\section{Equal-degree separation and unequal-degree windows}\label{sec:windows}
\subsection{A complete result at each fixed degree}
We first use the established fixed-sum, fixed-numerator and fixed-denominator
orderings of Markov numbers. Section~\ref{sec:snakes} proves them by
matching counts and records their earlier sources.
\begin{theorem}[Equal-degree unicity]\label{thm:equaldegree}
Distinct canonical integer-phase marks of the same Meridian degree have
different maxima. The marks may belong to the same or to different Meridians.
\end{theorem}
\begin{proof}
For $D\ge4$, Lemma~\ref{lem:dictionary} partitions the labels into three
classes. Helms $r/s$ have fixed sum $D$; near-helms $r/D,s/D$ have fixed
denominator $D$; outer phases have fixed numerator $D$. Each class is
strictly ordered by its corresponding monotonicity, and the dictionary
ensures that different marks have different labels.

The largest helm is $m_{1/(D-1)}$. The near-helm extrema are
$m_{1/D},m_{(D-1)/D}$; the smallest outer value is $m_{D/(D+1)}$.
They are separated by
\[
m_{1/(D-1)}<m_{1/D},\qquad
m_{(D-1)/D}<m_{(D-1)/(D+1)}<m_{D/(D+1)}.
\]
The intermediate fraction is reduced because $D$ is even. The first two
inequalities hold at fixed numerator and the last at fixed denominator.
For $D=2$, the unique seed gives $2,5$ and then the increasing family
$m_{2/(2n-1)}$ for $n\ge2$, starting at 29. These values are also distinct.
\end{proof}
Thus the remaining Meridian comparison has unequal degrees. To bound it,
we need exact leading terms and an explicit positive remainder.

\subsection{From Fibonacci blocks to a Meridian profile}
The ordering argument above uses the Fibonacci and six-piece comparisons
from Section~\ref{sec:snakes}. We now use the same run matrices to bound
growth.
Put $\rho=(3+\sqrt5)/2$ and $\lambda=\log\rho$.
The odd-index Fibonacci formula extends \eqref{eq:fibrun} to every real $t$:
\begin{equation}\label{eq:fibpositive}
\mathcal A_t=\frac1{\sqrt5}\left[
\rho^{t-1/2}\mat{\rho^2}{\rho}{\rho}{1}
+\rho^{-t+1/2}\mat{\rho^{-2}}{\rho^{-1}}{\rho^{-1}}{1}
\right].
\end{equation}
For integer $t$, its entries are exactly
$F_{2t+3},F_{2t+1},F_{2t+1},F_{2t-1}$, including negative odd indices.
Each displayed matrix is the outer product of a positive column with
itself: respectively $(\rho,1)^{\mathsf T}$ and
$(\rho^{-1},1)^{\mathsf T}$. Thus every choice of terms in a product has
a positive contribution to its upper-left entry.

Let the helm counts be $(r,s)$ and $D=r+s$. The phase dictionary and the
run-count formula \eqref{eq:fibword} give, for every integer $n\ge2$,
\[
M_\alpha(n)=m_{D/((n-1)D+r)}.
\]
The first run index is $n-1$; each later index is
$n-1+\lceil jr/D\rceil-\lceil(j-1)r/D\rceil$.
Consequently the full real profile is
\begin{equation}\label{eq:meridianfib}
M_\alpha(x)=\left[
\mathcal A_{x-1}\prod_{j=2}^{D}
\mathcal A_{x-1+\lceil jr/D\rceil-\lceil(j-1)r/D\rceil}
\right]_{11}.
\end{equation}
To justify the extension from integers to reals, observe that both sides
are finite Laurent polynomials in $\rho^x$. For the left side this follows
from the seed formula and the polynomial recurrence \eqref{eq:extend};
for the right side it follows from \eqref{eq:fibpositive}. Their difference
vanishes at every $x=n\ge2$. Multiplying by a sufficiently large power of
$\rho^x$ gives a polynomial with infinitely many roots, so it is zero.
For $M_\alpha(-x)$, replace $r$ by $s$ in the right side of
\eqref{eq:meridianfib}; the negative-arm dictionary gives the same argument.
For example, the seed has $D=2,r=s=1$ and
$M_1(x)=[\mathcal A_{x-1}^{\,2}]_{11}$, giving $2,5,29,194,\ldots$.

