# 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.

## A quick aside on the height function

Which height function is natural is itself a natural question. Perhaps the most obvious height function would come from looking at the magnitude of a point on the circle after homogenization. Rational points on $X^2 + Y^2 = 2$ are the same as integer points on $X^2 + Y^2 = 2Z^2$. If $p = (x : y : z)$ is such a point, then perhaps the most natural definition of the size of this point would be $\lvert p \rvert = x^2 + y^2 + z^2$.

But since this point satisfies $x^2 + y^2 = 2z^2$, we have that
$$\lvert p \rvert = 3z^2,$$
which is essentially the same as the height we actually use. Ordering rational points $(x, y)$ on $X^2 + Y^2 = 2$ by the size of the denominator is the same ordering of points $p$ on $X^2 + Y^2 = 2Z^2$ by $\lvert p \rvert$.

Thus this actually is a natural notion of height.

## Strategy of proof

To prove this equidistribution, we do the following steps.

1. Parametrize rational points by projection,
2. Study the problem in the parametrization space,
3. Translate this problem into a lattice-point problem in regions,
4. Then we execute this lattice-point counting argument.

## Parametrization of points

Rational points on $X^2 + Y^2 = 2$ are the same as integer points on $X^2 + Y^2 = 2Z^2$. These points are parametrized by
$$(x:y:z) = (m^2 + 2mn – n^2 : m^2 – 2mn – n^2 : m^2 + n^2).$$
This is a classical argument. The idea is to find a rational point (like $(-1, -1)$) and then consider lines with rational slope $t$ through this point. Every rational point on the circle will appear from such a slope (with the exception of the single point $(-1, 1)$, which would have infinite slope).

Projecting from this point gives rational points in terms of the rational parameter $t = m/n$. Clearing denominators gives the result.

Although this is not a complicated argument, it can be done in sage.

var('x y t')
show(solve([
x^2 + y^2 == 2,
y == t*(x+1) + 1
], [x, y]))


## Understanding primitive points

When counting rational points, we do not want to count (for example) $(-1, -1)$ and $(-2/2, -2/2)$, as these are the same point. Stated projectively, we want to count each point $(x : y : z) \in \mathbb{P}^2(\mathbb{Z})$ precisely once (and use its simplest form as its height, of course). These are primitive points.

We need to find conditions for $m,n$ such that the parametrization of points in terms of $m$ and $n$ yields only the primitive points. Each primitive point corresponds to a pair $(\pm m,\pm n)$ such that $\gcd(m,n)=1$ and the parity of $m$ is opposite from the parity of $n$.

It is very easy to see that if either $\gcd(m,n) \neq 1$ or if the parities of $m$ and $n$ are the same, then the resulting point $(x : y : z)$ is not primitive.

Conversely, suppose that the resulting point $(x : y : z)$ is not primitive. Then $\gcd(m^2 – 2mn – n^2, m^2 + 2mn – n^2, m^2 + n^2) > 1$, and thus
$$\gcd(4mn, m^2 + n^2) = g > 1.$$
The prime factors of $4mn$ are exactly $2$ and the primes diving $m$ and $n$. If $p \mid g$ divides $m$ or $n$, then it divides $m^2 + n^2$, and thus $p \mid \gcd(m, n)$. If $2 \mid g$, then $2 \mid m^2 + n^2$, and thus the parities of $m$ and $n$ are the same.

Thus we see that primitive points are precisely those obtained from $(m, n)$ with $\gcd(m,n) = 1$, $m, n$ having opposite parity, and $m \geq 0$.

## Equidistribution in terms of the parametrization

In terms of the parametrization $$(x : y : z) = (m^2 – 2mn – n^2, m^2 + 2mn – n^2, m^2 + n^2),$$ one can explicitly check that
$$(1+i)(m-in)^2 = x + iy.$$
(Noam Elkies pointed this out to me). Thus points $(x, y)$ are equidistributed (when ordered by denominator) if and only if the numbers $m + in$ are angularly equidistributed (when ordered by $m^2 + n^2$).

To prove this, we study the following question: given an arc $I = [\theta_1, \theta_2]$, how many pairs $(m, n)$ (with $\gcd(m, n) = 1$, where $m$ and $n$ have opposite parity, and where $m^2 + n^2 \leq H$) project to the arc $I$ under $m+in \mapsto (m+in)/\lvert m + in \rvert$? We want to show that the answer to this question is linear in the length of the arc $I$; this would give the required equidistribution for $m + in$, and thus for our initial question.

## A lattice point problem

The points $m + in$ form the typical $\mathbb{Z}^2$ lattice when $m$ and $n$ are unconstrained. We can handle the gcd condition later, but the opposite parity condition is annoying. Points $m+in$ when $m$ and $n$ are constrained to have opposite parity do not form a lattice. For example, $(1, 2) + (1, 2) = (2, 4)$, which is unfortunate.

But points $m + in$ where $m$ and $n$ are constrained to have the same parity do form a lattice. Let $\Lambda_1$ be the typical $\mathbb{Z}^2$ lattice formed from all $m + in$ as $m, n$ range across $\mathbb{Z}$. Let $\lambda_2$ be the lattice formed from $m + in$ where $m, n$ have the same parity. The points we are interested in are precisely those points in $\Lambda_1 \backslash \Lambda_2$.

