\documentclass[12pt]{article}
\include{macros}
\title{Lecture 6: Congruences, Part II}
\begin{document}
\maketitle
\hd{Key Ideas Today}
\begin{itemize}
\item Wilson's theorem
\item Chinese Remainder Theorem
\item Multiplicativity of $\vphi$
\end{itemize}

\section{Wilson's Theorem}
\begin{theorem}[John Wilson's theorem, from the 1770s]
An integer $p>1$ is prime if and only if
$$(p-1)! \con -1 \pmod{p}.$$
\end{theorem}
\begin{example}
\begin{verbatim}

? p=3
%1 = 3
? (p-1)! % 3
%2 = 2
? p=17
%3 = 17
? (p-1)!
%4 = 20922789888000
? (p-1)! % p
%5 = 16
\end{verbatim}
\end{example}

\begin{proof}
We first {\bf assume that~$p$ is prime} and prove that 
$(p-1)! \con -1\pmod{p}$.  If $a\in\{1,2,\ldots,p-1\}$ then
the equation 
$$
  ax\con 1\pmod{p}
$$
has a unique solution $a'\in\{1,2,\ldots,p-1\}$.
If $a=a'$, then $a^2\con 1\pmod{p}$, so 
$p\mid a^2-1 = (a-1)(a+1)$, so 
$p\mid (a-1)$ or $p\mid (a+1)$, so $a\in\{1,-1\}$.  
We can thus pair off the elements of 
$\{2,3,\ldots,p-2\}$,
each with its inverse. 
Thus
$$
  2\cdot 3 \cdot \cdots \cdot (p-2) \con 1\pmod{p}.
$$
Multiplying both sides by $p-1$ proves that
$(p-1)! \con -1\pmod{p}$.


Next we {\bf assume that $(p-1)! \con -1\pmod{p}$} and
prove that~$p$ must be prime.  Suppose not, so that~$p$ 
is a composite number $\geq 4$.  Let $\ell$ be a prime divisor
of~$p$.  Then $\ell<p$, so $\ell\mid (p-1)!$.  Also,
$$
  \ell \mid p \mid ((p-1)! - 1).
$$
This is a contradiction, because a prime can't divide a number $a$ and
also divide $a-1$, since it would then have to divide $a-(a-1)=1$.
\end{proof}

\begin{example}
When $p=17$, we have
$$
2\cdot 3\cdot \cdots \cdot 15
  = (2\cdot 9)\cdot(3\cdot 6)\cdot(4\cdot 13)\cdot
    (5\cdot 7)\cdot(8\cdot 15)\cdot(10\cdot 12)\cdot(14\cdot 11)
  \con 1\pmod{17},$$
where we have paired up the numbers $a, b$ for 
which $ab\con 1\pmod{17}$.
\end{example}

Let's test Wilson's Theorem in PARI:
\begin{verbatim}
? wilson(n) = Mod((n-1)!,n) == Mod(-1,n)   
? wilson(5)
%9 = 1
? wilson(10)
%10 = 0
? wilson(389)
%11 = 1
? wilson(2001)
%12 = 0
\end{verbatim}

\hd{Warning:} In practice, this is a horribly inefficient way to check
whether or not a number is prime.

\section{The Chinese Remainder Theorem}
Sun Tsu Suan-Ching (4th century AD):
\begin{quote}
There are certain things whose number is unknown. Repeatedly divided
by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the
remainder is 2. What will be the number?
\end{quote}
In modern notation, Sun is asking us to solve the following system of
equations:
\begin{align*}
x &\con 2 \pmod{3}\\
x &\con 3 \pmod{5}\\
x &\con 2 \pmod{7}
\end{align*}
The Chinese Remainder Theorem asserts that a solution to Sun's
question exists, and the proof gives a method to find a solution.

\begin{theorem}[The Chinese Remainder Theorem]
Let $a, b\in\Z$ and $n,m\in\N$ such that
$\gcd(n,m)=1$.  Then there exists $x\in\Z$ such that
\begin{align*}
x&\con a\pmod{m}\\
x&\con b\pmod{n}
\end{align*}
\end{theorem}
\begin{proof}
The equation
$$
  tm \con b-a\pmod{n}
$$
has a solution $t$ since $\gcd(m,n)=1$.
Set $x=a+tm$.  We next verify that $x$ is a solution to the two equations.  
Then
$$
 x \con a + (b-a) \con b \pmod{n},
$$
and 
$$
 x = a+tm \con a\pmod{m}.
$$
\end{proof}

