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\hypersetup{pdftitle={Hankel determinants and the length of finite Pierce expansions},pdfauthor={Jason Davies}}
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\title{Hankel determinants and the length of finite Pierce expansions}
\author{Jason Davies\thanks{\href{mailto:jason@jasondavies.com}{\nolinkurl{jason@jasondavies.com}}}}
\date{October 27, 2026}
\begin{document}
\maketitle
\begin{abstract}
For integers $a>b>0$, let $P(a,b)$ be the number of steps the iteration
$x\mapsto a\bmod x$ takes to reach $0$ from $b$; it is the length of the
Pierce expansion of $b/a$. We prove
\[
 P(a,b)\ll_\varepsilon a^{\beta_*+\varepsilon},\qquad
 \beta_*=\frac{41301-3\sqrt{82849}}{137480}=0.2941336553\ldots,
\]
improving the exponent $1/3-2/177$ of Chase and Pandey. The tail of an orbit
carries a sequence of positive integers, the Hankel determinants of a natural
moment sequence, and these are small where the drops of the orbit are small.
Grouping orbit states by such integers places the quotient differences of each group on
curves of bounded degree, whose integer points we count with the uniform
bounds of Castryck, Cluckers, Dittmann and Nguyen. The resulting exponent
approaches $5/17$ when these curves have large degree relative to the
logarithmic size of the box. We raise the degree by
adjoining further integer invariants, by an effective Hilbert irreducibility
sieve, and by an exact identity for the direction of the curvature vector,
and come within $1.61\cdot10^{-5}$ of $5/17$. We also give an
elementary proof of the exponent $3/10+\varepsilon$ and a bound for orbits
whose quotient increments rarely rise. All results are formalised in Lean~4
with Mathlib. Shallit's conjectured $O((\log a)^2)$ bound remains open.
\end{abstract}

\noindent\textbf{AI transparency statement.} This paper was produced entirely
by artificial intelligence, including the discovery of the main results, their
formalisation in Lean, and the writing of the manuscript. The human author
initiated the research and directed the AI systems, but did not write the text.
Details are given at the end of the paper.

\section{Introduction}\label{sec:intro}

Let $a>b>0$ be integers; we call $a$ the \emph{dividend}. Put $b_0=b$ and
$b_{i+1}=a\bmod b_i$ while $b_i>0$.
The sequence decreases until it reaches $0$; if $b_n=0$ we write $P(a,b)=n$.
For example, $P(35,22)=7$, from $22,13,9,8,3,2,1,0$. This is the length of
the Pierce expansion of $b/a$~\cite{Pierce,Shallit86}, since
\[
 \frac ba=\frac1{q_0}-\frac1{q_0q_1}+\frac1{q_0q_1q_2}-\cdots,
 \qquad q_i=\lfloor a/b_i\rfloor.
\]
Shallit~\cite{OPG} asked whether
\begin{equation}\label{eq:conj}
 P(a,b)=O\bigl((\log a)^2\bigr).
\end{equation}

Shallit~\cite{Shallit86} proved $P(a,b)\ll a^{1/2}$ by ``Archimedean'' arguments.
Erd\H{o}s and Shallit~\cite{ES} combined these with a divisor argument to obtain
$P(a,b)\ll_\varepsilon a^{1/3+\varepsilon}$, and showed that
$\max_bP(a,b)/\log a$ does not tend to $0$. Chase and Pandey~\cite{CP} improved
the upper exponent to $1/3-2/177$, using exponential sum estimates, and proved
$\max_bP(a,b)\gg\log a/\log\log a$ for all large $a$.
Baraskar and Vukusic~\cite{BV} study the related sets of iterated remainders
at fixed iteration depth. Shallit's talk pages~\cite{ShallitTalks} report a
bound $n^{3/10}$, described as ``LLM, unpublished so far''. We have not seen
that argument; our elementary bound below has the same exponent up to
$\varepsilon$, and we make no claim about how the arguments are related.

Our main result is the following uniform bound.
\begin{theorem}\label{thm:polar}
For every $\varepsilon>0$ there is $C_\varepsilon>0$, independent of $a,b$, such that
\[
 P(a,b)\le C_\varepsilon a^{\beta_*+\varepsilon}\qquad(0<b<a),\qquad
 \beta_*=\frac{41301-3\sqrt{82849}}{137480}.
\]
\end{theorem}
Here $\beta_*=0.2941336553\ldots$. The proof depends on two fixed positive
integers $m,n$, the powers of the adjoined coordinates (see ``Raising the
degree'' below), and $\beta_*$ is the limit of the resulting exponents as
$n/m$ tends to an irrational ratio. For instance $(m,n)=(3,2)$ gives the
rational exponent $4317/14677=0.2941336785\ldots$
(Theorem~\ref{thm:polarconditional}).

The curve-counting bounds of this paper (Theorems~\ref{thm:polar},
\ref{thm:joint} and~\ref{thm:lattice}) all lie just above
$5/17=0.2941176\ldots$, and Theorem~\ref{thm:polar} is within
$1.61\cdot10^{-5}$ of it. We first introduce the integer invariants of an
orbit, and then explain where $5/17$ comes from, since it organises the whole
argument.

\subsection*{Integer invariants of an orbit}
Write $k_i=q_{i+1}-q_i$ for the quotient increments and
$d_i=b_i-b_{i+1}$ for the drops. The \emph{curvature}
$D_i=d_ik_{i+1}-d_{i+1}k_i$ of four consecutive terms satisfies
\begin{equation}\label{eq:intro-curv}
 D_ib_{i+2}=d_id_{i+2}+k_id_id_{i+1}+(k_i-1)d_{i+1}^2.
\end{equation}
Thus $D_i$ is a positive integer, although its defining terms largely cancel,
and it is small when the drops are small. A six-term identity supplies a
second positive integer $L$, and with it the bound
$q_i\le12K^3\max(D_{i+1},D_{i+2})$ when the next four increments are at
most $K$ (Section~\ref{sec:six}).

These integers belong to one hierarchy. Each tail carries a finite positive
measure $\nu_i$ of mass $b_i/a$, and
\[
 I_r(i)=a\det\left(\int x^{j+l}\,d\nu_i\right)_{j,l<r}
\]
is a positive integer when the tail has at least $r$ positive terms.
The first four invariants are $b_i,d_i,D_i,L_i$, in the index ranges of
Theorem~\ref{thm:hankel}, which stay away from the end of the orbit.

\subsection*{Where the exponent $5/17$ comes from}
We count the $T$ orbit terms in a remainder window $[A,2A)$, then sum over
dyadic $A$. The total drop across the window is less than $A$, and
Erd\H{o}s and Shallit's divisor argument shows that a drop value can repeat
only $a^{o(1)}$ times; hence $A\ge T^{2-o(1)}$. Similarly the quotients
increase by at most $a/A$ across the window. Call an orbit index $i$ a
\emph{start} when we regard it as the first index of the block
$b_i,b_{i+1},\dots$. A fixed proportion of starts in the window then have
their next few increments bounded by $H\asymp a/(AT)\le a^{1+o(1)}/T^3$.
For such a start, form the $3\times5$ matrix whose rows are the next five
drops, the increments, and a quadratic expression in the increments. Its ten
maximal minors, among them $I_4(i)$, $I_4(i+1)$ and $I_4(i+2)$, are integers.
To leading order the drop row is a combination of the other two rows, since
the Pierce expansion gives $d_j/a=k_j/q_i^2-(r_j+k_j)/q_i^3+O(H^3/q_i^4)$ with
$r_j$ the quadratic row (Proposition~\ref{prop:allminors}). Subtracting that
combination leaves entries of size $O(aH^3/q_i^4)$; so the
minors have absolute value at most
\[
 X\asymp\frac{A^4H^6}{a^3}\asymp\frac{a^3}{A^2T^6}\le\frac{a^{3+o(1)}}{T^{10}}.
\]
Their vector is the \emph{pattern} of the start. It is an integral point of
the affine cone over the Grassmannian $\mathrm{Gr}(3,5)$, and there are
$X^{5+o(1)}$ such points (Section~\ref{sec:patterns}); the exponent $5$ is
the Fano index of $\mathrm{Gr}(3,5)$, so this count of all integral points of
the cone cannot be improved (orbit patterns could in principle be sparser). If
every pattern were shared by $O(1)$ starts, then $T\ll X^5$ would give
$T^{51}\le a^{15+o(1)}$, that is, $T\le a^{5/17+o(1)}$.

