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

A Hilbert cube of dimension $d$ is the set of integers $$H(a_0;a_1,\dots ,a_d)=a_0+\{0,a_1\}+\cdots +\{0,a_d\}=\{a_0+\sum _{i=1}^d{\epsilon }_i{a}_i:{\epsilon }_i\in \{0,1\}\}\mathrm{.}$$ Brown, Erdős and Freedman asked whether the maximal dimension of a Hilbert cube in the set $\mathcal{S}=\{{\mathcal{n}}^{\mathcal{2}}:\mathcal{n}\in \mathbb{N}\}$ of integer squares is absolutely bounded or not. Dietmann and Elsholtz proved that if $H(a_0;a_1,\dots ,a_d)\subset \mathcal{S}\cap [\mathcal{0},\mathcal{N}]$, then $d\le 7\log \log N$ for all sufficiently large values of $N$. Here we prove that there exist at least $\gg N^{1\mathrm{/}8}$ 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 $$\{\frac{a_i}{a_j}:H(a_0;a_1,a_2,a_3)\subset S\}$$ is dense in the set of positive real numbers (in the Euclidean topology).