The polynomial $x^{2k} + 1 + (x+1)^{2k}$ is not divisible by $x^2 + x +1$. Find value of $k$ such that it belongs to natural numbers.

Question

I did some progress by doing this-

I thought if it would be a factor, then

$x^2 +x +1 = 0 $

$x+1=-x^2$

Putting this in the expression, $$x^{4k}+x^{2k}+1$$ Then I tried to solve it further, but I am stuck here...

Answer 1

$$x^{2k}+(x+1)^{2k}+1\equiv x^{2k}+(x^2)^{2k}+1=x^{4k}+x^{2k}+1\equiv0$$ for any $k$, which is not divisible by $3$.

If $k$ is divisible by $3$, so since $x^3\equiv1$, we obtain: $$x^{4k}+x^{2k}+1\equiv3.$$

Can you end it now?