The rest of the argument bounds the number of starts sharing a pattern. For
a fixed pattern, three nodes $s_j=q_{i+j}-q_i$, $j=1,2,3$, determine the
increments (the pattern supplies two linear relations, which give $k_3$ and
$k_4$). The pattern also supplies a quadratic equation in the nodes. A second
equation, of degree at most twelve, says that $I_4(i)$ times the resultant of
the orthogonal quartic of state $i$ and $x$ times that of state $i+1$ equals
$a$ times the fourth power of the squared norm of the first quartic; it
involves the dividend $a$. The nodes therefore lie on a union of curves of total degree at most $24$. The
coefficient-uniform bound of Castryck, Cluckers, Dittmann and
Nguyen~\cite{CCDN} says that an integral curve over $\mathbb Q$ of degree
$\delta$ has at most $B^{1/\delta+o(1)}$ integer points in a box of size
$B$, whatever its coefficients. So a pattern carries $B^{1/\delta+o(1)}$
starts. If the degree could be made large with the box fixed, the factor
$B^{1/\delta}$ would become $a^{o(1)}$ and the exponent would tend to $5/17$.
A component has degree at most $24$, and the components of smallest degree
govern the bound; but the curves carry further integer-valued functions.
Adjoining powers of them as coordinates raises $\delta$ and $\log B$ together,
and what matters is the ratio $\log B/\delta$ (Remark~\ref{rem:jointscales}). Every refinement below either gains more degree per
unit of height or shrinks $B$, for almost all patterns.

\subsection*{Raising the degree}
We adjoin three integer curvature minors $W=(W_0,W_1,W_2)$ and the fifth
Hankel invariant $J=I_5(i)$, whose sizes satisfy
\[
 |W_j|\ll S:=A^3H^3/a^2,\qquad |J|\ll Y:=A^5H^{10}/a^4.
\]
On each component they are rational functions of the nodes: $J$ is a
polynomial, and $W$ is obtained by recovering the moments of $\nu_i$ from the
two orthogonal polynomials and the division equations, which gives the
curvature $I_3(i)$ as a rational function of the nodes (its
\emph{reconstruction}).
On a component of degree $d$, $J$ has a pole of total order $5d/2$ at
infinity, while the $W_j$ vanish there to total order $3d/2$. A suitable
combination of the $W_j$ also vanishes at the node triple of one orbit
state, so it has at least $3d/2+1$ zeros, hence as many poles, all affine.
For fixed positive integers $m,n$, consider the graph obtained by adjoining
$W_0^m,W_1^m,W_2^m,J^n$ to the nodes. The degree of an affine curve is the
pole degree of a generic linear combination of its coordinates, and on this
graph $J^n$ has poles only at infinity and the $W_j^m$ only at affine
points, so their pole degrees add. The graph therefore has
degree at least $\frac{5n}2d+m\bigl(\frac{3d}2+1\bigr)$, and the box grows
only to $\max(H,S^m,Y^n)$. Conics and components on which the reconstruction
fails are counted separately. This gives Theorem~\ref{thm:joint} below.

The pattern also determines a rank-three lattice (the rows satisfy
congruences modulo the largest minor). If $\lambda$ is the
smallest sup norm of a nonzero lattice vector, a reduced basis gives integer
increment and curvature coordinates of sizes $O(H/\lambda)$ and
$O(S/\lambda^2)$. Averaged bounds for short lattice vectors shrink the box
for most patterns without changing the graph degree
(Section~\ref{sec:lattice}, Theorem~\ref{thm:lattice}).

\subsection*{The polar improvement}
Two further facts give Theorem~\ref{thm:polar}. First, an effective Hilbert
irreducibility sieve (the \emph{Hilbert sieve}) shows that, outside a sparse set
of patterns, the pattern curve is irreducible over $\mathbb Q$, of the maximal
degree $24$. Second, an exact
identity, $JW=N_0^2(G\mathbf k+\ell)$, expresses the curvature vector through
three affine-linear forms in the increments $\mathbf k=(k_0,k_1,k_2)$. Here $N_0$
is a pattern coordinate, and $G\mathbf k+\ell$ arises by polarising the
pattern quadratic $f(\mathbf k)=\mathbf k\cdot(G\mathbf k+\ell)$. Since
$f(\mathbf k)=0$, the increment vector $\mathbf k$ is orthogonal to
$G\mathbf k+\ell$; the division equations say exactly that the drop vector
$\mathbf d$ is orthogonal to it too, so that $W=\mathbf d\times\mathbf k$ is
parallel to it.

On a component of degree $24$, $J$ has $60$ zeros,
all affine. Since $W=N_0^2(G\mathbf k+\ell)/J$, each zero of $J$ is a pole
of $W$ unless the three linear forms also vanish there, and outside a fixed
proper algebraic set of \emph{degenerate} patterns this happens at most four times,
counted with multiplicity. So $W$ has at least $60-4=56$ affine poles, and
the powered graph has degree
\[
 \deg\Gamma\ge\frac{5n}{2}\cdot24+m(60-4)=60n+56m.
\]
Without the sieve the general bound can use only $d\ge4$, which gives
$10n+7m$; forcing $d=24$ would give $60n+37m$, and the polar identity raises
the affine pole count from $37$ to $56$.

The sieve and the
degenerate set together discard $O_\eta(a^\eta X^{14/3})$ patterns. These are
counted with the unpowered graph of Section~\ref{sec:joint}, of degree at least
$17$, which gives the exponents $717/2440$ and $732/2491$, both below $5/17$. Applying the curve bound in the reduced
lattice coordinates and optimising the ratio $n/m$ gives $\beta_*$
(Section~\ref{sec:polar}).

\subsection*{Intermediate bounds}
The two curve-counting stages before the polar improvement give useful
bounds without the Hilbert sieve. Put
\[
 \alpha_* = \frac{1310-\sqrt{655}}{4365}=0.2942513\ldots,\qquad
 \gamma_* = \frac{471}{1559}-\frac{6\sqrt{105}}{7795}
          =0.2942294\ldots.
\]
\begin{theorem}[Joint-invariant bound]\label{thm:joint}
For every $\varepsilon>0$, $P(a,b)\ll_\varepsilon a^{\alpha_*+\varepsilon}$
uniformly for $0<b<a$.
\end{theorem}
\begin{theorem}[Exterior lattice bound]\label{thm:lattice}
For every $\varepsilon>0$, $P(a,b)\ll_\varepsilon a^{\gamma_*+\varepsilon}$
uniformly for $0<b<a$.
\end{theorem}
Fixed powers give the rational instances $128/435$ and $2201/7480$ of the
joint-invariant bound, and $153/520$ of the exterior lattice bound; see
Corollaries~\ref{cor:jointexponent} and~\ref{cor:latticeexponent}.
The table records how the exponent approached $5/17$ during this work. The
first four rows are superseded stages, recorded for orientation only; their
proofs are formalised in the repository but omitted here.
\begin{center}\small
\begin{tabular}{@{}lll@{}}
\toprule
Exponent & Exponent${}-5/17$ & Main input\\
\midrule
$23/77$ & $4.6\cdot10^{-3}$ & $I_4$ as an integer label\\
$11/37$ & $3.2\cdot10^{-3}$ & a ternary quadratic and norm equations on slices\\
$8/27$ & $2.2\cdot10^{-3}$ & the $X^{5+o(1)}$ count of minor vectors\\
$45/152$ & $1.9\cdot10^{-3}$ & lattices of pivot increments and their short vectors\\
$\alpha_*$ & $1.3\cdot10^{-4}$ & adjoining curvature and rank-five invariants\\
$\gamma_*$ & $1.1\cdot10^{-4}$ & reducing the joint coordinates in a lattice basis\\
$\beta_*$ & $1.6\cdot10^{-5}$ & the Hilbert sieve and the polar identity\\
\bottomrule
\end{tabular}
\end{center}

\subsection*{An elementary bound and a rise-sensitive bound}
The initial identities already yield an elementary improvement over the
exponent $1/3$.
\begin{theorem}\label{thm:main}
For every $\varepsilon>0$, $P(a,b)\ll_\varepsilon a^{3/10+\varepsilon}$
uniformly for $0<b<a$.
\end{theorem}
The proof uses only division equations and the divisor estimate
$\tau(n)\ll_\eta n^\eta$. For a window of $T$ remainders in $[A,2A)$,
\begin{align}
 T(T-1)&\le4A\Delta,\label{eq:intro1}\\
 T^6A^2&\ll a^3,\label{eq:intro3}
\end{align}
where $\Delta\ll_\eta a^\eta$ is a maximum of the divisor function. The first
is the Erd\H{o}s--Shallit drop argument. The second follows from the six-term
identity: some block of six terms has total drop at most $10A/T$, which bounds
its increments and curvatures, and these bound the quotient $q\asymp a/A$,
with no divisor loss. Multiplying the second
bound by the square of the first eliminates $A$ and gives
$T^{10}\ll a^3\Delta^2$ (Section~\ref{sec:proof310}). The six-term bound is
also used in all the later arguments.

