In the Euclidean space $\mathbb R^n$ :
What is the maximum cardinal of a family of equidistant points from each other?
How to build all these maximum families?
How to translate the condition of equidistance in terms of linear algebra ?
PS : There was a sequel to this exercise :
What is the maximum cardinal of a family of (vector) straight lines whose two by two angles are all equal?
Fix some distance $d$. Now fix your favorite point $x_1 \in \mathbb{R}^n$. Every point on the sphere $S(x_1,d) := \{ x\in \mathbb{R}^n : |x-x_1| = d\}$ is distance $d$ from $x_1$ (by definition of the sphere). It is also worth noting that $S(x_1,d)$ is an $n-1$ dimensional subset of $\mathbb{R}^n$ (hypersurface? manifold?—choose your linguistic and/or notational poison).
Next, fix your favorite point $x_2$ on the sphere $S(x_1,d)$, and note that every point in the intersection $$ S(x_1, d) \cap S(x_2, d) $$ is equidistant from both $x_1$ and $x_2$. Moreover, this intersection is an $n-2$ dimensional subset of $\mathbb{R}^n$.
We can continue inductively: in the $k$-th step, you will have $k$ spheres of radius $d$. The center of each sphere will be contained in all of the other spheres, and the $k$-fold intersection of the spheres will be an $n-k$ dimensional subset of $\mathbb{R}^n$. As long as $k \le n$, I can find a point in this intersection that is equidistant from all of the previously constructed points.
It therefore follows that I can find, via this construction, a collection of $n+1$ points that are mutually equidistant. Additionally, I cannot find any other points that are equidistant from every point in the collection. It remains only to show that no other construction is possible, but this can be done by actually being careful about the induction argument.