Proof: For all integers $x$ and $y$, if $x^2+ y^2= 0$ then $x =0$ and $y =0$ I need help proving the following statement:
For all integers $x$ and $y$, if $x^2+ y^2= 0$ then $x =0$ and $y =0$
The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, but I don't know where to begin.
Any help would be much appreciated.

An answer that avoids contradiction:
If $x^2 + y^2 = 0$ then $(x+y)^2 = 2xy$ and then $(x-y)^2 = -(x+y)^2$. But $(x-y)^2\geq 0$ for every $x,y$ and the same is true for $(x+y)^2$, hence we have
(1) $(x-y)^2 \geq 0$
(2) $ -(x+y)^2 \leq 0$
So $(x-y)^2 = (x+y)^2=0$. Therefore $x=y$ and $x=-y$, so the only solution is $x=0$ and $y=0$
