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Copy path11-00-ProofOfWilsonsTheorem.tex
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11-00-ProofOfWilsonsTheorem.tex
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\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{ProofOfWilsonsTheorem}
\pmcreated{2013-03-22 12:09:09}
\pmmodified{2013-03-22 12:09:09}
\pmowner{CWoo}{3771}
\pmmodifier{CWoo}{3771}
\pmtitle{proof of Wilson's theorem}
\pmrecord{9}{31343}
\pmprivacy{1}
\pmauthor{CWoo}{3771}
\pmtype{Proof}
\pmcomment{trigger rebuild}
\pmclassification{msc}{11-00}
\endmetadata
% This is Cosmin's preamble version 1.01 (planetmath.org version)
% Packages
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{mathrsfs}
%\usepackage[]{graphicx}
%%%%\usepackage{xypic}
%\usepackage[]{babel}
% Theorem Environments
\newtheorem*{thm}{Theorem}
\newtheorem{thmn}{Theorem}
\newtheorem*{lem}{Lemma}
\newtheorem{lemn}{Lemma}
\newtheorem*{cor}{Corollary}
\newtheorem{corn}{Corollary}
\newtheorem*{prop}{Proposition}
\newtheorem{propn}{Proposition}
% New Commands
% Other Commands
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\renewcommand{\leq}{\leqslant}
\newcommand{\vect}[1]{\boldsymbol{#1}}
\renewcommand{\div}{\!\mid\!}
% Local
%\DeclareMathOperator{}{}
\begin{document}
\PMlinkescapeword{inverse}
\PMlinkescapeword{pairing}
\PMlinkescapeword{satisfy}
\PMlinkescapeword{equation}
We first show that, if $p$ is a prime, then $(p-1)! \equiv -1 \pmod p.$ Since $p$ is prime, $\mathbb{Z}_p$ is a field and thus, pairing off each element with its inverse in the product $(p-1)! = \prod_{x=1}^{p-1}x,$ we are left with the elements which are their own inverses (i.e. which satisfy the equation $x^2 \equiv 1 \pmod p$), $1$ and $-1$, only. Consequently, $(p-1)! \equiv -1 \pmod p.$
To prove that the condition is necessary, suppose that $(p-1)! \equiv -1 \pmod p$ and that $p$ is not a prime. The case $p=1$ is trivial. Since $p$ is composite, it has a divisor $k$ such that $1 < k < p$, and we have $(p-1)! \equiv -1 \pmod k$. However, since $k\leq p-1$, it divides $(p-1)!$ and thus $(p-1)! \equiv 0 \pmod k,$ a contradiction.
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\end{document}