\begin{proposition}[Laurent profile]\label{prop:laurent}
For helm counts $(r,s)$, degree $D=r+s$, and $t=(r-s)/2$,
\begin{equation}\label{eq:laurent}
M_\alpha(x)=\sum_{j=-D,-D+2,\ldots,D}c_j\rho^{jx},\quad
c_j\ge0,\quad c_{\sigma D}=\frac13(9/5)^{D/2}\rho^{\sigma t}.
\end{equation}
The logarithm is strictly convex, and the real profile has one minimum.
\end{proposition}
\begin{proof}
Expand the $D$ Fibonacci factors in \eqref{eq:meridianfib} using
\eqref{eq:fibpositive}. Each factor supplies one power $\rho^x$ or
$\rho^{-x}$, with positive coefficients. This gives the stated exponents
and coefficient signs.

The run indices sum to $Dx+r-D-1$. Selecting the first term in every
factor gives the upper-left entry
\[
\frac{\rho^{Dx+r-D-1-D/2}\rho^2(3\rho)^{D-1}}{(\sqrt5)^D}
=\frac13(9/5)^{D/2}\rho^{Dx+(r-s)/2},
\]
because $\rho^2+1=3\rho$. Selecting every second term gives the other
extreme coefficient. The second derivative of the logarithm is
$\lambda^2$ times the variance of the exponents under their positive
normalized weights. Both extremes occur, so that variance is positive;
they also force growth at both ends.
\end{proof}

\begin{lemma}[Uniform Fibonacci-block remainder]\label{lem:tail}
For a Meridian of degree $D$, helm counts $(r,s)$, $n\ge0$ and
$\sigma=\pm1$,
\begin{equation}\label{eq:tail}
M_\alpha(\sigma n)
=\frac13(9/5)^{D/2}\rho^{\sigma(r-s)/2}\rho^{Dn}
 (1+\epsilon_{\alpha,\sigma}(n)),
\qquad
0\le\epsilon_{\alpha,\sigma}(n)\le(1+\rho^{1-2n})^D-1.
\end{equation}
\end{lemma}
\begin{proof}
Use \eqref{eq:meridianfib}, replacing $r$ by $s$ on the negative arm.
Every block index is either $n-1$ or $n$. Call the first and second
outer products in \eqref{eq:fibpositive} the $+$ and $-$ choices.
When two neighboring columns are multiplied, their inner products are
\[
\begin{array}{c|cc}
 &+&-\\\hline
+&3\rho&2\\
-&2&3/\rho
\end{array}.
\]
Relative to the all-$+$ choice, a mixed join has factor
$2/(3\rho)<\rho^{-1}$, and a $--$ join has factor $\rho^{-2}$.
At either outside boundary a minus choice has factor $\rho^{-2}$,
which is also at most $\rho^{-1}$. Each selected minus block touches two
joins or boundaries. Thus $k$ minus choices reduce these column and
boundary factors by at least $\rho^{-2k}$.

At a block of index $t$, replacing the scalar coefficient of the plus
term by that of the minus term multiplies it by $\rho^{1-2t}$.
Since $t\ge n-1$, the complete relative contribution of any specified
$k$ minus choices is at most
$\rho^{-2k}(\rho^{3-2n})^k=\rho^{k(1-2n)}$.
Sum this bound over all nonempty subsets of the $D$ positions.
The binomial theorem gives the displayed remainder bound.
\end{proof}

