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.

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.

  • 3
    $\begingroup$ By the same argument, there are no degenerate simplices where no edge length is a multiple of three except possibly in $\mathbb{R}^{3k-1}$ for some $k$. Maybe solving this related problem for e.g. $\mathbb{R}^2$ or $\mathbb{R}^5$ could give some intuition as to what happens in the odd length case. $\endgroup$
    – pregunton
    Commented Aug 28, 2021 at 10:08

1 Answer 1


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?". 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.

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    $\begingroup$ Thanks! To be clear, their $n$ is one less than I’ve used in the question here; it’s the dimension of the degenerate subspace. $\endgroup$
    – Greg Egan
    Commented Aug 29, 2021 at 4:56

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