Determine all real polynomials $P$ such that $P\left(X^{2}\right)=\left(X^{2}+1\right) P(X)$
Suppose $P$ is non-zero and a solution of the equation, let $n$ be its degree. The degree of $P\left(X^{2}\right)$ is $2n$, and that of $\left(X^{2}+1\right) P(X)$ is $2+n$, so $2n=n+2$, which gives $n=2$. We then set $P=a X^{2}+b X+c$ with $a, b, c$ real.

The equation can be rewritten as $a X^{4}+b X^{2}+c=\left(a X^{2}+b X+c\right)\left(X^{2}+1\right)=a X^{4}+b X^{3}+(a+c) X^{2}+b X+c$, by identification the equation implies the following 5 equations: $a=a, b=0, a+c=b, b=0$, $c=c$, which is equivalent to $b=0$ and $c=-a$, thus in all cases $P$ is of the form $a\left(X^{2}-1\right)$ with $a \in \mathbb{R}$.

Let's verify: if $P$ is of the form $a\left(X^{2}-1\right), P\left(X^{2}\right)=a\left(X^{4}-1\right)=a\left(X^{2}+1\right)\left(X^{2}-1\right)=$ $a\left(X^{2}+1\right) P(X)$. Thus, the polynomial solutions are the polynomials of the form $a\left(X^{2}-1\right)$ with $a \in \mathbb{R}$.