This question is inspired by a [riddle](https://math.stackexchange.com/questions/8101/iterated-polynomial-problem) 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:


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



When are infinitely many points in the orbit of a polynomial integers?
$P(x)= \frac{x(x+1)}{2} +1$. 


It is easy to see that $P^{n+1}(0) > P^n(0)$ and $P$ maps the integers into the integers.


But I think (didn't check it, might be one of these facts which are obvious but wrong) that 


$$P^{(n)}(x) = \frac{1}{2^{m}} x^{2^n}+....\notin \mathbb{Z} $$


where $m$ is probably $m=2^n+1$.


The right question to ask might be if $f$ maps the integers into the integers....


**Disregard the following part**, as it was pointed in the comments, it only works if for each $k$ we can find an $l$ and $n\_1,..., n\_k$ so that $f^{(n\_i)}(0)$ and $f^{(n\_i+l)}(0)$ are integers(or rational for the second question). 


EDIT: P.S. The answer with the rationals turns out to be true, I think (my algebra is rusty):


Let $P$ be such a polynomial, and let $m$ be the degree of $P$. Then using the Lagrange interpolation formula, you can reconstruct $P(x)$ from $m+1$ distinct integer values of the type $P^{(k)}(0)$, and since all of these are rational, all the coefficients are rational. Actually this way one can prove the following Lemma:


