, 2015

*Answer: $P(x) = a x - 1$, $a \neq 0$.*

First note that $P$ cannot be a constant. Now let $n = 1$ and $P(x) = a x + b$. Then
$$\begin{cases} P(P(x) + x) = a(a x + b + x) + b = (a^2 + a)x + a b + b, \\ P(P(x)) + P(x) + 1 = a(a x + b) + b + a x + b + 1 = (a^2 + a)x + a b + 2b + 1. \end{cases}$$
Hence
$$P(P(x) + x) = P(P(x)) + P(x) + 1 \iff b = -1.$$
Thus all the polynomials of the form $P(x) = a x - 1$, $a \neq 0$, satisfy the condition of the problem.

Finally, suppose $n \ge 2$. Let $a_n \neq 0$ denote the leading coefficient of $P(x)$. Then the leading coefficient of $P(P(x)+x)$ is $a_n^{n+1}$ and the leading coefficient of $P(P(x)) + P(x) + 1$ is $(a_n^{n+1} + a_n^n)$. Since $P(P(x)+x) = P(P(x)) + P(x) + 1$ we get that $a_n^{n+1} = a_n^{n+1} + a_n^n$. Hence $a_n = 0$, which is a contradiction.