Somebody asked about the Markov Diophantine equation recently http://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.