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.

The answer is yes. This is most easily seen via stereographic projection, as described here.
