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Markov number

From Wikipedia, the free encyclopedia
Solution to x*x + y*y + z*z = 3xyz
Not to be confused withMarkov constant orMarkov theorem.

Markov number orMarkoff number is a positiveintegerx,y orz that is part of a solution to the MarkovDiophantine equation

x2+y2+z2=3xyz,{\displaystyle x^{2}+y^{2}+z^{2}=3xyz,\,}

studied byAndrey Markoff (1879,1880).

The first few Markov numbers are

1,2,5,13,29,34,89,169,194,233, 433, 610, 985, 1325, ... (sequenceA002559 in theOEIS)

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

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The first levels of the Markov number tree

There are two simple ways to obtain a new Markov triple from an old one (xyz). First, one maypermute the 3 numbersx,y,z, so in particular one can normalize the triples so thatx ≤ y ≤ z. Second, if (xyz) is a Markov triple then so is (xy, 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,OEISA001653), and all the Markov numbers on the regions adjacent to 1's region are odd-indexed Fibonacci numbers (OEISA001519). Thus, there are infinitely many Markov triples of the form

(1,F2n1,F2n+1),{\displaystyle (1,F_{2n-1},F_{2n+1}),\,}

whereFk is thekthFibonacci number. Likewise, there are infinitely many Markov triples of the form

(2,P2n1,P2n+1),{\displaystyle (2,P_{2n-1},P_{2n+1}),\,}

wherePk is thekthPell number.[2]

Other properties

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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, ifm{\displaystyle m} is a Markov number then none of the prime divisors of9m24{\displaystyle 9m^{2}-4} 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

mn=13eCn+o(1)with C=2.3523414972.{\displaystyle m_{n}={\tfrac {1}{3}}e^{C{\sqrt {n+o(1)}}}\quad {\text{with }}C=2.3523414972\ldots \,.}

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

Error in the approximation of large Markov numbers

Moreover, he pointed out thatx2+y2+z2=3xyz+4/9{\displaystyle x^{2}+y^{2}+z^{2}=3xyz+4/9}, an approximation of the original Diophantine equation, is equivalent tof(x)+f(y)=f(z){\displaystyle f(x)+f(y)=f(z)} withf(t) =arcosh(3t/2).[11] The conjecture was proved[disputeddiscuss] byGreg McShane andIgor Rivin in 1995 using techniques fromhyperbolic geometry.[12]

ThenthLagrange number can be calculated from thenth Markov number with the formula

Ln=94mn2.{\displaystyle L_{n}={\sqrt {9-{4 \over {m_{n}}^{2}}}}.\,}

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

Markov's theorem

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Markoff (1879,1880) showed that if

f(x,y)=ax2+bxy+cy2{\displaystyle f(x,y)=ax^{2}+bxy+cy^{2}}

is anindefinitebinary quadratic form withreal coefficients anddiscriminantD=b24ac{\displaystyle D=b^{2}-4ac}, then there are integersxy for whichf takes a nonzero value ofabsolute value at most

D3{\displaystyle {\frac {\sqrt {D}}{3}}}

unlessf is aMarkov form:[13] a constant times a form

px2+(3p2a)xy+(b3a)y2{\displaystyle px^{2}+(3p-2a)xy+(b-3a)y^{2}}

such that

{0<a<p/2,aq±r(modp),bpa2=1,{\displaystyle {\begin{cases}0<a<p/2,\\aq\equiv \pm r{\pmod {p}},\\bp-a^{2}=1,\end{cases}}}

where (pqr) is a Markov triple.

Matrices

[edit]

Let tr denote thetrace function overmatrices. IfX andY are inSL2(C{\displaystyle \mathbb {C} }), then

tr(X)tr(Y)tr(XY)+tr(XYX1Y1)+2=tr(X)2+tr(Y)2+tr(XY)2{\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 iftr(XYX1Y1)=2{\textstyle \operatorname {tr} (XYX^{-1}Y^{-1})=-2} then

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

In particular ifX andY also have integer entries then tr(X)/3, tr(Y)/3, and tr(XY)/3 are a Markov triple. IfXYZ = I then tr(XtY) = tr(Z), so more symmetrically ifX,Y, andZ are in SL2(Z{\displaystyle \mathbb {Z} }) withXYZ = I and thecommutator of two of them has trace −2, then their traces/3 are a Markov triple.[14]

See also

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Notes

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  1. ^Cassels (1957) p.28
  2. ^OEISA030452 lists Markov numbers that appear in solutions where one of the other two terms is 5.
  3. ^Cassels (1957) p.27
  4. ^Frobenius, G. (1913). "Über die Markoffschen Zahlen".S. B. Preuss Akad. Wiss.:458–487.
  5. ^Guy (2004) p.263
  6. ^Aigner (2013)
  7. ^Rabideau, Michelle; Schiffler, Ralf (2020). "Continued fractions and orderings on the Markov numbers".Advances in Mathematics.370: 107231.arXiv:1801.07155.doi:10.1016/j.aim.2020.107231.
  8. ^Lee, Kyungyong; Li, Li; Rabideau, Michelle; Schiffler, Ralf (2023)."On the ordering of the Markov numbers".Advances in Applied Mathematics.143: 102453.doi:10.1016/j.aam.2022.102453.
  9. ^Aigner (2013) p. 55
  10. ^Zhang, Ying (2007)."Congruence and Uniqueness of Certain Markov Numbers".Acta Arithmetica.128 (3):295–301.arXiv:math/0612620.Bibcode:2007AcAri.128..295Z.doi:10.4064/aa128-3-7.MR 2313995.S2CID 9615526.
  11. ^Zagier, Don B. (1982)."On the Number of Markoff Numbers Below a Given Bound".Mathematics of Computation.160 (160):709–723.doi:10.2307/2007348.JSTOR 2007348.MR 0669663.
  12. ^Greg McShane; Igor Rivin (1995). "Simple curves on hyperbolic tori".Comptes Rendus de l'Académie des Sciences, Série I.320 (12).
  13. ^Cassels (1957) p.39
  14. ^Aigner (2013) Chapter 4, "The Cohn Tree", pp. 63–77

References

[edit]
Markoff, A. (1879)."First memoir".Mathematische Annalen.15 (3–4):381–406.doi:10.1007/BF02086269.S2CID 179177894.
Markoff, A. (1880)."Second memoir".Mathematische Annalen.17 (3):379–399.doi:10.1007/BF01446234.S2CID 121616054.
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