Locally nilpotent polynomials over $\mathbb{Z}$

Sayak Sengupta

Published 2022-11-12Version 1

For a polynomial $u(x)$ in $\mathbb{Z}[x]$ and $r\in\mathbb{Z}$, we can construct a dynamical sequence $u(r),u^{(2)}(r),\ldots$. Let $P(u^{(n)}(r)):=\{p \mathrm{\ prime\ number\ }|~p \mathrm{\ divides\ }u^{(n)}(r) \mathrm{\ for\ some\ } n\}$ and $\mathcal{P}:=\{p~|~p \mathrm{\ is\ a\ prime\ number}\}$. For which polynomials $u(x)$ and $r\in \mathbb{Z}$ can we say that $P(u^{(n)}(r))=\mathcal{P}$? If $0$ appears in this sequence, i.e., for some $m\in\mathbb{Z}$ $u^{(m)}(r)=0$, then we obviously have the equality $P(u^{(n)}(r))=\mathcal{P}$. In this paper we study these polynomials. Although the complete classification of such $u(x)'$s for arbitrary $r$ is open, several partial results are proven including the complete classification of $u$ when $r$ is $\pm 1$ and complete classification of linear polynomials $u(x)$ for arbitrary $r's$ with the extra condition that for no $m\in\mathbb{N}$, we get $u^{(m)}(r)=0.$ For the special case when $r$ is a prime we also classify almost all, i.e., all but finitely many linear polynomials $u(x)$ for which we have the equality $P(u^{(n)}(r))=\mathcal{P}$ and we also give a necessary condition for the remaining finitely many polynomials.