99-Graph: Is there a graph with 99 vertices in which every edge (i.e. pair of joined vertices) belong to a unique triangle and every nonedge (pair of unjoined vertices) to a unique quadrilateral? 



Is there a graph with 99 vertices in which every edge belong to a unique triangle and every nonedge to a unique quadrilateral?
First we will prove the graph is regular.


Let $x,y$ be two non-adjacent vertices, and let $a,b$ be their common neighbours. Define $X$ to be the neighbourhood of $x$ other than $a,b$, and $Y$ to be the neighbourhood of $y$ other than $a,b$. 


Considering the edge $ax$, there is a unique vertex $u\in X$ adjacent to both of them. Considering the non-edge $yu$, there must be exactly one edge from $u$ to $Y$. Similarly for the common neighbourbour of $b$ and $x$. For a vertex in $v\in X$ not adjacent to $a$ or $b$, the two common neighbours of $v$ and $y$ must lie in $Y$.


Consider the bipartite graph with parts $X,Y$ and the edges between them. We have proved that each part has 2 vertices of degree 1 and the others of degree 2. This is only possible if $|X|=|Y|$, which proves that $x$ and $y$ have the same degree. This proves the graph is regular. A simple count shows the degree must be 14.


Now you are looking for a strongly-regular graph of order 99, degree 14, $\lambda=1$ and $\mu=2$. According to [Andries Brouwer's table](http://www.win.tue.nl/~aeb/graphs/srg/srgtab51-100.html), the existence is unknown.


