Solve $x^2+y^2=2$ for $x,y\in\mathbb Q$.

Solve $x^2+y^2=2$ for $x,y\in\mathbb Q$.

Solve $x^2+y^2=2$ for $x,y\in\mathbb Q$.
I think the answer should be in terms of 1 integer variable $\in\mathbb Z$
only. I rewrite the equation to $(x+y)^2+(x-y)^2=2^2$, then by the formula
of pythagorean triples, $x+y=u^2-v^2,x-y=2uv,2=u^2+v^2$. How can I
proceed? Thanks.