A question about pairwise inner products What is the largest $m$ such that there exist  $v_1,\dots,v_m \in \mathbb{C}^n$ all with unit norm and all distinct such that for all $i$ and $j$,  $v_i \cdot v_j = x$ for some fixed constant value $x$.
If $x=0$ we know the answer is $n$ as we are then requiring that all the vectors are orthogonal.  
It was shown in the comments by Krassotkin that $m= n+1$ is possible at least for $n=1,2$. 

Is this maximum $m = n + 1$ for all $n$?

There is a related question at https://mathoverflow.net/q/31436/106623 but I don't know if the proof technique carries over.
 A: Yes, $n+1$ is the maximal value of $m$ for any $n$. Let the vectors $v_i$ span a $k$-dimensional subspace in $\mathbb{C}^n$. Then there are indices $a_1, \dots, a_k$ such that the vectors $v_{a_1}, \dots, v_{a_k}$ form a basis of this subspace. In particular, any $v_i$ can be represented as a linear combination of $v_{a_1}, \dots, v_{a_k}$. If $v_i$ and $v_j$ are both not among $v_{a_1}, \dots, v_{a_k}$, then $(v_i-v_j, v_{a_1}) = \dots = (v_i-v_j, v_{a_k}) = 0$. As $v_i-v_j$ is a linear combination of $v_{a_1}, \dots, v_{a_k}$, it means that $v_i - v_j = 0$, i.e., there are at most $k+1\leq n+1$ vectors.
And, as Ethan Bolker noted, the regular $n+1$-simplex gives an example of $m=n+1$.
A: The regular $d+1$-simplex does the job for dimension $d$. Just put its center at the origin. (This is the analogue of the equilateral triangle, the regular tetrahedron, ...)
It's probably the maximum but I haven't tried to prove it.
A: Actually this is possible, but I only have an example and I use real vectors, but of course they are also complex. Take the vectors
$$(1,0),\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) $$
Their pairwise inner product is always $-\frac{1}{2}$ and they are distinct vectors on the unit circle. So no is not the right answer.