Now we can solve Sun's problem:
\begin{align*}
x &\con 2 \pmod{3}\\
x &\con 3 \pmod{5}\\
x &\con 2 \pmod{7}.
\end{align*}
First, we use the theorem to find a solution to the pair
of equations 
\begin{align*}
x &\con 2 \pmod{3}\\
x &\con 3 \pmod{5}.
\end{align*}
Set $a=2$, $b=3$, $m=3$, $n=5$.
Step 1 is to find a solution to $t\cdot 3 \con 3-2\pmod{5}$.
A solution is $t=2$.  Then $x=a+tm=2+2\cdot 3 = 8$.
Since any $x'$ with $x'\con x\pmod{15}$ is also a solution to
those two equations, we can solve all three equations by
finding a solution to the pair of equations
\begin{align*}
x &\con 8 \pmod{15}\\
x &\con 2 \pmod{7}.
\end{align*}
Again, we find a solution to $t\cdot 15 \con 2-8\pmod{7}$.
A solution is $t = 1$, so 
$$x=a+tm=8+15=23.$$
Note that there are other solutions.  Any $x'\con x\pmod{3\cdot 5\cdot 7}$
is also a solution; e.g., $23+3\cdot 5\cdot 7 = 128$.

We can also solve Sun's problem in PARI:
\begin{verbatim}
? chinese(Mod(2,3),Mod(3,5))
%13 = Mod(8, 15)
? chinese(Mod(8,15),Mod(2,7))
%14 = Mod(23, 105)
\end{verbatim}


\section{Multiplicative Functions}
\begin{definition}
A function $f:\N\ra \Z$ is 
{\em multiplicative} if, whenever $m, n\in\N$ and $\gcd(m,n)=1$, we have
$$
   f(mn) = f(m)\cdot f(n).
$$
\end{definition}
Recall that the {\em Euler $\vphi$-function} is
$$
  \vphi(n) = \#\{a : 1\leq a \leq n\text{ and }\gcd(a,n)=1\}.
$$
\begin{proposition}
$\vphi$ is a multiplicative function.
\end{proposition}
\begin{proof}
Suppose that $m,n\in\N$ and $\gcd(m,n)=1$.
Consider the map
\begin{align*}\{c : & \,1 \leq c \leq mn \text{ and }\gcd(c,mn)=1\}
  \xrightarrow{\quad f\quad }\\
  &\{a : 1 \leq a \leq m \text{ and }\gcd(a,m)=1\}\cross
  \{b : 1 \leq b \leq n \text{ and }\gcd(b,n)=1\}\}
\end{align*}
defined by 
$$
 f(c) = (c\text{ mod } m, \quad{}c \text{ mod }n).
$$

\hd{The map~$f$ is injective:}  If $f(c)=f(c')$, then
$m\mid c-c'$ and $n\mid c-c'$, so, since $\gcd(n,m)=1$,
$nm\mid c-c'$, so $c=c'$.

\hd{The map~$f$ is surjective:} Given $a, b$ with $\gcd(a,m)=1$,
$\gcd(b,n)=1$, the Chinese Remainder Theorem implies that there
exists $c$ with $c\con a\pmod{m}$ and $c\con b\pmod{n}$.  We
may assume that $1\leq c\leq nm$, ans since $\gcd(a,m)=1$ and
$\gcd(b,n)=1$, we must have $\gcd(c,nm)=1$. Thus $f(c)=(a,b)$.

Because $f$ is a bijection, the set on the left has the same
size as the product set on the right.  Thus
$$
  \vphi(mn) = \vphi(m)\cdot \vphi(n).
$$
\end{proof}


\begin{example}
The proposition makes it easier to compute $\vphi(n)$. 
For example,
$$
   \vphi(12) = \vphi(2^2)\cdot \vphi(3) = 2\cdot 2 = 4.
$$
Also, for $n\geq 1$, we have
$$
   \vphi(p^n) = p^n - \frac{p^n}{p},
$$  
since $\vphi(p^n)$ is the number of numbers less than $p^n$
minus the number of those that are divisible by $p$.
Thus, e.g.,
$$
   \vphi(389\cdot 11^2) = 388 \cdot (11^2 - 11) = 388\cdot 110 = 42680.
$$
The $\vphi$ function is also available in PARI:
\begin{verbatim}
  ? eulerphi(389*11^2)
  %15 = 42680
\end{verbatim}
\end{example}

\begin{question}
Is computing $\vphi(\text{1000 digit number})$ really easy or really hard?
\end{question}

\end{document}