\documentclass[11pt,a4paper]{article} \usepackage[margin=25mm]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[T1]{fontenc} \usepackage{lmodern,microtype} \usepackage[dvipsnames]{xcolor} \usepackage[hyperfootnotes=false,colorlinks=true,linkcolor=blue,urlcolor=blue,citecolor=blue]{hyperref} \usepackage{indentfirst} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \DeclareMathOperator{\arcosh}{arcosh} \newcommand{\w}{\mathbf{w}} \title{Rational limits in Markov--Euclid asymptotics} \author{Qiyue Tang} \date{} \begin{document} \maketitle \begingroup \renewcommand{\thefootnote}{} \footnotetext{\textit{2020 Mathematics Subject Classification.} Primary 11D25; Secondary 11J81.\newline \textit{Keywords.} Markov equation, Euclid tree, irrationality, Fock's function.} \endgroup \begin{abstract} We construct an infinite reduced mutation sequence for which all three logarithmic Markov--Euclid ratios converge to $9/10$. This disproves the irrationality conjecture of Zhichao Chen and Zelin Jia (Conjecture~6.12). \end{abstract} \begingroup\color{blue} \section*{AI use and scholarly credit} The main results and their proofs were developed in a ChatGPT-6-Pro conversation lasting approximately 34 minutes. Codex subsequently carried out an adversarial review of the arguments, checked the supporting computations and literature, and prepared the manuscript using the mathematical writing skill. By posting this manuscript on this platform, I do not seek to assert ownership of the results or claim priority. I welcome anyone interested in preparing a finished account or checking the arguments; the eventual scholarly credit should go to those who undertake that work. \par\endgroup \section{Introduction} The classical Markov equation \begin{equation}\label{eq:markov} x_1^2+x_2^2+x_3^2=3x_1x_2x_3 \end{equation} arose in Markov's study of indefinite binary quadratic forms and Diophantine approximation \cite{Markov}. Its positive integer solutions are called Markov triples. Replacing one coordinate by the other root of the corresponding quadratic equation gives the Vieta mutation \[ x_i\longmapsto 3x_jx_k-x_i,\qquad \{i,j,k\}=\{1,2,3\}, \] with the other two coordinates fixed. These operations generate the positive solutions from $(1,1,1)$ and organize them into a tree. Zagier's counting argument related the Markov tree to a lattice-point problem; Baragar \cite{Bar} developed logarithmic comparisons with Euclid trees for higher-dimensional Markoff--Hurwitz equations. In these comparisons, the product appearing in a Vieta mutation becomes a sum after taking logarithms. Spalding and Veselov \cite{SpV} subsequently studied Lyapunov exponents of Markov and Euclid dynamics along paths. Their work also connects the Farey parametrization of the Markov tree with hyperbolic lengths and Fock's function. Cluster algebras, introduced by Fomin and Zelevinsky \cite{FZ}, provide another setting for these recursions. On a Markov triple, the equation gives \[ 3x_jx_k-x_i=\frac{x_j^2+x_k^2}{x_i}, \] which is the exchange relation of the Markov cluster pattern. Gyoda and Matsushita \cite{GM} studied the generalized equations \begin{equation}\label{eq:generalized} \begin{split} &x_1^2+x_2^2+x_3^2+\lambda_3x_1x_2+\lambda_1x_2x_3+\lambda_2x_3x_1\\ &\qquad=(3+\lambda_1+\lambda_2+\lambda_3)x_1x_2x_3, \end{split} \end{equation} where $\lambda_1,\lambda_2,\lambda_3$ are nonnegative integers. They gave recursive descriptions of the positive integer solutions and related