$m$-fold composite $p^{(m)}(x) \in \mathbb{Z}[x]$ implies $p(x) \in \mathbb{Z}[x]$

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edited Nov 18, 2021 at 0:50

Joe Silverman

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asked Dec 19, 2020 at 13:21

jack

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$p^{(m)}(x) \in \mathbb{Z}[x]$ implies $p(x) \in \mathbb{Z}[x]$

Let $p(x)$ be a polynomial, $p(x) \in \mathbb{Q}[x]$, and $p^{(m+1)}(x)=p(p^{(m)}(x))$ for any positive integer $m$.
If $p^{(2)}(x) \in \mathbb{Z}[x]$ it's not possible to say that $p(x) \in \mathbb{Z}[x]$.
Is it possible to conclude that $p(x) \in \mathbb{Z}[x]$ if $p^{(2)}(x) \in \mathbb{Z}[x]$ and $p^{(3)}(x) \in \mathbb{Z}[x]$?

More general, suppose there exist positive integers $k_1 <k_2$, such that $p^{(k_1)}(x) \in \mathbb{Z}[x]$ and $p^{(k_2)}(x) \in \mathbb{Z}[x]$. Does it follow that $p(x) \in \mathbb{Z}[x]$?

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$m$-fold composite $p^{(m)}(x) \in \mathbb{Z}[x]$ implies $p(x) \in \mathbb{Z}[x]$

$p^{(m)}(x) \in \mathbb{Z}[x]$ implies $p(x) \in \mathbb{Z}[x]$