$29 \cdot 61$ The polynomial $x^{2 n}+1+(x+1)^{2 n}$ is not divisible by $x^{2}+x+1$ if $n$ is equal to
(A) 17 .
(B) 20 .
(C) 21 .
(D) 64 .
(E) 65 .
(31st American High School Mathematics Examination, 1980)
［Solution］Let $f(x)=x^{2}+x+1$, and let $g_{n}(x)=x^{2 n}+1+(x+1)^{2 n}$.
Also, $f(x)=(x-r)\left(x-r^{\prime}\right)$,
where $r=\frac{-1+\sqrt{3} i}{2}$ and $r^{\prime}=\frac{-1-\sqrt{3} i}{2}$ are the two roots of $f(x)=0$.
Thus, $f(x)$ can divide $g_{n}(x)$ if and only if both $x-r$ and $x-r^{\prime}$ can divide $g_{n}(x)$, i.e., $g_{n}(r)=g_{n}\left(r^{\prime}\right)=0$.

Since $r$ and $r^{\prime}$ are conjugate complex numbers, it is sufficient to determine the $n$ for which $g_{n}(r)=0$. Note that
$$
g_{n}(x)=\left[x^{2}\right]^{n}+\left[(x+1)^{2}\right]^{n}+1,
$$

and note that $r$ and $r+1=\frac{1+\sqrt{3} i}{2}$ are both 6th roots of 1 (this can be easily seen from the unit circle in the complex plane).
Thus, $r^{2}$ and $(r+1)^{2}$ are both cube roots of 1.
Therefore, the value of $g_{n}(r)$ depends only on the remainder when $n$ is divided by 3. Direct calculation (or from the unit circle in the complex plane) yields
$$
g_{0}(r)=3, \quad g_{1}(r)=0, \quad g_{2}(r)=0 .
$$

That is, when $3 \mid n$, $f(x) \nmid g_{n}(x)$.
Hence, the answer is $(C)$.