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

TL;DR

This paper characterizes when numbers of the form a^n + 1 can be expressed as the sum of two squares, using Fermat's theorem, cyclotomic polynomials, and Aurifeuillian factorization, revealing conditions based on properties of a and n.

Contribution

It provides new necessary and sufficient conditions for a^n + 1 to be a sum of two squares, extending classical results with novel factorization techniques and modular considerations.

Findings

a^n + 1 is sum of two squares iff a is a perfect square

For certain a, if a^n + 1 is sum of two squares, then so are a^δ + 1 for divisors δ of n

If a is prime and ≡ 1 mod 4, then a^n + 1 is sum of two squares for infinitely many odd n

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 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.