There is also a bound sensitive to the order of the increments. Let
\[
 R=1+\#\{i:0\le i\le P(a,b)-3,\ k_i<k_{i+1}\}.
\]
\begin{theorem}\label{thm:rises}
$P(a,b)\ll a^{1/4}\sqrt R$ uniformly for $0<b<a$. In particular,
nonincreasing quotient increments give $P(a,b)\ll a^{1/4}$.
\end{theorem}
Curvature positivity makes the drops decrease between rises. The six-term
identity limits repeated equal increments, and summing the resulting
window bounds proves the theorem. This argument is independent of the
curve-counting refinements.

\subsection*{Reading guide and formal verification}
Sections~\ref{sec:curv}--\ref{sec:proof310} establish the elementary identities and
prove Theorem~\ref{thm:main}. Sections~\ref{sec:hankel}--\ref{sec:patterns} construct the
integer invariants and their patterns. Section~\ref{sec:joint} builds the enlarged
curves and proves Theorem~\ref{thm:joint}, and Section~\ref{sec:lattice} reduces their
lattice coordinates and proves Theorem~\ref{thm:lattice}. Section~\ref{sec:polar} proves
Theorem~\ref{thm:polar} by rerunning the window bound of Section~\ref{sec:lattice} with a
larger graph degree, so it depends on all of
Sections~\ref{sec:hankel}--\ref{sec:lattice}. The independent rise bound,
Theorem~\ref{thm:rises}, is proved in Section~\ref{sec:rises} and uses only
Sections~\ref{sec:curv} and~\ref{sec:six}. The conic count, graph-degree
upper bound, and three-dimensional lattice reduction are proved in
Appendices~\ref{app:conic}, \ref{app:degree} and~\ref{app:reduction}.
Appendix~\ref{app:hilbert} proves the effective Hilbert irreducibility estimate,
Appendix~\ref{app:linecertificates} records the degree certificates used to exclude
lines, and Appendix~\ref{app:polarcertificates} gives the certificates for the polar bound.

All the unconditional results above, including their curve-counting and
specialisation inputs, are formalised in Lean~4~\cite{Lean} with
Mathlib~\cite{Mathlib}. Section~\ref{sec:lean} identifies the proof artefact
and its verification scope. Section~\ref{sec:limits} explains the present
$5/17$ barrier.

\section{Preliminaries}

\subsection{Notation}
Throughout, $a>b>0$ are integers, $b_0=b$, $b_{i+1}=a\bmod b_i$, and $P=P(a,b)$, so that
$a>b_0>b_1>\dots>b_{P-1}>0=b_P$. Write $q_i=\lfloor a/b_i\rfloor$, so
\[
a=q_ib_i+b_{i+1},\qquad 0\le b_{i+1}<b_i\qquad(0\le i<P).
\]
An index $i$ of the orbit is also called a \emph{state}; quantities ``at state $i$'' are
computed from $b_i,b_{i+1},\dots$ and $q_i,q_{i+1},\dots$, and are defined whenever the
orbit terms they use exist. When we count blocks $b_i,b_{i+1},\dots$ in a window, we call
their first index $i$ a \emph{start}; a start is thus just a state, named for its role in the
count. We write
$k_i=q_{i+1}-q_i$ for the quotient \emph{increments} and $d_i=b_i-b_{i+1}$ for the
\emph{drops}.
Let $\tau(n)$ be the number of positive divisors of $n$, and
$\tau_{\max}(Y)=\max_{1\le j\le Y}\tau(j)$.
We write $F\ll_\eta G$ when $F\le C_\eta G$, with the constant uniform in the
dividend and orbit. A factor denoted $a^{o(1)}$ is bounded by
$C_\eta a^\eta$ for every fixed $\eta>0$; such factors include the divisor
losses below. We write $F\asymp G$ when both $F\ll G$ and $G\ll F$ hold.
Absolute constants play no role in the exponents, and we do not track them:
implied constants are absolute unless their dependence is indicated, and a
letter $c$ or $C$ with a descriptive subscript denotes a suitable absolute
constant introduced where it is needed. Explicit values of all
constants are recorded in the Lean development; a few of them are sharper than
the ones derived here.

\begin{lemma}[Quotients increase]\label{lem:quot}
If $b_{i+1}>0$ then $q_{i+1}\ge q_i+1$.
\end{lemma}

\begin{proof}
Since $b_{i+1}<b_i$ we have $(q_i+1)b_{i+1}\le q_ib_i+b_{i+1}=a$, so
$q_{i+1}=\lfloor a/b_{i+1}\rfloor\ge q_i+1$.
\end{proof}

\begin{lemma}[Divisor bound]\label{lem:div}
For every integer $k\ge1$ there is $C_k$ with $\tau(n)^k\le C_k\,n$ for all $n\ge1$.
\end{lemma}

\begin{proof}
This is standard: $\tau(n)^k/n=\prod_{p^\alpha\| n}(\alpha+1)^k/p^\alpha$. Each factor with
$p\ge 2^k$ is at most $(\alpha+1)^k/2^{k\alpha}\le1$. Each of the fewer than $2^k$ factors with
$p<2^k$ is at most $\sup_\alpha(\alpha+1)^k/2^\alpha<\infty$.
\end{proof}

\subsection{Window runs}

\begin{definition}
Let $A\ge1$ be an integer. A \emph{window run} of length $T$ is a sequence $x_0,x_1,\dots,x_{T-1}$ of
consecutive terms of the sequence $(b_i)$, so $x_{j+1}=a\bmod x_j$, all lying in $[A,2A)$.
\end{definition}

In a window run, $x_0>x_1>\dots>x_{T-1}$.

\begin{lemma}[Quotients in a window]\label{lem:fourAq}
If $0<b_i<2A$ then $a<4Aq_i$.
\end{lemma}

\begin{proof}
$a=q_ib_i+b_{i+1}<(q_i+1)b_i<2A(q_i+1)\le4Aq_i$, since $q_i\ge1$.
\end{proof}

\section{The curvature of three consecutive steps}\label{sec:curv}

Fix four consecutive terms
\[
x=b_i,\quad y=b_{i+1}=a\bmod x,\quad z=b_{i+2}=a\bmod y,\quad w=b_{i+3}=a\bmod z,
\]
with $x,y,z>0$ (and $w\ge0$). Write
\[
q=\left\lfloor \frac ax\right\rfloor,\quad
u=\left\lfloor \frac ay\right\rfloor-q,\quad
v=\left\lfloor \frac az\right\rfloor-\left\lfloor \frac ay\right\rfloor,\qquad
d=x-y,\quad e=y-z,\quad f=z-w .
\]
In the notation of the introduction, $q=q_i$, $u=k_i$, $v=k_{i+1}$ and
$d,e,f=d_i,d_{i+1},d_{i+2}$; the short letters keep the identities below
readable. By Lemma~\ref{lem:quot}, $u,v\ge1$, and clearly $d,e,f\ge1$. The division equations read
\begin{equation}\label{eq:div}
xq+y=a,\qquad y(q+u)+z=a,\qquad z(q+u+v)+w=a .
\end{equation}

\begin{definition}
The \emph{curvature} of the block is $D=dv-eu$.
\end{definition}

\begin{lemma}[Curvature identity]\label{lem:curv}
\[
D\,z=df+ude+(u-1)e^2 .
\]
In particular $D\ge1$.
\end{lemma}

\begin{proof}
Subtracting the first two equations of~\eqref{eq:div} gives $uy=qd+e$, hence
$uz=uy-ue=qd+(1-u)e$. Subtracting the last two gives $vz=(q+u)e+f$. Therefore
\[
(dv-eu)z=d\bigl((q+u)e+f\bigr)-e\bigl(qd+(1-u)e\bigr)=df+ude+(u-1)e^2 .
\]
The right-hand side is at least $df\ge1$, and $z>0$, so $D>0$.
\end{proof}

\begin{lemma}[Size of the parameters]\label{lem:size}
We have
\[
u\,yx\le a(d+e),\qquad v\,zy\le a(e+f),\qquad D\,z\le (u+1)(d+e+f)^2 .
\]
\end{lemma}

\begin{proof}
From $uy=qd+e$ we get $uyx=(qx)d+ex\le ad+ae$, since $qx\le a$ and $x<a$. Similarly
$vz=(q+u)e+f$ with $(q+u)y\le a$ and $y\le a$ gives the second bound. For the third, by
Lemma~\ref{lem:curv},
$Dz=df+ude+(u-1)e^2\le df+u\,e(d+e)\le (1+u)(d+e+f)^2$.
\end{proof}

