Another factor of integer polynomials with minimal integrals

TL;DR

This paper studies integer polynomials with minimal integrals over [0,1], showing infinitely many are divisible by a specific polynomial, improving previous results on polynomial divisibility related to prime distributions.

Contribution

It proves the existence of infinitely many polynomials in S_N divisible by (x^3(1-x)^2)^{⌊N/6⌋}, advancing prior work with a higher-degree polynomial.

Findings

Infinitely many polynomials in S_N are divisible by (x^3(1-x)^2)^{⌊N/6⌋}.

Improves previous divisibility results by using a higher-degree polynomial.

Connects polynomial divisibility to prime number distribution.

Abstract

Let $N$ be a positive integer and let $S_N$ be the set of polynomials with integer coefficients, degree less than $N$, and minimal positive integral over $[0,1]$. D. Bazzanella initiated the study of $S_N$ because of its relation to the distribution of prime numbers. Indeed, it is possible to prove that $\sum_{p^m \leq N} \log p = -\log \int_0^1 P(x) \mathrm{d} x$ for every $P \in S_N$, where the sum runs over prime numbers $p$ and positive integers $m$ such that $p^m \leq N$. For each real number $t$, let $\lfloor t \rfloor$ denote the maximal integer not exceeding $t$. The main result of this paper states that there exist infinitely many polynomials $P \in S_N$ such that $\big(x^3(1 - x)^2\big)^{\lfloor N / 6 \rfloor}$ divides $P(x)$ in $\mathbb{Z}[x]$. This improves upon a similar result of Sanna, who proved the same claim but with the lower-degree polynomial $\big(x(1-x)\big)^{\lfloor N / 3 \rfloor}$ in place of $\big(x^3(1 - x)^2\big)^{\lfloor N / 6 \rfloor}$.