What is the closed form of this sum? $$ S = \sum_{k\ge1, r>s\ge 1}\frac{1}{k^2(r^2 + s^2)^2} $$

Note: Though originally posted for Pythagorean triplets, their was an flaw in the question which changed the meaning of the question and was answered accordingly. I will post a separate question with the original question on Pythagorean triangles.

Update: I have posted the related question of Pythagorean triangles in the link belwo

What is the sum of the reciprocal of the square of hypotenuse of Pythagorean triangles?

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    $\begingroup$ @PaulSinclair Anything with a square root is not a Pythagorean triangle where all sides must be natural numbers. $\endgroup$ Oct 21, 2019 at 17:00
  • $\begingroup$ This $S$ is an interesting sum. But it is unclear what it should have to do with the geometric explanation given in the question. $\endgroup$ Oct 21, 2019 at 18:34
  • $\begingroup$ @ChristianBlatter Actually I came up with this sum geometrically while wondering what values of the sum of the reciprocal of hypotenuse squares would converge to so to be fair, without the geometrical interpretation, I would have no reason to think about this sum in the first place. $\endgroup$ Oct 21, 2019 at 19:05
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    $\begingroup$ @NilotpalKantiSinha - Sorry, I was thinking just "right triangle". But that just means I picked a poor example. The problem I'm pointing out is still there. So a better example would be the triangle $(12, 16, 20)$, which is included for both $k = 1, r = 16, s = 12$ and for $k = 4, r =4, s = 3$. You need to restrict to $r, s$ being relatively prime, and one or the other should be even. $\endgroup$ Oct 21, 2019 at 23:08
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    $\begingroup$ @Blue Fair point. I will post a separate question for Pythagorean triplets and make this question for the sum $S$ as defined in the original question. $\endgroup$ Oct 22, 2019 at 17:50

1 Answer 1


\begin{align*} S &= \sum_{k \geq 1} \frac{1}{k^2} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2} \\ &= \frac{\pi^2}{6} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2} \end{align*}

Note that $\frac{1}{(r^2 + s^2)^2}$ is unchanged by $\{r \mapsto -r\}$, $\{s \mapsto -s\}$, and $\{r \leftrightarrow s\}$. Applying these symmetries, we obtain eight copies of the region $r > s \geq 1$ in the $r$-$s$ plane. It lacks the four diagonals, each with value $\displaystyle \sum_{s \geq 1} \frac{1}{(s^2 + s^2)^2} = \frac{\pi^4}{360}$, the four coordinate rays starting at $\pm 1$, each with value $\displaystyle \sum_{s \geq 1} \frac{1}{(0^2 + s^2)^2} = \frac{\pi^4}{90}$, and the origin, which we will continue to exclude. Placing a prime on the summation sign to indicate that we omit the origin, $$ \sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 8 \sum_{r > s \geq 1} \frac{1}{(r^2 + s^2)^2} + 4 \cdot \frac{\pi^4}{360} + 4 \cdot \frac{\pi^4}{90} $$ Borwein and Borwein [1] (p. 291) (see also (38) at Double Series on MathWorld) show $$ \sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 4 \beta(2) \zeta(2) = \frac{2}{3} \pi^2 K \text{,} $$ where $\beta$ is Dirichlet's Beta function, $\zeta$ is the Riemann Zeta function, and $K$ is Catalan's constant.

Utilizing these facts, $$ S = \frac{\pi^2}{6} \cdot \frac{12 K \pi^2 - \pi^4}{144} = 0.1264945807\dots \text{.} $$

[1] Borwein, Jonathan M. and Peter B. Borwein, "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity", New York: Wiley, 1987.

P.S. Mathematica 11.3 does not evaluate the triple sum symbolically. It numerically estimates $0.126494$, which is comfortably close to the above result.

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    $\begingroup$ @NilotpalKantiSinha : Of course. The sum would be smaller, approximately $0.11687$. $\endgroup$ Oct 22, 2019 at 14:33
  • $\begingroup$ @NilotpalKantiSinha - you also have to require that either $r$ or $s$ is even. If they are both odd, then all three sides of the triangle will be divisible by $2$, which will duplicate another triangle in the sum. IIRC, every pythagorean triple is uniquely representable by $(k(r^2-s^2), 2krs, k(r^2 + s^2))$ with $\gcd(r,s) = 1$ and one of $r,s$ even. $\endgroup$ Oct 22, 2019 at 16:21
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    $\begingroup$ @NilotpalKantiSinha : Well, I don't currently see how to do the sum with the constraint $\gcd(r,s) = 1$, nor for Paul Sinclair's more complete set of constraints "$\gcd(r,s) = 1$ and one of $r,s$ is even". $\endgroup$ Oct 22, 2019 at 16:22
  • $\begingroup$ @NilotpalKantiSinha : With $\gcd(r,s) = 1$ and one of $r,s$ is even, the sum seems to be near $0.0935$ (and I don't trust the last digit). $\endgroup$ Oct 22, 2019 at 16:36
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    $\begingroup$ @NilotpalKantiSinha : I may have an idea for the restricted sum, but it will have to wait until tomorrow. $\endgroup$ Oct 22, 2019 at 21:45

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