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.

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


$$(p-2)(p-1)=2\pmod p\;\;$$

  • $\begingroup$ Could you please explain how can you say that? $\endgroup$ – user58905 Jan 19 '13 at 17:47
  • 3
    $\begingroup$ You know some modulo arithmetic? Well $$(p-2)(p-1)=p^2-3p+2=2\pmod p...$$ Or simpler: $\,(p-2)(p-1)=(-2)(-1)\pmod p=2\pmod p\,$ . $\endgroup$ – DonAntonio Jan 19 '13 at 17:50

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