; the latter is +the case when $a$ is a quadratic residue modulo $p$. We consider the +two cases separately. + +In the first case the numbers of the set (A) go in pairs such that +the product of the numbers in the pair is congruent to a modulo $p$. +Hence, taking the product of all $\tfrac{1}{2}(p - 1)$ pairs, we +have +\begin{align*} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &\equiv + +a^{\tfrac{1}{2}(p-1)} \bmod p. \\ +\intertext{But} +1 \cdot 2 \cdot 3 \ldots \overline{p-1} &= -1 \bmod p. \\ +\intertext{Hence} +a^{\tfrac{1}{2}(p-1)} \equiv -1 \bmod p, +\end{align*} +whence the truth of one part of the theorem. + +In the other case, namely that in which some $r$ and corresponding +$s$ are equal, we have for this $r$ +\begin{gather*} +r^{2} \equiv a \bmod p \\ +\intertext{and} +(p - r)^{2} \equiv a \bmod p. +\end{gather*} +Since $x^{2} \equiv a \bmod p$ has at most two solutions it follows +that all the integers in the set (A) except $r$ and $p - r$ fall in +pairs such that the product of the numbers in each pair is congruent +to a modulo $p$. Hence, taking the product of all these pairs, which +are $\frac{1}{2}(p - 1) - 1$ in number, and multiplying by $r(p-r)$ +we have +\begin{align*} +1 \cdot 2 \cdot 3 \cdots \overline{p -1} + &\equiv (p - r) r a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -r^{2} a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a a^{\frac{1}{2}(p -1) - 1} \bmod p \\ + &\equiv -a^{\frac{1}{2}(p -1)} \bmod p. \\ +\intertext{Since $1 \cdot 2 \cdot 3 \cdots \overline{p - 1} \equiv -1 +\bmod p$ we have} +a^{\frac{1}{2}(p -1)} &\equiv + 1 \bmod p +\end{align*} +whence the truth of another part of the theorem. + +Thus the proof of the entire theorem is complete.% +\index{Quadratic residues|)}\index{Wilson's theorem|)} + +\chapter{PRIMITIVE ROOTS MODULO $m$.} + +\section{Exponent of an Integer Modulo $m$}\label{s32}% +\index{Exponent of an integer|(}\index{Primitive roots|(} + +Let +\begin{equation*} +a_{1},\ a_{2},\ \cdots,\ a_{\phi(m)} \tag{A} +\end{equation*} +be the set of $\phi(m)$ positive integers not greater than $m$ and +prime to $m$; and let $a$ denote any integer of the set (A). Now any +positive integral power of $a$ is prime to $m$ and hence is +congruent modulo $m$ to a number of the set (A). Hence, among all +the powers of a there must be two, say $a^{n}$ and $a^{\nu}$, $n > +\nu$, which, are congruent to the same integer of the set (A). These +two powers are then congruent to each other; that is, +\begin{equation*} +a^{n} \equiv a^{\nu} \bmod m +\end{equation*} +Since $a^{\nu}$ is prime to $m$ the members of this congruence may +be divided by $a^{\nu}$. Thus we have +\begin{equation*} +a^{n - \nu} \equiv 1 \bmod m. +\end{equation*} +That is, among the powers of $a$ there is one at least which is +congruent to $1$ modulo $m$. + +\smallskip Now, in the set of all powers of $a$ which are congruent +to $1$ modulo $m$ there is one in which the exponent is less than in +any other of the set. Let the exponent of this power be $d$, so that +$a^{d}$ is the lowest power of $a$ such that +\begin{equation} +a^{d} \equiv 1 \bmod m. \tag{1} +\end{equation} + +We shall now show that if $a^{\alpha} \equiv 1 \bmod m$, then +$\alpha$ is a multiple of $d$. Let us write +\begin{gather} +\alpha = d\delta + \beta, \quad 0 \leqq \beta < d. \notag \\ +\intertext{Then} +a^{\alpha} \equiv 1 \bmod m, \tag{2} \\ +a^{d\delta} \equiv 1 \bmod m, \tag{3} \\ +\intertext{the last congruence being obtained by raising (1) to the +power $\delta$. From (3) we have} +a^{d\delta + \beta} \equiv a^{\beta} \bmod m; \notag \\ +\intertext{or} +a^{\beta}\equiv 1 \bmod m. \notag +\end{gather} +Hence $\beta = 0$, for otherwise $d$ is not the exponent of the +lowest power of $a$ which is congruent to 1 modulo $m$. Hence $d$ is +a divisor of $\alpha$. + +\smallskip These results may be stated as follows: + +\smallskip I.~\emph{If $m$ is any integer and $a$ is any integer +prime to $m$, then there exists an integer $d$ such that} +\begin{gather*} +a^d\equiv 1 \bmod m \\ +\intertext{\emph{while there is no integer $\beta$ less than $d$ for +which}} +a^\beta\equiv 1 \bmod m. \\ +\intertext{\emph{Further, a necessary and sufficient condition +that}} +a^\nu \equiv 1 \bmod m +\end{gather*} +\emph{is that $\nu$ is a multiple of $d$.} + +\smallskip \textsc{Definition.} The integer $d$ which is thus +uniquely determined when the two relatively prime integers $a$ and +$m$ are given is called the exponent of $a$ modulo $m$. Also, $d$ is +said to be the exponent to which $a$ belongs modulo $m$. + +Now, in every case we have +\begin{equation*} +a^{\phi(m)} \equiv 1,\quad a^{\lambda(m)} \equiv 1 \bmod m, +\end{equation*} +if $a$ and $m$ are relatively prime. Hence from the preceding +theorem we have at once the following: + +\smallskip II.~\textit{The exponent $d$ to which $a$ belongs modulo +$m$ is a divisor of both $\phi(m)$ and $\lambda(m)$.}% +\index{Exponent of an integer|)} + +\section{Another Proof of Fermat's General Theorem}\label{s33} + +In this section we shall give an independent proof of the theorem +that the exponent $d$ of $a$ modulo $m$ is a divisor of $\phi(m)$; +from this result we have obviously a new proof of Fermat's theorem +itself. + +We retain the notation of the preceding section. We shall first +prove the following theorem: + +\smallskip I.~\textit{The numbers} +\begin{equation} +1,\ a,\ a^2,\ \ldots,\ a^{d-1} \tag{A} +\end{equation} +\textit{are incongruent each to each modulo $m$.} + +For, if $a^\alpha \equiv a^\beta \bmod m$, where $0 \leqq \alpha < +d$ and $0 \leqq \beta < d$, $\alpha > \beta$, we have +$a^{\alpha-\beta} \equiv 1 \bmod m$, so that $d$ is not the exponent +to which $a$ belongs modulo $m$, contrary to hypothesis. + +\smallskip Now any number of the set (A) is congruent to some number +of the set +\begin{equation} +a_1,\ a_2,\ \ldots,\ a_{\phi(m)}. \tag{B} +\end{equation} +Let us undertake to separate the numbers (B) into classes after the +following manner: Let the first class consist of the numbers +\begin{equation} +\alpha_1,\ \alpha_2,\ \ldots,\ \alpha_{d-1}, \tag{I} +\end{equation} +where $\alpha_i$ is the number of the set (B) to which $a^i$ is +congruent modulo $m$. + +If the class (I) does not contain all the numbers of the set (B), +let $a_i$ be any number of the set (B) not contained in (I) and form +the following set of numbers: +\begin{equation} +\alpha_0 a_i,\ \alpha_1 a_i,\ \alpha_2 a_i,\ \ldots,\ + \alpha_{d-1}a_i. \tag{II'} +\end{equation} +We shall now show that no number of this set is congruent to a +number of class (I). For, if so, we should have a congruence of the +form +\begin{gather*} +a_i \alpha_j \equiv \alpha_k \bmod m; \\ +\intertext{hence} +a_i a^j \equiv a^k \bmod m, \\ +\intertext{so that} +a_i a^d \equiv a^{k+d-j} \bmod m; \\ +\intertext{or} +a_i \equiv a^{k+d-j} \bmod m, +\end{gather*} +so that $a_i$ would belong to the set (I) contrary to hypothesis. + +Now the numbers of the set (II$'$) are all congruent to numbers of +the set (B); and no two are congruent to the same number of this +set. For, if so, we should have two numbers of (II') congruent; that +is, $\alpha_k a_i \equiv \alpha_j a_i \bmod m,$ or $\alpha_k \equiv +\alpha_j \bmod m;$ and this we have seen to be impossible. + +Now let the numbers of the set (B) to which the numbers of the set +(II') are congruent be in order the following: +\begin{equation} +\beta_0,\ \beta_1,\ \beta_2,\ \ldots,\ \beta_{d-1}. \tag{II} +\end{equation} +These numbers constitute our class (II). + +If classes (I) and (II) do not contain all the numbers of the set +(B), let $a_j$ be a number of the set ($B$) not contained in