A Markov number or Markoff number is a positive integer x, y or z that is part of a solution to the Markov Diophantine equation studied by . The first few Markov numbers are 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325, ... appearing as coordinates of the Markov triples (1, 1, 1), (1, 1, 2), (1, 2, 5), (1, 5, 13), (2, 5, 29), (1, 13, 34), (1, 34, 89), (2, 29, 169), (5, 13, 194), (1, 89, 233), (5, 29, 433), (1, 233, 610), (2, 169, 985), (13, 34, 1325), ... There are infinitely many Markov numbers and Markov triples. Markov tree There are two simple ways to obtain a new Markov triple from an old one (x, y, z). First, one may permute the 3 numbers x,y,z, so in particular one can normalize the triples so that x ≤ y ≤ z. Second, if (x, y, z) is a Markov triple then by Vieta jumping so is (x, y, 3xy − z). Applying this operation twice returns the same triple one started with. Joining each normalized Markov triple to the 1, 2, or 3 normalized triples one can obtain from this gives a graph starting from (1,1,1) as in the diagram. This graph is connected; in other words every Markov triple can be connected to by a sequence of these operations. If we start, as an example, with we get its three neighbors , and in the Markov tree if z is set to 1, 5 and 13, respectively. For instance, starting with and trading y and z before each iteration of the transform lists Markov triples with Fibonacci numbers. Starting with that same triplet and trading x and z before each iteration gives the triples with Pell numbers. All the Markov numbers on the regions adjacent to 2's region are odd-indexed Pell numbers (or numbers n such that 2n2 − 1 is a square, ), and all the Markov numbers on the regions adjacent to 1's region are odd-indexed Fibonacci numbers (). Thus, there are infinitely many Markov triples of the form where Fk is the kth Fibonacci number. Likewise, there are infinitely many Markov triples of the form where Pk is the kth Pell number. Proof that this generates all possible triples Start with some solution (x, y, z), and assume all three are distinct. Now consider the quadratic