Wilson's Theorem

Theorem 1.1 (John Wilson's theorem, from the 1770s)   An integer $p>1$ is prime if and only if

$$(p-1)! \equiv -1 \pmod{p}.$$

Example 1.2  

? p=3
%1 = 3
? (p-1)! % 3
%2 = 2
? p=17
%3 = 17
? (p-1)!
%4 = 20922789888000
? (p-1)! % p
%5 = 16

Proof. We first assume that $p$ is prime and prove that $(p-1)! \equiv -1\pmod{p}$. If $a\in\{1,2,\ldots,p-1\}$ then the equation

$$ax\equiv 1\pmod{p}$$

has a unique solution $a'\in\{1,2,\ldots,p-1\}$. If $a=a'$, then $a^2\equiv 1\pmod{p}$, 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

$$2\cdot 3 \cdot \cdots \cdot (p-2) \equiv 1\pmod{p}.$$

Multiplying both sides by $p-1$ proves that $(p-1)! \equiv -1\pmod{p}$.

Next we assume that $(p-1)! \equiv -1\pmod{p}$ 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 $a-(a-1)=1$. $\qedsymbol$

Example 1.3   When $p=17$, we have

$$2\cdot 3\cdot \cdots \cdot 15 = (2\cdot 9)\cdot(3\cdot 6)\cdot(4\... ...(5\cdot 7)\cdot(8\cdot 15)\cdot(10\cdot 12)\cdot(14\cdot 11) \equiv 1\pmod{17},$$

where we have paired up the numbers $a, b$ for which $ab\equiv 1\pmod{17}$.

Let's test Wilson's Theorem in PARI:

? 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

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