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

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).