#### Vol. 12, No. 4, 2019

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When is $a^{n} + 1$ the sum of two squares?

### Greg Dresden, Kylie Hess, Saimon Islam, Jeremy Rouse, Aaron Schmitt, Emily Stamm, Terrin Warren and Pan Yue

Vol. 12 (2019), No. 4, 585–605
DOI: 10.2140/involve.2019.12.585
##### Abstract

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 $n\in ℕ$ if and only if $a$ is a square. We also prove that if $a\equiv 0,1,2\phantom{\rule{0.3em}{0ex}}\left(mod\phantom{\rule{0.3em}{0ex}}4\right)$, $n$ is odd, and ${a}^{n}+1$ is the sum of two squares, then ${a}^{\delta }+1$ is the sum of two squares for all $\delta \phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}n$, $\delta >1$. Using Aurifeuillian factorization, we show that if $a$ is a prime and $a\equiv 1\phantom{\rule{0.3em}{0ex}}\left(mod\phantom{\rule{0.3em}{0ex}}4\right)$, then there are either zero or infinitely many odd $n$ such that ${a}^{n}+1$ is the sum of two squares. When $a\equiv 3\phantom{\rule{0.3em}{0ex}}\left(mod\phantom{\rule{0.3em}{0ex}}4\right)$, we define $m$ to be the least positive integer such that $\left(a+1\right)∕m$ is the sum of two squares, and prove that if ${a}^{n}+1$ is the sum of two squares for $n$ odd, then $m\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}n$, and both ${a}^{m}+1$ and $n∕m$ are sums of two squares.

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##### Keywords
cyclotomic polynomials, Fermat's two squares theorem
##### Mathematical Subject Classification 2010
Primary: 11E25
Secondary: 11C08, 11R18