Polynomials whose divisors are enumerated by $SL_2(N_0)$

TL;DR

This paper classifies polynomials with integer coefficients that admit a certain invertible action related to divisor pairs, introduces a sequence linked to the polynomial n^2+1, and explores its properties and connections to prime numbers.

Contribution

It classifies enumerable polynomials under a specific monoid action and constructs a sequence with properties related to divisor functions and prime characterization.

Findings

The polynomial f(n) = n^2 + 1 is enumerable under the defined action.

The sequence S(k) has a fiber size equal to the divisor count of n^2+1.

n^2 + 1 is prime iff the fiber of n under S has exactly two elements.

Abstract

We consider a certain left action by the monoid $SL_2(\mathbf{N}_0)$ on the set of divisor pairs $\mathcal{D}_f := \{ (m, n) \in \mathbf{N}_0 \times \mathbf{N}_0 : m \lvert f(n) \}$ where $f \in \mathbf{Z}[x]$ is a polynomial with integer coefficients. We classify all polynomials in $\mathbf{Z}[x]$ for which this action extends to an invertible map $\hat{F}_f: SL_2(\mathbf{N}_0) \rightarrow \mathcal{D}_f$. We call such polynomials $\textit{enumerable}$. One of these polynomials happens to be $f(n) = n^2 + 1$. It is a well-known conjecture that there exist infinitely many primes of the form $p = n^2 + 1$. We construct a sequence $\mathcal{S}$ on the naturals defined by the recursions $$ \begin{cases} \mathcal{S}(4k) = 2\mathcal{S}(2k) - \mathcal{S}(k) \\ \mathcal{S}(4k+1) = 2\mathcal{S}(2k) + \mathcal{S}(2k+1) \\ \mathcal{S}(4k+2) = 2\mathcal{S}(2k+1) + \mathcal{S}(2k) \\ \mathcal{S}(4k+3) = 2\mathcal{S}(2k+1) - \mathcal{S}(k) \\ \end{cases} $$ with initial conditions $\mathcal{S}(1) = 0$, $\mathcal{S}(2) = 1$, $\mathcal{S}(3) = 1$. $$\{ \mathcal{S}(k) \}_{k \in \mathbf{N}} = \{0,1,1,2,3,3,2,3,7,8,5,5,8,7,3, \cdots \}$$ $\mathcal{S}$ is shown to have the properties $1.$ For all $n \in \mathbf{N}_0$, we have $\mathcal{S}(2^n) = \mathcal{S}(2^{n+1} - 1) = n$. $2.$ For all $n \in \mathbf{N}_0$, the size of the fiber of $n$ under $\mathcal{S}$ satisfies $|\mathcal{S}^{-1}(\{n\})| = \tau(n^2 + 1)$ where $\tau$ is the divisor counting function. $3.$ For all $n \in \mathbf{N}_0$, the integer $n^2 + 1$ is prime if and only if $\mathcal{S}^{-1}(\{n\}) = \{2^n, 2^{n+1} - 1\}$. $4.$ $\mathcal{S}(k)$ is a $2$-regular sequence.