(i) Prove Wilson's theorem: if p is prime then (p−1)!≡−1(modp).
Deduce that if p≡1(mod4) then
((2p−1)!)2≡−1(modp)
(ii) Suppose that p is a prime of the form 4k+3. Show that if x4≡1(modp) then x2≡1(modp).
(iii) Deduce that if p is an odd prime, then the congruence
x2≡−1(modp)
has exactly two solutions ( modulop) if p≡1(mod4), and none otherwise.