Using the results of Problems 15 and 16. prove that if p is an odd prime number,
ID: 3077017 • Letter: U
Question
Using the results of Problems 15 and 16. prove that if p is an odd prime number, then (p - 1)! = -1 mod p. (This is known as Wilson's Theorem.) It is, of course, also true if p = 2.Explanation / Answer
It is easy to check the result when p is 2 or 3, so let us assume p > 3. If p is composite, then its positive divisors are among the integers 1, 2, 3, 4, ... , p-1 and it is clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p). However if p is prime, then each of the above integers are relatively prime to p. So for each of these integers a there is another b such that ab = 1 (mod p). It is important to note that this b is unique modulo p, and that since p is prime, a = b if and only if a is 1 or p-1. Now if we omit 1 and p-1, then the others can be grouped into pairs whose product is one showing 2.3.4.....(p-2) = 1 (mod p) (or more simply (p-2)! = 1 (mod p)). Finally, multiply this equality by p-1 to complete the proof.
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