Factorization on an arbitrary field

Question

I have been asked:

Now, it seems quite clear that $n=$1 is an answer for any arbitrary field.

If the field was the reals, it seems that no other value for $n$ would statisfy the statement (that's mostly a hunch I have not been able to prove that to be the case)

If instead the field is such that $x^k = x$ and $a^l=a$ for some $k,l$ such that $gcd(k,l)=1$. Then we can find an integer $n'$ such that $(x+a)^{n'} = x^{n'}+a^{n'} = x+a$ by the chinese remainder theorem.

These are all the intuitions that I have had so far, but the question is proving challening. I wanted to know If I am on the correct track.

Answer 1

Hint: $a^{2k+1}+b^{2k+1}=a^2 (a^{2k-1}+b^{2k-1} )-b^{2k-1} (a^2-b^2 )$