these recursions to generalized cluster algebras. The choice $\lambda_1=\lambda_2=\lambda_3=0$ recovers \eqref{eq:markov}. Chen and Jia \cite{CJ} investigated the tropicalization and asymptotic behavior of \eqref{eq:generalized}. They constructed a deformed Fock--Goncharov tropicalization, compared generalized Euclid trees with classical Euclid trees, and established logarithmic comparison limits along infinite reduced mutation sequences. Their Theorem~6.10 distinguishes paths in which all three directions recur infinitely often from paths that eventually use only two directions. In the first case, the three coordinate ratios have a common limit; in the second, the comparison concerns the two changing coordinates. Their Fibonacci example, Example~6.11, motivates Conjecture~6.12, also stated as Conjecture~1.5: every such limit is a positive irrational number. Chen, Jia and Wu \cite{CJW} extended the logarithmic asymptotic comparison to a generalized Markov--Hurwitz family in arbitrarily many variables. For the classical equation, we use the following normalization. For a positive triple $(e_1,e_2,e_3)$, the Euclid operation in direction $i$ replaces $e_i$ by $e_j+e_k$ and fixes the other coordinates. A word $\w=(w_n)_{n\ge1}$ with $w_n\in\{1,2,3\}$ is reduced if $w_{n+1}\ne w_n$ for every $n\ge1$. Starting both triples at $(1,1,1)$, apply the first $n$ letters of the same word by Markov mutations and Euclid operations, respectively. Denote the resulting coordinate $i$ by $M_i(n)$ and $E_i(n)$, for $i\in\{1,2,3\}$ and integers $n\ge0$. The limit in Chen and Jia's conjecture, for a path with all three directions recurring, is \[ \lim_{n\to\infty}\frac{\log M_i(n)}{E_i(n)},\qquad i=1,2,3. \] Here $\log$ is the natural logarithm. We write $\arcosh:[1,\infty)\to[0,\infty)$ for the inverse of $\cosh$ restricted to $[0,\infty)$. \begin{theorem}\label{thm:main} Let \[ I=\left(\frac12\arcosh 3,\ \arcosh\frac32\right). \] For every real algebraic $c\in I$, there is a computable infinite reduced word $\w$ in which each of $1,2,3$ occurs infinitely often, such that \begin{equation}\label{eq:target} \lim_{n\to\infty}\frac{\log M_i(n)}{E_i(n)}=c \qquad (i=1,2,3). \end{equation} In particular, $c=9/10$ is a counterexample to Chen--Jia Conjecture 6.12. \end{theorem} The arithmetic input is the Hermite--Lindemann theorem: the exponential of a nonzero algebraic number is transcendental. It implies that the normalized lengths at rational Farey vertices are transcendental. Their continuous interpolation takes every intermediate value, so an algebraic target in the displayed interval must occur at an irrational Farey parameter. We construct its path by a strict sign comparison and identify every vertex denominator with the corresponding Euclid coordinate. The interpolation used here is Fock's function. Its continuous convex extension is recorded in Sorrentino--Veselov \cite[Theorem~1]{SV} and Spalding--Veselov \cite[Theorem~10]{SpV}. We prove the required continuity directly from the integer recursion, with a uniform error bound. The proof of Theorem~\ref{thm:main} concludes in Section~\ref{sec:bridge}. \section{An integer-labeled Farey tree} An active interval has reduced rational endpoints $a/bu. \end{equation} At depth $n$ of the interval tree, $xy\ge2^n$. \end{lemma} \begin{proof} The invariant assertions hold at the root. Since $\max(x,y)\le xy$, $z=3xy-u\ge2xy$. The identity \[ x^2+y^2+(3xy-u)^2-3xy(3xy-u) =x^2+y^2+u^2-3xyu=0 \] shows