Somebody asked about the Markov Diophantine equation recently https://math.stackexchange.com/questions/94394/diophantine-equation-in-positive-integers and I was reminded of some unfinished business. In 1907, A. Hurwitz wrote Uber eine Aufgabe der unbestimmten Analysis, in Archiv der Mathematik und Physik, pages 185-196. I am going to abbreviate...with integers $x_1, \ldots, x_n,$ define $$ T = x_1^2 + \cdots + x_n^2 $$ and let $$ P = x_1 \ldots x_n.$$ For some positive integer $a,$ consider the Markov-Hurwitz equation in positive integers, $$ T = a P.$$ For any solution to M-H, we can create a new solution with some $x_j$ replaced by $$ x_j' = \frac{aP}{x_j} - x_j.$$ So as with the Markov tree, any solution can be transformed into a "fundamental solution" (Grundlosung) by a finite sequence of such transformations. That is, eventually, no $x_j$ can be reduced by this process, which is what happens when all $$ 2 x_j^2 \leq aP. $$ The stuff that happens next is easier to type if we order the entries, so an OFS, or Ordered Fundamental Solution, is $$ x_1 \geq x_2 \geq x_3 \geq \cdots \geq x_n, \; T = a P, \; 2 x_1^2 \leq aP.$$ Hurwitz shows many things. In an OFS, we have $$ a x_3 \cdots x_n \leq n. $$ We get $$ a \leq n.$$ Also, most entries are 1, indeed if $ k > 2 + \left\lfloor \log_2 n \right\rfloor$ then $x_k = 1.$ Whenever $a=n,$ the only OFS has all $x_j = 1.$ Hurwitz gives a table of all OFS for $n \leq 10.$ There can be more than one value of $a$ for each $n.$

Kap and I investigated a bit for larger $n.$ For example, the first time there is more than one OFS for a single $(n,a)$ pair is $n = 14, a = 1.$ One OFS is $(3,3,2,2,1,1,\ldots,1).$ There is another, $(6,4,3,1,1,\ldots,1).$

We are finally getting to my conjecture. Whenever $n = 5 w^2 - 6,$ there is an OFS with $$a=1, \; x_1 = 3 w, \; x_2 = 2 w, \; x_3 = 3, \; x_4 = x_5 = \cdots = x_n = 1.$$ The common value of $T=aP$ is $18 w^2.$ Considerable computation suggests that this is the largest that $x_1$ ever gets, and the only occurrence of equality, at least once $n \geq 3.$ So we have

CONJECTURE: In an OFS with $n \geq 3,$ we always have $$ x_1 \leq \sqrt{\frac{9(n+6)}{5}}$$ and equality occurs only when $ n = 5 w^2 - 6.$

So that is the question. I can make no guarantee that any advanced mathematics is invloved in any way, only that I put a ton of effort into proving this and never got it. Let me know if more information would be helpful, I've got tons. Pantloads.

ADDED: as far as I know the second best OFS is related to the best, $$n = 5 w^2 - 7, \; \; a=1, \; x_1 = 3 w -1, \; x_2 = 2 w, \; x_3 = 3, \; x_4 = x_5 = \cdots = x_n = 1.$$ The common value of $T=aP$ is $18 w^2 -6w.$

SEPTEMBER 2013: there have been two recent articles on Apollonian Circle Packing. Motion between quadruples of integers in an ACP is accomplished by exactly the same trick as Hurwitz and Markov before him, which is called Vieta Jumping in high-school competitions. If we have the integer equation $$ x_1^2 + \cdots + x_n^2 = F(x_1, x_2, \cdots, x_n), $$ where $F$ is a symmetric polynomial with each term "squarefree," each exp[onent of a variable is either $0$ or $1,$ then we may "jump" to another $n$-tuple by fixing $n-1$ variables and switching just one. That is, write the relation as $x_j^2 - B(\mbox{other}) x_j + C(\mbox{other}) = 0.$ One root is the current value of $x_j,$ the other is also an integer $x_j' =B(\mbox{other}) - x_j. $ The ACP equation is $$ w^2 + x^2 + y^2 + z^2 = 2 wx + 2 w y + 2 w z + 2 x y + 2 x z + 2 y z. $$ It turns out that there are infinitely many distinct orbits of the Apollonian (Vieta) Group acting on integer quadruples solving the equation.