\section{Six consecutive terms}\label{sec:six}

Fix six consecutive terms $x_0>x_1>\dots>x_5\ge0$ of the sequence $(b_i)$ with $x_4>0$.
Write $q=\lfloor a/x_0\rfloor$ and let $k_0,\dots,k_3$ be the next four quotient
increments, so that
\[
\lfloor a/x_j\rfloor=q+k_0+\dots+k_{j-1}\qquad(0\le j\le4).
\]
By Lemma~\ref{lem:quot}, $k_0,\dots,k_3\ge1$. Let $f_j=x_j-x_{j+1}\ge1$ be the drops, and let
\[
D=f_0k_1-f_1k_0,\qquad E=f_1k_2-f_2k_1,\qquad F=f_2k_3-f_3k_2
\]
be the curvatures of the blocks starting at $x_0$, $x_1$ and $x_2$. Write $\mathcal E_j$ for the division
equation $x_jq_j+x_{j+1}-a=0$, where $q_j=\lfloor a/x_j\rfloor$. Define polynomials
\begin{align*}
\sigma(y_1,y_2,y_3)&=y_1^2y_2+(y_1-1)y_2^2+y_1y_3,\\
\omega(y_1,y_2,y_3,y_4)&=\bigl(y_1^2+(y_1-1)y_2\bigr)y_4+y_3\bigl(y_1^3+y_1y_2(2y_1-1)+(y_1-1)y_2^2\bigr)\\
&\qquad+y_3^2\bigl((y_1-1)(y_1+y_2)+1\bigr),
\end{align*}
and put
\[
L=(k_1k_2+k_2^2+k_3)\,D-\bigl(k_0^2+(k_0-1)k_1\bigr)E-k_2\,(k_2f_0-k_0f_2).
\]

\begin{lemma}[The six-term identity]\label{lem:six}
\begin{enumerate}
\item $\sigma(k_1,k_2,k_3)\,D-\sigma(k_0,k_1,k_2)\,E=k_1L$.
\item $q\,k_2L=\omega(k_0,k_1,k_2,k_3)\,E+\sigma(k_0,k_1,k_2)\,F$.
\end{enumerate}
\end{lemma}

\begin{proof}
Part~(1) is a polynomial identity in the drops and increments. Modulo $k_1$
we have $\sigma(k_1,k_2,k_3)\equiv-k_2^2$ and $\sigma(k_0,k_1,k_2)\equiv k_0k_2$,
and $k_2D+k_0E=k_1(k_2f_0-k_0f_2)$; so the left side is divisible by $k_1$, and
expanding the quotient gives $L$. Part~(2) holds modulo the division
equations: $qk_2L-\omega E-\sigma(k_0,k_1,k_2)F=k_2(c_0\mathcal E_0-c_1\mathcal E_1+c_2\mathcal E_2-c_3\mathcal E_3)$, where
\begin{align*}
c_0&=k_1^2k_2+k_1k_2^2+k_1k_3-k_2^2,\\
c_1&=k_0^2k_2+2k_0k_1k_2+k_0k_2^2+k_0k_3+k_1^2k_2+k_1k_2^2-k_1k_2+k_1k_3-k_2^2,\\
c_2&=k_0^2k_1+k_0^2k_2+k_0k_1^2+2k_0k_1k_2+k_0k_2^2+k_0k_2+k_0k_3-k_1^2-k_1k_2,\\
c_3&=k_0^2k_1+k_0k_1^2+k_0k_2-k_1^2.
\end{align*}
This is an identity of polynomials in $a$, $q$, the $x_j$ and the increments.
\end{proof}

\begin{corollary}\label{cor:six}
Suppose $k_0,\dots,k_3\le K$. Then $E,F\ge1$, $L\ge1$, and
\[
q\le 9K^3E+3K^3F .
\]
\end{corollary}

\begin{proof}
Lemma~\ref{lem:curv}, applied to the blocks starting at $x_1$ and $x_2$, gives $E,F\ge1$.
Clearly $\omega>0$ and $\sigma\ge0$, so by Lemma~\ref{lem:six}(2) $qk_2L>0$. Since $q\ge1$ (as
$x_0<a$) and $L$ is an integer, $L\ge1$ and $qk_2\le \omega E+\sigma F$. Bounding each monomial, using $k_2\ge1$, gives $\omega\le 9K^3k_2$ and
$\sigma(k_0,k_1,k_2)\le 3K^3k_2$. Dividing by $k_2$ gives the claim.
\end{proof}

So the quotient $q$, which is about $a/A$ when the six terms lie in a window $[A,2A)$, is controlled by the small parameters of a single block
of six terms. No counting over many blocks is needed.

\begin{proposition}[Six-term window bound]\label{prop:six}
For a window run of length $T\ge10$,
\[
T^6A^2\ll a^3 .
\]
\end{proposition}

\begin{proof}
For $0\le j\le T-6$ let $s_j=x_j-x_{j+5}$. By telescoping,
$\sum_{j\le T-6}s_j\le 5(2A-1)-5A=5A-5$. So some $j$ has $s_j(T-5)\le 5A-5$, and since
$T\le 2(T-5)$, we get $s:=s_j$ with $sT\le 10A$. Apply Corollary~\ref{cor:six} to
$x_j,\dots,x_{j+5}$, all of which lie in $[A,2A)$, with $K=\max(k_0,\dots,k_3)$.

By Lemma~\ref{lem:size} at $x_j$, $x_{j+1}$ and $x_{j+2}$, each increment $k$ satisfies
$kA^2\le a s$. Multiplying by $T$ gives $KAT\le 10a$. By
Lemma~\ref{lem:size} again, $EA\le Ex_{j+3}\le(k_1+1)s^2\le 2Ks^2$, and likewise $FA\le 2Ks^2$.
Hence $qA\le 24K^4s^2$. Also $a<4Aq$ by Lemma~\ref{lem:fourAq}. Therefore
\[
qA\cdot A^4T^6\le 24\,(KAT)^4(sT)^2\ll a^4A^2,
\]
so $qA^3T^6\ll a^4$, and $aA^2T^6\le 4qA^3T^6\ll a^4$.
\end{proof}


\section{Proof of Theorem~\ref{thm:main}}\label{sec:proof310}

We bound the number $T$ of orbit terms in a window $[A,2A)$ in two ways.
Write $B=a/A$, so that the quotients in the window are about $B$. The drop
argument (Proposition~\ref{prop:drop}) gives $T\ll a^{1/2+o(1)}B^{-1/2}$,
which is good when the quotients are large, and the six-term bound
(Proposition~\ref{prop:six}) gives $T\ll a^{1/6}B^{1/3}$, which is good when
they are small. The two agree at $B=a^{2/5}$, where both equal
$a^{3/10+o(1)}$. Multiplying suitable powers of the two bounds eliminates $A$
(Corollary~\ref{cor:window}), so no case split is needed. Throughout,
\[
 \Delta=\Delta(a)=\tau_{\max}(2a)\ll_\eta a^\eta
\]
by Lemma~\ref{lem:div}. In this section $x_0,\dots,x_{T-1}$ is a window run in
$[A,2A)$, and $q_j=\lfloor a/x_j\rfloor$ are its quotients.

\begin{proposition}[Drop bound; after Erd\H{o}s--Shallit]\label{prop:drop}
If $T\ge2$ then $T(T-1)\le 4A\Delta$.
\end{proposition}

\begin{proof}
Let $h_j=x_j-x_{j+1}\ge1$ for $0\le j\le T-2$. The total drop is
$\sum_jh_j=x_0-x_{T-1}\le A-1$. Thus fewer than $(T-1)/2$ drops exceed
$2A/(T-1)$. The remaining set $\mathcal S$ has $2|\mathcal S|\ge T$.

