# For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$?

For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$ (or equivalently, in $\mathbb{Q}^n$)?

Some easy observations:

• Such an $n$-simplex exists with vertices in $\mathbb{Z}^{n+1}$: just take the $n+1$ points $(0, \ldots, 0, 1, 0, \ldots, 0)$.

• If $n$ is even and $n+1$ is not a perfect square, then no regular $n$-simplex with vertices in $\mathbb{Z}^n$ exists: on the one hand, the side length $x$ is the square root of an integer, and thus the volume, namely $\frac{\sqrt{n+1}}{n! 2^{n/2}} x^n$, is irrational; on the other hand the volume must be rational by the formula using determinants.

• For $n=3$ a regular tetrahedron with vertices in $\mathbb{Z}^3$ does exist: $\{(0,0,0),(1,1,0),(1,0,1),(0,1,1)\}$.

• If there exists a Hadamard matrix of order $n+1$, we can normalize it so the first column is all ones and then the remaining $n$ columns give the coordinates of the $n+1$ vertices of a regular $n$-simplex with vertices in $\{1,-1\}^n$.

As shown in the accepted answer to this question, the Hadamard matrix condition is necessary and sufficent, so you have answered your own question...

EDIT As pointed out by Noam, I misread the linked-to question, and to atone: Robin Chapman gives the answer to the actual OP question, and here is his 1998 answer (slightly reformatted):

This is a good problem. It reduces to a question in the theory of rational quadratic forms. Let's ask for which $n$ an $n$-simplex can be embedded $n$-space with integral coordinates. The answer is if and only if

• $n + 1$ is an odd square, or

• $n + 1$ is a sum of $2$ odd squares, or

• $n + 1 \equiv 0 \mod 4.$

It's equivalent to consider rational coordinates, as we can scale, and we can also translate to put one of the vertices at the origin. Let $v_1,\dots, v_n$ be the other vertices. Then for some rational number m we have $v_i\cdot v_i = 2m,$ and $v_i\cdot v_j = m$ for $i \neq j.$ This means that the quadratic forms $Q_1 = x_1^2 + x_2^2 + ... + x_n^2$ and $2m Q_2$ where $Q_2 = x_1^2 + x_1 x_2 + x_1 x_3 + \dots + x_1 x_n + x_2^2 + x_2 x_3 + \dots + x_n^2$ are equivalent over the rationals. Indeed this is a necessary and sufficient condition. One can use the Hasse-Minkowski theory of rational quadratic forms to determine when an m exists so that this is the case, and doing so leads, after some effort, to the stated condition.

• Thanks, I somehow failed to find that question! But the paper by Adams, Zvengrowski and Laird mentioned there proves that there is a regular $n$-simplex in $\{0,1\}^n$ if and only if there exists a Hadamard matrix of order $n+1$. Is it easy to see that if there is any regular $n$-simplex in $\mathbb{Z}^n$ there must also be one in $\{0,1\}^n$? – Omar Antolín-Camarena Aug 15 '17 at 3:43
• The accepted answer answers the question for simplices inscribed in The accepted answer answers the question for simplices inscribed in a cube of the same dimension. Robin Chapman's answer claims a criterion for simplices with arbitrary integer coordinates: "The answer is yes iff $n+1$ is the sum of one, two, four or eight odd squares.". So for example $n=8,9$ are possible even though there's no Hadamard matrix of order $9$ or $10$ (any Hadamard matrix has order $1$, $2$, or a multiple of $4$). – Noam D. Elkies Aug 15 '17 at 4:39
• @NoamD.Elkies Oops, you are right! – Igor Rivin Aug 15 '17 at 4:43
• @OmarAntolín-Camarena see the edit, after Noam's sage commentary. – Igor Rivin Aug 15 '17 at 4:50