The Cayley-Menger determinant gives the squared volume of a simplex in $\mathbb{R}^n$ as a function of its $n(n+1)/2$ edge lengths:


$$v\_n^2 = \frac{(-1)^{n+1}}{(n!)^2 2^n}
\begin{vmatrix}
0&d\_{01}^2&d\_{02}^2&\dots&d\_{0n}^2&1\\
d\_{01}^2&0&d\_{12}^2&\dots&d\_{1n}^2&1\\
d\_{02}^2&d\_{12}^2&0&\dots&d\_{2n}^2&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
d\_{0n}^2&d\_{1n}^2&d\_{2n}^2&\dots&0&1\\
1&1&1&\dots&1&0
\end{vmatrix}$$
where $d\_{i j}$ is the distance from vertex $i$ to vertex $j$ of the simplex.


For the regular simplex with all edge lengths 1, i.e. $d\_{i j}=1$ for all $i,j$, we have:


$$(n!)^2 2^n v\_n^2 = (-1)^{n+1}
\begin{vmatrix}
0&1&1&\dots&1&1\\
1&0&1&\dots&1&1\\
1&1&0&\dots&1&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
1&1&1&\dots&0&1\\
1&1&1&\dots&1&0
\end{vmatrix}
= n+1$$


This follows from the fact that the $(n+2)\times(n+2)$ matrix here has eigenvector $(1,1,1,\dots,1)$ with eigenvalue $n+1$, along with $n+1$ eigenvectors of the form $(1,0,0,\dots,-1,0,0,\dots)$ with eigenvalue $-1$.


Any odd integer $2k+1$ has a square that is equal to 1 modulo 8, since:


$$(2k+1)^2 = 4k(k+1)+1$$


and either $k$ or $k+1$ must be even. It follows that, for any simplex whose edge lengths are all odd integers, the quantity $(n!)^2 2^n v\_n^2$ must equal $n+1$ modulo 8.


When $n+1$ is *not* a multiple of 8, this means a simplex in $\mathbb{R}^n$ whose edge lengths are all odd integers cannot be degenerate, i.e. it cannot have zero volume.


However, this does not settle the same question when $n=8k-1$.


Are there known examples of degenerate simplices in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths? Or is there a proof for their nonexistence that completes the proof that applies in other dimensions?


Acknowledgement: This question arose from [a discussion on Twitter between Thien An and Ian Agol](https://twitter.com/thienan496/status/1430773058218827778).


**Edited to add**: For all cases where $n = 7$ mod 16, it is possible to rule out a degenerate simplex by working modulo 16, where any squared odd integer must equal either 1 or 9. Computing the determinant when $x$ is added to any single squared edge length gives a quadratic in $x$ that has **even** coefficients for $x$ and $x^2$ (given that all the original entries are integers), from which it follows that adding 8 to any squared edge length preserves the determinant modulo 16. Since the determinant when all squared edge lengths are equal to 1 is $n+1$, changing any number of the squared edge lengths from 1 to 9 can never yield a determinant divisible by 16.



Is there a degenerate simplex in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths?
This is answered in a paper by R. L. Graham, B. L. Rothschild & E. G. Straus ["Are there $n+2$ Points in $E\_n$ with Odd Integral Distances?"](https://sci-hub.ru/https://www.tandfonline.com/doi/abs/10.1080/00029890.1974.11993491). Such simplexes exist iff $n+2 \equiv 0 \pmod {16}$. They also consider the related problem of integral distances relatively prime to 3 and 6.