For $j\in\mathcal S$ we have $1\le h_j\le\bar h:=\lfloor 2A/(T-1)\rfloor$. Also
$a+h_j=q_jx_j+x_{j+1}+x_j-x_{j+1}=(q_j+1)x_j$, so $x_j\mid a+h_j$. Moreover
$a+h_j\le 2a$. Since $j\mapsto x_j$ is injective, at most $\tau(a+h)\le\Delta$ indices
in $\mathcal S$ have $h_j=h$. Hence $|\mathcal S|\le\bar h\Delta$, and
\[
T(T-1)\le 2|\mathcal S|(T-1)\le 2\Delta\,\bar h(T-1)\le 4A\Delta. \qedhere
\]
\end{proof}

\begin{corollary}[Window bound]\label{cor:window}
If $T\ge10$ then $T^{10}\ll a^3\Delta^2$.
\end{corollary}

\begin{proof}
The drop bound and $T\le2(T-1)$ give $T^4\le64A^2\Delta^2$.
Multiply by the six-term bound $T^6A^2\ll a^3$ and cancel $A^2$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Let $0<b<a$. For $0\le k\le\log_2 a$ let $T_k$ be the number of indices $i<P$ with
$2^k\le b_i<2^{k+1}$. Every $b_i$ with $i<P$ lies in $[1,a)$, so
$P=\sum_{0\le k\le \log_2 a}T_k$. Since $(b_i)$ is decreasing, the indices counted by $T_k$
are consecutive. So when $T_k>0$ they form a window run in $[A,2A)$ with $A=2^k$.
Inserting $\Delta\ll_\eta a^\eta$ with $\eta=5\varepsilon/2$ into
Corollary~\ref{cor:window} yields
$T_k\ll_\varepsilon a^{3/10+\varepsilon/2}$; the windows with $T_k<10$
satisfy the same estimate. There are $O(\log a)\ll_\varepsilon a^{\varepsilon/2}$
windows, so summing proves the theorem.
\end{proof}

\section{The Hankel hierarchy}\label{sec:hankel}

From now on we write $N=P(a,b)$, freeing the letter $P$ for polynomials, so that
$b_0>\dots>b_{N-1}>0=b_N$, and $k_i=q_{i+1}-q_i$. The division equation
gives $b_i/a=(1-b_{i+1}/a)/q_i$, and iterating it expresses $b_i/a$ as the
alternating Pierce series in the tail quotients $q_i,q_{i+1},\dots$. The
construction below rewrites this series as a Stieltjes transform,
\[
 \frac{b_i}a=\int\frac{d\mu_i(x)}{q_i+x},
\]
of a probability measure $\mu_i$ on the nodes $q_{i+j}-q_i$
(Lemma~\ref{lem:moments}(2)). The measure $\nu_i=\mu_i/(q_i+x)$ then has total
mass $b_i/a$, and the Hankel determinants of its moments are the integer
invariants we need.

Put $p_i(j)=q_{i+j}-q_i$ for $0\le j<N-i$; the symbols $p_i,w_i$
are local to this construction. Define the weights backwards from
$w_{N-1}(0)=1$ by
\[
w_i(0)=1-\sum_{j=0}^{N-i-2}\frac{w_{i+1}(j)}{p_i(j+1)},\qquad
w_i(j+1)=\frac{w_{i+1}(j)}{p_i(j+1)}
\quad(0\le j<N-i-1).
\]
Let $\mu_i$ be the measure with mass $w_i(j)$ at $p_i(j)$, and let
\[
 d\nu_i(x)=\frac{d\mu_i(x)}{q_i+x},\qquad
 h_i(s)=\int x^s\,d\nu_i=\sum_{j=0}^{N-i-1}
 \frac{w_i(j)p_i(j)^s}{q_{i+j}}.
\]
The recursion is chosen to reproduce the division equation
$q_ih_i(0)+h_{i+1}(0)=1$ for $i+1<N$.

\begin{lemma}[The moment model]\label{lem:moments}
\begin{enumerate}
\item The weights are positive and sum to $1$.
\item $h_i(0)=b_i/a$, that is, $b_i/a=\mathbb E\bigl[1/(q_i+X_i)\bigr]$ for $X_i\sim\mu_i$; and
$h_{N-1}(s)=0$ for $s\ge1$.
\item $h_i(s+1)=\sum_{t\le s}\binom st k_i^{s-t}h_{i+1}(t)$ for $i+1<N$.
\item $h_i(s)=(-q_i)^sb_i/a+z$ with $z\in\mathbb Z$.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) Since $p_i(j+1)\ge1$, the sum defining $w_i(0)$ is at most $1$. It is strictly less: if
$N-i-1\ge2$, the term $j=1$ has $p_i(2)\ge2$; if $i=N-2$, the last increment satisfies
$q_{N-1}\ge q_{N-2}+2$, because $b_{N-1}\mid a$ gives
$q_{N-1}-q_{N-2}-1=q_{N-2}(b_{N-2}-b_{N-1})/b_{N-1}\ge1$. (2) Since $p_i(j+1)=q_{i+j+1}-q_i$, the definitions give
$q_ih_i(0)+h_{i+1}(0)=1$, which is the normalised division equation; and
$h_{N-1}(0)=1/q_{N-1}=b_{N-1}/a$; since $\mu_{N-1}=\delta_0$, also $h_{N-1}(s)=0$ for $s\ge1$.
(3) Use $p_i(j+1)=k_i+p_{i+1}(j)$ and the binomial theorem.
(4) For $i=N-1$ this holds with $z=-(-q_{N-1})^sb_{N-1}/a=-(-1)^sq_{N-1}^{s-1}$ for $s\ge1$, since
$q_{N-1}b_{N-1}=a$. Then induct backwards on $i$ using (3), $b_{i+1}/a=1-q_ib_i/a$ and
$k_i-q_{i+1}=-q_i$.
\end{proof}

The weights also explain the role of the Newton polynomials below. Write
$\mu_i(g)=\int g\,d\mu_i$. For $i+1<N$ and any polynomial $g$, the zero
atom contributes nothing to $\mu_i(xg)$, and
$p_i(j+1)w_i(j+1)=w_{i+1}(j)$ gives
\[
 \mu_i(xg(x))=\mu_{i+1}\bigl(g(k_i+x)\bigr).
\]
With Lemma~\ref{lem:moments}(1) this gives the low moments
\begin{equation}\label{eq:lowmoments}
 \mu_j(1)=\mu_j(x)=1,\qquad \mu_j(x^2)=k_j+1,\qquad
 \mu_j(x^3)=k_j^2+2k_j+k_{j+1}+1\qquad(j+3<N).
\end{equation}
Once the state $i$ is fixed we drop it from the notation and write
$s_j=p_i(j)=q_{i+j}-q_i$; put $F_j(x)=\prod_{0\le l<j}(x-s_l)$, with
$F_0=1$. Iterating the identity $\mu_i(xg(x))=\mu_{i+1}(g(k_i+x))$ gives the \emph{shift rule}
\begin{equation}\label{eq:moment-shift}
 \mu_i(F_j(x)g(x))=\mu_{i+j}\bigl(g(x+s_j)\bigr)
 \qquad(i+j<N).
\end{equation}
Indeed, each step removes the first factor of $F_j$ and translates the
remaining nodes by the next increment. Thus multiplication by a Newton
polynomial advances along the orbit; this is why this basis is useful for
bounding its Hankel determinants.

\begin{theorem}[Hankel determinants]\label{thm:hankel}
For $1\le r\le N-i$, the number $I_r(i)=a\det\bigl(h_i(j+l)\bigr)_{0\le j,l<r}$ is a positive
integer. Moreover $I_1(i)=b_i$, $I_2(i)=d_i$ (if $i+1<N$), $I_3(i)=D_i$ (if $i+3<N$), and
$I_4(i)=L_i$ (if $i+5<N$), where $D_i$ and $L_i$ are the curvature and the six-term integer
of the blocks starting at $b_i$.
\end{theorem}

\begin{proof}
\emph{Integrality.} By Lemma~\ref{lem:moments}(4),
$h_i(s+1)+q_ih_i(s)\in\mathbb Z$. Multiply the Hankel matrix on the left by the unit lower
triangular matrix that adds $q_i$ times row $j-1$ to row $j$. This makes rows
$1,\dots,r-1$ integral and leaves the determinant unchanged. Expand along row $0$, whose entries
are $(-q_i)^lb_i/a+(\text{integer})$; then $a\det$ is an integer.

