proof of Wilson’s theorem result

set of primesWe 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 $\mathrm{1}\mathrm{\le }k\mathrm{\le }p\mathrm{-}\mathrm{1}\mathrm{,}p\mathrm{\in }\mathrm{P}\mathrm{\iff }\mathrm{(}p\mathrm{-}k\mathrm{)}\mathrm{!}\mathrm{(}k\mathrm{-}\mathrm{1}\mathrm{)}\mathrm{!}\mathrm{\equiv }{\mathrm{(}\mathrm{-}\mathrm{1}\mathrm{)}}^k\mathrm{(}\mathrm{mod}p\mathrm{)}$

Proof.

If $p$ is a prime, then:

$$\begin{array}{c}(p-k)!\equiv (p-1)!\overline{(p-1)}\mathrm{\cdots }\overline{(p-k+1)}\equiv (p-1)!\overline{(-1)}\mathrm{\cdots }\overline{(1-k)}=\hfill \\ \hfill =(p-1)!{(-1)}^{k-1}\overline{(k-1)!}(modp),\end{array}$$

and since $(p-1)!\equiv -1(modp)$ (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 , and so $p$ would not be relatively prime to $(p-1)!(k-1)!$ as the initial hypothesis implies.∎