Let's determine those $n$-degree real-coefficient polynomials, which when divided by their derivative, result in a polynomial again.
I. solution. Let the sought polynomial be

$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+\alpha_{n}
$$

if $n>0$, its derivative is

$$
p^{\prime}(x)=n a_{0} x^{n-1}+(n-1) a_{1} x^{n-2}+\ldots+a_{n-1}
$$

and if $n=0$, then $p^{\prime}(x)$ is identically 0. We are looking for a $p(x)$ for which there exists a polynomial $q(x)$ such that

$$
p(x)=q(x) p^{\prime}(x)
$$

From this, if $n=0$, it follows that $p(x)=0$, which is not a solution because division by 0 is not allowed. For the rest, we assume that $n>0$, in which case we can also assume that $\alpha_{n} \neq 0$, i.e., $p$ is exactly of degree $n$. Then $p^{\prime}$ is of degree $(n-1)$, so (3) can only hold with a first-degree $q$:

$$
q(x)=b x+c
$$

Substituting this into (3) and using the fact that the equality of two polynomials is equivalent to the equality of the coefficients of the corresponding terms, we get the following relationships:

$$
\begin{aligned}
& a_{0}=n b a_{0} \\
& a_{1}=(n-1) b a_{1}+n c a_{0} \\
& \vdots \\
& a_{k}=(n-k) b a_{k}+(n-k+1) c a_{k-1} \\
& \vdots \\
& a_{n}=0+c a_{n-1}
\end{aligned}
$$

From these, on the one hand, $n b=1$, and on the other hand,

$$
a_{k}=\frac{n-k+1}{k} n c a_{k-1}, \quad(k=1,2, \ldots, n)
$$

follows. From this, after some calculation, we get that

$$
a_{k}=\binom{n}{k}(n c)^{k} a_{0}
$$

thus

$$
p(x)=a_{0}(x+d)^{n}
$$

where $d=n c$. Moreover,

$$
p^{\prime}(x)=n a_{0}(x+d)^{n-1}
$$

which indeed satisfies (3) if we choose $q$ to be the polynomial

$$
q(x)=\frac{1}{n}(x-d)
$$

Thus, (6) gives all solutions to the problem.

Strommer Pál (Budapest, Piarista Gimn., IV. o. t.)

II. solution. In the first solution, we showed that

$$
p(x)=\frac{1}{n}(x+d) p^{\prime}(x)
$$

i.e., $p$ is divisible by $(x+d)$. Let $k$ be the highest power of $(x+d)$ by which $p(x)$ is divisible:

$$
p(x)=(x+d)^{k} h(x)
$$

where $h$ is no longer divisible by $(x+d)$, i.e., $h(-d) \neq 0$. Then

$$
p^{\prime}(x)=k(x+d)^{k-1} h(x)+(x+d)^{k} h^{\prime}(x)
$$

Substituting this together with (8) into (7) and dividing by $(x+d)^{k}$, we get

$$
h(x)=\frac{k}{n} h(x)+(x+d) h^{\prime}(x)
$$

Substituting $x=(-d)$ into this, we get $k=n$ because $h(-d) \neq 0$, i.e., $h(x)$ is constant. In this case, (8) is equivalent to (6), so (6) gives all solutions.

III. solution. If in (3) $q$ is not identically 0, then (3) implies $\frac{p^{\prime}}{p}=\frac{1}{q}$. Since $\frac{p^{\prime}}{p}$ is the derivative of the function $\log p(x)$, and the function $\frac{1}{q}=\frac{n}{x+d}$ is the derivative of the function $\log (x+d)^{n}$, it follows that the difference between the functions $\log p(x)$ and $\log (x+d)^{n}$ is constant:

$$
\log p(x)-\log (x+d)^{n}=A
$$

i.e.,

$$
p(x)=B(x+d)^{n}
$$

where $A=\log B$, again giving us (6).