Define:
K(A,B,C) = (A^2 + B^2 + C^2) / (A*B*C).
Suppose you have the triple of Markov polynomials (A,B,C)., and let
C' = (A^2 + B^2)/C.
You can easily verify that K(A,B,C) = K(A,B,C').
Therefore, by induction,
we can prove that K(A,B,C) = K(x,y,z), if A,B and C are Markov polynomials
derived from x,y, and z.
But, C' = (A^2 + B^2)/C = (K(A,B,C)*A*B*C - C^2)/C
= K(x,y,z)*A*B - C.
Suppose that we substitute the graph B in the triple (A,B,C) by the graph D :
Then we need to prove that P(B) + P(D) = K(x,y,z) * P(A) * P(C)
But by simple algebra we can simplify this expression to:
Note that this proofs Laurentness of the Markov polynomials. This proof
can also be generalized for n-tuples instead of triples.