When are infinitely many points in the orbit of a polynomial integers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:40:04Z http://mathoverflow.net/feeds/question/43924 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43924/when-are-infinitely-many-points-in-the-orbit-of-a-polynomial-integers When are infinitely many points in the orbit of a polynomial integers? Yuval Filmus 2010-10-28T01:45:34Z 2010-10-28T16:50:14Z <p>This question is inspired by a <a href="http://math.stackexchange.com/questions/8101/iterated-polynomial-problem" rel="nofollow">riddle</a> in math.stackexchange.</p> <p>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?</p> <p>We can ask the same question replacing integers with rationals.</p> <p>EDIT: Nick and David gave simple counterexamples for the first question. Still open:</p> <ol> <li>In the setting of the original question, is it true that some composition power of $P$ takes integers to integers?</li> <li>The original question with rationals.</li> </ol> http://mathoverflow.net/questions/43924/when-are-infinitely-many-points-in-the-orbit-of-a-polynomial-integers/43929#43929 Answer by Nick S for When are infinitely many points in the orbit of a polynomial integers? Nick S 2010-10-28T02:28:37Z 2010-10-28T16:50:14Z <p>$P(x)= \frac{x(x+1)}{2} +1$. </p> <p>It is easy to see that $P^{n+1}(0) > P^n(0)$ and $P$ maps the integers into the integers.</p> <p>But I think (didn't check it, might be one of these facts which are obvious but wrong) that </p> <p>$$P^{(n)}(x) = \frac{1}{2^{m}} x^{2^n}+....\notin \mathbb{Z}$$</p> <p>where $m$ is probably $m=2^n+1$.</p> <p>The right question to ask might be if $f$ maps the integers into the integers....</p> <p><strong>Disregard the following part</strong>, 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). </p> <p>EDIT: P.S. The answer with the rationals turns out to be true, I think (my algebra is rusty):</p> <p>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:</p>