Baragar-Bourgain-Gamburd-Sarnak conjecture

$latex M$ is the markov triple $(x,y,z)$:

$x^2+y^2+z^2=xyz$ 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 solutions of  (*) could be generated from $latex (1,1,1)$. I got a similar result for a similar algebraic equation 1 half years ago when considering a $latex Q$ version of problem about the 1-form given by Xu Bin.

Now we know the graph with root $latex (1,1,1)$ and with node generated 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 $latex p$ sufficed large?