$P(x)\in\mathbb Z[x]$. There exist $n\in\mathbb Z^+$, $y\in\mathbb Z$ such that $\underbrace{P(P(\ldots P(y)\ldots))}_{n}=y$. Prove $P(P(y))=y$. 
$P(x)\in\mathbb Z[x]$. There exist $n\in\mathbb Z^+$, $y\in\mathbb Z$ such that $$\underbrace{P(P(\ldots P(y)\ldots))}_{n}=y$$
Prove $P(P(y))=y$.

I think the notation $P^n(y)=y$ is also standard somewhere. If $n\in\{1,2\}$, then it's trivial. I don't know the exact source of the problem. It could be an olympiad problem, but I'm not sure.
 A: In view of comments: Extracting from Olympiad (it was also shown some time ago by W. Narkiewics and possibly before...?). There are various generalizations to other rings and certain algebraic number fields. That tends to be very complicated. The integer case is easy and neat:
Let $y_0=y$ be an integer of period $p$ and not a fixed point. Set inductively $y_{k+1}=P(y_k)$ so that $y_p=y_0$. The trick is to look at succesive differences (all non-zero):
$$ y_{k+2}-y_{k+1}= P(y_{k+1}) - P(y_k)= R_k (y_{k+1}-y_k) .$$
By factorization, simply use that $y_{k+1}^n-y_k^n=(y_{k+1}-y_k)(y_{k+1}^{n-1} + \cdots y_k^{n-1})$, we see that $R_k$ must be an integer. Then by recursion (and using $y_p=y_0$):
$$ y_1-y_0= y_{p+1}-y_p=R_{p-1} ... R_0 (y_1-y_0)$$
shows that each $R_k=\pm 1$. If $R_j=-1$ then $y_{j+2}-y_{j+1} = y_j - y_{j+1}$ so $y_{j+2}=y_j$ and we have a 2-cycle. If every $R_j=1$ then every increment is the same and $0=y_0-y_0=y_p - y_0 = p (y_1-y_0)$. Impossible.
A: If
$$P^n(y)=y$$
Then it should be clear that
$$y=P^n(y)=P^{2n}(y)=\dots$$
And we may take the following limit:
$$y=\lim_{a\to\infty}P^{an}(y)$$
And if the limit exists, it is simple that
$$\lim_{a\to\infty}P^{an}(y)=\lim_{a\to\infty}P^{an+b}(y)$$
For any $b\in\mathbb Z$
And that actually implies
$$y=P^2(y)$$
for $b=1,2$
