proof of Wilson’s theorem
We first show that, if is a prime, then Since is prime, is a field and thus, pairing off each element with its inverse in the product we are left with the elements which are their own inverses (i.e. which satisfy the equation ), and , only. Consequently,
To prove that the condition is necessary, suppose that and that is not a prime. The case is trivial. Since is composite, it has a divisor such that , and we have . However, since , it divides and thus a contradiction.
|Title||proof of Wilson’s theorem|
|Date of creation||2013-03-22 12:09:09|
|Last modified on||2013-03-22 12:09:09|
|Last modified by||CWoo (3771)|