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

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 $n\in \mathbb{N}$ if and only if $a$ is a square. We also prove that if $a\equiv 0,1,2\left(mod4\right)$, $n$ is odd, and $a^n+1$ is the sum of two squares, then $a^{\delta }+1$ is the sum of two squares for all $\delta |n$, $\delta >1$. Using Aurifeuillian factorization, we show that if $a$ is a prime and $a\equiv 1\left(mod4\right)$, then there are either zero or infinitely many odd $n$ such that $a^n+1$ is the sum of two squares. When $a\equiv 3\left(mod4\right)$, 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 $m|n$, and both $a^m+1$ and $n∕m$ are sums of two squares.