proof of Wilson’s theorem

We first show that, if p is a prime, then (p-1)!-1(modp). Since p is prime, p is a field and thus, pairing off each element with its inversePlanetmathPlanetmathPlanetmath in the productPlanetmathPlanetmath (p-1)!=x=1p-1x, we are left with the elements which are their own inverses (i.e. which satisfy the equation x21(modp)), 1 and -1, only. Consequently, (p-1)!-1(modp).

To prove that the condition is necessary, suppose that (p-1)!-1(modp) and that p is not a prime. The case p=1 is trivial. Since p is composite, it has a divisor k such that 1<k<p, and we have (p-1)!-1(modk). However, since kp-1, it divides (p-1)! and thus (p-1)!0(modk), a contradictionMathworldPlanetmathPlanetmath.

Title proof of Wilson’s theorem
Canonical name ProofOfWilsonsTheorem
Date of creation 2013-03-22 12:09:09
Last modified on 2013-03-22 12:09:09
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 9
Author CWoo (3771)
Entry type Proof
Classification msc 11-00