Markov number

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 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 (1,1,1) by a sequence of these operations.^[1] If one starts, as an example, with (1, 5, 13) we get its three neighbors (5, 13, 194), (1, 13, 34) and (1, 2, 5) in the Markov tree if z is set to 1, 5 and 13, respectively. For instance, starting with (1, 1, 2) 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, 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

${\displaystyle (1,F_{2n-1},F_{2n+1}),\,}$

where F_k is the kth Fibonacci number. Likewise, there are infinitely many Markov triples of the form

${\displaystyle (2,P_{2n-1},P_{2n+1}),\,}$

where P_k is the kth Pell number.^[2]

Aside from the two smallest singular triples (1, 1, 1) and (1, 1, 2), every Markov triple consists of three distinct integers.^[3]

The unicity conjecture, as remarked by Frobenius in 1913,^[4] states that for a given Markov number c, there is exactly one normalized solution having c as its largest element: proofs of this conjecture 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 ${\displaystyle m}$ is a Markov number then none of the prime divisors of ${\displaystyle 9m^{2}-4}$ is congruent to 3 modulo 4. An even Markov number is 2 more than a multiple of 32.^[10]

In his 1982 paper, Don Zagier conjectured that the nth Markov number is asymptotically given by

${\displaystyle m_{n}={\tfrac {1}{3}}e^{C{\sqrt {n+o(1)}}}\quad {\text{with }}C=2.3523414972\ldots \,.}$

The error ${\displaystyle o(1)=(\log(3m_{n})/C)^{2}-n}$ is plotted below.

Moreover, he pointed out that ${\displaystyle x^{2}+y^{2}+z^{2}=3xyz+4/9}$, an approximation of the original Diophantine equation, is equivalent to ${\displaystyle f(x)+f(y)=f(z)}$ with f(t) = arcosh(3t/2).^[11] The conjecture was proved ^{[disputed – discuss]} by Greg McShane and Igor Rivin in 1995 using techniques from hyperbolic geometry.^[12]

The nth Lagrange number can be calculated from the nth Markov number with the formula

${\displaystyle L_{n}={\sqrt {9-{4 \over {m_{n}}^{2}}}}.\,}$

The Markov numbers are sums of (non-unique) pairs of squares.

Markoff (1879, 1880) showed that if

${\displaystyle f(x,y)=ax^{2}+bxy+cy^{2}}$

is an indefinite binary quadratic form with real coefficients and discriminant ${\displaystyle D=b^{2}-4ac}$, then there are integers x, y for which f takes a nonzero value of absolute value at most

${\displaystyle {\frac {\sqrt {D}}{3}}}$

unless f is a Markov form:^[13] a constant times a form

${\displaystyle px^{2}+(3p-2a)xy+(b-3a)y^{2}}$

such that

${\displaystyle {\begin{cases}0<a<p/2,\\aq\equiv \pm r{\pmod {p}},\\bp-a^{2}=1,\end{cases}}}$

where (p, q, r) is a Markov triple.

Let tr denote the trace function over matrices. If X and Y are in SL₂(${\displaystyle \mathbb {C} }$), then

${\displaystyle \operatorname {tr} (X)\operatorname {tr} (Y)\operatorname {tr} (XY)+\operatorname {tr} (XYX^{-1}Y^{-1})+2=\operatorname {tr} (X)^{2}+\operatorname {tr} (Y)^{2}+\operatorname {tr} (XY)^{2}}$

so that if ${\textstyle \operatorname {tr} (XYX^{-1}Y^{-1})=-2}$ then

${\displaystyle \operatorname {tr} (X)\operatorname {tr} (Y)\operatorname {tr} (XY)=\operatorname {tr} (X)^{2}+\operatorname {tr} (Y)^{2}+\operatorname {tr} (XY)^{2}}$

In particular if X and Y also have integer entries then tr(X)/3, tr(Y)/3, and tr(XY)/3 are a Markov triple. If X⋅Y⋅Z = I then tr(XtY) = tr(Z), so more symmetrically if X, Y, and Z are in SL₂(${\displaystyle \mathbb {Z} }$) with X⋅Y⋅Z = I and the commutator of two of them has trace −2, then their traces/3 are a Markov triple.^[14]

• Markov spectrum