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, Erdős and Freedman asked whether the maximal dimension of a Hilbert cube in the set $\mathcal{S}=\{n^2:n\in \mathbb{N}\}$ of integer squares is absolutely bounded or not. Dietmann and Elsholtz proved that if $H(a_0;a_1,\dots ,a_d)\subset \mathcal{S}\cap [0,\mathcal{N}]$, then $d\le 7\log \log N$ 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).

24 pages, submitted