either +of the classes (I) and (II): and form the set of numbers +\begin{equation} +\alpha_0 a_j,\ \alpha_1 a_j,\ \alpha_2 a_j,\ \ldots,\ + \alpha_{d-1} a_j. \tag{III'} +\end{equation} +Just as in the preceding case it may be shown that no number of this +set is congruent to a number of class (I) and that the numbers of +(III') are incongruent each to each. We shall also show that no +number of (III') is congruent to a number of class (II). For, if so, +we should have $a_k a_j \equiv \beta_l \bmod m$. Hence $a^k a_j +\equiv a^l a_i \bmod m$; or $a_j \equiv a^{l+d-k} a_i \bmod m$, from +which it follows that $a_j$ is of class (II), contrary to +hypothesis. + +Now let the numbers of the set (B) to which the numbers of the set +(III') are congruent be in order the following: +\begin{equation} +\gamma_0,\ \gamma_1,\ \gamma_2,\ \ldots,\ \gamma_{d-1}. \tag{III} +\end{equation} +These numbers form our class (III). + +It is now evident that the process may be continued until all the +numbers of the set (B) have been separated into classes, each class +containing $d$ integers, thus: +\begin{equation*} +\begin{matrix} +(\text{I}) & \alpha_0, & \alpha_1, & \alpha_2, + & \ldots, & \alpha_{d-1}, \\ +(\text{II}) & \beta_0, & \beta_1, & \beta_2, + & \ldots, & \beta_{d-1}, \\ +(\text{III}) & \gamma_0, & \gamma_1, & \gamma_2, + & \ldots, & \gamma_{d-1}, \\ +&\hdotsfor{5} \\ +(\quad ) & \lambda_0, & \lambda_1, & \lambda_2, + & \ldots, & \lambda_{d-1}. +\end{matrix} +\end{equation*} +The set (B), which consists of $\phi(m)$ integers, has thus been +separated into classes, each class containing $d$ integers. Hence we +conclude that $d$ is a divisor of $\phi(m)$. Thus we have a second +proof of the theorem: + +\smallskip II.~\emph{If $a$ and $m$ are any two relatively prime +integers and $d$ is the exponent to which $a$ belongs modulo $m$, +then $d$ is a divisor of $\phi(m)$.} + +In our classification of the numbers (B) into the rectangular array +above we have proved much more than theorem II; in fact, theorem II +is to be regarded as one only of the consequences of the more +general result contained in the array. + +If we raise each member of the congruence +\begin{equation*} +a^d \equiv 1 \bmod m +\end{equation*} +to the (integral) power $\phi(m)/d$, the preceding theorem leads +immediately to an independent proof of Fermat's general theorem. + +\section{Definition of Primitive Roots}\label{s34} + +\textsc{Definition.} Let $a$ and $m$ be two relatively prime +integers. If the exponent to which $a$ belongs modulo $m$ is +$\phi(m)$, $a$ is said to be a primitive root modulo $m$ (or a +primitive root of $m$). + +In a previous chapter we saw that the congruence +\begin{equation*} +a^{\lambda(m)} \equiv 1 \bmod m +\end{equation*} +is verified by every pair of relatively prime integers $a$ and $m$. +Hence, primitive roots can exist only for such a modulus $m$ as +satisfies the equation +\begin{equation*} +\phi(m) = \lambda(m). \tag{1} +\end{equation*} +We shall show later that this is also sufficient for the existence +of primitive roots. + +From the relation which exists in general between the +$\phi$-function and the $\lambda$-function in virtue of the +definition of the latter, it follows that (1) can be satisfied only +when $m$ is a prime power or is twice an odd prime power. + +Suppose first that $m$ is a power of $2$, say $m = 2^\alpha$. Then +(1) is satisfied only if $\alpha = 0,\ 1,\ 2$. For $\alpha = 0$ or +$1$, $1$ itself is a primitive root. For $\alpha = 2$, $3$ is a +primitive root. We have therefore left to examine only the cases +\begin{equation*} +m = p^\alpha,\quad m = 2p^\alpha +\end{equation*} +where $p$ is an odd prime number. The detailed study of these cases +follows in the next sections. + +\section{Primitive roots modulo $p$.}\label{s35} + +We have seen that if $p$ is a prime number and $d$ is the exponent +to which $a$ belongs modulo $p$, then $d$ is a divisor of $\phi(p) = +p - 1$. Now, let +\begin{gather*} +d_1,\ d_2,\ d_3,\ \ldots,\ d_r \\ +\intertext{be all the divisors of $p-1$ and let $\psi(d_i)$ denote +the number of integers of the set} +1,\ 2,\ 3,\ \ldots,\ p-1 +\end{gather*} +which belong to the exponent $d_i$. If there is no integer of the +set belonging to this exponent, then $\psi(d_i) = 0$. + +Evidently every integer of the set belongs to some one and only one +of the exponents $d_1, d_2, \ldots, d_r$. Hence we have the relation +\begin{gather} +\psi(d_1) + \psi(d_2) + \ldots + \psi(d_r) = p-1. \tag{1} \\ +\intertext{But} +\phi(d_1) + \phi(d_2) + \ldots + \phi(d_r) = p-1. \tag{2} \\ +\intertext{If then we can show that} +\psi(d_i) \leqq \phi(d_i) \tag{3} \\ +\intertext{for $i = 1, 2, \ldots, r$, it will follow from a +comparison of (1) and (2) that} +\psi(d_i) = \phi(d_i). \notag +\end{gather} +Accordingly, we shall examine into the truth of (3). + +Now the congruence +\begin{equation} +x^{d_i} \equiv 1 \mod p \tag{4} +\end{equation} +has not more than $d_i$ roots. If no root of this congruence belongs +to the exponent $d_i$, then $\psi(d_i) = 0$ and therefore in this +case we have $\psi(d_i) < \phi(d_i)$. On the other hand if $a$ is a +root of (4) belonging to the exponent $d_i$, then +\begin{equation} +a, a^2, a^3, \ldots, a^{d_i} \tag{5} +\end{equation} +are a set of $d_i$ incongruent roots of (4); and hence they are the +complete set of roots of (4). + +But it is easy to see that $a^k$ does or does not belong to the +exponent $d_i$ according as $k$ is or is not prime to $d_i$; for, if +$a^k$ belongs to the exponent $t$, then $t$ is the least integer +such that $kt$ is a multiple of $d_i$. Consequently the number of +roots in the set (5) belonging to the exponent $d_i$ is $\phi(d_i)$. +That is, in this case $\psi(d_i) = \phi(d_i)$. Hence in general +$\psi(d_i) \leqq \phi(d_i)$. Therefore from (1) and (2) we conclude +that +\begin{equation*} +\psi(d_i) = \phi(d_i), \quad i = 1,\ 2,\ \ldots,\ r. +\end{equation*} +The result thus obtained may be stated in the form of the following +theorem: + +\smallskip I.~\emph{If $p$ is a prime number and $d$ is any divisor +of $p-1$, then the number of integers belonging to the exponent $d$ +modulo $p$ is $\phi(d)$.} + +In particular: + +\smallskip II.~\emph{There exist primitive roots modulo $p$ and their +number is $\phi(p-1)$.} + +\section{Primitive Roots Modulo $p^\alpha$, $p$ an Odd +Prime}\label{s36} + +In proving that there exist primitive roots modulo $p^\alpha$, where +$p$ is an odd prime and $\alpha > 1$, we shall need the following +theorem: + +I.~\emph{There always exists a primitive root $\gamma$ modulo $p$ +for which $\gamma^{p-1}$ is not divisible by $p^2$.} + +Let $g$ be any primitive root modulo $p$. If $g^{p-1}-1$ is not +divisible by $p^2$ our theorem is verified. Then suppose that +$g^{p-1}-1$ is divisible by $p^2$, so that we have +\begin{gather*} +g^{p-1}-1 = kp^2 \\ +\intertext{where $k$ is an integer. Then put} +\gamma \equiv g + xp \\ +\intertext{where $x$ is an integer. Then $\gamma \equiv g \mod p$, and +hence} +\gamma^h \equiv g^h \mod p; +\end{gather*} +whence we conclude that $\gamma$ is a primitive root modulo $p$. But +\begin{align*} +\gamma^{p-1}-1 &= + g^{p-1} - 1 + \frac{p-1}{1!}g^{p-2}xp + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p^2 + \ldots \\ + &= p\left(kp + \frac{p-1}{1!}g^{p-2}x + + \frac{(p-1)(p-2)}{2!}g^{p-3}x^2p + \ldots\right). +\end{align*} +Hence +\begin{equation*} +\gamma^{p-1}-1 \equiv p(-g^{p-2}x) \mod p^2. +\end{equation*} +Therefore it is evident that $x$ can be so chosen that +$\gamma^{p-1}-1$ is not divisible by $p^2$. Hence there exists a +primitive root $\gamma$ modulo $p$ such that $\gamma^{p-1}-1$ is not +divisible by $p^2$. Q.