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 $x^2+y^2=1$.

Generally, if a conic curve with rational coefficients has one rational point $\rm\:P\:$ then it has infinitely many, since any rational line through $\rm\:P\:$ will intersect the curve in another point, necessarily rational, since if one root of a rational quadratic is rational then so is the other. Thus by sweeping lines of varying rational slopes through $\rm\:P\:$ we obtain infinitely many rational points on the conic. Further, projecting these points onto a line leads to a rational parametrization of the conic. For a very nice exposition see Chapter 1 of Silverman and Tate: Rational Points on Elliptic Curves.

For Pythagorean Triples there is even more structure. For example one may employ ascent in the Ternary Tree of Pythagorean Triples to simply and beautifully generate them all. Picking one simple ascending path yields this formula:

$\rm\quad\ (x,\:x+1,\:z) \to (X,\:X+1,\:Z),\ \ \ X\ =\ 3\:x+2\:z+1,\ \ \ Z\ =\ 4\:x+3\:z+2\:.\ \ $ For example

$\rm\quad (3,4,5)\to (20, 21, 29)\to (119, 120, 169)\to (696, 697, 985)\to (4059, 4060, 5741)\to\cdots$

Since $\rm\ a^2 + b^2 =\: c^2\ \Rightarrow\ (a/c)^2 + (b/c)^2 =\: 1\ $ this yields infinitely many solutions to your equation.