\section{Introduction}
A Markov triple consists of three positive integers satisfying
\begin{equation}\label{eq:markov}
 a^2+b^2+M^2=3abM,\qquad 1\le a\le b\le M.
\end{equation}
For example, $(1,2,5)$ and $(1,5,13)$ are Markov triples. The largest
entry of such a triple is a Markov number. The \emph{unicity problem}
asks a simple question: does that largest entry determine the other two?
A counterexample would consist of two different ordered triples with
the same largest entry. We call such an equality a \emph{collision}.
For the history and wider setting of the problem, see Aigner~\cite{Aigner}.
An early computational benchmark is due to Baragar~\cite[Section~4,
p.~9]{Baragar1996}, who reported 18,906 Markoff numbers below $10^{140}$
and no counterexample to unicity among all triples with maximal element
less than $10^{140}$.

Our starting point is to let the entries vary continuously. We build a
curve on which the Markov equation holds automatically, then ask which
positions give integer entries. This separates the equation itself from
the requirement that its solutions be integers. It also gives a picture
of what a collision would have to look like.

A second aim is practical: understand where a family of triples begins
and how its values grow. The Meridian viewpoint guided our choice to
compare candidates in limited size ranges, so that completed batches
could be cleared from working storage. The resulting verifier uses
Farey addresses and certified matrix bounds to select those batches.
It compares more than 1.1 trillion configurations and certifies the first
one trillion values in numerical order.

\subsection{The host and its catenary profile}
The \emph{host} $h$ is the fixed third entry around which we build the
curve. It determines a profile $f_h$: two heights on this profile, taken
one horizontal unit apart, will satisfy the Markov equation together
with $h$. We first allow positive real entries and call any such solution
a \emph{real Markov state}. The host need not yet be an integer or the
largest entry; those requirements will be imposed when we return to
the unicity question.

For $h>2/3$, define
\begin{equation}\label{eq:profile}
\begin{gathered}
q_h=\frac{3h}{2},\qquad R_h=\sqrt{q_h^2-1},\qquad Q_h=q_h+R_h,\\
f_h(x)=\frac{h}{2R_h}\bigl(Q_h^x+Q_h^{-x}\bigr).
\end{gathered}
\end{equation}
The restriction $h>2/3$ makes $R_h$ positive and $Q_h>1$; it includes
every positive integer host. The graph of $f_h$ is a rescaled catenary:
a smooth, symmetric U-shaped curve, with its minimum at $x=0$.
In particular, $f_h(-x)=f_h(x)$. In \eqref{eq:profile}, $x$ is a
free position on this profile. We use $c$ below for a particular selected
position.

The constants in \eqref{eq:profile} set the spacing and height scale so
that one horizontal unit has the required arithmetic meaning. The
calculation is short. First,
\[
(q_h+R_h)(q_h-R_h)=q_h^2-R_h^2=1,
\]
so
\[
Q_h^{-1}=q_h-R_h,\qquad
Q_h+Q_h^{-1}=3h,\qquad Q_h-Q_h^{-1}=2R_h.
\]
These three identities follow directly from the definitions.

\subsection{Why two heights one unit apart satisfy the equation}
Compare $f_h(x+1)$ with $Q_hf_h(x)$. The terms containing
$Q_h^{x+1}$ cancel, leaving
\[
\begin{aligned}
f_h(x+1)-Q_hf_h(x)
 &=\frac{h}{2R_h}\bigl(Q_h^{-x-1}-Q_h^{1-x}\bigr)\\
 &=\frac{h}{2R_h}Q_h^{-x}(Q_h^{-1}-Q_h)
 =-hQ_h^{-x}.
\end{aligned}
\]
Multiplying $f_h(x)$ by $Q_h^{-1}$ instead cancels the terms
containing $Q_h^{-x-1}$:
\[
\begin{aligned}
f_h(x+1)-Q_h^{-1}f_h(x)
 &=\frac{h}{2R_h}\bigl(Q_h^{x+1}-Q_h^{x-1}\bigr)\\
 &=\frac{h}{2R_h}Q_h^x(Q_h-Q_h^{-1})
 =hQ_h^x.
\end{aligned}
\]
Now multiply these two equalities. The powers on the right cancel:
\[
\bigl(f_h(x+1)-Q_hf_h(x)\bigr)
\bigl(f_h(x+1)-Q_h^{-1}f_h(x)\bigr)=-h^2.
\]
Expanding the left side gives
\[
f_h(x+1)^2-(Q_h+Q_h^{-1})f_h(x)f_h(x+1)+f_h(x)^2=-h^2.
\]
Finally, use $Q_h+Q_h^{-1}=3h$ and move $h^2$ to the left:
\begin{equation}\label{eq:edgeconic}
\boxed{f_h(x)^2+f_h(x+1)^2+h^2
       =3f_h(x)f_h(x+1)h.}
\end{equation}
Thus \emph{every} real position $x$ gives a real Markov state. No choice
of an integer value or a special position was used in this calculation.

