Solutions of $x_1^2+x_2^2-x_3^2=n^2$ with small $x_3$, revisited

Abstract

Let $r(N)$ be the number of representations of $N$ as a sum of two squares. For every integer $n\ge 2$ we prove an asymptotic formula for ${\sum }_{x\equiv e\mathrm{mod}2}\Phi (x/M)r(n^2+x^2)$, that is, for the number of integral solutions of $x_1^2+x_2^2-x_3^2=n^2$ with $x_3\equiv e\mathrm{mod}2$ in a range $|x_3|\ll M$, with smooth weight $\Phi$. The formula has an explicit secondary term and a power saving in the error term, and it is non-trivial as soon as $M\ge n^{5/6+\epsilon }$, that is, $M\ge D^{1/2-1/12+\epsilon }$ with $D=n^2$. The leading term is $\frac{4}{\pi }(\log n)\int \Phi (x/M)dx$, independently of the arithmetic of $n$; this corrects the main term stated in the published version (Ramanujan J. 63 (2024), 293-337). The proof factorises $n+i{x}_3$ in $\mathbb{Z}[i]$, which leads to Kloosterman sums. These are estimated by Weil's bound, combined with a Plancherel argument that exploits the curvature of the underlying quadric. This paper was written in collaboration with Claude (Opus 5.5, High), an AI model developed by Anthropic.

arXiv commentv2: completely rewritten. Corrects the main term of the published version (Ramanujan J. 63 (2024) 293-337; v1 is an earlier preprint), see Remark 1.1. New elementary proof, range M >= n^{5/6+eps}, all n. Spectral treatment: arXiv:2610.15706. 13 pages. Written with Claude (Opus 5.5), an AI model by Anthropic; see Subsection 1.1