“Let me say a few words about the solutions in positive integers of the equation


This equation is symmetric with respect to unknown terms x, y, z; therefore, knowing one of its solutions

x=\alpha, \quad y=\beta, \quad z=\gamma,

it is easy to find the following five:

x=\alpha, \quad y=\gamma, \quad z=\beta;
x=\beta, \quad y=\gamma, \quad z=\alpha;
x=\beta, \quad y=\alpha, \quad z=\gamma;
x=\gamma, \quad y=\alpha, \quad z=\beta;
x=\gamma, \quad y=\beta, \quad z=\alpha.

Although these six solutions may be different, we will consider them as one, denoted by


The above is a quote from A.A.Markov’s paper “Sur les formes quadratiques binaires indéfinies” (part 2). Back in 1880 people were patient enough to write out all permutations of three symbols… If we fix x=\alpha and y=\beta, the equation for z becomes

z^2-3\alpha \beta z +(\alpha^2+\beta^2)=0

which admits the second integer root \gamma'=3\alpha\beta-\gamma. We can also write \gamma\gamma'=\alpha^2+\beta^2 which does not immediately tell that \gamma' is an integer, but it does tell us that \gamma' is positive. For example, from the obvious solution (1,1,1) we get (1,1,2). Now it would not do us any good to mutate in the third variable again, for it would bring us back to (1,1,1). But we can mutate in the second variable, replacing it with 6-1=5. Having understood so well the symmetry of the equation, we write this new solution as (1,2,5), in the increasing order. Now we can mutate either 1 or 2, and so it goes…

Markov number tree, from http://en.wikipedia.org/wiki/File:MarkoffNumberTree.png

All triples, except for the two at the beginning, consist of distinct numbers (thus, they do generate six distinct solutions if order matters). The tree contains all solutions of the Markov equation. The Wikipedia article also points out the occurrence of Fibonacci numbers along the top branch, as well as a curious identity discovered by Don Zagier: let f(t)=\cosh^{-1}(3t/2); then (for triples written in increasing order)

\displaystyle x^2+y^2+z^2=3xyz+\frac{4}{9} \quad \Leftrightarrow \quad f(x)+f(y)=f(z)

Looks like a fun problem on simplification of inverse hyperbolic trigonometric functions.

Also, it’s still unknown whether two distinct Markov triples can have the same maximum \max(\alpha,\beta,\gamma). Looks like a fun problem for amateur number theorists.

To wrap this up, I will describe how Markov (the one of Markov chains fame, not his identically-named son of 4-manifold undecidability fame) came across the equation. Let Q(m,n)=am^2+2bmn+cn^2 be a quadratic form with real coefficients a,b,c normalized by b^2-ac=1. What is the best upper bound on

\displaystyle \min_{(m,n)\in \mathbb{Z}^2\setminus \lbrace(0,0)\rbrace} |Q(m,n)|?

In 1873 Korkin and Zolotarev published a paper showing that the best bound is 2/\sqrt{5}, attained by the form \displaystyle Q_1=\frac{2}{\sqrt{5}}(m^2-mn-n^2). Looks like the case is closed. But what if Q is not Q_1 (precisely, not equivalent to Q_1 under the action of SL(2,\mathbb Z))? Then the best bound improves to 1/\sqrt{2}, attained by \displaystyle Q_2=\frac{1}{\sqrt{2}}(m^2-2mn-n^2) (this is also due to KZ). Well, what if Q is not equivalent to either Q_1 or Q_2? Then the bound improves to \sqrt{\frac{100}{221}} (found by Markov), and we could go on…

But rather than continue in this fashion, Markov looked for the threshold \mu at which the number of inequivalent forms with minimum \ge \mu becomes infinite. And he found it: \mu =2/3 (for comparison, \sqrt{\frac{100}{221}}=0.67267\dots). Specifically, there are only finitely many forms with minimum above 2/3+\epsilon, for every \epsilon>0. But there are infinitely many forms with minimum exactly 2/3, such as \frac{2}{3}(x^2-\sqrt{5}xy-y^2). It was the iterative process of getting more and more of these forms that led Markov to the Diophantine equation x^2+y^2+z^2=3xyz.

The number 2/3 and its square also appear in Zagier’s identity with f(t)=\cosh^{-1}(3t/2)… But enough numerology for today.

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