I have encountered the following problem:
If $\Gamma$ is a graph on at least $3$ vertices and containing no triangles and with two non-neighbours vertices having exactly $b \geq 2$ common neighbours then $\Gamma$ is $k$-regular.
Edit: This is my worked realised so far:
The idea is basically to take to consider two vertices , and show that if they are not neighbours then the numbers of vertices that are nb to but not to is the same as the numbers of vertices that are nb to but not to . Then do the same for , being neighbours.
Case i: If , are not nb then let $_1$ be nb of but not , so there exists a set $Z_1$ of size of vertices nb to and . They clearly can't be nb to . I guess that the idea is to repeat this for all $z_i$ nb of but of but how can I make sure that the union of all this $Z_i$ is of the same as the nb of ?