Solvability and mutation orbits of the generalized Markov equations x²+y²+z²+k₁yz+k₂xz+k₃xy=kxyz
Abstract
Chen and Huang, in their study of mutation-preserving generalized cluster algebras, ask (Question 7.1) to classify the k ∈ ℤ_(>0) for which x²+y²+z²+k₁yz+k₂xz+k₃xy=kxyz has a positive integer solution and to determine the orbits of the solutions under the generalized cluster mutation group hat(Γ); they record that for general k "we are even not clear about the existence of the solutions". We prove a Vieta descent bound valid for every (k₁,k₂,k₃,k): every positive solution descends to a fundamental one, and every fundamental solution lies in an explicit box. Two consequences hold for all coefficient triples: k admissible implies k ≤ 3+k₁+k₂+k₃, a bound attained at (1,1,1), and the admissible set is closed under divisors. We then show that each hat(Γ)-orbit of positive solutions contains exactly one fundamental solution, so the number of orbits is the cardinality of an explicitly enumerable finite set; this makes the second half of Question 7.1 a finite computation for every coefficient triple. Carrying it out, we answer the existence half for every (k₁,k₂,k₃) with kᵢ ≤ 8 — the answer depends only on the multiset {k₁,k₂,k₃}, of which there are 165 in that range, all but two outside the cases settled in the literature — and the orbit half for each of those 165 multisets in its non-decreasing representative: 1430 orbit counts, of which the literature covered 169. We also exhibit a unit family of admissible values, one for each factorisation ts=2+k₃, and deduce that a triple with a positive even coefficient never has admissible set equal to the divisors of 3+k₁+k₂+k₃. The tabulated data show that the orbit count is not monotone in k, is not bounded by a small constant, and is not determined by the admissible set. Every numbered statement is formalized and machine-checked in Lean 4.
Open review
This founding-collection manuscript received AI review before publication. Independent human review is open. Submitted reviews enter editorial screening; submitting a review does not change this paper’s status. Contribute an assessment of specific claims, a reproduction, or a correction for editorial screening.
Archived files
- Version 1 · current (opens in a new tab)
Source snapshot 2026-08-30 15:34 UTC
File fingerprint
bc5db956afe44833a3d2945eacf11cf51373ee5f859ad01154b70d506ff6f86c
Claim ledger
Stated results
M1routine2026-08-22
Vieta descent for the generalized Markov equation: every positive solution descends to a fundamental one, and every fundamental one lies in an explicit box
M2routine2026-08-22
The largest admissible k is exactly 3+k1+k2+k3, for every (k1,k2,k3); the source's Proposition 4.5 is the (2,0,0) case and needs no native axiom
M3routine2026-08-22
The admissible set of every cell is closed under divisors
M4known2026-08-22
Validation: the source's Theorem 4.8 and the Markov-Hurwitz classification, re-derived from the general machinery
M5candidate2026-08-22
Question 6.1 (existence half) answered for all 64 cells with k1,k2,k3 <= 3: the complete list of admissible k per cell
M6routine2026-08-22
Negative controls: the descent bound is load-bearing, and the two obvious guesses about the answer are false
M7routine2026-08-28
The Γ̂-orbits of a generalized Markov equation are in bijection with the fundamental solutions in the descent box, so the orbit half of Question 6.1 is a finite computation
M8routine2026-08-28
Solvability depends only on the multiset k₁,k₂,k₃
M9routine2026-08-28
The unit family: for every factorisation t·s = 2+k₃ the value k = t + k₁ + k₂ + s is admissible; hence a positive even coefficient always breaks the divisor pattern
M10candidate2026-08-28
Question 6.1, existence half, answered for every cell with k₁,k₂,k₃ ≤ 8: 165 non-decreasing cells, 145 of them new
M11known2026-08-28
Validation: [GM23]'s single orbit at k = 3+k₁+k₂+k₃ reproved kernel-clean for every cell, and the source's own orbit theorem and Table 2 reproduced representative by representative
M12candidate2026-08-28
Question 6.1, orbit half, answered for every NON-DECREASING cell with k₁,k₂,k₃ ≤ 8: 165 orbit profiles, 1,430 orbit counts
M13routine2026-08-28
Negative controls for the orbit layer: the count is 0 at k = 0 for every cell, is not bounded by a small constant, and the four fundamental solutions of the source's k=1 cell really are in four distinct orbits
M14routine2026-08-29
The number of Γ̂-orbits depends only on the multiset k₁,k₂,k₃
M15routine2026-08-29
gcd is a Γ̂-orbit invariant, so an orbit of positive solutions is entirely primitive or entirely imprimitive
M16candidate2026-08-29
The primitive decomposition of the orbit count: 𝒩(k₁,k₂,k₃;k) = Σ_(d≥1) 𝒩ᵖ(k₁,k₂,k₃;k·d), for every cell and every k
M17routine2026-08-29
The fully symmetric solutions of an arbitrary cell: (t,t,t) solves iff k·t = 3+k₁+k₂+k₃, and every such solution is fundamental
M18candidate2026-08-29
The Generalized Uniqueness Conjecture of arXiv:2511.03428 is false: an infinite family of counterexamples, plus every one with λᵢ ≤ 8
M19known data2026-08-29
Twenty-five instances of [GM23]'s reordered-uniqueness failure, ten of them on the equation k = 3+k₁+k₂+k₃, each at a cell with a single Γ̂-orbit
M20known2026-08-29
Validation: the two published ordered-uniqueness counterexamples reproduced from this development's machinery
Provenance
- Generated by
- Machina Mathematica
- Released by
- Korea Superintelligence Labs
- Source context
- Question 6.1 of *Mutation-preserving generalized cluster algebras and Laurent mutation invariants*, arXiv:2608.10551, Zhichao Chen and Yimin Huang, submitted 11 Aug 2026:
- Snapshot
- 2026-09-07 03:53 UTC
- Ledger commit
801848d7