Proving the existence of a set of vectors Let $m \in ]0,\infty[$.
In the following, I will consider a product space on $R^n$ and its derived norm.
I'm asked to prove the existence of $(a_0,\ldots, a_n)$ a set of $n+1$ vectors in $\mathbb R^n$ with the following properties:
$$\forall i \in \{0, \ldots,n \}, ||a_i||=m $$
$$\sum_{i=0}^{n}a_i=0$$
$\forall (i,j) \in \{0, \ldots,n \} \ i \neq j$, $$\langle a_i,a_j \rangle=\frac{-m²}{n}$$
I think it can be done via induction, using the fact that $n+1$ vectors are linearly dependent. But how to deal with an extra coordinate (to keep a constant norm) ? I don't have any other enlightening thoughts on this.
Thanks for your suggestions.
 A: Note that you can assume without loss of generality that $m=1$. Then, what you are looking is for $n+1$ points on the unit sphere, such that they are equidistant.
Indeed, if $\|a_i-a_j\|=\|a_i-a_k\|$ whenever $i\ne j,k$ then
$$
2-2\langle a_,a_j\rangle=2-2\langle a_i,a_k\rangle,
$$
so $\langle a_i,a_j\rangle=\langle a_i,a_k\rangle$ whenever $i\ne j,k$.
From the fact that the points are regularly distributed you should get that $\sum a_i=0$. Finally,
$$
0=\langle a_1,\sum_{i=0}^na_i\rangle=1+\sum_{i=1}^n\langle a_1,a_i\rangle=1+n\langle a_1,a_2\rangle.
$$
So $\langle a_1,a_2\rangle=-1/n$, and we deduce that $\langle a_i,a_j\rangle=-1/n$ if $i\ne j$.
So let us construct such points. We proceed by induction. If $n=1$ then $\{1,-1\}$ is a set of two (trivially equidistant) points in the unit sphere (which is the same set!).

The picture above shows the idea of what comes below. The point $(1,0)$ represents the new point; the vertical dark line is the hyperplane, and the red part is the interior of $S_n$. The intersections of the red line with the circle are the equidistant points from the previous step.
Now assume that we can construct $n$ equidistant points, at distance $d_n$ in the unit sphere $S_{n-1}$. In the sphere $S_n$, fix a vector, say $e_1=(1,0,\ldots,0)$. In the hyperplane perpendicular to $e_1$ that crosses the axis of $e_1$ at $(2/d_n^2-1,0,\ldots,0)$, put a copy of the sphere $S_n$, shrunk so that its radius is $\frac2{d_n}\,\sqrt{1-\frac1{d_n^2}}$. The new $n$ points are at distance $1$ from the origin, because the distance to the origin is $$\sqrt{\left(\frac2{d_n^2}-1\right)^2+\frac4{d_n^2}\left(1-\frac1{d_n^2}\right)}=1
$$
The new $n$ points are at mutual distances of
$$
d_n\,\frac2{d_n}\sqrt{1-\frac1{d_n^2}}=2\sqrt{1-\frac1{d_n^2}},
$$
and the distance from any of these points to $e_1$ is
$$
\sqrt{\left(1-\left(\frac2{d_n^2}-1\right) \right)^2+\frac4{d_n^2}\left(1-\frac1{d_n^2}\right)}
=\sqrt{\left(2-\frac2{d_n^2} \right)^2+\frac4{d_n^2}\left(1-\frac1{d_n^2}\right)}\\
=2\sqrt{1-\frac1{d_n^2}}\,\sqrt{1-\frac1{d_n^2}+\frac1{d_n^2}}
=2\sqrt{1-\frac1{d_n^2}}.
$$
We have found $n+1$ points in the sphere $S_n$, at mutual distances of $d_{n+1}=2\sqrt{1-\frac1{d_n^2}}$.
