Zsigmondy's Theorem for Sums

Theorem
Let $a > b > 0$ be coprime positive integers.

Let $n \ge 1$ be a (strictly) positive integer.

Then there is a prime number $p$ such that
 * $p$ divides $a^n + b^n$
 * $p$ does not divide $a^k + b^k$ for all $k < n$

with the following exception:


 * $n = 3$, $a = 2$, $b = 1$

Outline of Proof
We apply Zsigmondy's Theorem to $a^{2 n} - b^{2 n}$.

Proof
By Zsigmondy's Theorem, there exists a prime divisor $p$ of $a^{2 n} - b^{2 n}$ which does not divide $a^k - b^k$ for all $k < 2 n$ unless:
 * $n = 1$ and $a + b$ is a power of $2$
 * $n = 3$, $a = 2$, $b = 1$

In particular, $p$ does not divide $a^{2 k} - b^{2 k} = \paren {a^k - b^k} \paren {a^k + b^k}$ for $k < n$.

It remains to check the case $n = 1$ and $a + b$ a power of $2$.

We have to show that $a^2 + b^2$ has an odd prime divisor.

Since $a$ and $b$ are coprime, both $a$ and $b$ are odd.

By Square Modulo 4, $a^2 + b^2 \equiv 2 \pmod 4$.

Because $a > b > 0$, $a^2 + b^2 > 2$.

But $4 \divides 2^k$ for $k > 1$.

Thus $a^2 + b^2$ is not a power of $2$.

Hence $a^2 + b^2$ has an odd prime divisor.