Cyclotomic ordering conjecture

TL;DR

This paper discusses a conjecture about the ordering of cyclotomic polynomials evaluated at integers greater than 1, exploring initial ideas towards proving the conjecture that either one polynomial is always greater or always lesser than the other.

Contribution

It introduces a conjecture on the ordering of cyclotomic polynomials and summarizes initial thoughts and related proofs regarding this conjecture.

Findings

The conjecture has been proved in a preprint by Pomerance and Rubinstein-Salzedo.

Initial thoughts towards a solution are presented.

The conjecture relates to the comparison of cyclotomic polynomials for different indices.

Abstract

This note describes a conjecture involving cyclotomic polynomials and some initial thoughts towards a solution. Given positive integers $m,n$, the conjecture is that either $\Phi_m(q)\leqslant\Phi_n(q)$ or $\Phi_m(q)\geqslant\Phi_n(q)$ holds for all integers $q\geqslant 2$. Pomerance and Rubinstein-Salzedo proved the conjecture in a preprint called `Cyclotomic coincidences' [arXiv:1903.01962].

Equations

$$q-1<q+1<q^2-q+1⩽q^2+1<q^2+q+1<q^4-q^3+q^2-q+1$$

$${\Phi }_m(q)<c^{-1}{q}^{\varphi (m)}⩽c^{-1}{q}^{\varphi (n)-2}⩽c{q}^{\varphi (n)}<{\Phi }_n(q).$$

$$c{q}^{\varphi (m)}<{\Phi }_m(q)<q^{\varphi (m)}<{\Phi }_{2m}(q)<c^{-1}{q}^{\varphi (m)}$$

Full text

We define a partial ordering $\preceq$ on the set $\mathcal{P}$ of positive integers. Recall that $t^{n}-1=\prod_{d\mid n}\Phi_{d}(t)$ where the roots of the $d$th cyclotomic polynomial $\Phi_{d}(t)$ are primitive roots of order $d$. Hence $\deg(\Phi_{d}(t))=\phi(d)$. For $m,n\in\mathcal{P}$ write $m\preceq n$ if $\Phi_{m}(q)\leqslant\Phi_{n}(q)$ for all integers $q\geqslant 2$, and write $m\prec n$ if $m\preceq n$ and $m\neq n$. (Clearly $a\preceq a$; $a\preceq b$ and $b\preceq a$ implies $a=b$; and $a\preceq b$ and $b\preceq c$ implies $a\preceq c$.) Since

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holds for all $q\geqslant 2$, we have $1\prec 2\prec 6\prec 4\prec 3\prec 10$. Similarly, one can show $10\prec 12\prec 8\prec 5\prec 14\prec 18\prec 9\prec 7\prec 15\prec 20\prec 24\prec 16\prec 30\prec 22\prec 11$.

Say that $m$ precedes $n$ (or $n$ succeeds $m$) if $m\prec n$ and there is no $x$ with $m\prec x\prec n$.

Proving $\Phi_{m}(q)<\Phi_{n}(q)$ for all $q$ is the same as proving $\Phi_{n}(q)-\Phi_{m}(q)>0$. After canceling any equal terms, this inequality can be written $A(q)>B(q)$ where $A(t)$ and $B(t)$ are integer polynomials whose nonzero coefficients are all positive. If the largest nonzero coefficient is $c$, then $A(q)>B(q)$ holds for all $q>c$ provided the leading monomial of $A$ is greater than the corresponding monomial of $B$. (The base-$q$ expansion of $A(q)$ is greater than $B(q)$.) The conjecture asserts that the inequality also holds for $2\leqslant q\leqslant c$.

This reasoning will determine a putative total ordering of $\mathcal{P}$ working for sufficiently large $q$ but maybe not for small $q$. I wrote a program in Magma that proved that the integers $\{1,2,\dots,2\cdot 10^{4}\}$ can be totally ordered. Since the coefficients of $\Phi_{n}(t)$ are unbounded as $n\to\infty$, and their maximum absolute value grows slowly, one might suspect that the conjecture is false and the smallest incomparable pair $(m,n)$ is large. What is positive evidence?

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The cases when $m<3$ or $n<3$ are easily handled. ∎

Thus it suffices to consider whether distinct $m,n\in\mathcal{P}$ with $\phi(m)=\phi(n)$ are comparable, i.e. $m\prec n$ or $n\prec m$. Clearly $\phi(m)=\phi(2m)$ if $m$ is odd.

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where $c=1-q^{-1}$. Similar inequalities (with $m\leftrightarrow 2m$) hold if $\mu(m_{0})=-1$, i.e. $m_{0}$ is a product of an odd number of primes. ∎