There are two simple ways to obtain a new Markov triple from an old one (x, y, z). First, one maypermute the 3 numbersx,y,z, so in particular one can normalize the triples so thatx ≤ y ≤ z. Second, if (x, y, z) is a Markov triple then 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 isconnected; in other words every Markov triple can be connected to(1,1,1) by a sequence of these operations.[1] If one starts, as an example, with(1, 5, 13) we get its threeneighbors(5, 13, 194),(1, 13, 34) and(1, 2, 5) in the Markov tree ifz is set to 1, 5 and 13, respectively. For instance, starting with(1, 1, 2) and tradingy andz before each iteration of the transform lists Markov triples withFibonacci numbers. Starting with that same triplet and tradingx andz before each iteration gives the triples withPell numbers.
All the Markov numbers on the regions adjacent to 2's region areodd-indexed Pell numbers (or numbersn such that 2n2 − 1 is asquare,OEIS: A001653), and all the Markov numbers on the regions adjacent to 1's region are odd-indexed Fibonacci numbers (OEIS: A001519). Thus, there are infinitely many Markov triples of the form
whereFk is thekthFibonacci number. Likewise, there are infinitely many Markov triples of the form
Aside from the two smallestsingular triples (1, 1, 1) and (1, 1, 2), every Markov triple consists of three distinct integers.[3]
Theunicity conjecture, as remarked byFrobenius in 1913,[4] states that for a given Markov numberc, there is exactly one normalized solution havingc as its largest element:proofs of thisconjecture have been claimed but none seems to be correct.[5]Martin Aigner[6] examines several weaker variants of the unicity conjecture. His fixed numerator conjecture was proved by Rabideau and Schiffler in 2020,[7] while the fixed denominator conjecture and fixed sum conjecture were proved by Lee, Li, Rabideau and Schiffler in 2023.[8]
None of the prime divisors of a Markov number is congruent to 3 modulo 4, which implies that an odd Markov number is 1 more than a multiple of 4.[9] Furthermore, if is a Markov number then none of the prime divisors of is congruent to 3 modulo 4. Aneven Markov number is 2 more than a multiple of 32.[10]
In his 1982 paper,Don Zagier conjectured that thenth Markov number is asymptotically given by
The error is plotted below.
Error in the approximation of large Markov numbers
Let tr denote thetrace function overmatrices. IfX andY are inSL2(), then
so that if then
In particular ifX andY also have integer entries then tr(X)/3, tr(Y)/3, and tr(XY)/3 are a Markov triple. IfX⋅Y⋅Z = I then tr(XtY) = tr(Z), so more symmetrically ifX,Y, andZ are in SL2() withX⋅Y⋅Z = I and thecommutator of two of them has trace −2, then their traces/3 are a Markov triple.[14]
Aigner, Martin (2013-07-29).Markov's Theorem and 100 Years of the Uniqueness Conjecture: A Mathematical Journey from Irrational Numbers to Perfect Matchings. Cham Heidelberg: Springer.ISBN978-3-319-00887-5.MR3098784.