# Wilson's Theorem

## Statement

If and only if p is a prime, then $(p-1)! + 1$ is a multiple of p. Written more mathematically, $(p-1)! \equiv -1 \pmod{p}$

## Proof

Wilson's theorem is easily verifiable for 2 and 3, so let's consider $p>3$. If p is composite, then its positive factors are among $1, 2, 3, \dots, p-1$ Hence, $gcd((p-1)!, p) > 1$, so $(p-1)! \neq -1 \pmod{p}$.

   However if $p$ is prime, then each of the above integers are relatively prime to $p$. So for each of these integers $a$ there is another $b$ such that $ab \equiv 1 \pmod{p}$. It is important to note that this $b$ is unique modulo $p$, and that since $p$ is prime, $a = b$ if and only if $a$ is $1$ or $p-1$. Now if we omit $1$ and $p-1$, then the others can be grouped into pairs whose product is congruent to one,


$2*3*4*\dots*(p-2) \equiv 1\pmod{p}$ Finally, multiply this equality by p-1 to complete the proof. Insert non-formatted text here