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# proof of Wilson’s theorem result

We denote by $\mathbb{P}$ the set of primes and by $\overline{x}$ the multiplicative inverse of $x$ in $\mathbb{Z}_{p}$.

###### Theorem (Generalisation of Wilson’s Theorem).

For all integers $1\leq k\leq p-1,\;p\in\mathbb{P}\Leftrightarrow(p-k)!(k-1)!\equiv(-1)^{k}\;\;(% \mathop{{\rm mod}}p)$

###### Proof.

If $p$ is a prime, then:

$(p-k)!\equiv(p-1)!\overline{(p-1)}\cdots\overline{(p-k+1)}\equiv(p-1)!% \overline{(-1)}\cdots\overline{(1-k)}=\\ =(p-1)!(-1)^{{k-1}}\overline{(k-1)!}\;\;(\mathop{{\rm mod}}p),$ |

and since $(p-1)!\equiv-1\;\;(\mathop{{\rm mod}}p)$ (Wilson’s Theorem, simply pair up each number — except $p-1$ and $1$, the only numbers in $\mathbb{Z}_{p}$ which are their own inverses — with its inverse), the first implication follows.

Now, if $p\!\mid\!(p-1)!(k-1)!-(-1)^{k}$, then $p\in\mathbb{P}$ as the opposite would mean that $p=ab$, for some integers $1<a,b<p$, and so $p$ would not be relatively prime to $(p-1)!(k-1)!$ as the initial hypothesis implies. ∎

## Mathematics Subject Classification

11-00*no label found*

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