~E.~D. + +\smallskip We shall now prove that this integer $\gamma$ is a +primitive root modulo $p^\alpha$, where $\alpha$ is any positive +integer. + +If +\begin{equation*} +\gamma^k \equiv 1\mod p, +\end{equation*} +then $k$ is a multiple of $p-1$, since $\gamma$ is a primitive root +modulo $p$. Hence, if +\begin{equation*} +\gamma^k \equiv 1 \mod p^\alpha, +\end{equation*} +then $k$ is a multiple of $p-1$. + +Now, write +\begin{equation*} +\gamma^{p-1} = 1 + hp. +\end{equation*} +Since $\gamma^{p-1}-1$ is not divisible by $p^2$, it follows that $h$ +is prime to $p$. If we raise each member of this equation to the +power $\beta p^{\alpha-2}$, $\alpha \stackrel{=}{>}2$, we have +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} = + 1 + \beta p^{\alpha-1}h + p^\alpha I, +\end{equation*} +where $I$ is an integer. Then if +\begin{equation*} +\gamma^{\beta p^{\alpha-2}(p-1)} \equiv 1 \mod p^\alpha, +\end{equation*} +$\beta$ must be divisible by $p$. Therefore the exponent of the +lowest power of $\gamma$ which is congruent to $1$ modulo $p^\alpha$ +is divisible by $p^{\alpha-1}$. But we have seen that this exponent +is also divisible by $p-1$. Hence the exponent of $\gamma$ modulo +$p^\alpha$ is $p^{\alpha-1}(p-1)$ since $\phi(p^\alpha) = +p^{\alpha-1}(p-1)$. That is, $\gamma$ is a primitive root modulo +$p^\alpha$. + +It is easy to see that no two numbers of the set +\begin{equation} +\gamma, \gamma^2, \gamma^3, \ldots, \gamma^{p^{\alpha-1}(p-1)} +\tag{A} +\end{equation} +are congruent modulo $p^\alpha$; for, if so, $\gamma$ would belong +modulo $p^\alpha$ to an exponent less than $p^{\alpha-1}(p-1)$ and +would therefore not be a primitive root modulo $p^\alpha$. Now every +number in the set (A) is prime to $p^\alpha$; their number is +$\phi(p^\alpha) = p^{\alpha -1}(p-1)$. Hence the numbers of the set +(A) are congruent in some order to the numbers of the set (B): +\begin{equation} +a_1,\ a_2,\ a_3,\ \ldots ,\ a_{p^{\alpha-1}(p-1)}, \tag{B} +\end{equation} +where the integers (B) are the positive integers less than +$p^\alpha$ and prime to $p^\alpha$. + +But any number of the set (B) is a solution of the congruence +\begin{equation} +x^{p^{\alpha-1} (p-1)} \equiv 1 \bmod p^\alpha. \tag{1} +\end{equation} +Further, every solution of this congruence is prime to $p^\alpha$. +Hence the integers (B) are a complete set of solutions of (1). +Therefore the integers (A) are a complete set of solutions of (1). +But it is easy to see that an integer $\gamma^k$ of the set (A) is +or is not a primitive root modulo $p^\alpha$ according as $k$ is or +is not prime to $p^{\alpha-1} (p-1)$. Hence the number of primitive +roots modulo $p^\alpha$ is $\phi \{p^{\alpha-1} (p-1) \}.$ + +The results thus obtained may be stated as follows: + +\smallskip II.~\emph{If $p$ is any odd prime number and $\alpha$ is +any positive integer, then there exist primitive roots modulo +$p^\alpha$ and their number is $\phi \{ \phi(p^\alpha) \}$}. + +\section{Primitive Roots Modulo $2p^\alpha$, $p$ an Odd +Prime}\label{s37} + +In this section we shall prove the following theorem: + +\emph{If $p$ is any odd prime number and $\alpha$ is any positive +integer, then there exist primitive roots modulo $2p^\alpha$ and +their number is $\phi \{\phi(2 p^{\alpha} )\}.$} + +Since $2 p^\alpha$ is even it follows that every primitive root +modulo $2 p^\alpha$ is an odd number. Any odd primitive root modulo +$p^\alpha$ is obviously a primitive root modulo $2p^\alpha$. Again, +if $\gamma$ is an even primitive root modulo $p^\alpha$ then $\gamma ++ p^\alpha$ is a primitive root modulo $2 p^\alpha$. It is evident +that these two classes contain (without repetition) all the +primitive roots modulo $2 p^\alpha$. Hence the theorem follows as +stated above. + +\section{Recapitulation}\label{s38} + +The results which we have obtained in \S\S \ref{s34}--\ref{s37} +inclusive may be gathered into the following theorem: + +\emph{In order that there shall exist primitive roots modulo $m$, it +is necessary and sufficient that $m$ shall have one of the values} +\begin{equation*} +m = 1, 2, 4, p^\alpha, 2p^\alpha +\end{equation*} +\emph{where $p$ is an odd prime and $\alpha$ is a positive integer.} + +\emph{If $m$ has one of these values then the number of primitive +roots modulo $m$ is $\phi\{\phi(m)\}$.} + +\section{Primitive $\lambda$-roots}\label{s39}% +\index{Primitive roots!$\lambda$-roots|(} + +In the preceding sections of this chapter we have developed the +theory of primitive roots in the way in which it is usually +presented. But if one approaches the subject from a more general +point of view the results which may be obtained are more general and +at the same time more elegant. It is our purpose in this section to +develop the more general theory. + +\smallskip We have seen that if $a$ and $m$ are any two relatively +prime positive integers, then +\begin{equation*} +a^{\lambda(m)} \equiv 1 \mod m. +\end{equation*} +Consequently there is no integer belonging modulo $m$ to an exponent +greater than $\lambda(m)$. It is natural to enquire if there are any +integers $a$ which belong to the exponent $\lambda(m)$. It turns out +that the question is to be answered in the affirmative, as we shall +show. Accordingly, we introduce the following definition: + +\smallskip \textsc{Definition.} If $a^{\lambda(m)}$ is the lowest +power of $a$ which is congruent to $1$ modulo $m$, $a$ is said to be +a primitive $\lambda$-root modulo $m$. We shall also say that it is +a primitive $\lambda$-root of the congruence $x^{\lambda(m)} = 1 +\mod m$. To distinguish we may speak of the usual primitive root as +a primitive $\phi$-root modulo $m$.% +\index{Primitive roots!$\phi$-roots} + +From the theory of primitive $\phi$-roots already developed it +follows that primitive $\lambda$-roots always exist when $m$ is a +power of any odd prime, and also when $m = 1,\ 2,\ 4$; for, for such +values of $m$ we have $\lambda(m) = \phi(m)$. + +We shall next show that primitive $\lambda$-roots exist when $m = +2^{\alpha}$, $a > 2$, by showing that 5 is such a root. It is +necessary and sufficient to prove that $5$ belongs modulo +$2^{\alpha}$ to the exponent $2^{\alpha-2} = \lambda(2^{\alpha})$. +Let $d$ be the exponent to which $5$ belongs modulo $2^{\alpha}$. +Then from theorem II of \S \ref{s32} it follows that $d$ is a +divisor of $2^{\alpha-2} = \lambda(2^{\alpha})$. Hence if $d$ is +different from $2^{\alpha-2}$ it is $2^{\alpha-3}$ or is a divisor +of $2^{\alpha-3}$. Hence if we can show that $5^{2^{\alpha-3}}$ is +not congruent to $1$ modulo $2^{\alpha}$ we will have proved that +$5$ belongs to the exponent $2^{\alpha-2}$. But, clearly, +\begin{gather*} +5^{2^{\alpha-3}} = (1+2^2)^{2^{\alpha-3}} + = 1+2^{\alpha-1}+ I\cdot 2^{\alpha}, \\ +\intertext{where $I$ is an integer. Hence} +5^{2^{\alpha-3}} \not\equiv 1 \bmod 2^{\alpha}. +\end{gather*} +Hence 5 belongs modulo $2^{\alpha}$ to the exponent +$\lambda(2^{\alpha})$. + +By means of these special results we are now in position to prove +readily the following general theorem which includes them as special +cases: + +\smallskip I.~\emph{For every congruence of the form} +\begin{gather*} +x^{\lambda(m)} \equiv 1 \bmod m +\end{gather*} +\emph{a solution $g$ exists which is a primitive $\lambda$-root, and +for any such solution $g$ there are $\phi\{\lambda(m)\}$ primitive +roots congruent to powers of $g$.