I asked this question [in MSE](https://math.stackexchange.com/questions/3728967/quaternionic-and-octonionic-analogues-of-the-basel-problem) around 3 months ago but I have received no answer yet, so following the suggestion in the comments I decided to post it here.


It is a well-known fact that


$$\sum\_{0\neq n\in\mathbb{Z}} \frac{1}{n^k} = r\_k (2\pi)^k$$


for any integer $k>1$, where $r\_k$ are rational numbers which can be given explicitly in terms of Bernoulli numbers. For example, for $k=2$ the sum equals $\pi^2/3$ (this is essentially the [Basel problem](https://en.wikipedia.org/wiki/Basel_problem)), and for $k=4$ it equals $\pi^4/45$. Note that for odd $k$ the sum vanishes.


The theory of [elliptic curves with complex multiplication](https://en.wikipedia.org/wiki/Complex_multiplication) allows us to extend this result to systems of complex integers such as the [Gaussian integers](https://en.wikipedia.org/wiki/Gaussian_integer), or more generally the ring of integers in an imaginary quadratic number field of class number 1. Namely, for $k>2$ we have


$$\sum\_{0\neq \lambda\in\mathbb{Z[\omega]}} \frac{1}{\lambda^k} = r\_k \varpi^k,$$


where again $r\_k$ are rational constants and $\varpi \in \mathbb{R}$ (the "complex $2\pi$") depends only on the ring $\mathcal{O}=\mathbb{Z[\omega]}$ and is an algebraic multiple of a so-called [Chowla–Selberg period](https://en.wikipedia.org/wiki/Chowla%E2%80%93Selberg_formula), given by a product of powers of certain gamma factors (note that the sum is always a real number since it is invariant under conjugation). For example, for the [Eisenstein](https://en.wikipedia.org/wiki/Eisenstein_integer) ($\omega = (1+\sqrt{3} i)/2$), Gaussian ($\omega = i$) and [Kleinian](https://en.wikipedia.org/wiki/Kleinian_integer) ($\omega = (1+\sqrt{7} i)/2$) integers, we have respectively


$$\varpi\_3 = 3^{-1/4} \sqrt{2\pi} \left(\frac{\Gamma(1/3)}{\Gamma(2/3)}\right)^{3/2}, \quad \varpi\_4 = 4^{-1/4} \sqrt{2\pi} \left(\frac{\Gamma(1/4)}{\Gamma(3/4)}\right), \quad \varpi\_7 = 7^{-1/4} \sqrt{2\pi} \left(\frac{\Gamma(1/7)\Gamma(2/7)\Gamma(4/7)}{\Gamma(3/7)\Gamma(5/7)\Gamma(6/7)}\right)^{1/2}.$$


For higher class numbers there is a similar formula, though in that case $r\_k$ will in general not be rational but algebraic. A nice exposition of this result can be found in Section 6.3 of [Zagier - Elliptic Modular Forms and Their Applications](https://people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/978-3-540-74119-0_1/fulltext.pdf).




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My question is whether this is still true for hypercomplex number systems, such as the [Hurwitz integers](https://en.wikipedia.org/wiki/Hurwitz_quaternion) or the [octonionic integers](https://en.wikipedia.org/wiki/Octonion#Integral_octonions). Define $$S\_k[\mathcal{O}] = \sum\_{0\neq \lambda\in\mathcal{O}} \frac{1}{\lambda^k}$$ for $k>\operatorname{dim} \mathcal{O}$, where $\mathcal{O}$ is now an order in a totally definite rational quaternion/octonion algebra of class number 1. The restriction on $k$ is so that the sum converges absolutely.



> 
> **Subquestion 1:** Do we have $S\_k[\mathcal{O}] = r\_k \varpi^k$ for some rational sequence $r\_k$ and some real number $\varpi$ depending only on $\mathcal{O}$ (a "quaternionic/octonionic $2\pi$")?
> 
> 
> 


Obviously $\varpi$ will only be defined up to a nonzero rational factor. An equivalent question is whether $(S\_m[\mathcal{O}])^n/(S\_n[\mathcal{O}])^m$ is rational for any $m, n$ such that $S\_n[\mathcal{O}]\neq 0$.



> 
> **Subquestion 2:** If so, can (some fixed choice of) $\varpi$ be expressed in terms of
> known constants such as $\zeta'(-1)$ or $\zeta'(-3)$?
> 
> 
> 


The reason I'm mentioning these particular constants is that in the previous cases (real and complex) the period $\varpi$ turns out to be equal to $e^{-\zeta'(\mathcal{O},0)/\zeta(\mathcal{O},0)}$ up to an algebraic factor, where the zeta function attached to the ring of integers $\mathcal{O}=\mathbb{Z}$ or $\mathbb{Z[\omega]}$ is defined as


$$\zeta(\mathcal{O},s) = \sum\_{0\neq \lambda\in\mathcal{O}} |\lambda|^{-s}.$$


(This is in general not the same as the previous sums, note the absolute value.) In the case that $\mathcal{O}$ is instead a quaternionic or octonionic order, the logarithmic derivative of this zeta function at $s=0$ can be expressed in terms of $\zeta'(-1)$ or $\zeta'(-3)$ respectively, where $\zeta(s)$ is the ordinary Riemann zeta function.




