Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$? Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$?
I'm fairly certain it does, but I'm hoping to see a rigorous proof of this statement. Thanks.

Here is my motivation. I'm working on a strange little problem. I'm working in a geometry over an ordered field. Suppose I have a circle $\Gamma$ with center $A$ passing through a point $B$. I want to prove that there are infinitely many points on $\Gamma$. Up to a change of variable, I'm considering the unit circle centered on the origin over $\mathbb{Q}$. To show there are an infinite number of points on $\Gamma$, it suffices to show there are an infinite number of solutions to $a^2+b^2=1$ for $a,b\in\mathbb{Q}$. I could then extend this to showing there are infinite number of solutions to $a^2+b^2=r^2$ for some $r$, which proves that any circle over $\mathbb{Q}$ has an infinite number of points. Then since any ordered field has a subfield isomorphic to $\mathbb{Q}$, I would be finished.

Just to make explicit something which was left implicit (or hyperlinked) in the other answers: for any rational $t$ we have
$\left(\frac{1-t^2}{1+t^2}\right)^2 + \left(\frac{2t}{1+t^2}\right)^2 = 1$.
This is just what you get by projecting the $y$ axis through the point $(-1,0)$ on the circle $x^2+y^2=1$. 
