Find all natural integers $a$ such that $a^{2}+a+1$ divides $a^{7}+3 a^{6}+3 a^{5}+3 a^{4}+a^{3}+a^{2}+3$
$$
\begin{gathered}
a^{2}+a+1 \mid a^{7}+3 a^{6}+3 a^{5}+3 a^{4}+a^{3}+a^{2}+3 \\
a^{2}+a+1 \mid\left(a^{5}+2 a^{4}+a^{3}\right)\left(a^{2}+a+1\right) \\
a^{2}+a+1 \mid a^{7}+3 a^{6}+3 a^{5}+3 a^{4}+a^{3}+a^{2} \\
a^{2}+a+1 \mid 3
\end{gathered}
$$

So $a=0$ or $a=1$ and these are indeed the only solutions.