Let $a \neq 0$ in a field $F$. Determine the integers $n \geq 1$ such that $x+a$ is a factor of $x^n+a^n$

Question

Problem:Let $a \neq 0$ in a field $F$. Determine the integers $n \geq 1$ such that $x+a$ is a factor of $x^n+a^n$.

Attempt: If $n$ is an odd integer, say $n=2k+1$ for some $k \in \Bbb Z_{\geq 0}$, then

$x^{2k+1} + a^{2k+1}= (x+a)(x^{2k}-x^{2k-1}a+x^{2k-2}a^2 + ... +x^{2}a^{2k-2}-xa^{2k-1} +a^{2k})$, and so $x+a$ is a factor of $x^{2k+1} + a^{2k+1}$.

Now if $n$ is even say $n=2j$, I suspect that $x+a$ is not a factor of $x^n+a^n$, but I am not sure how to show this is true. Any tips appreciated.

Answer 1

$x+a$ is a factor of a polynomial $P(x)$ if and only if $P(-a)=0$.

If $P(x)=x^n+a^n$, then $P(-a)=(-a)^n+a^n$ which is zero for odd $n$, and for even $n$ it is $2a^n$. Can that be zero?