Markov Equation

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

:$x^2 + y^2 + z^2 = 3xyz,\,$

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'', 3''xy'' − ''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 2''n''² − 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

:$(1, F_, F_),\,$

where ''F''_''k'' is the ''k''th Fibonacci number. Likewise, there are infinitely many Markov triples of the form

:$(2, P_, P_),\,$

where ''P''_''k'' is the ''k''th 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

:$f(t) = t^2 - t(3xy) + (x^2 + y^2)$

Note that ''z'' is a root. By Vieta's formulas, the other root ''z′'' satisfies ''z + z′'' = 3''xy'' and ''zz′ = x''² + ''y''². Thus since ''z'' is positive, ''z′'' is also positive, we see that ''z′'' = 3''xy − z'' gives another solution.

Now, WLOG, assume ''x'' > ''y'', then take

:$f(x) = 2x^2 + y^2 - 3x^2 y = x^2 ( 2 - 3y ) + y^2$

Since ''y'' > 0, 2 − 3''y'' ≤ −1, so ''f''(''x'') < 0. Since ''f''(''t'') is an upward-facing parabola, that means min(''z'', ''z′'' ) < ''x'' < max(''z'', ''z′'' ).

That means that we can construct three new solutions: (''x'', ''y'', 3''xy − z''), (''x'', 3''xz − y'', ''z''), and (3''yz'' − ''x'', ''y'', ''z'') and these are ''distinct''. By our calculation above, exactly one of the three new solutions will have a smaller maximum element than (''x'', ''y'', ''z'') (and the other two larger).

Thus we proceed in this way, reducing the maximum element each time (this is Vieta jumping). Since we are working with only positive integers, we must eventually stop, which means we reach a solution that has not all elements distinct.

It's left for us to consider such a solution. WLOG assume ''x'' = ''y'', then 2''x''² + ''z''² = 3''x''²''z''. Thus ''x''² , ''z''² and ''x'' , ''z'', so write ''z'' = ''ax''. So we get

:$2x^2 + a^2 x^2 = 3a x^3 \implies 2 + a^2 = 3a x \implies 2 = a(3x - a)$

So we see ''a, 2'' so ''a'' = 1 or 2. If ''a'' = 1 then we get (1, 1, 1) and if ''a'' = 2 then we get (1, 1, 2). And from (1, 1, 2) we get to (1, 1, 1) by taking (''x'', ''y'', 3''xy − z'').

Thus we see that starting from an arbitrary solution we eventually come to (1, 1, 1), and so these are all the solutions.

Other properties

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

The ''unicity conjecture'' 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.

Odd Markov numbers are 1 more than multiples of 4, while even Markov numbers are 2 more than multiples of 32.

In his 1982 paper, Don Zagier conjectured that the ''n''th Markov number is asymptotically given by
:$m_n = \tfrac13 e^ \quad\text C = 2.3523414972 \ldots\,.$
The error $(\log(3m_n)/C)^2 - n$ is plotted below.

Moreover, he pointed out that $x^2 + y^2 + z^2 = 3xyz + 4/9$, an approximation of the original Diophantine equation, is equivalent to $f(x)+f(y)=f(z)$ with ''f''(''t'') = arcosh(3''t''  / 2). The conjecture was proved by Greg McShane and Igor Rivin in 1995 using techniques from hyperbolic geometry.

The ''n''th Lagrange number can be calculated from the ''n''th Markov number with the formula

:$L_n = \sqrt.\,$

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

Markov's theorem

showed that if

:$f(x,y) = ax^2+bxy+cy^2$

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

:$\frac$

unless ''f'' is a ''Markov form'': a constant times a form
:$px^2+(3p-2a)xy+(b-3a)y^2$
such that
:\begin 0
where (''p'', ''q'', ''r'') is a Markov triple.

Matrices

Let Tr denote the trace function over matrices. If ''X'' and ''Y'' are in SL₂( ℂ), then

:Tr(''X'') Tr(''Y'') Tr(''X⋅Y'') + Tr(''X''⋅''Y''⋅''X''⁻¹⋅''Y''⁻¹) + 2 = Tr(''X'')² + Tr(''Y'')² + Tr(''X⋅Y'')²

so that if Tr(''X''⋅''Y''⋅''X''⁻¹⋅''Y''⁻¹) = −2 then

: Tr(''X'') Tr(''Y'') Tr(''X⋅Y'') = Tr(''X'')² + Tr(''Y'')² + Tr(''X⋅Y'')²

In particular if ''X'' and ''Y'' also have integer entries then Tr(''X'')/3, Tr(''Y'')/3, and Tr(''X⋅Y'')/3 are a Markov triple. If ''X''⋅''Y''⋅''Z'' =  I then Tr(''X⋅Y'') = Tr(''Z''), so more symmetrically if ''X'', ''Y'', and ''Z'' are in SL₂( ℤ) with ''X''⋅''Y''⋅''Z'' = I and the commutator of two of them has trace −2, then their traces/3 are a Markov triple..

See also

*Markov spectrum

Notes

References

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:: {{cite journal , last1=Markoff , first1=A. , authorlink = Andrey Markov, title=Second memory, journal=Mathematische Annalen , year=1880 , doi=10.1007/BF01446234 , volume=17 , pages=379–399 , issue=3 , s2cid=121616054 , url=https://gdz.sub.uni-goettingen.de/id/PPN235181684_0017?tify=%7B%22view%22:%22info%22,%22pages%22:%5B394%5D%7D

Diophantine equations
Diophantine approximation
Fibonacci numbers