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?
 A: 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.
A: Since you already have a rational solution, one strategy is to find a line with rational slope which passes through that point. The line will also intersect the curve in a second spot, which gives another solution (unless the line is tangent to the curve.) This is similar to in elliptic curves when we find the line through 2 points to find a thrid point.
Let $m$ be a rational number, find the line with slope $m$ which passes through the point $(1,1)$:
$y=m*x+(1-m)$.
Plug this back into the curve to get
$x^2+(m*x+(1-m))^2=2$
Solve for $x$, to find $x=\frac{m^2-2m-1}{m^2+1}$.
Plug this back into the equation for the line to find $y=\frac{m^2+2m-1}{m^2+1}$
So, we get a rational solution for any rational number $m$.
Convince yourself that this generates every rational solution (except for the one with a vertical line, $(1,-1)$).
A: 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 $st\ne 0$ 
$$\begin{cases}x=\frac{s(2t+s)-t^2}{(r^2+t^2)}\\y=\frac{t(2s+t)-s^2}{(r^2+t^2)}\end{cases}$$