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I calculated a few sums numerically for the ring of Hurwitz quaternions. The result is $$S\_6[\mathcal{O}] \approx 10.76,\quad S\_8[\mathcal{O}] \approx 1.196,\quad S\_{12}[\mathcal{O}] \approx 23.9905.$$


Unfortunately the calculations take a lot of time, and the precision is not enough to determine whether e.g. $S\_{12}[\mathcal{O}]/(S\_6[\mathcal{O}])^2$ is rational to any degree of confidence.


I also found the recent paper [Period Relations for Quaternionic Elliptic Functions](https://arxiv.org/abs/2002.06140) by Z. Amir-Khosravi that refers to previous works by R. Fueter ([Über die Quaternionenmultiplikation regulärer vierfachperiodischer Funktionen](https://doi.org/10.1007/BF02153625)) and R. Krausshar ([Generalized Analytic Automorphic Forms in Hypercomplex Spaces](https://www.springer.com/gp/book/9783764370596)). A certain $3$-parameter family of quaternionic Eisenstein-like functions associated to a lattice in $\mathbb{R}^4$ is introduced, and shown to enjoy period-like relations resembling those in the complex case. Unfortunately, the form of these functions is restricted by quaternionic regularity to contain factors of the quaternionic norm (c.f. equations (2.5)-(2.7) in the paper), and as far as I can see they aren't directly related to the sums of pure powers I'm interested in.



Quaternionic and octonionic analogues of the Basel problem
This isn't really a full answer, but it's too long for a comment, and perhaps it's informative all the same.


Your sum $S\_k[\mathcal{O}]$ can be written as the value at $s = k$ of the sum
$$\sum\_{0 \ne \lambda \in \mathcal{O}} \frac{\lambda^k}{Nm(\lambda)^s} = \sum\_{n \ge 1} a^{(k)}\_n n^{-s},$$
where $a^{(k)}\_n := \sum\_{N(\lambda) = n} \lambda^k$.


Now, I claim that $\sum\_{n \ge 1} a^{(k)}\_n q^n$ is the $q$-expansion of a modular form -- or something slightly more general, namely a quasi-modular form [\*] -- of weight $k + 2$ and some level depending on $\mathcal{O}$; for the Hurwitz integers the level is $\Gamma\_0(2)$. This should follow from thinking about Brandt matrices, which are a way of computing modular forms using quaternion algebras; see e.g. [this article](http://www2.math.ou.edu/%7Ekmartin/papers/basis-prob.pdf) by Kimball Martin.


Anyway, once you know what to look for, it's now quite easy to recognise the sequences $(a^{(k)}\_n)\_{n \ge 1}$ for small $k$. For example, when $k = 6$, what you get is exactly the $q$-expansion of $12f\_8$, where $f\_8$ is [the unique normalised modular cusp form of weight 8 and level 2](https://beta.lmfdb.org/ModularForm/GL2/Q/holomorphic/2/8/a/a/). So $S\_6[\mathcal{O}]$ is a value of the $L$-series of a modular form. In fact, we have $S\_6[\mathcal{O}] = 12 L(f\_8, 6) = 10.758540419274832757072...$, which agrees with your computations above. Similarly, unless I've slipped up in my computations, we have
$$S\_8[\mathcal{O}] = 12 \big( L(f\_{10}, 8) - L(f\_8, 7) \big) = 1.18636076594110...$$
where $f\_{10}$ is the cusp form of weight 10. Since the periods of $f\_{10}$ and $f\_{8}$ have essentially nothing to do with each other, this strongly suggests that there is no tidy algebraic relation between $S\_6[\mathcal{O}]$ and $S\_8[\mathcal{O}]$.


[\*] Quasi-modular forms are not too scary: they're exactly the ring of functions you get by starting with genuine modular forms and throwing in the function $E\_2 = 1 - 24\sum \sigma(n) q^n$.




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EDIT. Further numerical experiments suggest the following explicit formula: if $\mathfrak{S}(m)$ denotes the set of normalised newforms of level 2 and weight $m$, then for every $k \ge 6$ we seem to have
$$S\_k[\mathcal{O}] = 12\left( \sum\_{f \in \mathfrak{S}(k+2)} L(f, k) - \sum\_{f \in \mathfrak{S}(k)} L(f, k-1) \right).$$


