Find all polynomials $P$ such that $P(0)=0$ and $P\left(X^{2}+1\right)=P(X)^{2}+1$
A fixed point of $P$ is a number $x$ such that $P(x)=x$. Let's prove by induction on $n \geq 1$ that $P$ has at least $n$ positive fixed points.

Initialization: 0 is a fixed point of $P$.

Hereditary: If $P$ has at least $n$ positive fixed points $0 \leq x_{1} \leq \ldots \leq x_{n}$, then: $P\left(x_{n}^{2}+1\right) = P\left(x_{n}\right)^{2}+1 = x_{n}^{2}+1$. Therefore, $x_{n+1} = x_{n}^{2}+1$ is also a fixed point of $P$. Since $x_{n+1} \geq 2 \times x_{n} \geq x_{n}$ according to the AM-GM inequality and $x_{n} \geq 0$, one of the two equalities is strict (otherwise $x_{n}=0=1$, which is absurd). Since $x_{n+1} > x_{n}$, $x_{n+1}$ is different from all the $x_{i}(i \in[1 ; n])$. Hence the hereditary property.

Thus, $P$ has an infinite number of fixed points, and therefore $Q(X) = P(X) - X$ has an infinite number of roots. This means $Q(X)$ is the zero polynomial. Therefore, $P(X) = X$ is the only solution (conversely, it works well).

Solution to Exercise 6Si: By contradiction, if $|X|$ were a polynomial, then $Q(X) = |X| - X$ would also be a polynomial. However, $Q(X)$ has an infinite number of roots (all positive real numbers), but is not zero (since $Q(-1) = 2$), which is a contradiction.