} + +If any primitive $\lambda$-root $g$ exists, $g^\nu$ is or is not a +primitive $\lambda$-root according as $\nu$ is or is not prime to +$\lambda(m)$; and therefore the number of primitive $\lambda$-roots +which are congruent to powers of any such root $g$ is +$\phi\{\lambda(m)\}$. + +The existence of a primitive $\lambda$-root in every case may easily +be shown by induction. In case $m$ is a power of a prime the theorem +has already been established. We will suppose that it is true when +$m$ is the product of powers of $r$ different primes and show that +it is true when $m$ is the product of powers of $r+1$ different +primes; from this will follow the theorem in general. + +Put $m = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_r^{\alpha_r} +p_{r+1}^{\alpha_{r+1}}, \quad n = p_1^{\alpha_1} p_2^{\alpha_2} +\ldots p_r^{\alpha_r}$, and let $h$ be a primitive $\lambda$-root of +\begin{gather} +x^{\lambda(n)} \equiv 1 \mod n. \tag{1} \\ +\intertext{Then} +h + ny \notag +\end{gather} +is a form of the same root if $y$ is an integer. + +Likewise, if $c$ is any primitive $\lambda$-root of +\begin{equation} +x^\lambda(p_{r+1}^{\alpha_{r+1}}) + \equiv 1 \mod p_{r+1}^{\alpha_{r+1}} \tag{2} +\end{equation} +a form of this root is +\begin{equation*} +c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +where $z$ is any integer. + +Now, if $y$ and $z$ can be chosen so that +\begin{equation*} +h+ny = c+p_{r+1}^{\alpha_{r+1}}z +\end{equation*} +the number in either member of this equation will be a common +primitive $\lambda$-root of congruences (1) and (2); that is, a +common primitive $\lambda$-root of the two congruences may always be +obtained provided that the equation +\begin{equation*} +p_1^{\alpha_1} \ldots p_r^{\alpha_r}y - p_{r+1}^{\alpha_{r+1}}z = c-h +\end{equation*} +has always a solution in which $y$ and $z$ are integers. That this +equation has such a solution follows readily from theorem III of \S +\ref{s9}; for, if $c-h$ is replaced by $1$, the new equation has a +solution $\bar{y}$, $\bar{z}$; and therefore for $y$ and $z$ we may +take $y = \bar{y}(c-h)$, $z = \bar{z}(c-h)$. + +Now let $g$ be a common primitive $\lambda$-root of congruences (1) +and (2) and write +\begin{equation*} +g^\nu \equiv 1 \mod m, +\end{equation*} +where $\nu$ is to be the smallest exponent for which the congruence +is true. Since $g$ is a primitive $\lambda$-root of (1) $\nu$ is a +multiple of $\lambda(p_1^{\alpha_1} \ldots p_r^{\alpha_r})$. Since +$g$ is a primitive $\lambda$-root of (2) $\nu$ is a multiple of +$\lambda\left(p_{r+1}^{\alpha_{r+1}} \right)$. Hence it is a +multiple of $\lambda(m)$. But $g^{\lambda(m)} \equiv 1 \bmod m$; +therefore $\nu = \lambda(m)$. That is, $g$ is a primitive +$\lambda$-root modulo $m$. + +The theorem as stated now follows at once by induction. + +\smallskip There is nothing in the preceding argument to indicate +that the primitive $\lambda$-roots modulo $m$ are all in a single +set obtained by taking powers of some root $g$; in fact it is not in +general true when $m$ contains more than one prime factor. + +By taking powers of a primitive $\lambda$-root $g$ modulo $m$ one +obtains $\phi\{\lambda(m)\}$ different primitive $\lambda$-roots +modulo $m$. It is evident that if $\gamma$ is any one of these +primitive $\lambda$-roots, then the same set is obtained again by +taking the powers of $\gamma$. We may say then that the set thus +obtained is the set belonging to $g$. + +\smallskip II.~\emph{If $\lambda(m)>2$ the product of the +$\phi\{\lambda(m)\}$ primitive $\lambda$-roots in the set belonging +to any primitive $\lambda$-root $g$ is congruent to $1$ modulo $m$.} + +These primitive $\lambda$-roots are +\begin{gather*} +g,\ g^{c_1},\ g^{c_2},\ \ldots,\ g^{c_\mu} \\ +\intertext{where} +1,\ c_1,\ c_2,\ \ldots,\ c_\mu \\ +\end{gather*} +are the integers less than $\lambda(m)$ and prime to $\lambda(m)$. +If any one of these is $c$ another is $\lambda(m)-c$, since +$\lambda(m) > 2$. Hence +\begin{gather*} +1 + c_1 + c_2 + \ldots + c_\mu \equiv 0 \bmod \lambda(m). \\ +\intertext{Therefore} +g^{1 + c_1 + c_2 + \ldots + c_\mu} \equiv 1 \bmod m. +\end{gather*} +From this the theorem follows. + +\smallskip \textsc{Corollary.}\emph{The product of all the +primitive $\lambda$-roots modulo $m$ is congruent to $1$ modulo $m$ +when $\lambda(m) > 2$.}\index{Primitive roots!$\lambda$-roots|)} + +\begin{center} +EXERCISES +\end{center} + +\small\begin{enumerate} +\item[1.] If $x_1$ is the largest value of $x$ satisfying the equation +$\lambda(x) = a$, where $a$ is a given integer, then any solution +$x_2$ of the equation is a factor of $x_1$. + +\item[2*.] Obtain an effective rule for solving the equation +$\lambda(x) = a$. + +\item[3*.] Obtain an effective rule for solving the equation +$\phi(x) = a$. + +\item[4.] A necessary and sufficient condition that $a^{P-1} \equiv 1 +\mod P$ for every integer $a$ prime to $P$ is that $P \equiv 1 \mod +\lambda(P)$. + +\item[5.] If $a^{P-1} \equiv 1\mod P$ for every a prime to $P$, then +(1) $P$ does not contain a square factor other than $1$, (2) $P$ +either is prime or contains at least three different prime factors. + +\item[6.] Let $p$ be a prime number. If $a$ is a root of the congruence +$x^q \equiv 1 \mod p$ and $\alpha$ is a root of the congruence +$x^\delta\equiv 1 \mod p$, then $a\alpha$ is a root of the +congruence $x^{d\delta}\equiv 1 \mod p$. If $a$ is a primitive root +of the first congruence and $\alpha$ of the second and if $d$ and +$\delta$ are relatively prime, then $a\alpha$ is a primitive root of +the congruence $x^{d\delta} \equiv 1\mod p$. +\end{enumerate} \normalsize\index{Primitive roots|)} + +\chapter{OTHER TOPICS} + +\section{Introduction}\label{s40} + +The theory of numbers is a vast discipline and no single volume can +adequately treat of it in all of its phases. A short book can serve +only as an introduction; but where the field is so vast such an +introduction is much needed. That is the end which the present +volume is intended to serve; and it will best accomplish this end +if, in addition to the detailed theory already developed, some +account is given of the various directions in which the matter might +be carried further. + +To do even this properly it is necessary to limit the number of +subjects considered. Consequently we shall at once lay aside many +topics of interest which would find a place in an exhaustive +treatise. We shall say nothing, for instance, about the vast domain +of algebraic numbers, even though this is one of the most +fascinating subjects in the whole field of +mathematics.\index{Algebraic numbers} Consequently, we shall not +refer to any of the extensive theory connected with the division of +the circle into equal parts.\index{Circle, Division of} Again, we +shall leave unmentioned many topics connected with the theory of +positive integers; such, for instance, is the frequency of prime +numbers in the ordered system of integers---a subject which contains +in itself an extensive and elegant theory.\index{Prime numbers} + +In \S\S \ref{s41}--\ref{s44} we shall speak briefly of each of the +following topics: theory of quadratic residues, Galois imaginaries, +arithmetic forms, analytical theory of numbers. Each of these alone +would require a considerable volume for its proper development. All +that we can do is to indicate the nature of the problem in each case +and in some cases to give a few of the fundamental results. + +In the remaining three sections we shall give a brief introduction +to the theory of Diophantine equations, developing some of the more +elementary properties of certain special cases. We shall carry this +far enough to indicate the nature of the problem connected with the +now famous Last Theorem of Fermat. The earlier sections of this +chapter are not required as a preliminary to reading this latter +part. + +\section{Theory of Quadratic Residues}\label{s41}% +\index{Quadratic residues|(} + +Let $a$ and $m$ be any two relatively prime integers. In \S +\ref{s31} we agreed to say that $a$ is a quadratic residue modulo +$m$ or a quadratic non-residue modulo $m$ according as the +congruence +\begin{equation*} +x^2 \equiv a \bmod m +\end{equation*} +has or has not a solution. We saw that if $m$ is chosen equal to an +odd prime number $p$, then $a$ is a quadratic residue modulo $p$ or +a quadratic non-residue modulo $p$ according as +\begin{equation*} +a^{\frac{1}{2} (p-1)} \equiv 1\quad \mathrm{or}\quad + a^{\frac{1}{2} (p-1)} \equiv -1 \bmod p. +\end{equation*} +This is known as Euler's criterion.\index{Euler's!criterion} + +It is convenient to employ the Legendre symbol +\begin{equation*} +\left( \frac{a}{p} \right ) +\end{equation*} +to denote the quadratic character of $a$ with respect to $p$.% +\index{Legendre symbol} This symbol is to have the value $+1$ or the +value $-1$ according as $a$ is a quadratic residue modulo $p$ or a +quadratic non-residue modulo $p$. We shall now derive some of the +fundamental properties of this symbol, understanding always that the +numbers in the numerator and the denominator are relatively prime. + +From the definition of quadratic residues and non-residues it is +obvious that +\begin{equation} +\left ( \frac{a}{p} \right ) = \left ( \frac{b}{p} \right ) + \quad \text{if}\quad a \equiv b \bmod p. \tag{1} +\end{equation} + +It is easy to prove in general that +\begin{equation} +\left ( \frac{a}{p} \right ) \left ( \frac{b}{p} \right ) = + \left (\frac {ab}{p} \right ). \tag{2} +\end{equation} +This comes readily from Euler's criterion. We have to consider the +three cases +\begin{align*} +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=+1; & +\left( \frac{a}{p} \right ) &=+1,& + \left( \frac{b}{p} \right ) &=-1; \\ +&& \left( \frac{a}{p} \right ) &=-1,& + \left( \frac{b}{p} \right ) &=-1. +\end{align*} +The method will be sufficiently illustrated by the treatment +of the last case. Here we have +\begin{gather*} +a^{\frac 12 (p-1)}\equiv -1 \bmod p,\quad + b^{\frac 12 (p-1)}\equiv -1 \bmod p. \\ +\intertext{Multiplying these two congruences together member by +member we have} +(ab)^{\frac 12 (p-1)} \equiv 1 \bmod p, \\ +\intertext{whence} +\left( \frac {ab}{p} \right ) = 1 = + \left( \frac ap \right ) \left( \frac bp \right ), +\end{gather*} +as was to be proved. + +If $m$ is any number prime to $p$ and we write $m$ as the product of +factors +\begin{equation*} +m = \epsilon \cdot 2^\alpha \cdot q' q'' q''' \cdots +\end{equation*} +where $q',\ q'',\ q''',\ \ldots$ are odd primes, $\alpha$ is zero or +a positive integer and $\epsilon$ is $+1$ or $-1$ according as $m$ +is positive or negative, we have +\begin{equation} +\left( \frac{m}{p} \right ) = +\left( \frac{\epsilon}{p} \right ) +\left( \frac{2}{p} \right ) ^\alpha +\left( \frac{q'}{p} \right ) +\left( \frac{q''}{p} \right ) +\left( \frac{q'''}{p} \right ) \ldots, \tag{3} +\end{equation} +as one shows easily by repeated application of relation (2). +Obviously, +\begin{equation*} +\left( \frac{1}{p} \right ) = 1. +\end{equation*} +Hence, it follows from (3) that we can readily determine the +quadratic character of $m$ with respect to the odd prime $p$, that +is, the value of +\begin{equation*} +\left( \frac{m}{p} \right ), +\end{equation*} +provided that we know the value of each of the expressions +\begin{equation} +\left( \frac{-1}{p} \right ),\quad + \left( \frac{2}{p} \right ),\quad + \left( \frac{q}{p} \right ),\tag{4} +\end{equation} +where $q$ is an odd prime. + +The first of these can be evaluated at once by means of Euler's +criterion; for, we have +\begin{gather*} +\left( \frac{-1}{p} \right ) \equiv + (-1)^{\frac{1}{2} (p-1)} \bmod p \\ +\intertext{and hence} +\left( \frac{-1}{p} \right ) = (-1)^{\frac{1}{2} (p-1)}. +\end{gather*} +Thus we have the following result: The number $-1$ is a quadratic +residue of every prime number of the form $4k + 1$ and a quadratic +non-residue of every prime number of the form $4k + 3$. + +The value of the second symbol in (4) is given by the formula +\begin{equation*} +\left( \frac{2}{p} \right ) = (-1)^{\frac{1}{8} (p^2 -1)}. +\end{equation*} +The theorem contained in this equation may be stated in the +following words: The number $2$ is a quadratic residue of every +prime number of either of the forms $8k + 1, 8k + 7$; it is a +quadratic non-residue of every prime number of either of the forms +$8k + 3, 8k + 5$. + +The proof of this result is not so immediate as that of the +preceding one. To evaluate the third expression in (4) is still more +difficult. We shall omit the demonstration in both of these cases. +For the latter we have the very elegant relation +\begin{equation*} +\left( \frac{p}{q} \right ) \left( \frac{q}{p} \right ) = + (-1)^{\frac{1}{4}(p-1)(q-1)}. +\end{equation*} +This equation states the law which connects the quadratic character +of $q$ with respect to $p$ with the quadratic character of $p$ with +respect to $q$. It is known as the Law of Quadratic Reciprocity. +About fifty proofs of it have been given. Its history has been a +very interesting one; see Bachmann's Niedere Zablentheorie, Teil I, +pp.\ 180--318, especially pp.\ 200--206.\index{Bachmann}% +\index{Law of quadratic reciprocity}\index{Quadratic reciprocity} + +For a further account of this beautiful and interesting subject we +refer the reader to Bachmann, loc.\ cit., and to the memoirs to +which this author gives reference.\index{Quadratic residues|)} + +\section{Galois Imaginaries}\label{s42}% +\index{Galois imaginaries}\index{Imaginaries of Galois} + +If one is working in the domain of real numbers the equation +\begin{equation*} +x^2 + 1 = 0 +\end{equation*} +has no solution; for there is no real number whose square is $-1$. +If, however, one enlarges the ``number system'' so as to include not +only all real numbers but all complex numbers as well, then it is +true that every algebraic equation has a root. It is on account of +the existence of this theorem for the enlarged domain that much of +the general theory of algebra takes the elegant form in which we +know it. + +The question naturally arises as to whether we can make a similar +extension in the case of congruences. The congruence +\begin{equation*} +x^2 = 3 \bmod 5 +\end{equation*} +has no solution, if we employ the term solution in the sense in +which we have so far used it. But we may if we choose introduce an +imaginary quantity, or mark, $j$ such that +\begin{equation*} +j^2 \equiv 3 \bmod 5, +\end{equation*} +just as in connection with the equation $x^2 + 1 = 0$ we would +introduce the symbol $i$ having the property expressed by the +equation +\begin{equation*} +i^2 = -1. +\end{equation*} + +It is found to be possible to introduce in this way a general set of +imaginaries satisfying congruences with prime moduli; and the new +quantities or marks have the property of combining according to the +laws of algebra. + +The quantities so introduced are called Galois imaginaries. + +We cannot go into a development of the important theory which is +introduced in this way. We shall be content with indicating two +directions in which it leads. + +In the first place there is the general Galois field theory which is +of fundamental importance in the study of certain finite groups. It +may be developed from the point of view indicated here. An excellent +exposition, along somewhat different lines, is to be found in +Dickson's \emph{Linear Groups with an Exposition of the Galois Field +Theory.