-1

Given prime $p>2$, prove that $2(p-3)! \equiv -1 \pmod{p}$.

I know that I have to use Wilson's theorem but I have no idea how to do so.

Wilson's Theorem : For prime $p$, $(p-1)! \equiv -1 \pmod{p}$.

Any help is appreciated.

Haran
  • 9,717
  • 1
  • 13
  • 47
Raj
  • 21
  • 1
    Raj, welcome to MSE. Users here prefer people who ask questions to also share their thoughts and attempts on solving the problem, rather than just a question statement or a related concept. Please edit your question and ask questions in the future accordingly. Thank you! – Haran Oct 20 '20 at 16:35

1 Answers1

0

You know that $(p-1)! \equiv -1 \pmod{p}$. Thus: $$(p-1)(p-2)(p-3)! \equiv -1 \pmod{p}$$ Writing $p-1 \equiv -1 \pmod{p}$ and $p-2 \equiv -2 \pmod{p}$ gives: $$(-1)(-2)(p-3)! \equiv -1 \pmod{p}$$ $$2(p-3)! \equiv -1 \pmod{p}$$ which proves the required.

Haran
  • 9,717
  • 1
  • 13
  • 47
  • Thank you! I have a question though - how were you able to replace the (p-1) and (p-2) with -1,-2 respectively. – Raj Oct 20 '20 at 19:57
  • Because $p \equiv 0 \pmod{p}$, we can see that $p-1 \equiv -1 \pmod{p}$. – Haran Oct 21 '20 at 02:54