easy to see that we can find infynitiv numbers such that $P(x+1)=x$ so $P(x+1)=x$ for all real $x$

knock knock... who's there? Baltic Way 2008 https://artofproblemsolving.com/community/c6h239261p1316656

We set $P (x) = a_nx^n +a_{n-1}x^{n-1} +..+a_0$ Then by comparing the coefficients we get $a_n = 1 and (a_0 + 1) ^ 2 = a_n + ... + a_0(1)$ Assume $k$ is the second degree after degree n of the polynomial $P(x)(k<n)$ Comparing the coefficients of $x^{n+k}$ on both sides we get $a_k=0$. From this we deduce $P(x)=x^n+a_0 $ Using (1) with $a_n = 1,a_i=0$ with $i=1->(n-1)$ we get $a_0 = 0$ or $a_0 = -1$ If $a_0 = 0$, then $P (x) = x ^ n$, then from $P(2017) = 2016$ we get $n = log_ {2017} {2016} $(Not satisfied) If $a_0 = -1$ then $P(x) = x^n-1$ then from $P(2017) = 2016$ we get $n = 1 \Rightarrow:P(x) = x-1$