## Points on X^2 + Y^2 = 2 equidistribute with respect to height

When you order rational points on the circle $X^2 + Y^2 = 2$ by height, these points equidistribute.

Stated differently, suppose that $I$ is an arc on the circle $X^2 + Y^2 = 2$. Then asymptotically, the number of rational points on the arc $I$ with height bounded by a number $H$ is equal to what you would expect if $\lvert I\rvert /2\sqrt{2}\pi$ of all points with height up to $H$ were on this arc. Here, $\lvert I\rvert /2\sqrt{2}\pi$ the ratio of the arclength of the arc $I$ with the total circumference of the circle.

This only makes sense if we define the height of a rational point on the circle. Given a point $(a/c, b/c)$ (written in least terms) on the circle, we define the height of this point to be $c$.

In forthcoming work with my frequent collaborators Chan Ieong Kuan, Thomas Hulse, and Alexander Walker, we count three term arithmetic progressions of squares. If $C^2 – B^2 = B^2 – A^2$, then clearly $A^2 + C^2 = 2B^2$, and thus a 3AP of squares corresponds to a rational point on the circle $X^2 + Y^2 = 2$. We compare one of our results to what you would expect from equidistribution. From general principles, we expected such equidistribution to be true. But I wasn’t sure how to prove it.

With helpful assistance from Noam Elkies, Emmanuel Peyre, and John Voight (who each immediately knew how to prove this), I learned how to prove this fact.

The rest of this note contains this proof.