Suppose that $\displaystyle p=x^2+y^2$ p is prime with x odd, y even. Show that $\displaystyle (\frac{x}{p})=1$

I know that $\displaystyle p \equiv 1(4)$, so I can use LQR for Jacobi symbol

$\displaystyle (\frac{x}{p})=(\frac{p}{x})=(\frac{x^2+y^2}{x})=(\ frac{y^2}{x})=1$, 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?