\section{Meridian residues: modulo 27 and Fibonacci moduli}\label{sec:residues}
\subsection{Transferring a seed congruence to every address}
The recurrence for the companions contains no division, but a seed period of
the \emph{maximum} need not visibly preserve both companions. The following
four-entry record resolves that issue:
\[
\mathcal S=(A,B,C,H)=(9a^2,9b^2,3ab,H).
\]
Forward and backward orbit steps and pole reflection are the polynomial maps
\begin{align}
\mathcal R(A,B,C,H)&=(B,A+9H^2B-18HC,HB-C,H),\notag\\
\mathcal R^{-1}(A,B,C,H)&=(B+9H^2A-18HC,A,HA-C,H),\label{eq:record}\\
\mathcal J(A,B,C,H)&=(A,B,C,C-H).\notag
\end{align}
For $h(x)=M_1(x)$ the seed record is
\[
\mathcal S_1(x)=
(8h(x)-h(x-1)-2,\ h(x)+h(x-1)+2,\ h(x)+1,\ h(x)).
\]
Every congruence preserving two consecutive seed maxima therefore propagates
along every signed address.

\begin{theorem}[Largest modulus with a common phase period]\label{thm:period}
Every canonical Meridian has phase period $T$ modulo $m$ if and only if
\begin{equation}\label{eq:maxperiod}
m\mid q(T),\qquad
q(T)=3F_{2T}F_{\gcd(2T,4)}=
\begin{cases}3F_{2T},&T\text{ odd},\\9F_{2T},&T\text{ even}.
\end{cases}
\end{equation}
In particular, $s$ is a common period modulo $F_{2s}$.
\end{theorem}
\begin{proof}
Fibonacci addition and \eqref{eq:seedfib} give
\[
h(x+T)-h(x)=3F_{2T}F_{4x+2T}.
\]
Fibonacci strong divisibility,
$\gcd(F_a,F_b)=F_{\gcd(a,b)}$, follows by the Euclidean algorithm from
the addition formula. The gcd of all displayed differences is already
the gcd of those at $x=0,1$, namely $q(T)$. Thus the seed has period $T$
modulo $m$ exactly when $m\mid q(T)$. The record \eqref{eq:record} proves
sufficiency at every address; the seed itself proves necessity.
\end{proof}

The theorem specifies a period common to all Meridians. It does not assert
that the period is the smallest for each individual sequence. Nor does a
large period exclude a scalar residue match at two earlier phases.
For powers of three it gives
\begin{equation}\label{eq:3period}
M_\alpha(x+2\cdot3^n)\equiv M_\alpha(x)\pmod{3^{n+3}},\qquad n\ge0.
\end{equation}
Indeed, $F_{3j}=F_j(5F_j^2+3)$ for even $j$, beginning with $F_4=3$,
gives $v_3(F_{4\cdot3^n})=n+1$. Figure~\ref{fig:fibresidues} shows the
common periods at the requested even-indexed Fibonacci moduli.

\paperfigure{fibresidues}{.97}{figures/fig_fibonacci_residues.pdf}{Five
Meridians modulo $F_{2s}$ for $s=2,\ldots,8$: the moduli are
$3,8,21,55,144,377,987$. Each column has common phase period $s$,
possibly a multiple of an individual least period. The shaded row is the
helm; blank cells omit the seed's negative phases. The displayed residues
use the reversed Meridian indexing of this edition.}{p}

\subsection{Why modulo 27 gives only ten classes}
No positive Markov entry is divisible by three: mutation negates the changed
coordinate modulo three and the integral tree starts at $(1,1,1)$.
Let $t,h$ be preceding and current poles. An extension by a nonzero index
has the particularly simple action
\begin{equation}\label{eq:mod27maps}
S(t,h)=(h,t)\quad(k\text{ even}),\qquad
R(t,h)=(h,7h-t)\quad(k\text{ odd})\pmod{27}.
\end{equation}
To see this, put $N_j=3u_ju_{j+1}-h$. Direct use of the orbit recurrence gives
\[
N_0=t,\quad N_1=(9u_1^2-2)h-t,\quad
N_{j+2}-N_j=9h(u_{j+2}^2-u_{j+1}^2).
\]
All orbit squares are one modulo three, so $N_{j+2}\equiv N_j$ and
$N_1\equiv7h-t\pmod{27}$. This works for negative indices as well.
Only the parity of each recursively applied index survives.

\begin{proposition}[Universal modulo-27 classes]\label{prop:tenclasses}
At integer phases every Meridian alternates between the two residues of
one of the following sets, allowing the two singleton cases:
\[
\{1\},\ \{2,5\},\ \{4,16\},\ \{7,13\},\ \{8,17\},\
\{10,19\},\ \{11,23\},\ \{14,20\},\ \{22,25\},\ \{26\}.
\]
The ten sets are pairwise disjoint. Meridians in different classes cannot
share an integer value, regardless of degree or phase.
\end{proposition}
\begin{proof}
The common phase period is two because $q(2)=27$. Track preceding and current
poles at both phase parities, starting from $(1,2,1,5)$. The simultaneous
maps satisfy $S^2=I$ and $SRS=R^{-1}$. Successive current-pole pairs on
the $R$ orbit are
\[
\begin{array}{rrrrrr}
(2,5)&(13,7)&(8,17)&(16,4)&(23,11)&(10,19)\\
(20,14)&(22,25)&(26,26)&(25,22)&(14,20)&(19,10)\\
(11,23)&(4,16)&(17,8)&(7,13)&(5,2)&(1,1).
\end{array}
\]
The preceding pair is the previous pair cyclically, and
$b_{j+1}=7b_j-b_{j-1}$ checks the return in 18 steps.
These states and their $S$ images form a closed set of 36 states.
Their current-pole images are exactly the ten listed sets. Congruence of
integer values would require membership in the same set.
\end{proof}

\begin{table}[htbp]\centering
\caption{The ten universal classes and their populations among the 519
Meridians of degree at most 100. Only the population column is bounded-degree
data.}\label{tab:tenclasses}
\begin{tabular}{cr@{\hspace{.35in}}cr}\toprule
Residue set&Population&Residue set&Population\\\midrule
$\{1\}$&42&$\{10,19\}$&47\\
$\{2,5\}$&88&$\{11,23\}$&47\\
$\{4,16\}$&61&$\{14,20\}$&38\\
$\{7,13\}$&78&$\{22,25\}$&35\\
$\{8,17\}$&65&$\{26\}$&18\\\bottomrule
\end{tabular}\end{table}
At that degree bound, the classification leaves $15{,}215$ of the
$134{,}421$ pairs for further comparison. Larger Fibonacci moduli and
compatible phases, rather than extrapolation from this table, supply the
degree-10,000 certificate in Section~\ref{sec:certificate}.