Why was the $243$-graph version of Conway's $99$-graph problem easy enough to be solved? I know that there is a thousand-dollar prize for Conway's 99-graph problem. I also know that there is a graph called the Berlekamp–Van Lint–Seidel graph that solves the 243-graph version of this problem. Why was the 243-graph problem easier than the 99-graph problem even though all the conditions for the existence of such a graph were true?
 A: Even for $99$ vertices, you can't just solve a problem like this by checking all approximately-$10^{1304}$ possible graphs to see if they satisfy the condition. Both of these cases of the problem are out of the reach of brute force.
So what do we do if we abandon brute force, because checking all graphs is hard? We can hope to find an answer that's a very nice graph with a simple and highly symmetric construction. Here, $243$ has an advantage over $99$, because $243 = 3^5$ is a high power of a small prime. We can define Berlekamp–Van Lint–Seidel graph in terms of modulo-$3$ arithmetic in $\mathbb F_3^5$. When we do this, it's so symmetric that we can check Conway's condition for it by checking it for just one pair of adjacent vertices and one pair of non-adjacent vertices.
With $99$ vertices, there is no hope for an equally-nice construction. It has already been checked (by Wilbrink in 1984) that there's no vertex-transitive solution. We cannot prove that there is no $99$-vertex solution at all, because it's hard to guarantee that a weird, asymmetric, messy graph will not spontaneously happen to satisfy Conway's condition. But we also don't know where to start looking for weird, asymmetric, messy solutions.
