A special type of Strongly regular graph Let $G$ be a strongly regular graph with parameters $(n,k,p,q)$. I am looking for results  where $p=q=1$.  In web, there are many theorems related to Strongly regular graphs but I wonder what happens it this specific case. Any reference and comment is welcome.
 A: There is no strongly regular graph with these parameters. 
Suppose $X$ is an srg with parameters $(n,k,a,c)$. Then the basic theory tells us that the eigenvalues of the graph are the valency $k$ and the two zeros of 
\[
  t^2-(a-c)t - (k-c),
\]
which in the case of hand is $t^2-(k-1)$. Again by the basic theory, this graph is not a conference graph ($k\ne 2c$), and so the eigenvalues are integers. Assume then that $k=\ell^2+1$ for some integer $\ell$.
Suppose $f$ and $g$ denote the respective multiplicities of $-\ell$ and $\ell$ as eigenvalues. Since $k$ must be simple and since the trace of the adjacency matrix is zero, we have
\[
  0= \ell^2+1 +(g-f)\ell,
\]
implying that $\ell$ divides 1. Therefore $k=2$ and $X$ is a 2-regular graph that contains a triangle ($a=1$). Hence $X=K_3$, which is not strongly regular.
Note that if $A$ is the adjacency matrix of $X$, then 
\[
  A^2 = (k-1)I + J
\]
(where $J$ is the all-ones matrix), whence $A$ is the incidence matrix of a projective plane. Since $A$ is symmetric and its diagonal is zero, this
plane has a polarity with no absolute points. By a theorem due to Baer, no such finite plane exists. (However the simplest proof of this is the one just given.)
And, finally, proving this result is a key step in the proof of the so-called Friendship Theorem, which is very likely where this question started.
