Determine the real polynomials satisfying the relation:

$$
P\left(x^{2}\right)=\left(x^{2}+1\right) P(x)
$$
The zero polynomial is obviously a solution.

We seek non-zero solutions by analysis-synthesis:

- Analysis. Suppose that $P$ is a solution, then necessarily:

$$
2 \operatorname{deg} P=\operatorname{deg} P+2 \Longleftrightarrow \operatorname{deg} P=2
$$

The candidate solutions are polynomials of degree $2: P(x)=a x^{2}+b x+c$ with $a, b$ and $c$ real.

- Synthesis. We have the following equivalences:

$$
\begin{aligned}
P(x)=a x^{2}+b x+c \text { is a solution } & \Longleftrightarrow P\left(x^{2}\right)=\left(x^{2}+1\right) P(x) \\
& \Longleftrightarrow a x^{4}+b x^{2}+c=a x^{4}+b x^{3}+(a+c) x^{2}+b x+c \\
& \Longleftrightarrow b=0 \text { and } c=-a \\
& \Longleftrightarrow P(x)=a x^{2}-a=a\left(x^{2}-1\right)
\end{aligned}
$$

The polynomial solutions of the functional equation are exactly the $P(x)=a\left(x^{2}-1\right)$ with $a$ any real number.

& Explanation Some combinatorial formulas can be proven using the principle of identification: