Assume $P$ is non-constant. Then, there is an $n_0$ such that $P(n)>1$ for every $n\ge n_0$. Next, if $a,b\in S$ are distinct (such elements exist), then $a+b\in S$, $2a+b\in S$ and continuing, we find $S$ is infinite. Denote $F= \{k\in S:k\ge n_0\}$. Take $a,b\in F$ and observe that $a+b\in F$ and $(P(a),P(b))=1$. Since $P(a),P(b)>1$, there is a prime $r_1$ such that $r_1\mid P(a)$. Notice that $a+b\in F$, $a+2b\in F$, and continuing, $a+r_1b\in F$. But now, $a+r_1>a$, $a+r_1,a\in S$, so $P(a+r_1)$ and $P(a)$ must be coprime. However, $P(a+r_1)\equiv P(a)\equiv 0\pmod{r_1}$, yielding a contradiction.

Iran MO 2016