}\index{Dickson} + +Again, the whole matter may be looked upon from the geometric point +of view. In this way we are led to the general theory of finite +geometries, that is, geometries in which there is only a finite +number of points. For a development of the ideas which arise here +see Veblen and Young's \emph{Projective Geometry} and the memoir by +Veblen and Bussey in the Transactions of the American Mathematical +Society, vol.\ 7, pp.\ 241--259.\index{Bussey}\index{Veblen}% +\index{Young} + +\section{Arithmetic Forms}\label{s43}% +\index{Arithmetic forms|(}\index{Forms|(} + +The simplest arithmetic form is $ax + b$ where $a$ and $b$ are fixed +integers different from zero and $x$ is a variable integer. By +varying $x$ in this case we have the terms of an arithmetic +progression. We have already referred to Dirichlet's celebrated +theorem which asserts that the form $ax + b$ has an infinite number +of prime values if only $a$ and $b$ are relatively +prime.\index{Dirichlet} This is an illustration of one type of +theorem connected with arithmetic forms in general, namely, those in +which it is asserted that numbers of a given form have in addition a +given property.\index{Prime numbers} + +Another type of theorem is illustrated by a result stated in \S +\ref{s41}, provided that we look at that result in the proper way. +We saw that the number $2$ is a quadratic residue of every prime of +either of the forms $8k + 1$ and $8k + 7$ and a quadratic +non-residue of every prime of either of the forms $8k + 3$ and $8k + +5$. We may state that result as follows: A given prime number of +either of the forms $8k + 1$ and $8k + 7$ is a divisor of some +number of the form $x^2 - 2$, where $x$ is an integer; no prime +number of either of the forms $8k + 3$ and $8k + 5$ is a divisor of +a number of the form $x^2 - 2$, where $x$ is an integer. + +The result just stated is a theorem in a discipline of vast extent, +namely, the theory of quadratic forms. Here a large number of +questions arise among which are the following: What numbers can be +represented in a given form? What is the character of the divisors +of a given form? As a special case of the first we have the question +as to what numbers can be represented as the sum of three squares. +To this category belong also the following two theorems: Every +positive integer is the sum of four squares of integers; every prime +number of the form $4n + 1$ may be represented (and in only one way) +as the sum of two squares.\index{Prime numbers} + +For an extended development of the theory of quadratic forms we +refer the reader to Bachmann's Arithmetik der Quadratischen Formen +of which the first part has appeared in a volume of nearly seven +hundred pages.\index{Bachmann} + +It is clear that one may further extend the theory of arithmetic +forms by investigating the properties of those of the third and +higher degrees. Naturally the development of this subject has not +been carried so far as that of quadratic forms; but there is a +considerable number of memoirs devoted to various parts of this +extensive field, and especially to the consideration of various +special forms. + +Probably the most interesting of these special forms are the +following: +\begin{equation*} +\alpha^n + \beta^n , \quad + \frac{\alpha^n - \beta^n}{\alpha - \beta} = + \alpha^{n-1} + \alpha^{n-2} \beta + \cdots + \beta^{n-1}, +\end{equation*} +where $\alpha$ and $\beta$ are relatively prime integers, or, more +generally, where $\alpha$ and $\beta$ are the roots of the quadratic +equation $x^2 - ux + v = 0$ where $u$ and $v$ are relatively prime +integers. A development of the theory of these forms has been given +by the present author in a memoir published in 1913 in the Annals of +Mathematics, vol.\ 13, pp.\ 30--70.% +\index{Arithmetic forms|)}\index{Carmichael}\index{Forms|)}% +\index{Quadratic forms} + +\section{Analytical theory of numbers}\label{s44}% +\index{Analytical theory of numbers|(} + +Let us consider the function +\begin{equation*} +P(x) = \frac{1}{\prod_{k=0}^\infty (1-x^{2^k} )} , \quad + |x|\leqq \rho < 1. +\end{equation*} +It is clear that we have +\begin{align*} +P(x) = \prod_{k=0}^\infty \frac{1}{(1-x^{2^k} )} &= + \prod_{k=0}^\infty + ( 1 + x^{2k} + x^{2\cdot 2^k} + x^{3\cdot 2^k} + \cdots ) \\ +&= \sum_{s=0}^\infty G(s) x^s, +\end{align*} +where $G(0) = 1$ and $G(s)$ (for $s$ greater than $0$) is the number +of ways in which the positive integer $s$ may be separated into like +or distinct summands each of which is a power of $2$. + +We have readily +\begin{equation*} +(1-x)\sum_{s=0}^\infty G(s) x^s = (1-x)P(x) = P(x^2) = + \sum_{s=0}^\infty x^{2^s}; +\end{equation*} +whence +\begin{equation} +G(2s + 1) = G(2s) = G(2s - 1) + G(s), \tag{A} +\end{equation} +as one readily verifies by equating coefficients of like powers of +$x$. From this we have in particular +\begin{gather*} +G(0) = 1, \quad G(1) = 1, \quad G(2) = 2, \quad G(3) = 2, \\ +G(4) = 4, \quad G(5) = 4, \quad G(6) = 6, \quad G(7) = 6. +\end{gather*} +Thus in (A) we have recurrence relations by means of which we may +readily reckon out the values of the number theoretic function +$G(s)$. Thus we may determine the number of ways in which a given +positive integer $s$ may be represented as a sum of powers of $2$. + +We have given this example as an elementary illustration of the +analytical theory of numbers, that is, of that part of the theory of +numbers in which one employs (as above) the theory of a continuous +variable or some analogous theory in order to derive properties of +sets of integers. This general subject has been developed in several +directions. For a systematic account of it the reader is referred to +Bachmann's Analytische Zahlentheorie.% +\index{Analytical theory of numbers|)}\index{Bachmann} + +\section{Diophantine equations}\label{s45}% +\index{Diophantine equations}\index{Equations!Diophantine} + +If $f(x, y, z, \ldots)$ is a polynomial in the variables $x, y, z, +\ldots$ with integral coefficients, then the equation +\begin{equation*} +f(x, y, z, \ldots) = 0 +\end{equation*} +is called a Diophantine equation when we look at it from the point +of view of determining the integers (or the positive integers) $x, +y, z, \ldots$ which satisfy it. Similarly, if we have several such +functions $f_i(x, y, z, \ldots)$, in number less than the number of +variables $x, y, z, \ldots$, then the set of equations +\begin{equation*} +f_i(x, y, z, \ldots) = 0,\quad i = i, 2, \ldots, +\end{equation*} +is said to be a Diophantine system of equations. Any set of integers +$x, y, z, \ldots$ which satisfies the equation [system] is said to +be a solution of the equation [system]. + +We may likewise define Diophantine inequalities by replacing the +sign of equality above by the sign of inequality. But little has +been done toward developing a theory of Diophantine inequalities. +Even for Diophantine equations the theory is in a rather fragmentary +state. + +In the next two sections we shall illustrate the nature of the ideas +and the methods of the theory of Diophantine equations by developing +some of the results for two important special cases. + +\section{Pythagorean triangles}\label{s46}% +\index{Pythagorean triangles|(} + +\textsc{Definitions.} If three positive integers $x, y, z$ satisfy +the relation +\begin{equation} +x^2 + y^2 = z^2 \tag{1} +\end{equation} +they are said to form a Pythagorean triangle or a numerical right +triangle; $z$ is called the hypotenuse of the triangle and $x$ and +$y$ are called its legs. The area of the triangle is said to be +$\frac{1}{2} xy$.