Recall Helly theorem : If $X_i\subset \mathbb{R}^d,\ 1\leq i\leq n $ is a convex body and any $d+1$ sets of $X_i$ has a nonempty intersection, then $\bigcap_{i=1}^n\ X_i$ is nonempty.
Question : In $\mathbb{R}^d$, consider a finite number of unit cubes inscribed in a fixed sphere $S$. If any three of them share a common vertex, then all cubes share a common vertex.
Proof : For convenience, assume that $\{e_i\}$ is a standard basis and a center of the sphere $S$ is $\frac{\sum_{i=1}^d e_i}{2}$. That is $$V_1=\bigg\{ \sum_{i=1}^d \epsilon_ie_i\bigg| \epsilon_i\in \{0,1\} \bigg\}$$ is a vertex set of the first cube inscribed in $S$. Note that distance between two vertices in a cube inscribed is $\sqrt{i}$ for some $i\in \mathbb{Z}$.
Assume that the origin $v_4$ is in $V_1\bigcap V_2\bigcap V_3$, $v_2=\sum_{i=1}^k e_i$ is in $V_1\bigcap V_3\bigcap V_4$ and $ v_3 = \sum_{i=1}^{k+l} e_i$ is in $V_1\bigcap V_2\bigcap V_4$ where $k+l<d$.
Hence $v_1=\sum_i w_i e_i$ is in $V_2\bigcap V_3\bigcap V_4$. Hence we have a claim that all $w_i$ are integrer.
$|v_1-v_4|^2 + |v_1-\sum_{i=1}^d e_i|^2=d $ so that $$ \sum_i w_i^2=\sum_i w_i $$
$|v_2-v_1|^2$ is an integer so that $$-\sum_{i=1}^k w_i + \sum_{i=k+1}^d w_i \in \mathbb{Z} $$
I can not proceed more. How can we solve this ?
Thank you in advance.