Regular graph proof efficiently Suppose Γ is a non-complete graph, with every pair of joined vertices having exactly one common neighbor and every pair of non-joined vertices having exactly 2 common neighbors. How can I prove this is a regular graph? The proofs I have seen are considering joined and non joined vertices separately and are rather long. Is there a compact neat way to prove this ? Thank You

First note that $G$ must be connected, so it is sufficient to show that two adjacent vertices have the same degree. Let $x$ and $y$ be two adjacent vertices, with common neighbour $v$.

We define a mapping $f:N(x)\to N(y)$ as follows:

$f(y)=x$

$f(v)=v$

$f(t)$ is the unique common neighbour of $t$ and $y$ different from $x$ when $t\in N(x)-y-v$.

We show that $f$ is injective. The only nontrivial part is to show that $f(s)\
e f(t)$ for different $s,t\in N(x)-v-y$. Suppose $f(s)=f(t)=u$. Then $u$ and $x$ have three common neighbours, $s,t,y$. Contradiction.

By symmetry and the finiteness of our sets, $f$ must be bijective, which settles the case.

I am not sure if this classifies as neat and compact.