Fermat's Little Theorem
\usepackage{amsmath}% \setcounter{MaxMatrixCols}{30}% \usepackage{amsfonts}% \usepackage{amssymb}% \usepackage{graphicx} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} %%end-prologue%% \begin{theorem} Let $p$ be a prime and $a\in Z^{+}$ such that $a\operatorname{mod}p\neq0$. Then \begin{equation} a^{p-1}\equiv1\operatorname{mod}p \end{equation} \end{theorem} \begin{proof} [Fermat's Little Theorem]Consider $Z_{p}=\left\{ 0,1,...,p-1\right\} $, we know that multiplying each element of $Z_{p}$ by $a$ modulo $p$ just gives us $Z_{p}$ in some order. Since $0\times a\operatorname{mod}p=0$ we have the last $p-1$ numbers of $Z_{p}$ multiplied by $a$ as:% \begin{align} a\times Z_{p}\backslash\left\{ 0\right\} & =\left\{ a,2a,3a,...,\left( p-1\right) a\right\} \nonumber\\ & \equiv\left\{ a\operatorname{mod}p,2a\operatorname{mod}p,...,\left( p-1\right) a\operatorname{mod}p\right\} \end{align} If we multiply all the numbers in the set we have \begin{align} a\times2a\times3a\times...\times\left( p-1\right) a & =a^{p-1}\left( 1\times2\times...\times\left( p-1\right) \right) \nonumber\\ & =a^{p-1}\left( p-1\right) ! \end{align} and \begin{equation} a^{p-1}\left( p-1\right) !\equiv a\operatorname{mod}p\times 2a\operatorname{mod}p\times...\times\left( p-1\right) a\operatorname{mod}% p\label{eq:1}% \end{equation} Because we know that all the terms in (\ref{eq:1}) map to some unique element in $Z_{p}$ not $0$ we have the following% \begin{equation} a^{p-1}\left( p-1\right) !\equiv\left( 1\times2\times...\times\left( p-1\right) \right) \operatorname{mod}p \end{equation} and \begin{equation} a^{p-1}\left( p-1\right) !\equiv\left( p-1\right) !\operatorname{mod}% p\label{eq:2}% \end{equation} Since $\left( p-1\right) $ is relatively prime to $p$ (because $p$ is prime) we can divide out $\left( p-1\right) !$ from each side of (\ref{eq:2}) to get the result:% \[ a^{p-1}\equiv1\operatorname{mod}p \] \end{proof}