Math Help - Sum of two squares

1. Sum of two squares

Suppose that $p=x^2+y^2$ p is prime with x odd, y even. Show that $(\frac{x}{p})=1$
I know that $p \equiv 1(4)$, so I can use LQR for Jacobi symbol
$(\frac{x}{p})=(\frac{p}{x})=(\frac{x^2+y^2}{x})=(\ frac{y^2}{x})=1$, but for the last equality holds I need to have x,y are relatively prime. Are they relatively prime in this case? Can someone help me argue that they are relatively prime?

2. Originally Posted by namelessguy
Are they relatively prime in this case? Can someone help me argue that they are relatively prime?
Let $(x,y) = d$ then $d|x^2 + y^2 = p$ . Thus $d = 1 or p$ .
But $d = p \Rightarrow x,y \geq p$ but thats a contradiction