type: reasoning

text:
**Enumerating Markov triples below 100**
**Analyzing Markov tree mutations and mod4 residues**


text:
**Analyzing coordinate parity and coprimality constraints**
**Developing gcd-divisibility lemma with valuation contradictions**
**Examining divisibility impact on scaled variables**


text:
**Verifying coprimality preservation**
**Analyzing inequalities for mutations**


text: **Solving equation for variable bounds**


Referee report

Let
\[
M=\{(x,y,z)\in\mathbf Z_{>0}^3:x\le y\le z,\ x^2+y^2+z^2=3xyz\}.
\]
For fixed \(x,y\), the equation is quadratic in \(z\):
\[
z^2-3xyz+(x^2+y^2)=0.
\]
Hence the other root is
\[
z'=3xy-z=\frac{x^2+y^2}{z},
\]
so every Vieta mutation is integral and positive.

### Descent

Take \((x,y,z)\in M\), with \(x\le y\le z\). Write
\[
f(t)=t^2-3xyt+(x^2+y^2),
\]
whose roots are \(z,z'\). Since
\[
f(y)=x^2+y^2(2-3x)\le0,
\]
we have \(z'\le y\le z\). Equality \(f(y)=0\) occurs only when \(x=y=1\). Thus, except for the pair \((1,1,2)\), mutation of the largest coordinate gives
\[
0<z'<y<z
\]
or \(0<z'<y=z\). More precisely, when \(x=y=1\), the roots are \(1,2\), so the only triples with \(z'=y\) are \((1,1,1)\) and \((1,1,2)\); the latter descends to the former.

It remains to note that the largest coordinate cannot be repeated except in these cases. If \(y=z\), then
\[
x^2+2y^2=3xy^2.
\]
If \(x=1\), this gives \(y=1\); if \(x\ge2\), the right side is at least \(6y^2\), while the left side is at most \(y^2+2y^2=3y^2\), impossible. Hence \(y=z\) only for \((1,1,1)\). Similarly, \(x=y\) gives only \((1,1,1)\) and \((1,1,2)\).

Therefore repeated-coordinate exceptions are completely exhausted by those two vertices. Repeatedly mutating the largest coordinate strictly decreases the maximum until reaching \((1,1,1)\). This proves claim A.

### The rooted graph

Orient edges away from \((1,1,1)\). The root has one neighbor:
\[
(1,1,1)\longleftrightarrow(1,1,2).
\]
At \((1,1,2)\), mutating either occurrence of \(1\) gives the same vertex \((1,2,5)\), while mutating \(2\) returns to the root. Thus it has one child.

Every other vertex \((x,y,z)\) has three distinct coordinates. Mutation of \(z\) is its unique parent, by the descent just proved. Mutating \(x\) or \(y\) gives two distinct children:
\[
(3yz-x,y,z),\qquad (x,3xz-y,z),
\]
after reordering. Both new coordinates exceed \(z\): for example,
\[
3yz-x>z
\]
because \(y\ge2\) and \(3y z-x\ge6z-z>z\); the other case is identical. Consequently no child can coincide with an ancestor, and every vertex has a unique parent. Hence \(G\) is a tree consisting of a single edge from the root to \((1,1,2)\), followed by a full binary tree.

Thus claim B is false as stated. The exact repair is: the root and its child have one child each, and every subsequent vertex has exactly two children.

### Parity, residues modulo \(4\), and coprimality

Starting from \((1,1,1)\), every mutation preserves pairwise coprimality. Indeed, if \(\gcd(x,y)=\gcd(y,z)=\gcd(z,x)=1\), and \(z'=3xy-z\), then
\[
\gcd(x,z')=\gcd(x,z)=1,\qquad
\gcd(y,z')=\gcd(y,z)=1.
\]
Thus every vertex is pairwise coprime.

In particular, at most one coordinate is even. Reducing the equation modulo \(4\), the exact possibilities are:

1. all three coordinates are odd, and
   \[
   xyz\equiv1\pmod4;
   \]
   equivalently, among the three odd residues, the number congruent to \(3\pmod4\) is even;

2. exactly one coordinate is even, and it is congruent to \(2\pmod4\), while the other two are odd.

Indeed, three odd squares contribute \(3\pmod4\), forcing \(3xyz\equiv3\pmod4\), hence \(xyz\equiv1\pmod4\). If exactly one coordinate is even, the left side is \(2\pmod4\), so the even coordinate must be \(2\pmod4\). The other parity patterns are excluded by coprimality.

Therefore claim C is false only in its assertion that all coordinates are odd. Its correct form is: every vertex is pairwise coprime, and either all coordinates are odd with product \(1\pmod4\), or exactly one coordinate is \(2\pmod4\).

### Vertices with \(z\le100\)

The tree begins
\[
(1,1,1)\to(1,1,2)\to(1,2,5).
\]
From \((1,2,5)\), the two children are
\[
(1,5,13),\qquad(2,5,29).
\]
The next children are
\[
(1,13,34),\qquad(1,34,89),
\]
while the children of \((2,5,29)\) already have largest coordinates at least
\[
3\cdot2\cdot29-5=169>100.
\]
Likewise, every descendant of a vertex with largest coordinate \(>100\) also has largest coordinate \(>100\), since children strictly increase the maximum.

Hence the complete list with \(z\le100\) is
\[
\boxed{(1,1,1),\ (1,1,2),\ (1,2,5),\ (1,5,13),\ (2,5,29),\ (1,13,34),\ (1,34,89).}
\]
Completeness follows from the rooted-tree structure and the strict increase of the maximum along every downward edge, not from a computer search.