Question

wisons theorem proof?

Answer 1

Answer:

Let p be an odd prime number.  

Consider the group Up={equivalent classes of a|p>a>0, gcd(a,p)=1}  

(equivalent relation:a≡b(modp), binary operation:[a][b]=[ab]).  

p is a prime,so Up={[a]|1≤a≤p−1}.  

Since Up is a finite abelian group, (∏p−11[a])2=∏p−11[a]∗∏p−11[a−1]=[1],  

so [(p−1)!]2≡1(modp),  

therefore, either (p−1)!≡1(modp)(!) or (p−1)!≡−1(modp)(!!).  

Now we'll show that the first statement (!) is incorrect, thus forcing the second statement to be true.  

Assume that there exists an element [a],2≤a≤p−1, such that [a]2=[1],  

therefore a2≡1(modp),  

or p|(a−1)(a+1),  

so p|(a−1) and/or p|(a+1).  

so a−1≥p and/or a+1≥p  

Therefore, a=p−1  

So,only [1] and [p−1] are self-paired.  

Therefore, consider the product: x=[1]...[p−1],  

Apart from [1] and [p−1], all other elements are paired together with their inverses,  

so x=[(p−1)!]=[p−1] which doesn't equal to [1] since it requires p|p-2, which is true only when p=1,2.  

so [(p−1)!]≠[1].  

So it is false that (p−1)!≡1(modp).  

This forces (!!)to be true,so it must be true that (p−1)!≡−1(modp).  

This completes the proof.