## General properties

This page displays an applet showing the tree of Markov numbers. It gives the basic description of the tree
and the instructions for navigating in that tree. To display the tree, click on the bluish button, labeled "Display the Markov Tree".
Depending on your security setting, a yellow bar at the top of the page may appear, warning you that running this
applet ("ActiveX controls" it will say) maybe dangerous. In that case, click on this yellow bar and allow the applet to be run. Trust me. In any case,
after you click on the bluish button, the tree will appear in a separate window. You may resize the window, minimize it (in which case
its icon will appear in the bottom tray so you can restore it), or close it altogether. If this last action is taken, you must click on the bluish
button again to restore the applet.

The Markov numbers are the solutions of the diophantic equation:
(1)

Each node in the tree holds a solution triplet (*x, y, z*) with *x ≤ y ≤ z.* The two children of a node
are generated as follows: Suppose
(*x, y, z*), *x ≤ y ≤ z*

is a solution held at some node. If we hold two of the variables fixed,
this defines a quadratic equation in the third variable. That equation is monic, with integer coefficients, thus the other
root, denoted by *x', y'*, or *z'*, is also an integer, giving rise to another solution. This produces three
other triplets solving the equation (1):

(*y, z, x'*) denoted by *X*(*x, y, z*)

(*x, z, y'*) denoted by *Y*(*x, y, z*) and

(*z', x, y*) or (*x, z', y*) denoted by *Z*(*x, y, z*)

The smallest solution, strictly speaking, is (1, 1, 1), which gives rise to
three identical triplets (1, 1, 2). That triple, in turn, produces two triplets (1, 2, 5),
and one already accounted for: (1, 1, 1). The triplets (1, 1, 1) and
(1, 1, 2) are the only solutions in which *x = y* or *y = z*.
Every other node (*x, y, z*) in the tree has then two new distinct children:
The left child: (*x, z, y'*) and the right child: (*y, z, x'*), with *y' < x' *.
The triplet containing *z'*, be it (*z', x, y*) or (*x, z', y*), already
appears further up the tree, in fact it is the parent of (*x, y, z*).
The reason for this is that if *x < y < z* then:
*z < x'*, * z < y'*, *y' < x' *, and *z' < y*

Thus, if we display the tree with (1, 2, 5) at the root, the tree is quite regular, every node has two children
as described above.
The tree structure at the node
(*x, y, z*) can be described by this picture showing
the location of the nodes *X*(*x, y, z*), *Y*(*x, y, z*)
and *Z*(*x, y, z*):

Parent *Z*(*x, y, z*) = (*x, z', y*) or (*z', x, y*)

Left child *Y*(*x, y, z*) = (*x, z, y'*)

Right child *X*(*x, y, z*) = (*y, z, x'*)

See the mathematical discussion for all the proofs and further references.

It is a long standing conjecture, going back to Frobenius in 1913, whether the tree, as described, defines each Markov number
uniquely. More precisely, given a Markov number *m*, is there **only one** triplet of the form (*x, y, m*)?
(With *x ˂ y ˂ m*, of course.)
## Navigation of the tree

The tree in the applet has initially the triplet (1, 2, 5) at the root, and the children of all the nodes
are chosen according to the scheme above. The values of *x, y,* and *z* at the root are displayed constantly. At any given time,
the display shows all the 255 nodes in 7 levels below the root. You can place any node on the display in the root, by left clicking on that node.
To see the content at any of the nodes, **right click on the node and hold the mouse button down**.
Releasing the mouse button makes the display disappear. To move a given node to the root,
**left click on the node**. The applet will display the tree below the chosen node. To move the root one level up, or to reset the root to (1, 2, 5),
click the relevant button in the bottom panel. The numbers involved are quite large, as you can test yourself. When the numbers get to be 100+ digits long,
the response will be sluggish, there is a lot of stuff to compute. When the tree seems to be unresponsive, it does **not** mean the
program crashed, it is just doing its computation. It will eventually finish, give it some time.
There is at least one more Web page devoted to Markov numbers, it
is maintained by Tom Ace. For some mathematical discussion, references, proofs etc. see this link.

Return to Vladimir Drobot's home page.