  • $\begingroup$ Is there a form of the conjecture that depends on $a$? ie, as $a\rightarrow n$? $\endgroup$ Jan 5, 2012 at 2:46
  • $\begingroup$ Actually, it appears $a=n$ is isolated, the next largest value of $a$ seems never larger than $(n+3)/2.$ Indeed, for odd $n,$ there is always an OFS with $a = (n+3)/2, \; x_1 = 2, \; x_2 = x_3 = \cdots = 1$ $\endgroup$
    – Will Jagy
    Jan 5, 2012 at 3:51
  • 1
    $\begingroup$ Perhaps you will be interested in these experimental results : graphics and upper-bounds (note that I didn't know about works on Hurwitz equations 2 hours before now). $\endgroup$ Sep 19, 2013 at 20:30

2 Answers 2


Do the Apollonian and Markoff-Hurwtiz problems share the same "jumping" movement? That would depend on one's interpretation of "share." There is a link; one that my research followed.

I was first introduced to the Hurwitz equations via the Markoff equation; and the fractal nature of the exponent of growth lead me to the work of Boyd on the Apollonian Packing Problem. I later heard about a certain class of K3 surfaces ... those surfaces that are smooth $(2,2,2)$ forms in $P\times P\times P$. (Note that $P\times P\times P$ is not the same as $P^3$.) These are surfaces in three projective variables whose defining equation is quadratic in each variable. They therefore have the same "jumping" action as the Markoff equation. These surfaces have the advantage of being smooth, so the jumps end up being automorphisms of the surface. The automorphisms act on the group of divisors on the surface (the Picard group), and this action shares a lot of similarities with the Apollonian packing.

The Apollonian packing can be thought of this way: The bilinear form that you mention in your note is a Lorentz product. That is, we can write it as $X^tJX=0$, where $X=(w,x,y,z)$ and $J$ is a symmetric matrix with signature $(3,1)$. (That is, $J$ has $3$ positive eigenvalues and one negative eigenvalue.) Thus, a surface of the form $X^tJX=-1$ is a hyperboloid of two sheets, and one sheet is a model of 3D hyperbolic geometry, with distance defined by $\cosh(|XY|)=-X^tJY$. Three dimensional hyperbolic space has the Poincare upper half space model, and planes in this model are (Euclidean) hemi-spheres perpendicular to the boundary. A plane is therefore represented by a circle. The "jumps" used in the Apollonian packing is a subgroup of the group of all such jumps, and is a subgroup of isometries in the hyperbolic space. The packing is an orbit of one plane under the action of this group. Because the fundamental domain for the group of "jumps" has infinite volume, the packing has an associated fractal.

For the K3 surfaces, the Lorentz product is the intersection pairing, which has signature $(1,3)$ (if the Picard group has dimension $4$). The group of automorphisms on the K3 surface (analogs of the jumps for the Markoff equation) acts as isometries (analogs of jumps for the Apollonian packing) on the underlying hyperbolic space. The orbit of a circle for these cases are not always tangent, as they are in the Apollonian case, but are sometimes perpendicular. An example is pictured on my homepage (http://faculty.unlv.edu/baragar/). Note that one of the jumps in the Picard space came from an automorphism that isn't exactly a "jump" on the surface, though it has a very interesting interpretation.

There is no K3 surface, though, that is associated to the Apollonian packing.

  • $\begingroup$ Arthur, thank you for answering. I gather that all the interest in this, and the difficult part, has more to do with the growth of solutions in the same "orbit." So those examples with "jumping" that are null vectors of an indefinite quadratic form, or possibly $Q(\vec{x}) = b,$ will be rather special. Kaplansky and I mostly looked at the "root" or "fundamental" solutions. $\endgroup$
    – Will Jagy
    Sep 28, 2013 at 22:14

Not an answer, but a bit long for a comment. If you don't already know them, you might find the papers of Baragar to be of interest:

[1] The exponent for the Markoff-Hurwitz equations. Pacific J. Math. 182 (1998), no. 1, 1–21

[2] Asymptotic growth of Markoff-Hurwitz numbers. Compositio Math. 94 (1994), no. 1, 1–18

[3] Integral solutions of Markoff-Hurwitz equations. J. Number Theory 49 (1994), no. 1, 27–44

If one lets $N_{n,a}(X)$ be the number of solutions up to size $X$ (assuming the unicity conjecture, I guess), then [1,2] study the exponent $e(n)$ such that $N_{n,a}(X) \asymp (\log X)^{e(n)}$. Zagier showed that $e(3)=2$, but for $n\ge4$, Baragar shows that $e(n)$ isn't an integer, and indeed, probably isn't even rational.

  • $\begingroup$ Joe, thanks for answering. The name Baragar is familiar, but not these articles. This seems to be the main interest, also for the Apollonian circle packing, how many solutions with largest element up to some large $N.$ I haven't seen anything in print that pointed out that the Apollonian and Markov-Hurwitz problems share the same "jumping" movement, same possibility of "fundamental" solutions of some kind, and so on. $\endgroup$
    – Will Jagy
    Sep 19, 2013 at 20:59

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