2
$\begingroup$

I thought of using Wilson's theorem for the proof.

First we have by Wilson's theorem $$(p - 1)!+1 \equiv 0 \pmod p$$ We can write this as $$(p - 2)!(p-1)+1 \equiv 0 \pmod p$$ $$(p - 2)!(p-1)+1=p(p-2)!-[(p-2)!-1]$$ Hence the right side p(p-2)! is divisible by p also the other term is divisible by p. So we have $$(p - 2)!-1 \equiv 0 \pmod p$$.

Is this correct?

$\endgroup$
8
  • 1
    $\begingroup$ You know wilson's theorem? Use that. $\endgroup$ Dec 26 '15 at 18:04
  • 1
    $\begingroup$ An equivalent statement is widely (and probably unreasonably) known as Wilson's Theorem. A search will yield many hits. $\endgroup$ Dec 26 '15 at 18:05
  • 1
    $\begingroup$ see here primes.utm.edu/notes/proofs/Wilsons.html $\endgroup$ Dec 26 '15 at 18:10
  • $\begingroup$ The proof is trivial, you simply invoke that there is a primitive element x, so that the numbers you need to multiply will be given by the powers of x. You need to omit p-1, but that's -1 so, you multiply by -1. $\endgroup$ Dec 26 '15 at 18:13
  • $\begingroup$ For the direction that the congruence fails when $p$ is composite, see this earlier Question. $\endgroup$
    – hardmath
    Dec 26 '15 at 18:35
2
$\begingroup$

The proof is very simple: first by Wilson theorem: $$(p-1)! \equiv -1 \pmod p$$ with $p$ prime, now we add $p$ to the right side: $$(p-1)(p-2)! \equiv (p-1) \pmod p$$ since $gcd(p-1,p)=1$ by cancellation law we can cancel $(p-1)$ on both sides $$(p-2)! \equiv 1 \pmod p$$ $$(p-2)!-1 \equiv 0 \pmod p$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.