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

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 $a\equiv 0,1,2(\mathrm{m}\mathrm{o}\mathrm{d}4),$ if $a^n+1$ is the sum of two squares, then $a^{\delta }+1$ is the sum of two squares for all $\delta | n,
\delta>1$\mathrm{.}UsingAurifeuillianfactorization,weshowthatif$a$ is a prime and $a\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$, then there are either zero or infinitely many odd $n$ such that $a^n+1$ is the sum of two squares. When $a\equiv 3(\mathrm{m}\mathrm{o}\mathrm{d}4),$ 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 $m\mathrm{\mid }n$, and both $a^m+1$ and $\frac{n}{m}$ are sums of two squares.