It is true that it solves for a very few $n$. But it strikes me as a surprise that the statement that "no solution exists for any non-Wilson prime less one." is not reported even if it is obvious.

But as you go up in dimensions the gap between the cubes gets smaller and smaller. So while the 2D case was sort of a checkerboard square lattice (with gaps of the same width as the square), the 3D case has cubes with gaps two-thirds of the cube-edge length.

If $ry$ increases by $3(y/z)\ge 3$ wouldn't that imply an increment in $rx$ by $3(x/y)\ge 3$ as well?

Can Case $1$ and $2$ with $m\ne 1$ be given an inductive flavor to cover for all dimensions? Or do more cases arise when you have an increment in dimensions?

It was inspired by the View Obstruction paper by Cusick, though this one is on a completely different route, more like scaling n-cubes to cover the entire totally positive orthant of $\mathbb{R}^n$.

I am actually interested in a general version of this problem where you have $a_1, a_2, a_3,... a_{\lambda-2} \in \mathbb{N}$ such that $a_i\geq 1$ for all $1\leq i\leq \lambda-2$ and it is conjectured that the bound is $x=\lambda-1$ such that $n\in[1,x]$ to ensure that all $a_i$ are contained within $\lambda-2$-dimensional hypercubes generated by $[\lambda (m-1)+1,\lambda m -1]$ in all dimensions. It would be quite a task to solve this in general as apparently the problem increases in difficulty as you increase the dimensions.

"It is easy to check that, any point in this triangle, we can rescale it by a factor of $\leq 3$ to land in the square $[9,11]\times[5,7]$." Or precisely this?

@domotorp exactly.