Suppose p is an odd prime and a 2 Z is not divisible by p. It turns out that 1 always has exactly 2 square roots in Z/(p) for odd prime p: 1 and −1. (a) Please prove that a has a square root in Z/(p) if and only if a(p−1)/2 =1 mod p. Hint: The "only if" direction (a has a square root in Z/(p) =) a(p−1)/2=1) is very easy to prove if you remember FLIT. To prove the "if" direction you may assume the basic fact that (Z/(p)) always has a generator. How then, would you express a square root of a if there is one? (b) Please prove that if p=3 mod 4 and a has a square root mod p, then the square roots of a are exactly a(p+1)/4 and −a(p+1)/4. Hint: To show that a as no more than 2 square roots you’ll need to use the fact that a quadratic polynomial has at most 2 roots in Z/(p). We’ll say more about this fact later in class.