\index{Triangles, Numerical} + +We shall determine the general form of the integers $x$, $y$, $z$, +such that equation (1) may be satisfied. Let us denote by $\nu$ the +greatest common divisor of $x$ and $y$ in a particular solution of +(1). Then $\nu$ is a divisor of $z$ and we may write +\begin{equation*} +x = \nu u, \quad y = \nu v,\quad z = \nu w. +\end{equation*} +Substituting these values in (1) and reducing we have +\begin{equation} +u^2 + v^2 = w^2, \tag{2} +\end{equation} +where $u, v, w$ are obviously prime each to each, since $u$ and $v$ +have the greatest common divisor $1$. + +Now an odd square is of the form $4k + 1$. Hence the sum of two odd +squares is divisible by $2$ but not by $4$; and therefore the sum of +two odd squares cannot be a square. Hence one of the numbers $u$, +$v$ is even. Suppose that $u$ is even and write equation (2) in the +form +\begin{equation} +u^2 = (w - v)(w + v). \tag{3} +\end{equation} +Every common divisor of $w - v$ and $w + v$ is a divisor of their +difference $2v$. Therefore, since $w$ and $v$ are relatively prime, +it follows that $2$ is the greatest common divisor of $w - v$ and $w ++ v$. Then from (3) we see that each of these numbers is twice a +square, so that we may write +\begin{equation*} +w - v = 2b^2,\quad w + v = 2a^2 +\end{equation*} +where $a$ and $b$ are relatively prime integers. From these two +equations and equation (3) we have +\begin{equation} +w = a^2 + b^2, \quad v = a^2 -b^2,\quad u = 2ab. \tag{4} +\end{equation} +Since $u$ and $v$ are relatively prime it is evident that one of the +numbers $a$, $b$ is even and the other odd. + +The forms of $u$, $v$, $w$ given in (4) are necessary in order that +(2) may be satisfied. A direct substitution in (2) shows that this +equation is indeed satisfied by these values. Hence we have in (4) +the general solution of (2) where $u$ is restricted to be even. A +similar solution would be obtained if $v$ were restricted to be +even. Therefore \emph{the general solution of (1) is +\begin{gather*} +x = 2\nu ab,\quad y = \nu (a^2 - b^2),\quad z = \nu (a^2 + b^2)\\ +\intertext{and} +x = 2\nu (a^2 - b^2 ),\quad y = 2\nu ab,\quad z = \nu (a^2 + b^2) +\end{gather*} +where $a$, $b$, $\nu$ are arbitrary integers except that $a$ and $b$ +are relatively prime and one of them is even and the other odd.} + +By means of this general solution of (1) we shall now prove the +following theorem: + +\smallskip I.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero, such that} +\begin{equation} +q^2 + n^2 = m^2 , \quad m^2 + n^2 = p^2. \tag{5} +\end{equation} + +It is obvious that an equivalent theorem is the following: + +\smallskip II.~\emph{There do not exist integers $m$, $n$, $p$, $q$, +all different from zero such that} +\begin{equation} +p^2 + q^2 = 2m^2, \quad p^2 - q^2 = 2n^2. \tag{6} +\end{equation} + +Obviously, we may without loss of generality take $m$, $n$, $p$, $q$ +to be positive; and this we do. + +The method of proof is to assume the existence of integers +satisfying equations (5) and (6) and to show that we are thus led to +a contradiction. The argument we give is an illustration of Fermat's +famous method of ``infinite descent.''% +\index{Descent, Infinite}\index{Fermat}\index{Infinite descent} + +If any two of the numbers $p$, $q$, $m$, $n$ have a common prime +factor $t$, it follows at once from (5) and (6) that all four of +them have this factor. For, consider an equation in (5) or in (6) in +which these two numbers occur; this equation contains a third +number, and it is readily seen that this third number is divisible +by $t$. Then from one of the equations containing the fourth number +it follows that this fourth number is divisible by $t$. Now let us +divide each equation of system (6) through by $t^2$; the resulting +system is of the same form as (6). If any two numbers in this +resulting system have a common prime factor $t_1$, we may divide +through by $t_1^2$; and so on. Hence if a pair of simultaneous +equations (6) exists then there exists a pair of equations of the +same form in which no two of the numbers $m$, $n$, $p$, $q$ have a +common factor other than unity. Let this system of equations be +\begin{equation} +p_1^2 + q_1^2 = 2m_1^2, \quad p_1^2 - q_1^2 = 2n_1^2. \tag{7} +\end{equation} + +From the first equation in (7) it follows that $p_1$ and $q_1$ are +both even or both odd; and, since they are relatively prime, it +follows that they are both odd. Evidently $p_1 > q_1$. Then we may +write +\begin{equation*} +p_1 = q_1 + 2\alpha, +\end{equation*} +where $\alpha$ is a positive integer. If we substitute this value of +$p_1$ in the first equation of (7), the result may readily be put in +the form +\begin{equation} +(q_1 + \alpha)^2 + a^2 = m_1^2. \tag{8} +\end{equation} +Since $q_1$ and $m_1$ have no common prime factor it is easy to see +from this equation that $\alpha$ is prime to both $q_1$ and $m_1$, +and hence that no two of the numbers $q_1 + \alpha, \alpha, m_1$ +have a common factor. + +Now we have seen that if $a$, $b$, $c$ are positive integers no two +of which have a common prime factor, while +\begin{equation*} +a^2 + b^2 = c^2, +\end{equation*} +then there exist relatively prime integers $r$ and $s$, $r > s$, +such that +\begin{gather} +c = r^2 + s^2,\quad a = 2rs,\quad b = r^2 - s^2 \notag \\ +\intertext{or} +c = r^2 + s^2,\quad a = r^2 - s^2,\quad b = 2rs. \notag \\ +\intertext{Hence from (8) we see that we may write} +q_1 + \alpha = 2rs,\quad \alpha = r^2 - s^2 \tag{9} \\ +\intertext{or} +q_1 + \alpha = r^2 - s^2, \alpha = 2rs. \tag{10} \\ +\intertext{In either case we have} +p_1^2 - q_1^2 = (p_1 - q_1)(p_1 + q_1) = + 2\alpha \cdot 2(q_1 + \alpha) = 8rs(r^2 - s^2). \notag \\ +\intertext{If we substitute in the second equation of (7) and divide +by 2 we have} 4rs(r^2 - s^2) = n_1^2. \notag +\end{gather} + +From this equation and the fact that $r$ and $s$ are relatively +prime it follows at once that $r$, $s$, $r^2 - s^2$ are all square +numbers; say, +\begin{gather} +r = u^2,\quad s = v^2,\quad r^2 - s^2 = w^2. \notag \\ +\intertext{Now $r - s$ and $r + s$ can have no common factor other +than 1 or 2; hence from} +w^2 = (r^2-s^2) = (r-s)(r+s) = (u^2-v^2)(u^2+v^2) \notag \\ +\intertext{we see that either} +u^2 + v^2 = 2w_1^2,\quad u^2 - v^2 = 2w_2^2 \tag{11} \\ +\intertext{or} +u^2 + v^2 = w_1^2,\quad u^2 - v^2 = w_2^2. \notag \\ +\intertext{And if it is the latter case which arises, then} +w_1^2 + w_2^2 = 2u^2,\quad w_1^2 - w_2^2 = 2v^2. \tag{12} +\end{gather} +Hence, assuming equations of the form (6) we are led either to +equations (11) or to equations (12); that is, we are led to new +equations of the form with which we started. Let us write the +equations thus: +\begin{equation} +p_2^2 + q_2^2 = 2m_2^2,\quad p_2^2 - q_2^2 = 2n_2^2; \tag{13} +\end{equation} +that is, system (13) is identical with that one of systems (11), +(12) which actually arises. + +Now from (9) and (10) and the relations $p_1 = q_1 + 2\alpha, r +> s$, we see that +\begin{gather*} +p_1 = 2rs + r^2 - s^2 > 2s^2 + r^2 - s^2 = + r^2 + s^2 = u^4 + v^4. \\ +\intertext{Hence $u < p_1$. Also,} +w_1^2 \leqq w^2 \leqq r+s < r^2 + s^2. +\end{gather*} +Hence $w_1 < p_1$. Since $u$ and $w_1$ are both less than $p_1$ it +follows that $p_2$ is less than $p_1$. Hence, obviously, $p_2 < p$. +Moreover, it is clear that all the numbers $p_2, q_2, m_2, n_2$ are +different from zero. + +From these results we have the following conclusion: If we assume a +system of the form (6) we are led to a new system (13) of the same +form; and in the new system $p_2$ is less than $p$. + +Now if we start with (13) and carry out a similar argument +we shall be led to a new system +\begin{gather*} +p_3^2 + q_3^2 = 2m_3^2,\quad p_3^2 - q_3^2 = 2n_3^2, +\end{gather*} +with the relation $p_3 < p_2$, starting from this last system we +shall be led to a new one of the same form, with a similar relation +of inequality; and so on \emph{ad infinitum.