1. Find all polynomials $p$ with real coefficients for which

$$
p(p(x))=\left(x^{2}+x+1\right) p(x)
$$

for all $x \in \mathbb{R}$.
III/1. The zero polynomial is obviously a solution to the problem. Let $p$ be a non-zero polynomial and write $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$, where $a_{n} \neq 0$. The leading term on the left side of the equation is then equal to $a_{n}\left(a_{n} x^{n}\right)^{n}=a_{n}^{n+1} x^{n^{2}}$, and the leading term on the right side is equal to $x^{2} \cdot a_{n} x^{n}=a_{n} x^{n+2}$. Since these two terms must be of the same degree, it follows that $n^{2}=n+2$ or $(n-2)(n+1)=0$. Since $n$ is a non-negative integer, we must have $n=2$. Therefore, we write $p(x)=a x^{2}+b x+c$, where $a \neq 0$. Then we have

$$
\begin{gathered}
p(p(x))=a^{3} x^{4}+2 a^{2} b x^{3}+\left(a b^{2}+2 a^{2} c+a b\right) x^{2}+\left(2 a b c+b^{2}\right) x+\left(a c^{2}+b c+c\right) \\
\left(x^{2}+x+1\right) p(x)=a x^{4}+(b+a) x^{3}+(c+b+a) x^{2}+(c+b) x+c
\end{gathered}
$$

From the given equation, we obtain the system of equations

$$
\begin{aligned}
a^{3} & =a \\
2 a^{2} b & =b+a \\
a b^{2}+2 a^{2} c+a b & =c+b+a \\
2 a b c+b^{2} & =c+b \\
a c^{2}+b c+c & =c
\end{aligned}
$$

Since $a \neq 0$, from the first equation we get $a=1$ or $a=-1$. For $a=1$, the second equation gives $2 b=b+1$, so $b=1$. If we substitute both into the third equation, we get $2 c+2=c+2$ or $c=0$. We can verify that these three numbers satisfy the last two equations as well. For $a=-1$, we get a contradictory system. The solutions to the problem are thus two, $p(x)=0$ and $p(x)=x^{2}+x$.

Solution $p(x)=0$