{ "id": "2211.06760", "version": "v1", "published": "2022-11-12T22:46:14.000Z", "updated": "2022-11-12T22:46:14.000Z", "title": "Locally nilpotent polynomials over $\\mathbb{Z}$", "authors": [ "Sayak Sengupta" ], "comment": "23 pages, 0 figures", "categories": [ "math.NT" ], "abstract": "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.", "revisions": [ { "version": "v1", "updated": "2022-11-12T22:46:14.000Z" } ], "analyses": { "subjects": [ "11A41", "37P05", "11A05", "11A07", "37P25" ], "keywords": [ "locally nilpotent polynomials", "complete classification", "linear polynomials", "necessary condition", "special case" ], "note": { "typesetting": "TeX", "pages": 23, "language": "en", "license": "arXiv", "status": "editable" } } }