When is \(a^{n} + 1\) the sum of two squares?

• Published in: arXiv.org 2016-09
• Main Authors: Dresden, Greg, Hess, Kylie, Islam, Saimon, Rouse, Jeremy, Schmitt, Aaron, Stamm, Emily, Terrin Warren, Pan, Yue
• Format: Article
• Language: English
• Subjects: Polynomials
• ISSN: 2331-8422
• DOI: 10.48550/arxiv.1609.04391

Using Fermat's two squares theorem and properties of cyclotomic polynomials, we prove assertions about when numbers of the form \(a^{n}+1\) can be expressed as the sum of two integer squares. We prove that \(a^n + 1\) is the sum of two squares for all \(n \in \mathbb{N}\) if and only if \(a\) is a perfect square. We also prove that for \(a\equiv 0,1,2\pmod{4},\) if \(a^{n} + 1\) is the sum of two squares, then \(a^{\delta} + 1\) is the sum of two squares for all \(\delta | n, \ \delta>1\). Using Aurifeuillian factorization, we show that if \(a\) is a prime and \(a\equiv 1 \pmod{4}\), then there are either zero or infinitely many odd \(n\) such that \(a^n+1\) is the sum of two squares. When \(a\equiv 3\pmod{4},\) we define \(m\) to be the least positive integer such that \(\frac{a+1}{m}\) is the sum of two squares, and prove that if \(a^n+1\) is the sum of two squares for any odd integer \(n,\) then \(m | n\), and both \(a^m+1\) and \(\frac{n}{m}\) are sums of two squares.