\emph{Positivity.} The Hankel matrix is the Gram matrix of $1,x,\dots,x^{r-1}$ in $L^2(\nu_i)$.
Its quadratic form is $\sum_jw_i(j)q_{i+j}^{-1}\bigl(\sum_lc_lp_i(j)^l\bigr)^2$. If this
vanishes, the polynomial $\sum_lc_lx^l$ vanishes at the $N-i\ge r$ distinct points $p_i(j)$, so
it is zero, by the invertibility of the Vandermonde matrix. So the matrix is positive definite.

\emph{The first four.} First, $I_1(i)=ah_i(0)=b_i$.
For $i+1<N$, write $y_j=b_{i+j}/a$. Then
$h_i(0)=y_0$, $h_i(1)=y_1$ and
$h_i(2)=k_iy_1+y_2$. Hence
\[
 I_2(i)=a\bigl(y_0(k_iy_1+y_2)-y_1^2\bigr)
       =a(y_0-y_1)=d_i,
\]
using $q_iy_0+y_1=q_{i+1}y_1+y_2=1$.
For the larger determinants we use the Newton basis. By
Lemma~\ref{lem:moments}(3), $\nu_i(xg(x))=\nu_{i+1}\bigl(g(x+k_i)\bigr)$, so
$\nu_i$ obeys the same shift rule~\eqref{eq:moment-shift} as $\mu_i$, and for
$j\le l$
\[
 \nu_i(F_jF_l)=\nu_{i+l}\bigl(F_j(x+s_l)\bigr)
 =\nu_{i+l}\Bigl(\prod_{m<j}(x+s_l-s_m)\Bigr).
\]
Since $F_0,\dots,F_{r-1}$ are monic of degrees $0,\dots,r-1$, the Gram matrix
$\bigl(\nu_i(F_jF_l)\bigr)_{j,l<r}$ has the same determinant as the Hankel
matrix. Its entries are explicit: $\nu_{i+l}(1)=y_l$, and higher moments
of $\nu_{i+l}$ follow from the same rule. For $r=3$ the matrix is
\[
\begin{pmatrix}
 y_0&y_1&y_2\\
 y_1&y_2+k_iy_1&y_3+s_2y_2\\
 y_2&y_3+s_2y_2&y_4+(s_2+k_{i+1}+k_{i+2})y_3+s_2k_{i+1}y_2
\end{pmatrix}.
\]
Using $y_{j+1}=1-q_{i+j}y_j$ for $0\le j\le3$ to express every entry in
terms of $y_0$, $q_i$ and the increments, and expanding, gives $a$ times its
determinant as
$(b_i-b_{i+1})k_{i+1}-(b_{i+1}-b_{i+2})k_i=D_i$. At the end of the orbit
this computation uses $h_{N-1}(s)=0$ for $s\ge1$, that is, $b_N=0$.

For $r=4$ we avoid the analogous $4\times4$ expansion by passing through the
orthogonal cubic of $\nu_i$. Let $i+5<N$, and for $m\le2$ write
$k_m=k_{i+m}$, $d_m=d_{i+m}$, $r_m=k_m(s_m+s_{m+1})+k_{m+1}$ and
$G_m=F_{m+1}-F_m=F_m(x)(x-s_m-1)$. The shift rule, $\nu_{i+m}(1)=b_{i+m}/a$,
$\nu_{i+m}(x)=b_{i+m+1}/a$ and the low moments~\eqref{eq:lowmoments} at
$j=i+m$ give
\[
 a\,\nu_i(G_m)=-d_m,\quad \mu_i(G_m)=0,\quad \mu_i(xG_m)=k_m,\quad
 \mu_i(x^2G_m)=r_m+k_m\qquad(m\le2).
\]
Put $\mathbf d=(d_0,d_1,d_2)$, $\mathbf k=(k_0,k_1,k_2)$,
$\mathbf r=(r_0,r_1,r_2)$ and $W=\mathbf d\times\mathbf k$, whose last
coordinate is $W_2=d_0k_1-d_1k_0=D_i\ge1$. The monic cubic
$\pi=\sum_{m\le2}(W_m/W_2)G_m$ satisfies $\nu_i(\pi)=-\mathbf d\cdot W/(aW_2)=0$,
$\mu_i(\pi)=0$ and $\mu_i(x\pi)=\mathbf k\cdot W/W_2=0$. Since
$\mu_i=(q_i+x)\nu_i$, we have $\nu_i(xg)=\mu_i(g)-q_i\nu_i(g)$, so
$\nu_i(x\pi)=\nu_i(x^2\pi)=0$: $\pi$ is the monic orthogonal cubic of $\nu_i$.
Hence
\[
 \nu_i(\pi^2)=\nu_i(x^3\pi)=\mu_i(x^2\pi)-q_i\nu_i(x^2\pi)
 =\frac{\mathbf r\cdot W+\mathbf k\cdot W}{W_2}
 =\frac{\det(\mathbf d,\mathbf k,\mathbf r)}{D_i}.
\]
In the basis $1,x,x^2,\pi$ the Gram matrix of $\nu_i$ is block diagonal and has
the same determinant as the Hankel matrix, so
$I_4(i)=I_3(i)\,\nu_i(\pi^2)=\det(\mathbf d,\mathbf k,\mathbf r)$, using
$I_3(i)=D_i$.
Finally $\det(\mathbf d,\mathbf k,\mathbf r)=L_i$. Expanding, with the nodes
cancelling,
\[
 k_1r_2-k_2r_1=\sigma(k_1,k_2,k_3),\qquad k_0r_1-k_1r_0=\sigma(k_0,k_1,k_2),
\]
and $k_1(k_0r_2-k_2r_0)=k_0(k_1r_2-k_2r_1)+k_2(k_0r_1-k_1r_0)$. Multiplying
the expansion of $\det(\mathbf d,\mathbf k,\mathbf r)$ along its first row by
$k_1$ therefore gives
\[
 k_1\det(\mathbf d,\mathbf k,\mathbf r)
 =(d_0k_1-d_1k_0)\,\sigma(k_1,k_2,k_3)-(d_1k_2-d_2k_1)\,\sigma(k_0,k_1,k_2)
 =k_1L_i
\]
by Lemma~\ref{lem:six}(1), since $d_0k_1-d_1k_0$ and $d_1k_2-d_2k_1$ are the
curvatures $D$ and $E$ of that lemma.
\end{proof}

The ranges $i+3<N$ and $i+5<N$ cannot be relaxed to all $i$ at which $D_i$ or $L_i$ is defined.
For $a=926$ and $b=229$ the orbit is $229,10,6,2$, so $N=4$, and $I_3(1)=984$ while $D_1=988$.

\begin{remark}
In these index ranges, the positivity of $D$ and $L$, proved earlier by explicit identities, is an instance of the
positivity of Gram determinants. The integrality of $L$ (Lemma~\ref{lem:six}(1)) is an instance
of the integrality of $I_4$. A priori $I_r(i)$ involves the moments $h_i(s)$ for $s\le2r-2$,
and hence $b_i,\dots,b_{i+2r-2}$. In fact $I_3(i)$ depends only on $b_i,b_{i+1},b_{i+2}$ and
$q_i,q_{i+1},q_{i+2}$. Similarly $I_4(i)$ depends only on $b_i,\dots,b_{i+3}$ and
$q_i,\dots,q_{i+4}$: the later remainders cancel.
\end{remark}

\section{Patterns}\label{sec:patterns}

In this section we develop the patterns used in Section~\ref{sec:joint}, together with the
bounds on their size and number. We keep the notation of
Section~\ref{sec:hankel}: $b_0>b_1>\dots>b_{N-1}>0$ is the positive part of the orbit,
$N=P(a,b)$, and $q_i=\lfloor a/b_i\rfloor$.

\subsection{Heuristics}
The exponent $3/10$ of Theorem~\ref{thm:main} comes from a single balanced configuration:
$T\asymp a^{3/10}$ remainders in a window $[A,2A)$ with $A\asymp a^{3/5}$, quotients
$q\asymp a^{2/5}$ and typical increments $k\asymp a^{1/10}$. At this scale the rank-four Hankel
invariant $I_4(i)=L_i$ has size $\asymp aH^6/q^4\asymp 1$, where $H$ bounds the increments.
The idea is that a positive integer of bounded size carries information. The labels
$I_4(i),I_4(i+1),I_4(i+2)$, together with seven further $3\times3$ minors of the same size, form a
\emph{pattern} (made precise in Section~\ref{sec:minors}). States with a given pattern satisfy two linear equations and one quadratic
equation in five consecutive increments (Lemma~\ref{lem:pattern}), and the quotient $q_i$ is a
root of a quartic fixed by the pattern and the increments (Lemma~\ref{lem:multid}). So the
problem becomes one of counting patterns, and the states belonging to each pattern. Counting the
latter with these equations alone already improves on $3/10$ (the first rows of the table in
Section~\ref{sec:intro}). To go further, the linear and quadratic relations alone do not
suffice: for some patterns they admit constant increments $k_j=t$ for every $t$
(an example follows Lemma~\ref{lem:jointcurve}). Section~\ref{sec:joint} therefore adds
the fixed-dividend equation, an equation in the nodes alone that involves $a$;
it excludes such families.

