Is there a graph with 99 vertices in which every edge belong to a unique triangle and every nonedge to a unique quadrilateral? 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? 
 A: 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, the existence is unknown.
A: I claim that this paper contains a solution of the problem. Unfortunately, Conway's email account is inactive and I do not know how to reach him via the Internet.
01.09.2018 
To clarify, I notice that the problem has two different interpretations:
(1) It is not important whether the set of nonedges is empty or not; in particular, a solution without nonedges would be acceptable.
(2) The set of nonedges must be nonempty.
If we choose (1) - and that's what I did - then my paper does provide an affirmative answer to the problem. Indeed, from Proposition 1 it follows that the number $99$ is good, i.e. there exist $\frac{99\cdot 98}{6}=1617 $ triangles such that every edge in $K_{99}$ belongs to a unique triangle. Further, the set of nonedges in $K_{99}$ is empty and therefore there is no requirement about quadrilaterals. Thus the desired graph is the complete graph $K_{99}$.
On the other hand, if we choose (2) the situation is more complicated because the following question arises:

What does "belong to a quadrilateral" mean?

Please, give me a definition.
05.09.2018 
According to Gordon Royle's comment below, Conway's problem can be stated as follows:

Is there a graph with $99$ vertices in which every pair of joined vertices have one common neighbour and every pair of unjoined vertices have two common neighbours?

If so, I must acknowledge that I misread the problem and thus the question is still open.
P.S.: The problem that I solved is the following one:

Are there $n_1$ triangles and $n_2$ quadrilaterals such that every edge in the complete graph $K_{99}$ belongs either to a unique triangle or to a unique quadrilateral?

(In particular the equality $3n_1 +4n_2 =4851$ must hold.) My paper provides a solution of this problem with $(n_1,n_2)=(1617,0)$. One can also prove that it is possible to replace $4$ triangles by $3$ quadrilaterals and derive a solution with   $(n_1,n_2)=(1613,3)$.
