Find all rational solutions to $x^2+y^2=2$ 
Find all rational solutions to $x^2+y^2=2$

I rewrote the equation to $x= \sqrt{2-y^2}$ and thought that $x$ is rational if and only if $2-y^2$ is a square. So the only solution to the first problem is $x=1$ or $-1$ and $y=1$ or $-1$. 
Is there another approach? Thank you.
Edit: Another question: Is there maybe a way to show this using elliptic curves?

Given any rational $$ u^2 + v^2 = 1,$$ we find
$$ (u-v)^2 + (u+v)^2 = 2  $$
Meanwhile, given any integer Pythagorean triple $$  a^2 + b^2 = c^2,  $$ we get rational
$$ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1  $$
From the comments above, I guess I should add that all primitive solutions in integers to $x^2 + y^2 = 2 z^2,$ since this means $x \equiv y \pmod 2$ and $z$ odd, so we get
$$  \left( \frac{x-y}{2} \right)^2 + \left( \frac{x+y}{2} \right)^2 = z^2 $$
in integers and primitive. We can find all of these with the tandard parametrization for primitive Pythagorean triples.