The rest of this section builds the pattern matrix and identifies its
cofactors with the orthogonal quartic of $\nu_i$ (Section~\ref{sec:patternmatrix}), ties $q_i$ to
the pattern by a quartic equation (Section~\ref{sec:normid}), proves the signs and the
supply of good starts used throughout Section~\ref{sec:joint} (Sections~\ref{sec:signs}
and~\ref{sec:goodstarts}), and bounds the size and number of patterns
(Sections~\ref{sec:minors} and~\ref{sec:minorcount}).

\subsection{The pattern matrix}\label{sec:patternmatrix}
Fix a state $i$ with $i+6<N$ and write, for $m\ge0$,
\begin{gather*}
k_m=q_{i+m+1}-q_{i+m},\qquad d_m=b_{i+m}-b_{i+m+1},\\
s_m=k_0+\dots+k_{m-1},\qquad r_m=k_m(s_m+s_{m+1})+k_{m+1}.
\end{gather*}
All indices here are local to the chosen state, and the definitions apply
whenever the indicated orbit entries exist; the empty sum gives $s_0=0$.
So $s_m=q_{i+m}-q_i$, $r_0=k_0^2+k_1$, $r_1=k_1(2k_0+k_1)+k_2$, and so on.
In this local notation the low moments~\eqref{eq:lowmoments} at state $i+m$ read
$\mu_{i+m}(x^2)=k_m+1$ and $\mu_{i+m}(x^3)=k_m^2+2k_m+k_{m+1}+1$.

The \emph{pattern} $p(i)$ of the state is the vector of the ten $3\times3$
minors of the $3\times5$ matrix with rows $(d_0,\dots,d_4)$, $(k_0,\dots,k_4)$
and $(r_0,\dots,r_4)$ (Section~\ref{sec:minors}). Every object built from the
pattern below (the pattern integers, the normalised parameters, the Newton
coefficients $\gamma$ and, in Section~\ref{sec:polar}, $G$ and $\ell$) is a
function of $p(i)$. Consider the
$3\times4$ integer matrix
\[
\mathcal M_i=\begin{pmatrix} d_0&d_1&d_2&d_3\\ k_0&k_1&k_2&k_3\\ r_0&r_1&r_2&r_3\end{pmatrix},
\]
and let $U_m(i)=(-1)^{m+1}\det(\mathcal M_i\text{ with column $m$ deleted})$ for
$0\le m\le3$. The three rows are, up to sign and the factor $a$, the moments of
$F_{m+1}-F_m$ against $\nu_i$, $x\mu_i$ and $(x^2-x)\mu_i$ (proof of
Lemma~\ref{lem:orthquartic}); this is why the cofactor vector describes an
orthogonal polynomial.

\begin{lemma}[Cofactor relations]\label{lem:pattern}
\begin{enumerate}
\item $U_3(i)=L_i=I_4(i)$ and $U_0(i)=-L_{i+1}=-I_4(i+1)$, where $L$ is the six-term integer of
Lemma~\ref{lem:six}.
\item $\sum_m U_m(i)\,d_m=\sum_m U_m(i)\,k_m=\sum_m U_m(i)\,r_m=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
(2) is the Laplace expansion of a $4\times4$ determinant with a repeated row. For (1), the
identity $\det(\text{columns }0,1,2)=L_i=I_4(i)$ was proved in the case $r=4$ of
Theorem~\ref{thm:hankel}. Passing from $i$ to $i+1$ shifts the $d$- and $k$-rows by one
column, and, since $s_m(i+1)=s_{m+1}(i)-k_0$, it replaces $r_m$ by $r_{m+1}-2k_0k_{m+1}$. This
row operation, which subtracts $2k_0$ times the $k$-row, does not change the determinant, so
$\det(\text{columns }1,2,3)=L_{i+1}$.
\end{proof}

For example, the positive part of the orbit with $(a,b)=(95,61)$ is
$61,34,27,14,11,7,4,3,2,1$. Its first pattern matrix and cofactor vector are
\[
 \mathcal M_0=\begin{pmatrix}27&7&13&3\\1&1&3&2\\2&6&23&29\end{pmatrix},
 \qquad U(0)=(-81,833,-296,68).
\]
Thus $L_0=68$, $L_1=81$, and each row of $\mathcal M_0$ is orthogonal
to $U(0)$.

The cofactor vector has a meaning, which is the source of everything that
follows. Recall the Newton polynomials $F_m(x)=\prod_{l<m}(x-s_l)$ and the
measures $\mu_i,\nu_i$ of Section~\ref{sec:hankel}.

\begin{lemma}[The orthogonal quartic]\label{lem:orthquartic}
Let $i+6<N$ and $C_m=U_m(i)/U_3(i)$. Then
\[
 \pi_i=\sum_{m=0}^3C_m\,(F_{m+1}-F_m)
\]
is the monic orthogonal polynomial of degree four for $\nu_i$. Writing
$\pi_i=\sum_{l\le4}\gamma_lF_l$, with \emph{Newton coefficients} $\gamma_l$, one has $U_m(i)/I_4(i)=\gamma_{m+1}+\dots+\gamma_4$
for $0\le m\le3$, and $\gamma_0=I_4(i+1)/I_4(i)$.
\end{lemma}

\begin{proof}
Note that $F_{m+1}-F_m=F_m(x)(x-s_m-1)$. The shift rule~\eqref{eq:moment-shift}
holds for $\nu_i$ as well as for $\mu_i$ (proof of Theorem~\ref{thm:hankel}).
Together with $\nu_{i+m}(1)=b_{i+m}/a$, $\nu_{i+m}(x)=b_{i+m+1}/a$ and the low
moments~\eqref{eq:lowmoments} at $j=i+m$, it gives, for $m\le3$,
\begin{gather*}
 \nu_i\bigl(F_m(x)(x-s_m-1)\bigr)=\nu_{i+m}(x-1)=-d_m/a,\\
 \mu_i\bigl(F_m(x)(x-s_m-1)\bigr)=\mu_{i+m}(x-1)=0,\\
 \mu_i\bigl(xF_m(x)(x-s_m-1)\bigr)=\mu_{i+m}\bigl((x+s_m)(x-1)\bigr)=k_m,\\
 \mu_i\bigl(x^2F_m(x)(x-s_m-1)\bigr)=\mu_{i+m}\bigl((x+s_m)^2(x-1)\bigr)=r_m+k_m.
\end{gather*}
By Lemma~\ref{lem:pattern}(2), the three rows of $\mathcal M_i$ are orthogonal to
$(C_m)$. Hence $\nu_i(\pi_i)=\mu_i(\pi_i)=\mu_i(x\pi_i)=\mu_i(x^2\pi_i)=0$. Since
$\mu_i=(q_i+x)\nu_i$, we have $\nu_i(xg)=\mu_i(g)-q_i\nu_i(g)$, and therefore
$\nu_i(x^j\pi_i)=0$ for $j\le3$, inductively in $j$. The coefficient of $F_4$
in $\pi_i$ is $C_3=1$, so $\pi_i$ is monic of degree four. Finally
$\gamma_l=C_{l-1}-C_l$ with $C_{-1}=C_4=0$, which gives the tail sums, and
$\gamma_0=-C_0=I_4(i+1)/I_4(i)$ by Lemma~\ref{lem:pattern}(1).
\end{proof}

The \emph{pattern integers} of $i$ are the seven integers
\[
\theta(i)=\bigl(L_i,\,L_{i+1},\,L_{i+2},\,U_1(i),\,U_2(i),\,U_1(i+1),\,U_2(i+1)\bigr),
\]
and we write $\theta(i)=(N_0,N_1,N_2,u_1,u_2,v_1,v_2)$; the subscripted $N_j$ have
nothing to do with the orbit length $N$. By Lemma~\ref{lem:pattern},
$\theta(i)$ determines $U(i)=(-N_1,u_1,u_2,N_0)$ and $U(i+1)=(-N_2,v_1,v_2,N_1)$ completely,
and the $k$-relations at $i$ and $i+1$ read
\[
N_1k_0=u_1k_1+u_2k_2+N_0k_3,\qquad N_1k_4=N_2k_1-v_1k_2-v_2k_3 .
\]

