# Show the graph is regular

If $G$ is a connected finite graph which has no triangles, and $G$ has the property that if two vertices have a common neighbour then they have exactly two common neighbours, does $G$ have to be strongly regular?

Note: I showed it is regular, but how to prove/disprove it's strongly regular?

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Now I get a proof. The idea is to show each pair of adjacent vertices has the same degree.(The proof does not need the triangle free condition) –  Harry May 3 '12 at 23:25
The proof is for regularity, but for strongly regularity I have not found any counterexample so far, does the triangle-free condition play any role in this context? –  Harry May 3 '12 at 23:37
For the benefit of other users, and to get this question off the unanswered list, you could submit your proof of regularity as an answer. Answering your own questions is encouraged. –  Brian M. Scott May 3 '12 at 23:44
I can't answer my own question within 8 hours of asking~I will soon it allows me to do so. –  Harry May 4 '12 at 1:21
The title doesn't correspond to the question. –  joriki May 4 '12 at 14:18

The answer is "no". A counterexample is provided by the hypercube graph $Q_n$ for any $n\gt2$, in particular by the nearest-neighbour graph of the corners of a cube. The conditions are fulfilled, but some non-adjacent vertices have $2$ common neighbours while others have $0$.
$Q_n$ (for $n > 2$) does not fullfil the requirement that any two vertices with a common neighbor have exactly 2 common neighbors. Look at the drawing of $Q_3$ here: en.wikipedia.org/wiki/File:Hypercubeconstruction.png. The vertex in the lower left and the upper right have exactly one common neighbor. –  utdiscant Jul 8 '12 at 6:06