Now we ask: which points in $\Lambda_1$ and $\Lambda_2$ project to a given arc under the projection map $z \mapsto z/\lvert z \rvert$, and have $\lvert z \rvert \leq H$? These are precisely the points contained within the region
$$\Omega_H = \{ re^{i\theta} : 0 \leq r \leq H, \theta_1 \leq \theta \leq \theta_2 \}.$$
For a given lattice $\Lambda$, we can mimic Gauss’s heuristic for the Gauss Circle Problem to approximate the number of points $N(H; \Lambda)$ contained within the region $\Omega_H$. The Gaussian heuristic gives the estimate
$$N(H; \Lambda) = \mathrm{Area}(\Omega_H) \cdot \rho(\Lambda) + O(\lvert \delta \Omega_H \rvert) + O(1),$$
where $\rho(\Lambda)$ is the density of the lattice $\Lambda$ (how many lattice points there are per unit square, on average), and where $\lvert \delta \Omega_H \rvert = O(H)$ is the length of the boundary of the region $\Omega_H$.

The density of $\Lambda_1$ is $1$. The density of $\Lambda_2$ is $1/2$. Thus the number of points $(m, n)$ with $m, n$ having opposite parity that project to the arc $I$ and which have $m^2 + n^2 \leq H^2$ is given by
$$N(H) = \mathrm{Area}(\Omega_H) \cdot \tfrac{1}{2} + O(H) + O(1).$$

## Removing imprimitive points

Finally, we restrict attention to primitive points (equivalently, points where $\gcd(m, n) = 1$) through an inclusion-exclusion argument. Notice that if $(n, m)$ is a point in our region, then the points $(3n, 3m)$, $(5n, 5m)$, and so on are also in the region. (But not the point $(2n, 2m)$, as this violates the parity constraint).

To remove them, we can thus use inclusion-exclusion (e.g. remove points such that $(3n, 3m)$ are counted, and points such that $(5n, 5m)$ are counted, but then add back in points such that $(15n, 15m)$ should be counted, etc.). This is accomplished by the mobius sum $$\widetilde{N}(H) = \sum_{d \leq H, 2 \nmid d} \mu(d) N(h/d).$$

From the Gaussian heuristic and the fact that $\mathrm{Area}(\Omega_H) =A H^2$ for the constant $A = \pi \cdot (\theta_2 – \theta_1)/2\pi$, we have that
$$\widetilde{N}(H) = A H^2 \cdot \frac{1}{2} \sum_{d \leq H, 2 \nmid d} \frac{\mu(d)}{d^2} + O\Big( \sum_{d \leq H} \frac{H}{d} \Big) + O(H).$$
The error term is of size $O(H \log H)$. The main can be understood by extending the sum to all $d \geq 1$ with $2 \nmid d$. This introduces an error term bounded by
$$\sum_{d \geq H} \frac{H^2}{d^2} \ll H,$$
which is swallowed by the other error term. And thus we have that
$$\widetilde{N}(H) = \pi\frac{\theta_2 – \theta_1}{2\pi} \frac{H^2}{2} \sum_{d \geq 1, 2 \nmid d} \frac{\mu(d)}{d^2} + O(H \log H).$$

The sum is exactly $$\frac{1}{\zeta^{(2)}(2)} = \frac{8}{\pi^2},$$ where $\zeta^{(2)}(s)$ is the Riemann zeta function, but with the $2$-factor of the Euler product removed. This gives the evaluation $$\widetilde{N}(H) = \frac{\theta_2 – \theta_1}{2\pi} H^2 \frac{4}{\pi} + O(H \log H).$$

And thus the number of points that project to the arc $I$ is linear in the length of the arc, as expected. The points equidistribute as claimed.

## An alternate computation for the total number of points

From the relationship $(x + iy) = (1 + i)(m – in)^2$, an arc of length $\ell$ in $(m, n)$ coordinates will correspond to an arc of length $2\ell$ in $(x + iy)$ coordinates.

On the other hand, $N(H)$ as defined above counted all $n, m$ of opposite parity, with $n^2 + m^2 \leq H^2$, such that $m+in$ projected to the arc $I$. This includes points with $m$ coordinate positive and negative. But from the analysis of primitive parameters above, we see that this double counts actual lattice points $x + iy$ (as $(-m, -n)$ and $(m, n)$ correspond to the same lattice point).

These two facts are deeply related. The effect is that the asymptotic $\widetilde{N}(H)$ above counts lattice points in $(x, y)$ coordinates. (That is, the doubling of the arc length and the double-counting of points cancel each other).

It follows that if we were to take an arc of length $2\pi$, i.e. take the whole circle, we have computed that there will be
$$\widetilde{N}(H) = \frac{4}{\pi}H^2 + O(H \log H)$$
points in the circle.

We can compute this in another way.

Let $A(b)$ denote the number of rational points on $x^2 + y^2 = 2$ of the (reduced) form $(a/b, c/b)$. We see that
$$\sum_ {d \mid b} A(d)= \#\{(a, c) \in \mathbb{Z}^2 : a^2 + c^2 = 2b^2\} = r_ 2(2b^2) = r_ 2(b^2),$$
where $r_2(n)$ is the number of ways of writing $n$ as a sum of $2$ squares. The function $r_2(n)/4$ is multiplicative, and we can compute the Dirichlet series
$$\sum _{n \geq 1} \frac{A(n)}{n^s} = \frac{4 \zeta(s) L(s, \chi _4)}{(1 + 2^{-s})\zeta(2s)},$$
where $\chi _4 = (\frac{-1}{\cdot})$ is the non-trivial character of modulus $4$. This can be seen upon comparing Euler products.

A simple application of Perron’s formula then shows that the number of rational points on $x^2 + y^2 = 2$ of the (reduced) form $(a/b, c/b)$ with $b \leq X^2$ is
$$\sum _{b \leq X^2} A(b) = \frac{4}{\pi} X^2 + O(X^{\frac{4}{3} + \epsilon}).$$
The leading asymptotics agree! (And a more sophisticated analysis in the Perron estimate would allow one to lower the error there).

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