$latex M$ is the markov triple $latex (x,y,z)$:
$latex x^2+y^2+z^2=xyz$ and $latex (x,y,x)\in \mathbb Z^3 \ \ \ \ (*)$.
It is easy to see:
$latex R_1: (x,y,z)\to (3yz-x,y,z)$.
map markov triple to markov triple.
This is also true for $latex R_2,R_3$. and the transform $latex R_1,R_2,R_3$ and permutation a classical result of markov claim that all solution of (*) could be generated from $latex (1,1,1)$. I get a similar result for a similar algebraic equation 1 half years ago when consider a $latex Q$ version of problem about 1-form given by Xu Bin.
Now we know the graph with root $latex (1,1,1)$ and with node generate by transform $latex R_1\cup R_2 \cup R_3 \cup S_3$ is connected.
The B-B-G-S conjecture is is the connected property still true for prime $latex p$ surfficed large?