Austrian Mathematical Olympiad National Competition (Final Round, part 1)

Letting $Q(x) := P(x) + 1$ we get the two new conditions $Q(2017) = 2017$ and $Q(x^2+1) = Q(x)^2 + 1$, $x \in \mathbb{R}$.

We now define the sequence $\langle x_n \rangle_{n \ge 0}$ recursively by $x_0 = 2017$ and $x_{n+1} = x_n^2 + 1$, $n \ge 0$. A straightforward induction yields $Q(x_n) = x_n$, $n \ge 0$, because $Q(x_{n+1}) = Q(x_n^2+1) = Q(x_n)^2 + 1 = x_n^2 + 1 = x_{n+1}$.

Because of $x_0 < x_1 < x_2 < \dots$ the two polynomials $Q(x)$ and $\text{id}(x) = x$ coincide at infinitely many arguments $x$. Therefore, $Q(x) = x$ and thus the unique polynomial satisfying the two conditions of our problem is $P(x) = x - 1$.