For what $n$ is $x^2 + x+ 1\mid x^{2n} + x^n + 1$?

Question

For what $n \in \mathbb{N}$ is $x^2 + x+ 1\mid x^{2n} + x^n + 1$?

The only obvious thing that I could see was noticing that $(x^3 - 1) = (x-1)(x^2 + x+1)$. So, if $x^3 - 1\mid x^{2n} + x^n + 1$. I don't even know if this will help.

Answer 1

Hint: This happens precisely if the two non-real cube roots of unity are roots of the polynomial on the right. Things are much simplified on the right by noting that if $\omega$ is such a cube root of unity, then $w^{3k}=1$.

Answer 2

HINT:

If $\omega$ is a root of $x^2+x+1=0, \omega^2+\omega+1=0\implies \omega^3=1$

So if $3|n, n=3m$(say),

$$\omega^n=\omega^{3m}=(\omega^3)^m=1 \text{ and } \omega^{2n}=\omega^{6m}=(\omega^3)^{2m}=1$$

$$\implies \omega^{2n}+\omega^n+1=3\ne0$$

$\implies \omega$ is not a root of $x^{2n}+x^n+1=0$

If $n=3m+1,\omega^{2n}+\omega^n+1=\omega^{2(3m+1)}+\omega^{3m+1}+1=\omega^2+\omega+1=0$

$\implies \omega$ is a root of $x^{2n}+x^n+1=0$

$\implies (x-\omega)|(x^{2n}+x^n+1)$

Similarly, $(x-\omega^2)|(x^{2n}+x^n+1)$

and as $(x-\omega)(x-\omega^2)=x^2+x+1\implies (x^2+x+1)|(x^{2n}+x^n+1)$

Similarly, for $n=3m+2$

Answer 3

$x^2+x+1\mid x^{2n}+x^n+1$ iff the two complex roots $\omega:=\frac{-1+i\sqrt3}2$ and $\overline\omega$ of $x^2+x+1$ are also roots of $x^{2n}+x^n+1$, i.e. iff $\omega$ is (since $x^{2n}+x^n+1\in\Bbb R[x]$). Now, $\omega^3=1$ hence $$\omega^{2n}+\omega^n+1=\begin{cases}3&\text{if }3\mid n\\\omega^2+\omega+1=0&\text{else.}\end{cases}$$ Therefore,$$x^2+x+1\mid x^{2n}+x^n+1\iff3\not\mid n.$$