TITLE: Find all rational solutions to $x^2+y^2=2$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Let $\cal P$ be the set of rational points in the circle $\Gamma:x^2+y^2=1$. It is well known that $\cal P$ is dense in $\Gamma$ and that $$(a,b)\in \cal P \iff \left(\frac{a+b}{\sqrt 2}, \frac{a-b}{\sqrt 2}\right)\in \Gamma$$ The second fact is because of the parametrization of the Pythagorean triples and the simple verification of
$$\left(\frac{2st}{s^2+t^2}, \frac{ s^2-t^2}{ s^2+t^2}\right)\in \cal P \iff  \left(\frac{s(2t+s)-t^2}{\sqrt 2(s^2+t^2)}\right)^2+\left( \frac{t(2s+t)-s^2}{\sqrt 2(s^2+t^2)}\right)^2=1$$
This gives a parametrization of the solutions and shows that the set of these is dense in the circle $\Gamma$ also (because to each element of $\cal P$ it is associated a solution and reciprocally). The set of solutions is for $t,s$ rational with $st\ne 0$ 
$$\begin{cases}x=\frac{s(2t+s)-t^2}{(r^2+t^2)}\\y=\frac{t(2s+t)-s^2}{(r^2+t^2)}\end{cases}$$