Find all polynomials $\mathrm{P}$ with real coefficients such that

$$
P\left(x^{2}\right) P\left(x^{3}\right)=P(x)^{5}
$$

for all real numbers $x$.
(Solved by Yassine Hamdi)

Let $P(X)=a_{n} X^{n}+\cdots+a_{0}, a_{n} \neq 0$ be the solution of the problem, then we have

$$
\left(a_{n} X^{n}+\cdots+a_{0}\right)^{5}=\left(a_{n} X^{2 n}+\cdots+a_{0}\right)\left(a_{n} X^{3 n}+\cdots+a_{0}\right)
$$

By focusing on the term in $X^{5 n}$, we find $a_{n}^{5}=a_{n}^{2}$ and $a_{n} \in \mathbb{R}^{*}$, so $a_{n}=1$. Now let's prove by induction on $k \in \mathbb{N}^{*}$ that $a_{n-k}=0$. For the initialization, the term in $X^{5 n-1}$ is $5 a_{n} a_{n-1}$ on the left and 0 on the right, so $a_{n-1}=0$.

Suppose the rank $k, k \geqslant 1$, by the induction hypothesis:

$$
\begin{aligned}
\left(a_{n} X^{n}+a_{n-k-1} X^{n-k-1}+\cdots+a_{0}\right)^{5}= & \left(a_{n} X^{2 n}+a_{n-k-1} X^{2(n-k-1)}+\cdots+a_{0}\right) \\
& \times\left(a_{n} X^{3 n}+a_{n-k-1} X^{3(n-k-1)}+\cdots+a_{0}\right)
\end{aligned}
$$

The term in $X^{5 n-k-1}$ is $5 a_{n-k-1}$ on the left and 0 on the right, so $a_{n-(k+1)}=0$. This closes the induction. We conclude that $P(X)=X^{n}$, and conversely these polynomials are solutions for all $n \in \mathbb{N}$.