Suppose that the odd prime p = x^2 + y^2 with x > y > 0. Find integers u and v (in terms of x and y) such that 2p = u^2 + v^2 with u > v > 0.

Okay, I know that this means 2p = 2x^2 + 2y^2, and then u^2 + v^2 = 2x^2 + 2y^2. I'm not sure it helps, but since p is prime and it can be written as the sum of two squares, then p ≡ 1 mod 4. I'm not sure what I should do next.