This is a cross post from Math SE that no one seemed able to solve.

Here is a problem that came up in a conversation with a professor after I made a false assumption about the geometry of $\mathbb{Z}^n$. I do not know if he knew the answer (and told me none of it) and has since passed so I can no longer ask him about it.

Let $C$ be a lattice cube in $\mathbb{R}^n$. Characterize all possible volumes for $C$. A cube is called a *lattice cube* if and only if every vertex has integer coordinates.

I broke this proof into three cases, the last of which I am having trouble with in one direction. We will let $V(n)$ be the set of all numbers $V$ for which there exists a lattice cube of volume $V$ in dimension $n$. We will break into three cases based on the value mod 4.

\begin{align*} V(2k+1)&=\{a^n:a\in\mathbb{N}\} \\ V(4k)&=\{a^\frac{n}{2}:a\in\mathbb{N}\} \\ V(4k+2)&\supseteq\{(a^2+b^2)^\frac{n}{2}:a,b\in\mathbb{N}\} \end{align*}

These statements I have proven (none of them are hard), and conjecture that the last one is an equality. I've been trying to use a collapsing dimension argument to show if I can make a cube of side length $s$ in $\mathbb{R}^{4k+2}$ then I can in $\mathbb{R}^{4k-2}$, at which point the theorem follows since I have proven the special case of $n=2$, unfortunately this does not seem to be fruitful. After trying general collapsing arguments, I got into technical arguements about the matrices whose rows are the vectors that define the cubes, but again, no avail.