Sum of the inverse squares of the hypotenuse of Pythagorean triangles

What is the sum of the series

S=152+1132+1172+1252+1292+1372+

where the sum is taken over all hypotenuse of primitive Pythagorean triangles.

By numerical computation, I found the sum to be 0.056840308812554488 correct to 18 decimal places. I would like to know if this sum has a closed form.

Using the general formula for primitive Pythagorean triangles, S=r>s1,gcd(r,s)=11(r2+s2)2

Trivially, for all primitive and non primitive Pythagorean triangles, the sum will be ζ(2)=π2/6 times the corresponding sum for primitive Pythagorean triangles which turn out to be about 0.09349856033594433852.

Motivation: We have equated the sum of the square of the sides of a right triangle to the square of the hypotenuse, so I was curious to know what the sum of the reciprocal of the square of the hypotenuse would be. Also since ζ(2) converges, and the density of hypotenuse is smaller than that of natural numbers, this sum must trivially converge.

Related question: What is the sum of the reciprocal of the hypotenuse of Pythagorean triangles?

Answer

Let us use the notations from

http://mathworld.wolfram.com/DoubleSeries.html

We fix a positive definite binary quadratic form q given by q(m,n)=am2+bmn+cn2, a,b,c integers.
We use sumations over the index set J=Z×Z{(0,0)} .
We define
S(q;s)=S(a,b,c;s)=(m,n)Jq(m,n)s=(m,n)J(am2+bmn+cn2)s ,S1(q;s)=S1(a,b,c;s)=(m,n)J(1)mq(m,n)s ,S2(q;s)=S2(a,b,c;s)=(m,n)J(1)nq(m,n)s ,S12(q;s)=S12(a,b,c;s)=(m,n)J(1)m+nq(m,n)s .
The last three sums are “twisted versions” of the first sum,
the “twist” occurs by using a character for the first parameter, for the second one, for both.
In our case, q(m,n)=m2+n2, and (a,b,c)=(1,0,1), we have a symmetric case (w.r.t. the exchange ac).

We will drop q below from notations in S?(q,s), since we use only the above quadratic form q. I decided during the edit operation that should bring us quick to numbers we can compute that it is better for the check to introduce the versions S+ for all sums, where the plus index indicates a further restriction to (m,n)J with
(+)m,n>0 .

From loc. cit. we extract the following relations:
S(s)=(m,n)J(m2+n2)s=4β(s)ζ(s) ,S12(s)=(m,n)J(1)m+n(m2+n2)s=4β(s)η(s)=4β(s)(121s)ζ(s) . Then the plus versions are:S+(s)=β(s)ζ(s)ζ(2s) ,S+12(s)=β(s)η(s)η(2s)=β(s)(121s)ζ(s)(1212s)ζ(2s) , which givesS+(s)S+12(s)=2β(s)(12s)ζ(s)2(122s)ζ(2s) .

Now let us search for a linear combination of the above sums
that correspond to summing q(m,n)^{-s} over the set of K of all (m,n) with positive (components with) different parity.
This is
\frac 12(\ S^+(s)-S^+_{12}(s)\ )\ .
So far can we write:

\begin{aligned}
&\beta(s)\;(1-2^{-s})\zeta(s) – (1-2^{-2s})\zeta(2s)
\\
&\qquad=\frac 12(\ S^+(s)-S^+_{12}(s)\ ) \\
&\qquad=\sum_{\substack{(m,n)\in K\\m,n> 0}} q(m,n)^{-s}\\
&\qquad=2\sum_{\substack{(m,n)\in K\\m>n> 0}} q(m,n)^{-s}\\
&\qquad=2\sum_{\substack{(m,n)\in K\\m>n> 0\\ d=(m,n)\text{ odd}}} q(m,n)^{-s}\qquad\text{ and with }M=m/d,\ N=n/d\\
&\qquad=2\sum_{d>0\text{ odd}}d^{-2s}
\sum_{\substack{(M,N)\in K\\M>N> 0\\ (M,N)=1}} q(M,N)^{-s}\\
&\qquad=
2(1-2^{-2s})\; \zeta(2s)\;
\color{blue}{
\sum_{\substack{(M,N)\in K\\M>N> 0\\ (M,N)=1}} q(M,N)^{-s}
}
\ .
\end{aligned}

The isolated sum in the last expresson is the sum we need, let us take it for s=2.

The value we obtain is:

\color{brown}{
\frac{\beta(2)\;\zeta(2)}{2(1+2^{-2})\zeta(4)}
-\frac 12\ = \frac{6C}{\pi^2} – \frac 12.}


\color{brown}{
\frac{\beta(2)\zeta(2)}{2(1+2^{-2})\zeta(4)} – \frac{1}{2}
= \frac{6C}{\pi^2} – \frac 1 2.}

where C is the Catalan constant.
Numerically:

sage: E = catalan * zeta(2) / 2 / (1+2^-2) / zeta(4) - 1/2 
sage: E.n()
0.0568403090661582

Attribution
Source : Link , Question Author : Nilotpal Sinha , Answer Author : Nilotpal Sinha

Leave a Comment