The Linear Equation

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:
                     
                                                          M(B) *  z^(na/2 + 2) * y^(wa) * x^(ha) + M(D) = (x^2+y^2+z^2) * M(A) * M(C)


Note that this proofs Laurentness of the Markov polynomials. This proof can also be generalized for n-tuples instead of triples.