48th 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 $⟨x_n{⟩}_{n\ge 0}$ 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 $P(x)=x-1$.