\subsection{A finite rectangle depending only on the degrees}
\begin{theorem}[Unequal-degree collision window]\label{thm:window}
Let $D_\alpha\ne D_\beta$, and define
\[
\begin{gathered}
m=\frac{|D_\alpha-D_\beta|}{2},\qquad E=\max(D_\alpha,D_\beta),\\
h=\min\{j\in\Z_{\ge0}:\rho^j\ge E\},\qquad N=3m+h+3,\qquad T=E(N+2).
\end{gathered}
\]
An integer equality $M_\alpha(x)=M_\beta(y)$ requires
\begin{equation}\label{eq:window}
|x|\le\left\lceil\frac T{D_\alpha}\right\rceil,\qquad
|y|\le\left\lceil\frac T{D_\beta}\right\rceil,
\qquad M_\alpha(x)<\rho^T/3.
\end{equation}
The seed still uses only nonnegative phases.
\end{theorem}
\begin{proof}
The degrees are even, so $m$ is a positive integer. First,
\begin{equation}\label{eq:leadinggap}
|m\log(9/5)-j\log\rho|>\frac1{81^m+1}\qquad(j\in\Z).
\end{equation}
Choose $j_0\ge0$ with $1<(9/5)^m\rho^{-j_0}<\rho$; equality at an
endpoint is impossible by taking the field norm. The positive algebraic
integers $9^m-5^m\rho^{j_0}$ and $5^m\rho^{j_0+1}-9^m$ have nonzero
integer norms and conjugates of absolute value below $9^m$. Each is
therefore greater than $9^{-m}$. Dividing by the appropriate positive
denominators makes both adjacent multiplicative gaps greater than $81^{-m}$.
The inequality $\log(1+u)>u/(1+u)$ gives \eqref{eq:leadinggap}; all other
integers $j$ are farther away.

At phases $x=\sigma n$, $y=\tau n'$, the leading logarithms differ by
\[
\frac{D_\alpha-D_\beta}{2}\log(9/5)
+(D_\alpha n-D_\beta n'+\sigma t_\alpha-\tau t_\beta)\log\rho.
\]
The coefficient of $\log\rho$ is integral. Put $r_m=(81^m+1)^{-1}$.
Since $E\le\rho^h$, $\rho^6>81$ and $\rho^5>5$,
\[
E\rho^{1-2N}\le\rho^{-6m-5}
<\frac1{2\cdot81^m+3}=\frac{r_m}{r_m+2}.
\]
For a positive integer $g$ and $u\ge0$ with $gu<1$, the elementary bound
$(1+u)^g\le\sum_{j\ge0}(gu)^j=(1-gu)^{-1}$ makes each relative
remainder in Lemma~\ref{lem:tail} smaller than $r_m/2$ when its phase
magnitude is at least $N$. Their logarithmic corrections cannot cancel
\eqref{eq:leadinggap}. At least one collision phase therefore has magnitude
less than $N$.

The leading coefficient of $M_\alpha$ lies strictly between
$\rho^{-D_\alpha}/3$ and $\rho^{D_\alpha}/3$: use
$|t_\alpha|<D_\alpha/2$ and $1<3/\sqrt5<\sqrt\rho$ in
\eqref{eq:laurent}. Lemma~\ref{lem:tail}, with
$1+\rho^{1-2N}<\rho$, gives
\[
M_\alpha(\pm N)<\rho^{D_\alpha(N+2)}/3\le\rho^T/3.
\]
The same bound holds for $M_\beta$. Convexity bounds the value at any
phase inside either finite strip by its endpoint values, so the common
value is below $\rho^T/3$. Positivity also gives
$M_\alpha(\sigma n)>\rho^{D_\alpha(n-1)}/3$ and the corresponding
bound for $M_\beta(\tau n')$. Comparing the upper and lower bounds yields
$n<1+T/D_\alpha$ and $n'<1+T/D_\beta$, hence the integer ceilings in
\eqref{eq:window}.
\end{proof}

This theorem bounds the remaining candidates; it does not assert that any
candidate collides. It also does not bound the degrees of a hypothetical
collision. The next section supplies a different finite reduction, expressed
in Farey numerators and denominators rather than Meridian phases.