Observations
From the correspondence of Markov polynomials with the points {(a,b)| a>b, gcd(a,b)=1},
we make several observations characterizing the snake graph corresponding to the point (a,b).
As stated earlier, the graph corresponding to this point (a,b) [a better notation here??] can
be created by triangulating all the squares through which a vector from the origin passes in
order to get to the point (a,b). To the entire snake graph, each square not on the end of the
path then contributes a pair of x and y edges forming a 2-by-2 block, for convenience here called
a "unit," and 2 z-edges attached to opposite corners of the unit. Squares on the end of course
contribute only 1 z-edge as one does not consider the endpoints of the graph as earlier shown.
Distinctness of denominators in the Markov polynomials
The vector from the origin to (a,b) then has to pass through a squares horizontally to get to (a,b)
and b squares vertically. This corresponds to a width of 2*a and a height of 2*b in the original model.
The powers of x and y in the denominators are then
$x^{frac{h-2}{2}} = x^{b-1}$ and $y^{frac{w-2}{2}} = y^{a-1}$
As there is a correspondence between the the distinct points whose entries are
relatively prime and the powers of x and y in the denominators of the Markov polynomials,
it follows that the denominators are distinct.
For completeness, one may also calculate the power of z in the denominator to be $z^{a+b-1}$.
This follows from the fact that the vector from 0 to the point (a,b) passes through a+b-1 squares
thereby contributing 4*(a+b-1) vertices. The power of z in the denominator from our earlier model
is z^{n/4} where n is the number of vertices.
Distinctness of numerators of the Markov polynomials
From the proof of the combinatorial model for the Markov polynomials using cuts to rearrange the graphs,
it was noted that a given snake graph has a matching that contains all the x-edges and a matching that
contains all the y-edges.
Our earlier proposition (lemma? theorem?) noted that for a