\subsection{Choosing an integral state: the role of \texorpdfstring{$c$}{c}}
To look for an ordered triple with largest entry $h$, select
$0<c\le1/2$ and take the unit interval $[c-1,c]$ across the bottom
of the curve. Its two heights are $f_h(c-1)$ and $f_h(c)$.
By symmetry, $f_h(c-1)=f_h(1-c)$, so the state, with its companions
in increasing order, is
\[
\bigl(f_h(c),\ f_h(1-c),\ h\bigr).
\]
It solves the equation by \eqref{eq:edgeconic}, with $x=c-1$.
For it to be a normalized Markov triple, $h$ and both heights must
be positive integers, and $f_h(1-c)\le h$ must hold. The latter
condition keeps the chosen host as the largest entry. Restricting
$c$ to this half of the unit interval avoids counting the reflected pair twice.

Figure~\ref{fig:phasechoice} shows the distinction for $h=5$.
One choice, $c\approx0.351515$, gives the heights $1$ and $2$, hence
the integer triple $(1,2,5)$. A different choice, $d=0.44$, gives
heights approximately $1.207458$ and $1.602614$. It still solves the
same equation with host $5$, but its companion heights are not integers.

\begin{figure}[htbp]\centering
\includegraphics[width=\linewidth]{figures/fig0_catenary_choices.pdf}
\caption{Two choices on the same catenary profile, with host $h=5$.
In each panel the marked contacts are exactly one horizontal unit apart.
The left contact has height $f_5(1-c)$ or $f_5(1-d)$ by symmetry.
The choice $c$ gives an integer triple; the displayed choice $d$ gives
only a real state. A collision would require two distinct choices with
both pairs of heights integral and with the same host as their largest
entry. No collision is shown here.}
\label{fig:phasechoice}\end{figure}

This is the geometric form of the unicity question. For a fixed integer
host $h$, could two distinct choices $c$ and $d$ in $(0,1/2]$ both give
integer companion heights no larger than $h$? If so, they would give
different ordered triples with the same maximum. The curve makes these
possible comparisons visible. To exclude them requires information about
the integer values on the curve.

\subsection{How the paper organizes those comparisons}
The next step is \emph{descent}. Starting from two contact heights, we
look for a smaller host whose profile passes through the same heights
one unit apart. We then move inward along that profile. Repeating these
operations reduces a normalized integer Markov triple to $(1,1,1)$.
Section~\ref{sec:geometry} makes the procedure precise and explains how
its final host and final position recognize the integer starting states.

Reversing the steps organizes the triples into families called
\emph{Meridians}. Each family has an address recording the steps used
to construct it, and a phase specifying a position along it.
This Meridian phase is a separate coordinate from the selected $c$ on
a fixed catenary. Integer Meridian phases pick out integer triples;
the corresponding $c$ need not be an integer, as Figure~\ref{fig:phasechoice} shows.
Section~\ref{sec:indexing} proves that these addresses and integer phases
recover every normalized triple exactly once. This is a statement about
triples: whether two of them have the same largest entry remains the
question to be answered. The same section introduces the alternative
tree and the \emph{phasors} that connect its Meridians.

The remainders of the Markov values follow exact periodic patterns
(Section~\ref{sec:residues}). Section~\ref{sec:snakes} then develops the
ordering tools used in the collision arguments. It represents Markov
numbers by counts of perfect matchings on chains of square tiles, called
snake graphs; a perfect matching pairs all the vertices using edges of
the graph. The author's Fibonacci matrices and six-piece decomposition
provide concrete ways to compare these counts.

The following two sections use those constructions to control growth.
A Meridian's \emph{degree} measures the largest exponent in its formula.
We prove that distinct integer marks of the same degree cannot collide.
For different degrees, a positive expansion of the Fibonacci blocks gives
a finite range of phases that needs comparison (Section~\ref{sec:windows}).
\emph{Corridor walls} follow other paths through the same tree, using
both Meridian and phasor steps. Adding one to each Fibonacci run index
gives their recurrence and a second bound on possible collisions
(Section~\ref{sec:corridors}).

The resulting computer-checked certificate covers all $5{,}067{,}077$
Meridians of degree at most $10{,}000$, at every integer phase
(Theorem~\ref{thm:degreebound}). Distinct triples within that family
have different maxima. This bounds the complexity of the families,
while allowing their values to grow without limit; comparisons with
higher-degree Meridians remain outside this certificate.
Section~\ref{sec:verification} records the separate numerical-order
verification. Meridian helms and degree-dependent logarithmic spacing
motivated its organization by scale. The production code implements
that organization through groups of Farey addresses with shared,
certified size bounds; it does not schedule one Meridian at a time.
The same band algorithm can be formulated entirely in Farey and matrix
terms. Its finite-prefix conclusion does not use the Meridian collision
exclusions, and the timings do not isolate a speedup due to the change
of coordinates alone. It compared $1{,}101{,}388{,}071{,}945$ configurations and
certified uniqueness of the first one trillion values, including every
possible competitor at those values. Production finished on 10 November 2026
at 03:54:13 UTC (9 November, 21:54:13 CST), using 23.956 hours of retained
active band processing, including the inherited work, and at most
0.992 GiB of recorded temporary key files. Both final replays passed.
The section gives the exact timing and storage scopes, together with the earlier
fifty-billion result. The general unicity problem remains open within
this framework.

As the later sections introduce their coordinates, $M_\alpha(x)$ will
denote the value at Meridian address $\alpha$ and phase $x$, and
$m_{p/q}$ the value of the triple indexed by the reduced fraction $p/q$.
Both label triples before any assertion that their maxima are distinct.
Analytical logarithms are natural; plotted horizontal axes use $\log_{10}M$.