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](https://en.wikipedia.org/wiki/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 $\{1,-1\}^n$.



For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$?
As shown in the accepted answer to [this question](https://mathoverflow.net/questions/38724/coordinates-of-vertices-of-regular-simplex), 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.


