If the polynomial $x^{2 k}+1+(x+1)^{2 k}$ is not divisible by $x^{2}+x+1$, find $k$.

The polynomial $x^{2 k}+1+(x+1)^{2 k}$ is not divisible by $x^{2}+x+1$, find $k$.
Let $w$ be a complex cube root of 1, then $w^{3}=1, w^{2}+w+1=0$, so $w$ is a root of $x^{2}+x+1=0$.
When $k=3 l$ (where $l$ is a non-negative integer), $w^{2 k}+1+(w+1)^{2 k}=1+1+\left(-w^{2}\right)^{2 k}=3$;
When $k=3 l+1$, $w^{2 k}+1+(w+1)^{2 k}=w^{2}+1+w=0$;
When $k=3 l+2$, $w^{2 k}+1+(w+1)^{2 k}=w+1+w^{2}=0$.
That is, $w$ is not a root of $x^{2 k}+1+(x+1)^{2 k}=0$ if and only if $3 \mid k$, so the answer to this problem is $3 \mid k$.
／cjkstartComment: In the 1981 Austrian Mathematical Competition, there was a problem: For which positive integer $n$ does $\left(x^{2}+x+1\right) \mid \left[x^{2 n}+1+(x+1)^{2 n}\right]$?
From the solution of Example 6, we know that $\left(x^{2}+x+1\right) \mid \left[x^{2 n}+1+(x+1)^{2 n}\right]$ if and only if the positive integer $n$ is not divisible by 3.