1. Find all real polynomials $P$ such that $P\left(X^{2}\right)=P(X)^{2}$.
2. Find all real polynomials $P$ such that $P\left(X^{2}\right)=P(X) P(X+1)$.
3. Find all real polynomials $P$ such that $P(X+1)+P(X-1)=2 P(X)$.
1. If $x=\operatorname{rexp}(i \theta)$ is a root of $P$ and $y=\sqrt{r} \exp (i \theta / 2)$ is a square root of $x$ (with $0 \leqslant r$ and $0<\theta \leqslant 2 \pi)$, then $\rho$ is also a root of $P$. Thus, if $P$ has a non-zero root, then it has roots with an infinite number of different arguments, and therefore $P=0$. We can thus look for $P$ in the form $P(X)=c X^{n}$, which is suitable if and only if $x \in\{0,1\}$.
2. If $x$ is a root of $P$, then $x^{2}$ and $(x-1)^{2}$ are also roots: if $P$ is non-zero, then $|x| \in\{0,1\}$, so $|x-1| \in\{0,1\}$, which means that $x \in\{0,1, \omega, \bar{\omega}\}$, where $\omega=\exp \left(\frac{i \pi}{6}\right)$. Thus, neither $(\omega-1)^{2}$ nor $\left(\bar{\omega}-1^{2}\right)$ is a root of $P$, so $x \in\{0,1\}: P(X)=c X^{a}(X-1)^{b}$. The equation then becomes $c X^{2 a}(X-1)^{b}(X+1)^{b}=c^{2}(X-1)^{b} X^{a+b}(X+1)^{a}$: the solutions are therefore polynomials of the form $c X^{n}(X-1)^{n}$, where $x \in\{0,1\}$.
3. Let $a=P(1)-P(0)$ and $b=P(0)$. The equation gives $P(X+1)-P(X)=$ $P(X)-P(X-1)$, so $P(n+1)-P(n)=$ a for all integers $n \in \mathbb{N}$. In particular, the polynomial $P(X+1)-P(X)-$ a is therefore zero, and we know that $P(n)=b+$ na for all integers $n \in \mathbb{N}$, so the polynomial $P(X)-a X-b$ is also zero. Thus, $P$ is an affine polynomial, and it indeed fits.