There are infinitely many Hilbert cubes of dimension 3 in the set of squares

TL;DR

This paper proves that there are infinitely many 3-dimensional Hilbert cubes within the set of perfect squares, with a lower bound on their quantity and a density property of their ratios.

Contribution

It establishes the existence of many such cubes and demonstrates the density of ratios of their defining parameters, advancing understanding of Hilbert cubes in squares.

Findings

At least on the order of N^{1/8} Hilbert cubes of dimension 3 exist within the first N squares.

The ratios of parameters defining these cubes are dense in the positive real numbers.

The result provides a lower bound on the number of such cubes and their ratio density.

Abstract

A Hilbert cube of dimension $d$ is the set of integers \[ H(a_{0}; a_{1}, \ldots, a_{d})=a_{0}+\{0, a_{1}\}+\cdots+\{0, a_{d}\}=\left\{a_{0}+\sum_{i=1}^{d}\varepsilon_{i}a_{i}:\;\varepsilon_{i}\in\{0,1\}\right\}. \] Brown, Erd\H{o}s and Freedman asked whether the maximal dimension of a Hilbert cube in the set $\cal{S}=\{n^2:\;n\in\mathbb{N}\}$ of integer squares is absolutely bounded or not. Dietmann and Elsholtz proved that if $H(a_{0}; a_{1}, \ldots, a_{d})\subset \cal{S}\cap [0, N]$, then $d\leq 7 \log\log N$ for all sufficiently large values of $N$. Here we prove that there exist at least $\gg N^{1/8}$ Hilbert cubes $H(a_{0}; a_{1}, a_{2}, a_{3})$ with $a_{0}, a_{1}, a_{2}, a_{3}\in [0,N]$ in the set of squares. Moreover, we prove that for each $i, j\in\{0, 1, 2, 3\}$ with $i<j$, the set $$ \left\{\frac{a_{i}}{a_{j}}:\;H(a_{0}; a_{1}, a_{2}, a_{3})\subset S\right\} $$ is dense in the set of positive real numbers (in the Euclidean topology).