Given an arbitrary field $F$ and an arbitrary element $a\neq0$ of that field, I am sked to find all the integers $n \geq 1$ such that $x+a$ is a root of $x^n+a^n$ in $F[x]$
I proved it as follows:
Proof
$x+a$ is a factor of $x^n+a^n \iff (x+a)=0 \implies x^n+a^n=0$
Then we get $(x+a)=0 \iff x=-a$.
For $n$ odd we also have that $(-a)^n = -(a^n)$
Thus $x=-a \implies x^n = -(a^n) \implies x^n+a^n=0$
Thus whenever $n$ is odd $x+a$ is a root of $x^n+a^n$
For $n$ even we have that $(-a)^n = a^n$
Thus $x = -a \implies x^n = a^n \implies x^n+a^n = 2a^n$
Thus it's never the case that if $n$ is even $x+a$ is a root of $x^n+a^n$.
We have exhausted all possibilities, so $x+a$ is a root of $x^n+a^n$ if and only if $n$ is odd.
End of Proof
I am very new to working with arbitrary fields so I am not 100% confident in my proof.
First I am not sure if my initial statement:
$x+a$ is a factor of $x^n+a^n \iff (x+a)=0 \implies x^n+a^n=0$
Holds for any arbitrary field regardless of order
Second, I am not fully convinced that my seocnd statemnt is correct, because I am not confident that $a\neq 0 \implies 2a^n \neq 0$ for example in a field with characteristic 2, $2a^n$ is $2\cdot1^n=0$
So i think the proof is fallacious and incomplete.
It is true that $x+a$ is a factor of $x^n+a^n$ iff $x=-a$ is a root of $x^n+a^n$. This can be proved by polynomial long division over any field (or even over any commutative ring), just like the familiar case of $\mathbb{R}$.
However, as you suspect, the implication $a\neq 0 \implies 2a^n \neq 0$ is not valid in general. It is only valid in a field where $2\neq 0$ (in other words, a field of characteristic different from $2$). So your argument is correct if the characteristic of $F$ is different from $2$. But when the characteristic of $F$ is equal to $2$, we have $2a^n=0$ for all $a$, and so $x+a$ is a factor of $x^n+a^n$ for all $n$.