TITLE: When are infinitely many points in the orbit of a polynomial integers?
QUESTION [6 upvotes]: This question is inspired by a riddle in math.stackexchange.
Let $P$ be a polynomial, and $O = \{P^{(n)}(0) : n \geq 0\}$ be its orbit under zero (viewed as a set). Suppose that $O$ contains infinitely many integers. Is it true that for some $n$, $P^{(n)}$ is a polynomial with integral coefficients?
We can ask the same question replacing integers with rationals.
EDIT: Nick and David gave simple counterexamples for the first question.
Still open:

In the setting of the original question, is it true that some composition power of $P$ takes integers to integers?
The original question with rationals.

REPLY [5 votes]: A slight variant that turns out to be non-elemenatry is to replace the polynomial with a rational function. This leads to:
Theorem: Let $R(x)\in\mathbf{Q}(x)$ be a rational function of degree at least 2, let $\alpha\in\mathbf{Q}$ be an initial value, and suppose that the orbit $O_R(\alpha)=\{R^{(n)}(\alpha) : n\ge0\}$ contains infinitely many integers. Then the second iterate $R^{(2)}(x)$ of $R$ is a polynomial.
For specific $R$ there are often easy proofs, but in general the proof seems to require some non-trivial result on Diophantine approximation such as Thue's theorem. There's an exposition of the proof in The Arithmetic of Dynamical Systems (Springer 2007), Section 3.7. See Section 3.8 for a stronger result saying roughly that as $n$ gets large, then the numerator and denominator of $R^{(n)}(\alpha)$ have about the same number of digits. (There's one extra technical condition for this last result.)