that the Markov equation is preserved. In either child, the opposite label is one of $x,y$, whereas the new endpoint $z$ is larger than both. This proves the induction. The endpoint maximum at least doubles on each descent; the root maximum is one. Finally $xy\ge\max(x,y)\ge2^n$. \end{proof} \begin{lemma}\label{lem:farey} Each refinement is a partition into intervals satisfying \eqref{eq:det}. Every rational in $[0,1]$ appears at a finite level and has a well-defined inherited label. \end{lemma} \begin{proof} The two child determinants are \[ b(a+c)-a(b+d)=bc-ad=1,\qquad (b+d)c-(a+c)d=bc-ad=1. \] A common divisor of $a+c$ and $b+d$ would divide the first determinant, so the mediant is reduced. Its distances from the two endpoints are \[ \frac{a+c}{b+d}-\frac ab=\frac1{b(b+d)},\qquad \frac cd-\frac{a+c}{b+d}=\frac1{d(b+d)}. \] They are positive. Thus each level is a partition, and every new vertex lies in the interior of exactly one parent interval. Its label is assigned once and is inherited thereafter. Every reduced rational $p/q\in(0,1)$ occurs at a finite level. To prove this, suppose it is interior to $[a/b,c/d]$ and define the positive integers \[ \mu=cq-dp,\qquad \nu=bp-aq. \] The determinant identity yields \[ (p,q)=\mu(a,b)+\nu(c,d). \] Since \[ \mu-\nu=q(a+c)-p(b+d), \] if $p/q<(a+c)/(b+d)$, then $\mu>\nu$ and the coefficients relative to the left child are $(\mu-\nu,\nu)$. If $p/q>(a+c)/(b+d)$, the coefficients relative to the right child are $(\mu,\nu-\mu)$. In either case their positive integer sum decreases. Eventually $\mu=\nu$; then $(p,q)=\mu(a+c,b+d)$, and reducedness forces $\mu=1$. Hence $p/q$ is the inserted mediant. The endpoints $0,1$ occur at level zero. \end{proof} For each rational $r\in[0,1]$, let $m(r)$ denote its inherited positive integer label. \section{A quantitative length defect} For an integer $m\ge1$, define \begin{equation}\label{eq:h} h(m)=\arcosh\frac{3m}{2} =\log\frac{3m+\sqrt{9m^2-4}}2. \end{equation} \begin{lemma}\label{lem:defect} For every insertion \eqref{eq:insert}, \begin{equation}\label{eq:defectbound} 00$. As $z>u$ and $z+u=3xy$, we have $z=(3xy+\sqrt D)/2$. Since $\sinh h(x)=\sqrt{9x^2-4}/2$, the addition formula gives \[ \cosh(h(x)+h(y)) =\frac{9xy+\sqrt{(9x^2-4)(9y^2-4)}}4. \] Here $(9x^2-4)(9y^2-4)=9D+16$ and $\cosh h(z)=3z/2=(9xy+3\sqrt D)/4$. Subtracting and rationalizing, \[ \cosh(h(x)+h(y))-\cosh h(z) =\frac{\sqrt{9D+16}-3\sqrt D}{4} =\frac4{\sqrt{9D+16}+3\sqrt D}>0. \] The positive difference and strict increase of $\cosh$ on positive reals imply $h(x)+h(y)>h(z)$. Moreover, \[ \cosh(h(x)+h(y))-\cosh h(z) =\int_{h(z)}^{h(x)+h(y)}\sinh v\,dv \ge\bigl(h(x)+h(y)-h(z)\bigr)\frac{\sqrt{9z^2-4}}2. \] Consequently \[ 0<\bigl(h(x)+h(y)-h(z)\bigr)\le \frac8{\sqrt{9z^2-4}(\sqrt{9D+16}+3\sqrt D)}. \] Also \[ 9z^2-4\ge36x^2y^2-4\ge25x^2y^2, \] since $x^2y^2\ge1$. Thus $\sqrt{9z^2-4}\ge5xy$, while $\sqrt{9D+16}+3\sqrt D\ge6xy$. These inequalities prove \eqref{eq:defectbound}. \end{proof} \section{Continuous Farey interpolation} For each integer $n\ge0$, let $P_n$ be the ordered set of vertices after $n$ complete refinement levels. Thus $P_0=\{0,1\}$, and the gaps between consecutive elements of $P_n$ are exactly the depth-$n$ intervals. Define $F_n:[0,1]\to\mathbb R$ to be affine on each gap and to take, at every reduced vertex $p/d\in P_n$, the value \begin{equation}\label{eq:vertex} F_n(p/d)=\frac{h(m(p/d))}{d}. \end{equation} For a depth-$n$ interval $[a/b,c/d]$ with endpoint