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} [Fermat's Little 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} 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} \begin{theorem} An alternate form of Fermat's Little theorem: Let $p$ be prime and $a\in Z^{+}$ such that $a\operatorname{mod}p\neq0$. Then% \[ a^{p}=a\operatorname{mod}p \] \end{theorem} \begin{definition} Let $n\in Z^{+}$ then we define the totient function $\phi\left( n\right) $ is defined to be the number of positive integers less than $n$ that are relatively prime to $n$. That is \begin{equation} \phi\left( n\right) =\left\{ x:x\in Z^{+},x<n,\gcd\left( x,n\right) =1\right\} \end{equation} \end{definition} \begin{lemma} Let $n$ be a prime number. Then% \begin{equation} \phi\left( n\right) =n-1 \end{equation} \end{lemma} \begin{theorem} Given a composit number $n=p\times q$ where $p,q$ are prime then \begin{equation} \phi\left( n\right) =\phi\left( p_{1}\right) \times\phi\left( p_{2}\right) \end{equation} \end{theorem} \begin{proof} Consider that the set of residues in $Z_{n}$ is $\left\{ 0,1,...,pq-1\right\} $. Now the residues that are not relatively prime to $p$ are \begin{equation} \left\{ p,2p,...,\left( q-1\right) p\right\} \label{eq:3}% \end{equation} and the residues that are not relatively prime to $q$ are \begin{equation} \left\{ q,2q,...,\left( p-1\right) q\right\} \label{eq:4}% \end{equation} Clearly the size of the two sets of residues not relatively prime to $n$ are $\left( p-1\right) $ and $\left( q-1\right) $ plus the $0$ element. So we have \begin{align} \phi\left( n\right) & =pq-\left[ \left( p-1\right) +\left( q-1\right) +1\right] \nonumber\\ & =pq-\left( p-1\right) -\left( q-1\right) -1\nonumber\\ & =pq-p-q+1\nonumber\\ & =p\left( q-1\right) -\left( q-1\right) \nonumber\\ & =\left( p-1\right) \left( q-1\right) \end{align} \end{proof} \begin{theorem} [Euler's Theorem]Let $a$ and $n$ be relatively prime positive numbers, then \begin{equation} a^{\phi\left( n\right) }\equiv1\left( \operatorname{mod}n\right) \label{eq:5}% \end{equation} \end{theorem} \begin{proof} First we not that if $n$ is prime that the Theorem holds from Fermat's little Theorem. Namely (\ref{eq:5}) reduces to \begin{equation} a^{n-1}=1\operatorname{mod}n \end{equation} Consider the set of integers relatively prime to $n$: \[ R=\left\{ x_{1},...,x_{\phi\left( n\right) }\right\} \] Multiplying the set by $a$ modulo $n$ gives:% \[ S=\left\{ ax_{1}\operatorname{mod}n,...,ax_{\phi\left( n\right) }\operatorname{mod}n\right\} \] We claim that $S$ is a permutation of $R$. Consider that $a$ and $x_{i}\in R$ are relatively prime to $n$. Then $ax_{i}$ must also be relatively prime to $n$ and $ax_{i}\operatorname{mod}n\neq0$. Thus all the members of $S$ are relatively prime to $n$. There can be no duplicates in $S$ because \[ ax_{i}\operatorname{mod}n=ax_{j}\operatorname{mod}n\rightarrow x_{i}% \operatorname{mod}n=x_{j}\operatorname{mod}n \] because there exists a $a^{-1}$ in $Z_{n}$. But $x_{i}<n$ and $x_{j}<n$, so we must have that \[ x_{i}=x_{j}% \] and we have that there are no duplicates in $S$. Therefore $S$ is a permutation of $R$. Consider% \[ \prod_{i=1}^{\phi\left( n\right) }\left( ax_{i}\operatorname{mod}n\right) =\prod_{i=1}^{\phi\left( n\right) }x_{i}% \] Then% \[ \prod_{i=1}^{\phi\left( n\right) }ax_{i}\equiv\prod_{i=1}^{\phi\left( n\right) }x_{i}\left( \operatorname{mod}n\right) \] Which gives% \[ a^{\phi\left( n\right) }\times\prod_{i=1}^{\phi\left( n\right) }% x_{i}\equiv\prod_{i=1}^{\phi\left( n\right) }x_{i}\left( \operatorname{mod}% n\right) \] Now since $\prod_{i=1}^{\phi\left( n\right) }x_{i}$ is relatively prime to $n$, we can cancel on each side to get the result% \[ a^{\phi\left( n\right) }\equiv1\left( \operatorname{mod}n\right) \] \end{proof} \begin{theorem} Alternate form of Euler's Theorem: Let $a$ and $n$ be relatively prime positive numbers, then% \[ a^{\phi\left( n\right) +1}=a\left( \operatorname{mod}n\right) \] \end{theorem} \begin{corollary} Let $n=pq$ and $m$ be integers where $p$ and $q$ are prime numbers and $0<m<n$. Then \[ m^{\phi\left( n\right) +1}=m^{\left( p-1\right) \left( q-1\right) +1}\equiv m\left( \operatorname{mod}n\right) \] \end{corollary} \begin{theorem} [CRT]Let $S=\left\{ m_{1},...,m_{k}\right\} $ and% \[ M=\prod_{i=1}^{k}m_{i}% \] where for all $m_{i},m_{j}\in S,\gcd\left( m_{i},m_{j}\right) =1$ (that is they are pairwise relatively prime). We can represent any integer in $Z_{m}$ by the $k-$tuple whose elements are in $Z_{m_{i}}$. That is we have a bijection: \[ A\longleftrightarrow\left( a_{1},...,a_{k}\right) \] \end{theorem} \begin{proof} Define \[ a_{i}=A\operatorname{mod}m_{i}% \] Let \[ M_{i}=M/m_{i}\text{ for }1\leq i\leq k \] so that the following condition holds:% \[ M_{i}=0\left( \operatorname{mod}m_{i}\right) \] Since $M_{i}$ is relatively prime to $m_{i}$ we define the following:% \[ c_{i}=M_{i}\times\left( M_{i}^{-1}\operatorname{mod}m_{i}\right) \text{ for }1\leq i\leq k \] We claim that the following holds, but why we have no idea!% \[ A\equiv\left( \sum_{i=1}^{k}a_{i}c_{i}\right) \operatorname{mod}M \] \end{proof}