We know that given the Markov Diophantine equation $x^2+y^2+z^2=3xyz$ we have a solution $(x,y,z)$ then markov show that $(x,z,3xz-y)$ and $(y,z,3yz-x)$ are also solutions, using them we can build a tree which begin at $(1,1,1)$.
My question is the following, how can I prove that those are all the integral solution for the equation?
Thanks!
The best source on this is HURWITZ 1907. A proof for the Markov numbers is included. Lots of students on this site are aware of a contest technique called "Vieta Jumping." The improvement that Hurwitz gives is to emphasize the geometry of "fundamental solutions" of one of his diophantine equations, where we just need to show existence of solutions obeying some fairly tight inequalities. For this direction, see https://math.stackexchange.com/questions/tagged/vieta-jumping