coordinates of vertices of regular simplex

One interesting problem is to determine the $n$ such that there is a regular simplex in $\mathbb{R}^n$ with rational/integer coordinates. This is a well-known old problem which I discussed in this 1998 usenet thread, and is a nice application of the Hasse-Minowski theory of rational quadratic forms. The answer is yes iff $n+1$ is the sum of one, two, four or eight odd squares.

It is known that there is a regular simplex of side length $\sqrt{(d+1)/2}$ whose vertices are vertices of the cube $[-1,1]^d$ in $\Bbb{R}^d$ if and only if there exists a Hadamard matrix of order $d+1$; this is a square matrix of $\pm 1$-entries with pairwise orthogonal columns.

In particular, there exist Hadamard matrices of order $2^n$, one of which can be constructed using the recursive Sylvester's construction as explained on the above linked wikipedia page:

Let $H_0=[1]$ and $H_{n+1}=\left[\array{H_n & H_n \\\\ H_n & -H_n}\right].$

Note that the first column of $H_n$ consists only of ones. Delete it to obtain $2^n$ row vectors in $\Bbb{R}^{2^n-1}$. These are the coordinates of a regular simplex.

The $d$ points $(0,\dots,0,1,0,\dots,0)$ are the vertices of a regular $(d-1)$-simplex. If you want it to be centered at the origin, just substract their barycenter from them.