Czech-Slovak-Polish Match

The constant polynomial $P(x)=c$ is a solution if and only if $c=c^2$, thus the polynomials $P(x)=0$ and $P(x)=1$ are solutions of the problem. We claim that the only polynomial of a positive degree $n$ which solves the equation is of the form $P(x)=(x-1{)}^n$. In view of the identity $(x^2-1{)}^n=(x-1{)}^n(x+1{)}^n$, the latter is clearly a solution for any $n\ge 1$. If $a{x}^n$ ($a\ne 0$) is the leading term of a polynomial $P(x)$ of a positive degree $n$, then $a{x}^{2n}$ is the leading term of the polynomial $P(x^2)$ and $a^2{x}^{2n}$ is the leading term of the polynomial $P(x)P(x+2)$. If $P$ satisfies the given equality, comparing the leading order terms thus gives $a=a^2$, hence $a=1$. The polynomial $P$ can therefore be written in the form $P(x)=(x-1{)}^n+Q(x)$, where $Q$ is either identically zero, or is a nonzero polynomial of degree $k$, where $0\le k<n$. Comparing the polynomials $$\begin{gathered} P(x^2)=(x^2-1{)}^n+Q(x^2), \\ P(x)P(x+2)=[(x-1{)}^n+Q(x)][(x+1{)}^n+Q(x+2)] \end{gathered}$$ we obtain (upon multiplying out the brackets and cancelling the terms $(x^2-1{)}^n$ on both sides) the equality $$Q(x^2)=(x-1{)}^{n}Q(x+2)+(x+1{)}^{n}Q(x)+Q(x)Q(x+2).$$ The zero polynomial $Q$ clearly satisfies this relation. For a nonzero $Q$ of degree $k<n$, however, $Q(x^2)$ is a polynomial of degree $2k$, while on the right-hand side of the last equation there is a polynomial of degree $n+k$ (whose leading term is $2b{x}^{n+k}$, if $b{x}^k$ is the leading order term of the polynomial $Q(x)$). Since $2k<n+k$, this is not possible. Conclusion. The solutions are the constant polynomials $P(x)=0$ and $P(x)=1$ and the polynomial $P(x)=(x-1{)}^n$ for any natural number $n$.