Talk:Wilson's Theorem

$(p-1)!=-1 (\bmod\ p)\,\!$  ???

yes ... and? --Matt Westwood 12:51, 3 January 2009 (UTC)

i knw a few proof. i asked for a confiramtion that is the theorem i know... xD -- Gamma 16:42, 3 January 2009 (UTC)

It's true iff p is prime I believe. --Joe (talk) 18:12, 3 January 2009 (UTC)

And for any composite number $n \neq 4$, we have $(n-1)!\equiv 0~mod~n$--Grambottle 16:54, 4 January 2009 (UTC)

... which of course shows the converse of Wilson's Theorem. First assume $n$ is not square, so $n = pq$ and both $p$ and $q$ will be factors of $(n-1)!$. If $n = p^2$ then if $n>4$ you'll have both $p$ and $2p$ as factors. So that bit's easy. --Matt Westwood 19:55, 4 January 2009 (UTC)