Why do lattice cubes in odd dimensions have integer edge lengths?

This is a spinoff from Characterization of Volumes of Lattice Cubes. That question claims a number of facts as being proven, but doesn’t include the full proofs. That’s fine for the question as it stands, but I find the subject interesting enough to wonder about the details. So here I’m asking.

Let’s start by defining the set of squared lattice cube lengths in dimension $n$ as

$$S(n)=\left\{s\in\mathbb N\;\middle|\;\exists M\in\mathbb Z^{n\times n}:M^TM=s\,I_n\right\}$$

In other words, a number $s$ is part of that set iff there exists some $n\times n$ matrix $M$ with integer coordinates whose vectors are orthogonal to one another and all have Euclidean norm $\sqrt s$. They can be interpreted as edge vectors of a cube whose corners are all lattice points in $\mathbb Z^n$ and whose edges are of length $\sqrt s$. Hence the name lattice cube.

Notes about notation: I use $I_n$ to denote the $n\times n$ identity matrix, and I consider $0\in\mathbb N$.

The referenced question claims that for $k\in\mathbb N$,

\begin{align*}
S(2k+1)&=\bigl\{a^2\;\big|\;a\in\mathbb N\bigr\} \\
S(4k)&=\bigl\{a\;\big|\;a\in\mathbb N\bigr\}=\mathbb N \qquad\text{(}k>0\text{ of course)}\\
S(4k+2)&\supseteq\bigl\{a^2+b^2\;\big|\;a,b\in\mathbb N\bigr\} \\
\end{align*}

While that question asks for proof that the last point is in fact an identity as well, here I’m concerned with proofs for the rest of the points. Some of them are easy, others are harder.

  • $S(2k+1)\supseteq\bigl\{a^2\;\big|\;a\in\mathbb N\bigr\}$:
    Simply choose $M=a\,I_n$.
  • $S(2k+1)\subseteq\bigl\{a^2\;\big|\;a\in\mathbb N\bigr\}$:
    I don’t have a proof here yet. This is the core of my question.
  • $S(4k)\supseteq\bigl\{a\;\big|\;a\in\mathbb N\bigr\}$:
    According to Lagrange’s four-square theorem, any natural number $a$ can be described as the sum of four squares, i.e. $a=b^2+c^2+d^2+e^2$. For $n=4$ you can e.g. choose
    $$M=\begin{pmatrix}b&-c&-d&-e\\c&b&e&-d\\d&-e&b&c\\e&d&-c&b\end{pmatrix}$$
    For $k>1$ you can let $M$ be a block diagonal matrix with blocks shaped as above.
  • $S(4k)\subseteq\bigl\{a\;\big|\;a\in\mathbb N\bigr\}$:
    This one is easy: since the elements of $M$ are integers, the squared norm $s$ must be a non-negative integer.
  • $S(4k+2)\supseteq\bigl\{a^2+b^2\;\big|\;a,b\in\mathbb N\bigr\}$:
    Choose $\displaystyle M=\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ or a block diagonal matrix formed from blocks of this form.

So essentially what I’m asking for is a proof of the second point in the above list, i.e. a proof of $S(2k+1)\subseteq\bigl\{a^2\;\big|\;a\in\mathbb N\bigr\}$. Or, as the title of this question states, I want to know why every lattice cube in odd dimensions must have integer edge lengths.

If you have an idea for an alternate proof which also covers one of the other aspects, I’ll welcome alternate views on these as well. But make sure to include an answer for the main question as well, or post a comment instead.

Solutions Collecting From Web of "Why do lattice cubes in odd dimensions have integer edge lengths?"

OP of the linked post here. Your proofs are identical to mine. The one you are missing is as follows: A parallelepiped volume can be calculated as the determinant of a matrix for which the rows are the vectors defining the paralelipiped. Thus if all the coordinates are integers, then the determinant must be as well. Thus the volume of any paralelipiped in $\mathbb{R}^n$ for any $n$ must be an integer. If we let the side length of the cube be $\ell=\sqrt{s}$, then $V=\ell^n=s^\frac{n}{2}$ must be an integer which forces $s$ to be a perfect square when $n$ is odd, making $\ell$ an integer.