Prove that if a is a primitive root of odd prime p, then a^((p-1)/2) ≡ -1 (mod p). ?

Question

Prove that if a is a primitive root of odd prime p, then a^((p-1)/2) ≡ -1 (mod p). ?

Answer 1

This should be straightforward if you understand what primitive root means.  A primitive root mod p is a generator of the multiplicative group of the field Z/pZ, or, in other words, it is an element a in Z/pZ with order phi(p) = p-1.  Therefore,

a^(p-1) ≡ 1 (mod p)  

(we know by FLT that this is true for all 0 < a < p )  

BUT

a^x !≡ 1 (mod p) for any 0 < x < p-1.

Since a^(p-1) is congruent to 1 mod p, a^((p-1)/2) must be congruent to either 1 or -1 mod p.  (Do you see why?)  

Since 0 < (p-1)/2 < p-1, it follows that, because a is a primitive root, a^((p-1)/2) !≡ 1 (mod p) (otherwise it would contradict the minimality of the order of a).  But then, the only other possibility is:

a^((p-1)/2) ≡ -1 (mod p).

Ask me if that was confusing, because I might have used something that you haven't done...