Radical bound for Zaremba's conjecture

TL;DR

This paper proves a bound on partial quotients in continued fractions for most integers, confirming Zaremba's conjecture for many cases and extending previous results with a new bound related to the radical of the integer.

Contribution

The paper establishes a bound on partial quotients for integers not of the form 2^n or 3^n, confirming Zaremba's conjecture for these cases and generalizing prior work.

Findings

Zaremba's conjecture holds for q=2^n3^m with k=5.

Bound on partial quotients is given by the radical of q minus one.

Improves previous bounds for prime power integers.

Abstract

Famous Zaremba's conjecture (1971) states that for each positive integer $q\geq2$, there exists positive integer $1\leq a <q$, coprime to $q$, such that if you expand a fraction $a/q$ into a continued fraction $a/q=[a_1,\ldots,a_n]$, all of the coefficients $a_i$'s are bounded by some absolute constant $\mathfrak k$, independent of $q$. Zaremba conjectured that this should hold for $\mathfrak k=5$. In 1986, Niederreiter proved Zaremba's conjecture for numbers of the form $q=2^n,3^n$ with $\mathfrak k=3$ and for $q=5^n$ with $\mathfrak k=4$. In this paper we prove that for each number $q\neq 2^n,3^n$, there exists $a$, coprime to $q$, such that all of the partial quotients in the continued fraction of $a/q$ are bounded by $ \operatorname{rad}(q)-1$, where $\operatorname{rad}(q)$ is the radical of an integer number, i.e. the product of all distinct prime numbers dividing $q$. In particular, this means that Zaremba's conjecture holds for numbers $q$ of the form $q=2^n3^m, n,m\in\mathbb N \cup \{0\}$ with $\mathfrak k= 5$, generalizing Neiderreiter's result. Our result also improves upon the recent result by Moshchevitin, Murphy and Shkredov on numbers of the form $q=p^n$, where $p$ is an arbitrary prime and $n$ sufficiently large.