Is it known? A sum over lattice parallelograms of area one is equal to $\pi$

Question

I recently discovered a formula, my proof is really a high school proof in three lines.

$$4\sum_{x, \, y \, \in \, \mathbb Z_{\geq 0}^2, \, \det(x \ \ y) = 1} \frac{1}{\lVert x\rVert^2\cdot\lVert y\rVert^2\cdot \lVert x+y\rVert^2} = \pi$$

We sum over all pairs of lattice vectors in the first quadrant such that the oriented area of the parallelogram spanned by them is one. Note that each primitive vector in the first quadrant appears as $x+y$ for some unique pair $x$ and $y.$ So it can be seen as a kind of zeta function for Gaussian integers.

It should be known to humanity but I cannot find anything similar. Who can suggest a source where to look for such formulae?

I will later post my proof here but before I encourage you to suggest another proofs =) Maybe this way we can discover some nice connections!

Added: I wrote about this in https://arxiv.org/abs/2410.10884

Answer 1

This is not a proof, but since you are interested in connections I have a conjectural generalization. I have verified it numerically in a number of small cases.

Let $f(x,y)=ax^2+bxy+cy^2$ be a positive definite quadratic form with $a,b,c\in\mathbb{Z}$, and let $A$ be the area of the ellipse $f(x,y) \leq 1$. Then $$ \sum (f(x_1,y_1)f(x_2,y_2)f(x_3,y_3))^{-1} = \frac{A^3}{\pi^2}, $$ where the sum is over all sets $\{(x_1,y_1),(x_2,y_2),(x_3,y_3)\}$ such that $x_1+x_2+x_3=y_1+y_2+y_3=0$ and $\left|\det \begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}\right| = 1$ (which implies the same for the other two pairs).

The motivation for this comes from Conway's topograph for quadratic forms. The three factors in each term of the sum are the numbers at the three faces that meet at each node of the topograph for $f$. Note that each vertex of the topograph is counted twice since we count $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ and $(-x_1,-y_1),(-x_2,-y_2),(-x_3,-y_3)$ separately (in Conway's language, the sum is over strict superbases, not lax superbases).

Presumably this can be proved along the same lines as David E Speyer's answer, but I have not yet tried to do so.

Update: This version of your formula was published by Hurwitz in 1905 in a paper titled "Über eine Darstellung der Klassenzahl binärer quadratischer Formen durch unendliche Reihen" (see §4, equation (8)). It can also be found Dickson, "History of the Theory of Numbers," Vol. III, page 167. The proof given there is essentially the same as David E Speyer's. (As an aside, his proof can be simplified by observing that the length of the edge connecting the images of $a/b$ and $c/d$ is $$ \frac{2}{\sqrt{(a^2+b^2)(c^2+d^2)}} $$ and then applying the formula $K=\frac{abc}{4R}$ he mentions in the comments.)

I found these references by looking for papers that mention summing over the vertices of a topograph. This led me to a very recent preprint by Cormac O'Sullivan, which contains Hurwitz's formula phrased in topographical terms (Theorem 9.1). O'Sullivan's source is this paper by Duke, Imamoḡlu, and Tóth.

All of these references contain a number of related formulas and connections to the theory of quadratic forms (not just positive definite, and not just binary) that might be of interest to you.

Answer 2

I am very curious how this identity was discovered. Here is a proof. If I have time, I'll clean it up a bit.

As is well known, there is a triangulation of the hyperbolic plane, with vertices at the points $\mathbb{Q} \cup \{ \infty \} \subset \mathbb{RP}^1$, such that there is an edge between $\tfrac{x_1}{y_1}$ and $\tfrac{x_2}{y_2}$ iff and only if $x_1 y_2 - x_2 y_1 = \pm 1$. Moreover, this edge is in two triangles, whose the third vertices are $\tfrac{x_1+x_2}{y_1+y_2}$ and $\tfrac{x_1-x_2}{y_1-y_2}$. Summing over all nonnegative integer matrices with $x_1 y_2 - x_2 y_1=1$ is the same as summing over all triangles in the right half of the hyperbolic plane. I'd rather sum over the whole hyperbolic plane, so my goal is to show that the sum is $\pi/2$, not $\pi/4$, once all triangles are used.

Now, use the standard parametrization of the Pythagorean triples on the unit circle: $$\phi(x,y) := \left( \frac{x^2-y^2}{x^2+y^2}, \frac{2xy}{x^2+y^2} \right).$$ If we have a triangle in the hyperbolic plane with vertices at $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, then we view $\phi(x_1, y_1)$, $\phi(x_2, y_2)$, $\phi(x_3, y_3)$ as the vertices of an ordinary Euclidean triangle inscribed in the unit circle. Let's compute the area of that triangle. It is

$$\frac{1}{2} \det \begin{bmatrix} \frac{x_1^2-y_1^2}{x_1^2+y_1^2} & \frac{x_2^2-y_2^2}{x_2^2+y_2^2} & \frac{x_3^2-y_3^2}{x_3^2+y_3^2} \\ \frac{2 x_1 y_1}{x_1^2+y_1^2} & \frac{2 x_2 y_2}{x_2^2+y_2^2} & \frac{2 x_3 y_3}{x_3^2+y_3^2} \\ 1&1&1 \\ \end{bmatrix}=$$ $$ \frac{1}{2 (x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2)} \det \begin{bmatrix} x_1^2-y_1^2 & x_2^2-y_2^2 & x_3^2-y_3^2 \\ 2 x_1 y_1 & 2 x_2 y_2 & 2 x_3 y_3 \\ x_1^2+y_1^2 & x_2^2+y_2^2 & x_3^2+y_3^2 \\ \end{bmatrix}.$$ The matrix factors as $$\begin{bmatrix} 1&0&-1 \\ 0&2&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} x_1^2 & x_2^2 & x_3^2 \\ x_1 y_1 & x_2 y_2 & x_3 y_3 \\ y_1^2 & y_2^2 & y_3^2 \\ \end{bmatrix}.$$ The first factor has determinant $4$ and the second is a Vandermonde matrix with determinant: $$(x_1 y_2 - x_2 y_1) (x_1 y_3 - x_3 y_1) (x_2 y_3 - x_3 y_2).$$ But each of the Vandermonde factors is assumed to be $1$. So, putting it all together, the area of the triangle is $$\frac{1}{2} \cdot 4 \cdot \frac{1}{2 (x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2)} = \frac{2}{(x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2)}.$$

The triangles cover the whole circle, so their total area is $\pi$, and we obtain that the sum of $\frac{1}{(x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2)}$ over all triangles is $\pi/2$, as desired.