# Calculating that 13 is a Wilson Prime

I'm confused by the idea of a Wilson Prime. The theorem states that $$p^2=(p-1)!+1$$

This makes sense for $5$: $$5^2=(4\times3\times2)+1$$ so $5^2=25$

But it makes no sense to me for $13$: $$13^2=4790016001$$ Clearly I am as far from a mathematician as possible. If you can help in very simple terms...

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Which theorem states that $p^2=(p-1)!+1$? Wilson's theorem states that $p$ divides $(p-1)!+1$ for all primes $p$. A Wilson prime is a prime $p$ such that $p^2$ divides $(p-1)!+1$. – Hagen von Eitzen Dec 27 '12 at 18:48

A Wilson prime is a prime $p$ such that $p^2$ divides $(p-1)!+1$.

So you need to check that $169$ divides $12!+1$.

Remark: By Wilson's Theorem (called that despite the fact it was never proved by Wilson) $(p-1)!+1$ is always divisible by $p$ if $p$ is prime. The prime $p$ is called a Wilson prime if $(p-1)!+1$ has at least one "extra" factor of $p$.

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"...because it was never proved by Wilson" or "despite the fact that it was never proved by Wilson?" – Thomas Andrews Dec 27 '12 at 18:53

To prove that $13$ is a Wilson prime, you need to show that

$$12! \equiv -1 \pmod {13^2} \,.$$

To make the computations faster, observe that

$$12!= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot (2 \cdot 5)\cdot 11 \cdot (4 \cdot 3)$$

We split the product in three:

$$2 \cdot 3 \cdot 4 \cdot 7 =(2 \cdot 7)(3 \cdot 4)=(13+1)(13-1)\equiv -1 \pmod {13^2}$$ Also $$6 \cdot 11 \cdot 8 \cdot 2 \cdot 4 = (6 \cdot 11)(8 \cdot 8)=(5 \cdot 13+1)(5 \cdot 13-1) \equiv -1 \pmod{13^2}$$

and $$5 \cdot 9 \cdot \cdot 5 \cdot 3=25 \cdot 27=(2 \cdot 13-1)(2 \cdot 13+1) \equiv -1 \pmod {13^2} \,.$$

Multiplying we get

$$12! \equiv -1 \pmod{13^2} \,.$$

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