RagvaloD wrote: Are there such natural $n$, that exist polynomial of degree $n$ and with $n$different real roots, and a) $P(x)P(x+1)=P(x^2)$ b) $P(x)P(x+1)=P(x^2+1)$ a ) $ P(x) = x^m(x-1)^m$

All roots should be different

a) If $r$ is a root of $P$, then $r^2$ and $(r-1)^2$ are also roots. From this it is easy to see that the only possible roots are $0$ and $1$. Then we get $x(x-1)$ as the only (non-constant) solution. b) If $r$ is a root of $P$, then $r^2+1$ is a root as well. Since for all real $r_0=r$, the sequence defined by $r_{n+1}=r_n^2+1$ tends to infinity, $P$ cannot have any roots if it is nonzero. Thus there are only the obvious constant solutions.

For part 2, if we disregard the restriction of having only real roots and distinct roots, using similar arguments on roots of unity, we get $x^2-x+1$ and its powers.