Is equidistant points an open problem? This post asks whether for any $n$-dimensional (presumably real) normed vector space, you can find $n+1$ equidistant points. They receive two answers saying that it is possible, but neither give much detail. Wikipedia states that this is an open problem, though its source is from 1971. Neither of these seem particularly convincing as to whether the problem is open or not. They also seem to contradict each other.
So I would like to know whether it is an open problem whether for any $n$-dimensional normed vector space $(\mathbb{R}^n,\|\cdot\|)$ there is always a set of $n+1$ equidistant points.
 A: Take $n$  points at the unit directions, $(1,0,0...,0)$  and $(0,1,0,0..0)$ These are $\sqrt 2$ apart.
The final point, take as $(-t,-t,-t, ... , -t,-t)$   and find $t$  so the se diatnces are also $\sqrt 2$
Want $$  2 = (1+t)^2 + (n-1)t^2  $$
$$   2 =  n t^2 +2t + 1$$
$$ n t^2 +2t - 1 = 0$$
To get the same outcome  for a different inner product use matrices. By translating and shrinking my original example, I may put one of the points at the origin, all length one vectors and pairwise distances $1.$  That is, let my $n$ unit vectors make up the columns of a square matrix $P.$   The distance property says that $P^T P = W,$   where my $W$  will have all $1$   on the main diagonal and $1/2$  on all other elements.  If two unit vectors have dot product $\frac{1}{2}$  there is a $60^\circ$  angle between them, meaning distance $1$
You are now going to give me a positive symmetric matrix $G$   for which thenew inner product of $u,v$   will be $u^T G v.$  Now, we know that $G$ is a Gram matrix, meaning there is a matrix $R$   with $R^T R = G,$   while $R$  is also invertible.
Define $$   Q  = R^{-1} P  $$
Then  the (new)distances among the columns of $Q$
$$Q^T   GQ  = P^T R^{(-1)T}  R^T R  R^{-1}  P = P^T P = W $$
Your new $n+1$ points are the origin and the columns of $Q$
Next, here is the first page of a 2008 article settling many cases of the more general problem. But not all.

