If p is odd prime (!=2), how to proof that x^2 == 1 (mod p) has only 2 solutions

What theorems should I look into if not...

Thank you.

Since x^2 = 1 (mod p), we have p | (x^2 - 1).
==> p | (x - 1)(x + 1)
==> p | (x + 1) or p | (x - 1), since p is prime
==> x = -1 (mod p) or x = 1 (mod p), which are the two distinct solutions mod p.

I hope this helps!