labels $x,y$ and mediant label $z=3xy-u$, the identity \[ \frac{a+c}{b+d}=\frac b{b+d}\frac ab+\frac d{b+d}\frac cd \] gives \[ F_n\left(\frac{a+c}{b+d}\right)=\frac{h(x)+h(y)}{b+d}, \qquad F_{n+1}\left(\frac{a+c}{b+d}\right)=\frac{h(z)}{b+d}. \] On each parent interval, $F_n-F_{n+1}$ vanishes at both endpoints and is affine on each child. Its value at the mediant is \[ \frac{h(x)+h(y)-h(z)}{b+d}>0, \] which is therefore its maximum on that interval. Lemma~\ref{lem:invariants}, Lemma~\ref{lem:defect}, and $b+d\ge2$ give \begin{equation}\label{eq:onelevel} 0\le F_n(t)-F_{n+1}(t)\le\frac2{15}4^{-n} \qquad(0\le t\le1). \end{equation} \begin{proposition}\label{prop:continuous} The functions $F_n$ converge uniformly to a continuous function $F$ on $[0,1]$. For every integer $n\ge0$ and $t\in[0,1]$, \begin{equation}\label{eq:uniformtail} 0\le F_n(t)-F(t)\le\frac8{45}4^{-n}. \end{equation} At every reduced rational $p/d\in[0,1]$, \begin{equation}\label{eq:F} F(p/d)=\frac{\arcosh(3m(p/d)/2)}{d}. \end{equation} \end{proposition} \begin{proof} For integers $N>n\ge0$, telescoping \eqref{eq:onelevel} gives \[ 0\le F_n(t)-F_N(t) \le\frac2{15}\sum_{k=n}^{N-1}4^{-k} \le\frac8{45}4^{-n} \qquad(t\in[0,1]). \] Thus $(F_n)$ is uniformly Cauchy and has a uniform limit $F$; letting $N\to\infty$ gives \eqref{eq:uniformtail}. To check continuity at $t_0\in[0,1]$, fix $\varepsilon>0$ and choose $n$ with $\sup_t|F_n(t)-F(t)|<\varepsilon/3$. Continuity of $F_n$ gives a neighborhood of $t_0$ on which $|F_n(t)-F_n(t_0)|<\varepsilon/3$. The triangle inequality then gives $|F(t)-F(t_0)|<\varepsilon$. Every rational vertex appears at a finite level and keeps the same assigned value at all subsequent levels. Taking its limit proves \eqref{eq:F}. \end{proof} \section{Algebraic values along infinite Farey paths}\label{sec:path} At the two endpoints and the first mediant, \begin{equation}\label{eq:endpoints} F(0)=F(1)=\arcosh(3/2),\qquad F(1/2)=\tfrac12\arcosh3. \end{equation} The power series for $\cosh$ gives \[ \cosh(9/5)>1+\frac{(9/5)^2}{2}+\frac{(9/5)^4}{24} =\frac{15287}{5000}>3. \] For every integer $k\ge2$, \[ \frac{(9/10)^{2k+2}/(2k+2)!}{(9/10)^{2k}/(2k)!} =\frac{81}{100(2k+2)(2k+1)}\le\frac{27}{1000}. \] Bounding the tail beginning with $k=2$ by a geometric series gives \[ \cosh(9/10) \le1+\frac{81}{200}+\frac{(9/10)^4}{24(1-27/1000)} =\frac{557761}{389200}<\frac32. \] As $\cosh$ increases on positive reals, these prove \begin{equation}\label{eq:bracket} \frac12\arcosh3<\frac9{10}<\arcosh\frac32. \end{equation} \begin{lemma}\label{lem:transcendence} For every rational $r\in[0,1]$, $F(r)$ is transcendental. \end{lemma} \begin{proof} Write $r=p/d$ in reduced form with integers $0\le p\le d$ and $d\ge1$. If $F(r)$ were algebraic, then $dF(r)=h(m(r))$ would be a nonzero algebraic number, whereas \[ \exp(dF(r))=\frac{3m(r)+\sqrt{9m(r)^2-4}}2 \] is algebraic. This contradicts the Hermite--Lindemann theorem \cite{HL}. \end{proof} \begin{proposition}\label{prop:path} For each real algebraic $c\in I$, the following rule defines an infinite nested sequence of Farey intervals $J_k$ ($k\ge0$), with $J_0=[1/2,1]$: insert the mediant in $J_k$ and let $J_{k+1}$ be the child whose endpoint values lie on opposite sides of $c$. Their intersection is a single irrational number $\alpha$ satisfying $F(\alpha)=c$. Each child can be determined by a terminating comparison of rational bounds. \end{proposition} \begin{proof} The endpoint values of $J_0$ satisfy $F(1/2)