} But, since there is +only a finite number of positive integers less than the given +positive integer $p$ this is impossible. We are thus led to a +contradiction; whence we conclude at once to the truth of II and +likewise of I. + +By means of theorems I and II we may readily prove the following +theorem: + +\smallskip III.~\emph{The area of a numerical right triangle is +never a square number.} + +Let the sides and hypotenuse of a numerical right triangle be $u, v, +w$, respectively. The area of this triangle is $\frac{1}{2} uv$. If +we assume this to be a square number $t^2$ we shall have the +following simultaneous Diophantine equations +\begin{equation} +u^2 + v^2 = w^2,\quad uv = 2t^2. \tag{14} +\end{equation} +We shall prove our theorem by showing that the assumption of such a +system leads to a contradiction. + +If any two of the numbers $u, v, w$ have a common prime factor $p$ +then the remaining one also has this factor, as one sees readily +from the first equation in (14). From the second equation in (14) it +follows that $t$ also has the same factor. Then if we put $u = pu_1, +v = pv_1, w = pw_1, t = pt_1$, we have +\begin{equation*} +u_1^2 + v_1^2 = w_1^2,\quad u_1 v_1 = 2t_1^2, +\end{equation*} +a system of the same form as (14). It is clear that we may start +with this new system and proceed in the same manner as before, and +so on, until we arrive at a system +\begin{equation} +\bar{u}^2 + \bar{v}^2 = \bar{w}^2,\quad + \bar{u}\bar{v} = 2\bar{t}^2, \tag{15} +\end{equation} +where $\bar{u}$, $\bar{v}$, $\bar{w}$ are prime each to each. + +Now the general solution of the first equation (15) may be written +in one of the forms +\begin{gather*} +\bar{u} = 2ab,\quad \bar{v} = a^2 - b^2,\quad \bar{w} = a^2 + b^2 \\ +\bar{u} = a^2 - b^2,\quad \bar{v} = 2ab, \quad \bar{w} = a^2 + b^2. \\ +\intertext{Then from the second equation in (15) we have} +\bar{t}^2 = ab(a^2 - b^2 ) = ab(a-b)(a+b). +\end{gather*} +It is easy to see that no two of the numbers $a$, $b$, $a - b$, $a + +b$ in the last member of this equation have a common factor; for, if +so, $\bar{u}$ and $\bar{v}$ would have a common factor, contrary to +hypothesis. Hence each of these four numbers is a square. That is, +we have equations of the form +\begin{gather*} +a = m^2,\quad b = n^2,\quad a + b = p^2,\quad a - b = q^2; \\ +\intertext{whence} +m^2 - n^2 = q^2,\quad m^2 + n^2 = p^2. +\end{gather*} +But, according to theorem I, no such system of equations can exist. +That is, the assumption of equations (14) leads to a contradiction. +Hence the theorem follows as stated above.% +\index{Pythagorean triangles|)} + +\section{The Equation $x^n + y^n = z^n$.}\label{s47}% +\index{Equation $x^n + y^n = z^n$|(}\index{Fermat's!last theorem} + +The following theorem, which is commonly known as Fermat's Last +Theorem, was stated without proof by Fermat in the seventeenth +century: + +\smallskip\emph{If n is an integer greater than 2 there do not exist +integers x, y, z, all different from zero, such that} +\begin{equation} +x^n + y^n = z^n. \tag{1} +\end{equation} + +No general proof of this theorem has yet been given. For various +special values of $n$ the proof has been found; in particular, for +every value of $n$ not greater than 100. + +In the study of equation (1) it is convenient to make some +preliminary reductions. If there exists any particular solution of +(1) there exists also a solution in which $x$, $y$, $z$ are prime +each to each, as one may show readily by the method employed in the +first part of \S \ref{s46}. Hence in proving the impossibility of +equation (1) it is sufficient to treat only the case in which $x$, +$y$, $z$ are prime each to each. + +Again, since $n$ is greater than 2 it must contain the factor +4 or an odd prime factor $p$. If $n$ contains the factor $p$ we write +$n = mp$, whence we have +\begin{gather*} +(x^m)^p + (y^m)^p = (z^m)^p). \\ +\intertext{If $n$ contains the factor 4 we write $n = 4m$, whence we +have} +(x^m)^4 + (y^m)^4 = (z^m)^4. +\end{gather*} +From this we see that in order to prove the impossibility of (1) in +general it is sufficient to prove it for the special cases when $n$ +is 4 and when $n$ is an odd prime $p$. For the latter case the proof +has not been found. For the former case we give a proof below. The +theorem may be stated as follows: + +\smallskip I.~\emph{There are no integers $x, y, z$, all different +from zero, such that} +\begin{equation*} +x^4 + y^4 = z^4. +\end{equation*} + +This is obviously a special case of the more general theorem: + +\smallskip II.~\emph{There are no integers $p$, $q$, $\alpha$, all +different from zero, such that} +\begin{equation} +p^4 - q^4 = \alpha^2. \tag{2} +\end{equation} + +The latter theorem is readily proved by means of theorem III of \S +\ref{s46}. For, if we assume an equation of the form (2), we have +\begin{gather} +(p^4 - q^4)p^2 q^2 = p^2 q^2 \alpha^2. \tag{3} \\ +\intertext{But, obviously,} +(2p^2 q^2)^2 + (p^4 - q^4)^2 = (p^4 + q^4)^2. \tag{4} +\end{gather} +Now, from (3) we see that the numerical right triangle determined by +(4) has its area $p^2 q^2(p^4 - q^4)$ equal to the square number +$p^2 q^2 \alpha^2$. But this is impossible. Hence no equation of the +form (2) exists. + +\begin{center} +EXERCISES +\end{center} + +\begin{enumerate} +\item[1.] Show that the equation $\alpha^4 + 4\beta^4 = \gamma^2$ is +impossible in integers $\alpha$, $\beta$, $\gamma$ all of which are +different from zero. + +\item[2.] Show that the system $p^2 - q^2 = km^2$, $p^2 + q^2 = kn^2$ +impossible in integers $p$, $q$, $k$, $m$, $n$, all of which are +different from zero. + +\item[3*.] Show that neither of the equations $m^4 - 4n^4 = \pm t^2$ +is possible in integers $m$, $n$, $t$, all of which are different +from zero. + +\item[4*.] Prove that the area of a numerical right triangle is not +twice a square number. + +\item[5*.] Prove that the equation $m^4 + n^4 = \alpha^2$ is not +possible in integers $m$, $n$, $\alpha$ all of which are different +from zero. + +\item[6*.] In the numerical right triangle $a^2 + b^2 = c^2$, +not more than one of the numbers $a$, $b$, $c$ is a square. + +\item[7.] Prove that the equation $x^{2k} + y^{2k} = z^{2k}$ implies +an equation of the form $m^k + n^k = 2^{k-2} t^k$. + +\item[8.] Find the general solution in integers of the equation +$x^2 + 2y^2 = t^2$. + +\item[9.] Find the general solution in integers of the equation +$x^2 + y^2 = z^4$. + +\item[10.] Obtain solutions of each of the following Diophantine +equations: +\begin{align*} +x^3 + y^3 + z^3 &= 2t^3, \\ +x^3 + 2y^3 + 3z^3 &= t^3, \\ +x^4 + y^4 + 4z^4 &= t^4, \\ +x^4 + y^4 + z^4 &= 2t^4. +\end{align*} +\end{enumerate}\index{Equation $x^n + y^n = z^n$|)} + +\addcontentsline{toc}{chapter}{Index} +\printindex + + +\newpage +\chapter{PROJECT GUTENBERG "SMALL PRINT"} +\small +\pagenumbering{gobble} + +*** END OF THE PROJECT GUTENBERG EBOOK THE THEORY OF NUMBERS *** + +***** This file should be named 13693-pdf.pdf or 13693-pdf.zip ***** +This and all associated files of various formats will be found in: + https://www.gutenberg.org/1/3/6/9/13693/ + +Updated editions will replace the previous one--the old editions will +be renamed. + +Creating the works from print editions not protected by U.S. copyright +law means that no one owns a United States copyright in these works, +so the Foundation (and you!) can copy and distribute it in the +United States without permission and without paying copyright +royalties. 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