Which are the polynomials $p(x)$ for which $p(x) \cdot p(x+1)$ and $p(x+p(x))$ are identical?
Solution. Let $p(x)$ be an $n$-degree polynomial that satisfies the condition of the problem.

If $n=0$, then $p(x) \equiv c$ and $p(x+1) \equiv c$. According to the condition,

$$
c \cdot c=c, \quad \text { hence } \quad c=0 \quad \text { or } \quad c=1
$$

Thus, $p(x) \equiv 0$, or $p(x) \equiv 1$.

If $n=1$, then $p(x)=a x+b$, so $p(x+1)=a x+a+b$, where $a \neq 0$. Substituting,

$$
\begin{aligned}
(a x+b)(a x+a+b) & =a(x+a x+b)+b \\
a^{2} x^{2}+a b x+a^{2} x+a b+a b x+b^{2} & =a x+a^{2} x+a b+b \\
a^{2} x^{2}+\left(a^{2}+2 a b\right) x+b^{2} & =\left(a^{2}+a\right) x+b
\end{aligned}
$$

The two polynomials can only match if $a=0$, but $a \neq 0$. Therefore, $p(x)$ cannot be a first-degree polynomial.

If $n>1$, then $p(x)$ is an $n$-degree polynomial, $p(x+1)$ is an $n$-degree polynomial, $p(x) \cdot p(x+1)$ is a $2n$-degree polynomial. $x+p(x)$ is an $n$-degree polynomial, $p(x+p(x))$ is an $n^2$-degree polynomial. Since the two polynomials are equal, $n^2=2n$, so if $n>1$, then $n=2$, the polynomial is quadratic.

Let $p(x)=a x^{2}+b x+c$, where $a \neq 0$. Let's see what the condition means:

$$
\begin{aligned}
p(x+p(x)) & =a(x+p(x))^{2}+b(x+p(x))+c= \\
& =a x^{2}+2 a x p(x)+a p^{2}(x)+b x+b p(x)+c
\end{aligned}
$$

If this is identical to the $p(x) \cdot p(x+1)$ polynomial, then

$$
p(x) \cdot\left[a(x+1)^{2}+b(x+1)+c-2 a x-a p(x)-b\right] \equiv a x^{2}+b x+c \equiv p(x)
$$

that is, $p(x) \cdot\left[a x^{2}+2 a x+a+b x+b+c-2 a x-a p(x)-b-1\right]$ is the identically zero polynomial. Since $p(x)$ is quadratic, the second factor is identically zero:

$$
\underbrace{a x^{2}+b x+c}_{p(x)}+a-a p(x)-1=0
$$

Factoring, $(1-a)(p(x)-1)$ is identically zero. Since $p(x)$ is quadratic, this is only possible if $a=1$. Then $p(x)=x^{2}+b x+c$, where $b$ and $c$ are arbitrary real numbers. Since our steps are reversible, the equality holds in this case.

Therefore, the polynomials that satisfy the conditions of the problem are: $p(x)=0, p(x)=1$ and $p(x)=x^{2}+b x+c$.