The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals 
$\text{(A)} \ 17 \qquad  \text{(B)} \ 20 \qquad  \text{(C)} \ 21 \qquad  \text{(D)} \ 64 \qquad  \text{(E)} \ 65$
Let $h(x)=x^2+x+1$.
Then we have
\[(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,\]
where $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem).
Notice that
\[x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x            -x^{2n-1}-x^{2n-2}-x^{2n-3}          +...\]
$x^n = x^n+x^{n-1}+x^{n-2}          -x^{n-1}-x^{n-2}-x^{n-3}   +....$
Therefore, the left term from $x^2n$ is $x^{(2n-3u)}$

          the left term from $x^n$ is $x^{(n-3v)}$, 

If divisible by h(x), we need 2n-3u=1 and n-3v=2  or 

                             2n-3u=2 and n-3v=1

The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible
~~Wei