Solutions - Problem Set 2

Problem Set 2 4x − y ≡ b (mod p), andb(x + y) ≡ (x − y)(x + y) ≡ a (mod p).If b is such that b · b ≡ 1(mod p), this last system is satisfied if and only ifx − y ≡ b (mod p), andx + y ≡ b · a (mod p)if and only ifx ≡ 2(b + b · a) (mod p)y ≡ 2(−b + b · a) (mod p)Thus, the number of elements in S is the same as the number of choices of b, where 1 ≤ b ≤ p−1.There are p − 1 such choices.Finally, consider the case that a ≡ 0 (mod p). We want to know the number of solutions,modulo p, to the congruencex 2 − y 2 ≡ 0 (mod p).This is much simpler. Now x 2 − y 2 = (x − y)(x + y). Hence (x, y) is a solution if and only if x ≡ y(mod p) or x ≡ −y (mod p). The 2p − 1 solutions in this case are{(0, 0), (1, 1), (1, p − 1), (2, 2), (2, p − 2), . . . , (p − 1, p − 1), (p − 1, 1)}.