# Difference between revisions of "Wilson's Theorem"

## Statement

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

## 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, $2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}$

Finally, multiply this equality by $p-1$ to complete the proof.

## Example

Let ${p}$ be a prime number such that dividing ${p}$ by 4 leaves the remainder 1. Show that there is an integer ${n}$ such that $n^2 + 1$ is divisible by ${p}$.

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