TITLE: Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$
QUESTION [32 upvotes]: Find all polynomials $P$ such that 
$$P(x^2+1)=P(x)^2+1$$

REPLY [3 votes]: It depends on whether the coefficients of the polynomial come from a field (or ring) of characteristic $0$. [ http://en.wikipedia.org/wiki/Characteristic_%28algebra%29 ]
There are extra solutions in characteristic $p$, such as $P(x) = x^p$. It could be a hard problem to determine whether solutions exist that are not generated by compositions of $x^p$ and $x^2+1$.  The general problem of determining all commuting pairs of polynomials over a finite field is well-known and unsolved.
In characteristic $0$, solutions of $P(Q(x))=Q(P(x))$ are classified. The case $Q(x)=x^2+1$ does not fall into one of the families of nontrivial solutions (it is not linearly conjugate to $x^2$ or to the degree 2 Chebyshev polynomial), so that the only possibililty is $P(x) = Q^{\circ \hskip0.7pt n}$ for some $n$.