Restricted sums of four integral squares

TL;DR

This paper presents a straightforward quaternionic proof of a recent mathematical result concerning the representation of numbers as sums of four integral squares, focusing on restricted sums.

Contribution

It introduces a simple quaternionic approach to prove a recent theorem on restricted sums of four integral squares, offering an alternative to previous proofs.

Findings

Quaternionic proof simplifies understanding of the theorem

Provides new insights into sums of four squares

Validates the recent result through an alternative method

Abstract

We give a simple quaternionic proof of a recent result of Goldmakher and Pollack on restricted sums of four integral squares.

Full text

In [1] the authors prove the following result:

We give here a different proof using a very simple computation in the ring of integral quaternions.

Since $(1-i-j-k)^{-1}=(1+i+j+k)/4$ we see that $f(R)=\{\beta\in R\mid\beta(1+i+j+k)/4\in R\}=\{\beta=b_{0}+b_{1}i+b_{2}j+b_{3}k\in R\ \mid\sum_{\nu\neq\mu}b_{\nu}\equiv b_{\mu}\bmod 4\text{ for }0\leq\mu\leq 3\}$.

For $\beta^{\prime}=b_{0}^{\prime}+b_{1}^{\prime}i+b_{2}^{\prime}j+b_{3}^{\prime}k\in X$ one easily sees that the $b_{\nu}^{\prime}$ are all congruent modulo $2$ and if they are even, the number of $b_{\nu}^{\prime}\equiv 2\bmod 4$ is $0,2$ or $4$. In the latter case, the congruence condition $\sum_{\nu\neq\mu}b_{\nu}^{\prime}\equiv b_{\mu}^{\prime}\bmod 4\text{ for }0\leq\mu\leq 3$ is satisfied. If the $b_{\nu}^{\prime}$ are all odd, it is satisfied if either one or three of them are congruent to $-1$ modulo $4$. In both cases we find, changing a sign if necessary, a $\beta=b_{0}+b_{1}i+b_{2}j+b_{3}k\in f(R)$ with $b_{0}=b_{0}^{\prime},b_{\nu}\in\{b_{\nu}^{\prime},-b_{\nu}^{\prime}\}$ for all $\nu$.

Obviously, for $\alpha=a_{0}+a_{1}i+a_{2}j+a_{3}k\in R$ with $\sum_{\nu=0}^{3}a_{\nu}^{2}=n,\sum_{\nu=0}^{3}a_{\nu}=T$ and $f(\alpha)=:b_{0}+b_{1}i+b_{2}j+b_{3}k$ we have $\sum_{\nu=1}^{3}b_{\nu}^{2}=4n-T^{2}$, which establishes the “only if” part of the the assertion.

Conversely, let $n\equiv T\bmod 2$ be such that $4n-T^{2}=\sum_{\nu=1}^{3}(b_{\nu}^{\prime})^{2}$ is a sum of three squares. For $\beta^{\prime}=T+b_{1}^{\prime}i+b_{2}^{\prime}j+b_{3}^{\prime}k$ we have $N(\beta)\equiv 4\operatorname*{Re}(\beta^{\prime})\bmod 8$, hence $\beta^{\prime}\in X$. By what we have shown about elements of $X$ there exists $\beta=T+b_{1}i+b_{2}j+b_{3}k\in f(R)$ with $N(\beta)=N(\beta^{\prime})=4n$, and $\alpha=a_{0}+a_{1}i+a_{2}j+a_{3}k\in R$ with $\beta=f(\alpha)$ is as required.

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