I will claim a proof

 The unicity conjecture states that for a given Markov number c, there is exactly one normalized solution having c as its largest element: proofs of this conjecture have been claimed but none seems to be correct.^[4]

Are Markov numbers relatively prime? Yes they must else a cycle through the combinations would lead to an incongruent overlap and violate the Markov conditions.

So if Markov triples are relatively prime they they can be sorted, uniquely. I can count the total count along one path. This means that Tere must be a shortest path to any z in some Markov x,y,z.  

This also means I can reorder them systematically and still find the unique p[ath through all combinations.  There must exist a form of the Markov tree such that either of two branches lead to a Markov tuple with c as its large counter. Since only one tuple change per jump, there must exist a x,k,z and x,j,z and both satisfy Markov. That mean both triple can use a vieta jump to return to the common node. That can happen only if i = k. We have two linear equations in one unknown\ and the same answer from integer arithmetic. The two jumps must be the same, the value C is a unique Markov triple when z=C.

It is graph theory on the Markov triple tree.  The tree is a spanning tree, three links per node.  Because it is a spanning tree,  I can systematically swap, or rotate links as long as I apply the rotation to each node. I can always count direction left right or down on the tree, as long as I can recover the rotations.  Then you can see that some rotations will result in contradictions in adjacent nodes because of convergence to a root any any of the three directions.

The Bayes spanning tree, counts all unique paths in order. 

What and who is relatively prime?

We really want each of the Markov fields to contribute equally to the counting. hey have to be -iLog(i) all within an integer, they all need to use 'bandwidth' optimally.   That point, x^x, has an equivalent form in the mean value of a coin tossed x times. The number of combinstions and rate of happening is when (pq)^x and p equals q  That is the probability of getting the mean value of x coin tosses when p and q are 1/2.   In the  combinatorials  p = 1/x and q = 1-1/x  The effect of relative sampling is to convert that 1/x into a 1/2 by  setting rate to x/(x+y+z), or relative frequency of an arrival.   then seting bandwidth using -iLog(i) will center all the arrivals when seen as an adapted system, that is all subsets separable and ordered.  There is a compression process, Botzman once again.

Send me a banana.