Simple conjecture about rational orthogonal matrices and lattices The following conjecture grew out of thinking about topological phases of matter. Despite being very elementary to state, it has evaded proof both by me and by everyone I've asked so far. The conjecture is:

Let $R$ be an $N \times N$ rational orthogonal matrix. Define a sublattice $\Lambda \subseteq \mathbb{Z}^N$ by
  $$ \Lambda = \{ v \in \mathbb{Z}^N : Rv \in \mathbb{Z}^N \} = \mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N $$
  Then if $N$ is not a multiple of 4, $\Lambda$ contains a vector of odd length-squared.

Note that by this question, such matrices $R$ are in plentiful supply, so the statement is far from vacuous.
To motivate this conjecture, we can first look at several examples where it is easy to prove.


*

*$N = 1$: Here it is trivially true. The only possible $R$ are $(\pm 1)$, so $\Lambda = \mathbb{Z}$.

*$N = 2$: In this case, $R$ takes the form
$$ R = \frac{1}{c} \begin{pmatrix}a & \mp b \\ b & \pm a\end{pmatrix} $$
with $a^2 + b^2 = c^2$ a primitive Pythagorean triple, which in particular means that $c$ must be odd. Now it can be shown that $\Lambda$ admits a basis in which the inner product has matrix
$$ \begin{pmatrix}c & 0 \\ 0 & c\end{pmatrix} $$
So $\Lambda$ once again contains vectors of odd length-squared.

*$N = 3$: I haven't found an elegant proof for this case, but did manage to reduce it to checking a finite list of cases by computer. Sparing the messy details, the conjecture turns out to be true.

*$N = 4$: Here something new happens! Consider the matrix
$$ R = \frac{1}{2} \begin{pmatrix}+1 & -1 & -1 & -1 \\ -1 & +1 & -1 & -1 \\ -1 & -1 & +1 & -1 \\ -1 & -1 & -1 & +1\end{pmatrix} $$
Then a basis for $\Lambda$ consists of $(1,-1,0,0)$, $(0,1,-1,0)$, $(0,0,1,-1)$, $(0,0,1,1)$, which all have even length-squared. This 'counterexample' is the origin of the requirement $4 \nmid N$ in the statement of the conjecture.

*$N = 5$: The same strategy as for $N = 3$ works. However, it can't be pushed any further.
To gain further confidence, we can turn to Monte-Carlo experiments. For each matrix size $N = 1, \dots, 16$, the following table shows the result of generating 1 million random rational orthogonal matrices $R$, and counting the number of them for which $\Lambda$ is an even lattice:
$$ \begin{array}{r|cccccccccccccccc}
   N & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\
   \hline
   \#R & 0 & 0 & 0 & 95444 & 0 & 0 & 0 & 3299 & 0 & 0 & 0 & 136 & 0 & 0 & 0 & 5
\end{array} $$
Obviously, I haven't specified exactly what kind of "random" matrix generator is being used here. However, the important point is this: whatever the distribution, it has yielded many examples for each of $N = 4,8,12,16$. This suggests that, if there were examples to be found for $4 \nmid N$, we would have found some, which we haven't.
What's more, this pattern is exactly what one expects from considerations of the physics of topological phases of fermions. Together with the above checks, this makes a compelling case in favour of the conjecture.
So, assuming the conjecture to be true, my question is: why? If it can be related to any known result, then it would be nice simply to have a reference to it. And if not, it would be great to know of a proof all the same, in the hope that some of the ingredients may shed light on the physics, or vice-versa.
 A: Proof
Let $R$ be any matrix. We have the obvious exact sequence
$$ 0 \longrightarrow\mathbb{R}^N \xrightarrow[\left(\begin{matrix} I \\ R \end{matrix}\right)]{} \mathbb{R}^N \oplus \mathbb{R}^N \xrightarrow[\left(\begin{matrix} I & -R^{-1} \end{matrix}\right)]{} \mathbb{R}^N \longrightarrow 0 $$
This contains as a subsequence
$$ 0 \longrightarrow\mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N \longrightarrow \mathbb{Z}^N \oplus \mathbb{Z}^N \longrightarrow \mathbb{Z}^N + R^{-1} \mathbb{Z}^N \longrightarrow 0 $$
By definition, $\mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N = \Lambda$. Taking the dual of this equation, $\mathbb{Z}^N + R^T \mathbb{Z}^N = \Lambda^\star$. At this point, we need to invoke the fact that $R$ is orthogonal, so that $R^T = R^{-1}$. Then taking the quotient of the two sequences yields
$$ 0 \longrightarrow \frac{\mathbb{R}^N}{\Lambda} \longrightarrow \frac{\mathbb{R}^N}{\mathbb{Z}^N} \oplus \frac{\mathbb{R}^N}{\mathbb{Z}^N} \longrightarrow \frac{\mathbb{R}^N}{\Lambda^\star} \longrightarrow 0 $$
Suppose that every vector of $\Lambda$ has even length-squared. Then $(1, \dots, 1)$ has even inner product with every vector in $\Lambda$, so $\tfrac{1}{2} (1, \dots, 1)^T \in \Lambda^\star$. This tells us that $\tfrac{1}{2}(1, \dots, 1) \oplus 0$, viewed as an element of the middle group above, maps to zero. By exactness, it must therefore be the image of some $v \in \mathbb{R}^N$. So we have
$$ \begin{align} v &= \tfrac{1}{2}(1, \dots, 1) \mod \mathbb{Z}^N \\ Rv &= 0 \mod \mathbb{Z}^N \end{align} $$
Comparing length-squareds,
$$ \underbrace{v^2 \vphantom{)^2}}_{\frac{N}{4} \text{ mod } 2} = \; \underbrace{(Rv)^2}_{\vphantom{\frac{N}{4}} 0 \text{ mod } 1} $$
we immediately read off that $N$ is a multiple of $4$.
A connection
A closely related problem is considered here. For an $N \times N$ rational orthogonal matrix $R$, and a sublattice $L \subseteq \mathbb{Z}^N$, define the coincidence index
$$ \Sigma_L(R) := [ L : L \cap R L ] $$
It can be shown that
$$ \frac{\Sigma_{\mathbb{Z}^N}(R)}{\Sigma_{D_N}(R)} \in \{1, 2\} $$
where $D_N$ is understood to mean the sublattice of vectors with even component sum. The original conjecture is equivalent to saying that
$$ 4 \nmid N \; \implies \; \Sigma_{\mathbb{Z}^N}(R) = \Sigma_{D_N}(R) $$
The special case $N = 3$ is a known result, stated in the paper as Fact 3.
A: This is from papers about 1940 by Gordon Pall, one with B. W. Jones. I'm looking for statements about things being primitive, especially odd/even. Found it, also in "Rational Automorphs," in order to ge6t the gcd of the nine integer elements and $n$ to be $1,$ we have $n$ odd.  This is Theorem 1 on page 754
You did not mention this, so, in case this will make dimension 3 neater, all rational orthogonal matrices come from integers $a^2 + b^2 + c^2 + d^2 = n$ 
and the standard matrix describing quaternions,
$$
\frac{1}{n} \;
\left(
\begin{array}{ccc}
a^2 + b^2 - c^2 - d^2 & 2(-ad+bc) & 2(ac+bd)  \\
2(ad+bc) & a^2 - b^2 + c^2 - d^2 & 2(-ab+cd)  \\
2(-ac+bd) &  2(ab+cd) & a^2 - b^2 - c^2 + d^2  \\
\end{array}
\right)
$$
