# Wilson's theorem

Can you hint me on how to show that $2(p-3)!\equiv -1\pmod{p}$, for $p>2$ prime.

I that Wilson's theorem says that $(p-1)!\equiv-1\pmod{p}$, and that $(p-3)!=(p-3)(p-2)(p-1)!$, but I'm not seeing how to fit this together.

• Certainly not true that $(p-3)!=(p-3)(p-2)(p-1)!$. You mean $(p-1)!=(p-1)(p-2)(p-3)!$. – André Nicolas Jan 19 '13 at 17:31

$$(p-2)(p-1)=2\pmod p\;\;$$
• You know some modulo arithmetic? Well $$(p-2)(p-1)=p^2-3p+2=2\pmod p...$$ Or simpler: $\,(p-2)(p-1)=(-2)(-1)\pmod p=2\pmod p\,$ . – DonAntonio Jan 19 '13 at 17:50