\subsection{The normalisation identity and the label bound}\label{sec:normid}

\begin{lemma}[The normalisation identity]\label{lem:multid}
Let $i+6<N$, let $\pi_i$ be as in Lemma~\ref{lem:orthquartic}, and let
$H_4=\det\bigl(\mu_i(x^{j+l})\bigr)_{j,l<4}$, a polynomial in $k_0,\dots,k_4$
independent of $i$. Then
\[
I_4(i)\,\pi_i(-q_i)=a\,H_4(k_0,\dots,k_4).
\]
As a polynomial in $q$, $I_4(i)\pi_i(-q)$ has degree four and leading coefficient
$I_4(i)=L_i\ge1$.
\end{lemma}

This is Christoffel's formula for the modification $\mu_i=(q_i+x)\nu_i$ of
the measure; we include the short proof.

\begin{proof}
Write $h_s=h_i(s)=\nu_i(x^s)$ and $M_s=\mu_i(x^s)$, so that $h_{s+1}+q_ih_s=M_s$.
Expanding along the last row shows that
\[
 \det\bigl(h_{j+l}\bigr)_{j,l<4}\,\pi_i(x)=
 \det\begin{pmatrix}
 h_0&h_1&\cdots&h_4\\
 \vdots&&&\vdots\\
 h_3&h_4&\cdots&h_7\\
 1&x&\cdots&x^4
 \end{pmatrix},
\]
since the right side is a polynomial of degree four with leading coefficient
$\det(h_{j+l})_{j,l<4}$ that is orthogonal to $1,x,x^2,x^3$ (replacing the
last row by a moment row gives a repeated row). Put $x=-q_i$ and, for
$l=4,3,2,1$ in turn, add $q_i$ times column $l-1$ to column $l$. The last row
becomes $(1,0,0,0,0)$, and the entry in row $j<4$ and column $l\ge1$ becomes
$h_{j+l}+q_ih_{j+l-1}=M_{j+l-1}$. Expanding along the last row gives
$\det(h_{j+l})\,\pi_i(-q_i)=\det(M_{j+l})_{j,l<4}=H_4$. Multiply by $a$ and
use $a\det(h_{j+l})=I_4(i)$. The leading coefficient is that of the monic
$\pi_i$.
\end{proof}

\begin{lemma}[Labels]\label{lem:labels}
Let $i+6<N$ and put $\mathcal G=\prod_{j<4}\mu_i(F_j^2)$. Explicitly,
\[
\mathcal G(k_0,\dots,k_4)=(k_0+1)\bigl(k_1s_2+s_2+k_1+k_2+1\bigr)
\bigl(k_2(k_1+k_2)s_3+g_3\bigr),
\]
where $g_3=s_3(k_1+k_2)+s_3k_2+(k_1+k_2)k_2+(k_0+2k_1+3k_2)(k_3+1)+k_3^2+2k_3+k_4+1$. Then
$L_i\,q_i^4\le a\,\mathcal G(k_0,\dots,k_4)$.
\end{lemma}

\begin{proof}
The Newton polynomials $F_0,\dots,F_3$ are monic of degrees $0,\dots,3$. So the matrix
$\bigl(\nu_i(F_jF_l)\bigr)$ is $\mathcal U\mathcal H\mathcal U^{\mathsf T}$, where $\mathcal H$ is the Hankel matrix of $\nu_i$ and $\mathcal U$ is unit
lower triangular. Hence it has determinant $I_4(i)/a$, and it is positive definite, as in the
proof of Theorem~\ref{thm:hankel}. By Hadamard's inequality,
\[
\frac{I_4(i)}a\le\prod_{j<4}\nu_i(F_j^2)\le\prod_{j<4}\frac{\mu_i(F_j^2)}{q_i},
\]
since $\nu_i=\mu_i/(q_i+x)\le\mu_i/q_i$ on the nonnegative support. By the shift
rule~\eqref{eq:moment-shift} and the low moments~\eqref{eq:lowmoments}, the product
$\prod_{j<4}\mu_i(F_j^2)$ expands to the displayed formula.
\end{proof}

\subsection{Signs}\label{sec:signs}
Throughout this subsection the pattern and increments arise from the same state,
so the cofactor relations of Lemma~\ref{lem:pattern} remain in force.
Besides the polynomial $\sigma$ of Section~\ref{sec:six}, put
$\tilde\sigma(y_1,y_2,y_3,y_4)=y_1^2y_3+(2y_1-1)y_2y_3+y_1y_3^2+y_1y_4$. Both $\sigma$
and $\tilde\sigma$ are positive at positive integers.

\begin{lemma}[Signs]\label{lem:signs}
If $N_0,N_1\ge1$ and $k_0,\dots,k_4\ge1$, then $u_1>0>u_2$.
\end{lemma}

\begin{proof}
The $2\times2$ minors of the $k$- and $r$-rows do not involve the nodes: expanding
$r_m=k_m(s_m+s_{m+1})+k_{m+1}$ with $s_{m+1}=s_m+k_m$ gives
\[
 k_mr_{m+1}-k_{m+1}r_m=\sigma(k_m,k_{m+1},k_{m+2}),\qquad
 k_mr_{m+2}-k_{m+2}r_m=\tilde\sigma(k_m,k_{m+1},k_{m+2},k_{m+3}).
\]
The $k$- and $r$-relations of Lemma~\ref{lem:pattern} read
$u_1k_1+u_2k_2=N_1k_0-N_0k_3$ and $u_1r_1+u_2r_2=N_1r_0-N_0r_3$. Solving them for
$(u_1,u_2)$ by Cramer's rule and using these minors gives the exact identities
\begin{align*}
u_1\sigma(k_1,k_2,k_3)&=N_1\tilde\sigma(k_0,k_1,k_2,k_3)+N_0\sigma(k_2,k_3,k_4),\\
-u_2\sigma(k_1,k_2,k_3)&=N_1\sigma(k_0,k_1,k_2)+N_0\tilde\sigma(k_1,k_2,k_3,k_4),
\end{align*}
since the Cramer determinant is $k_1r_2-k_2r_1=\sigma(k_1,k_2,k_3)$.
\end{proof}


\subsection{Good starts}\label{sec:goodstarts}

\begin{lemma}[Good starts]\label{lem:good}
Let $b_{i_0},\dots,b_{i_0+T-1}$ be a window run in $[A,2A)$ with $T\ge36$, and put
$H=\lfloor 28a/(AT)\rfloor+1$. Call a start $i$ \emph{good} if $i_0\le i$, $i+9\le i_0+T$ and
$q_{j+1}-q_j\le H$ for $i\le j\le i+6$, that is, $k_0,\dots,k_6\le H$ at state $i$. Then at least
$T/2$ starts are good, and every good start satisfies $i+8<N$ and $b_i,b_{i+1},b_{i+2}<2A$.
\end{lemma}

\begin{proof}
Let $\mathcal B$ be the set of $j\in[i_0,i_0+T-1)$ with $q_{j+1}-q_j>H$. Then
$|\mathcal B|(H+1)\le q_{i_0+T-1}-q_{i_0}\le a/A$. Since $H\cdot AT>28a$, this gives
$28|\mathcal B|\le T$. Each $j\in\mathcal B$ spoils at most seven starts, so at least
$T-9-7|\mathcal B|\ge T/2$ starts are good, as $T\ge36$. The last assertion is immediate,
since the states $i,\dots,i+8$ lie in the window.
\end{proof}

\subsection{The full minor vector}\label{sec:minors}
The seven pattern integers are not independent: they are coordinates of a single vector of
$3\times3$ minors. If all ten minors are at most $X$ in absolute value, there are only
$X^{5+o(1)}$ such vectors (Proposition~\ref{prop:minorvec}), rather than the $X^7$ choices of seven
independent integers.
Let $\widehat{\mathcal M}_i$ be the $3\times5$ matrix with rows $(d_0,\dots,d_4)$, $(k_0,\dots,k_4)$
and $(r_0,\dots,r_4)$, and let $p_{c_1c_2c_3}$ be its minor on columns $c_1<c_2<c_3$. We define
the \emph{pattern} of a state $i$ to be the vector $p(i)$ of these ten minors. Expanding
the cofactors above shows that it determines the pattern integers:
\[
\theta(i)=(p_{012},\,p_{123},\,p_{234},\,p_{023},\,-p_{013},\,p_{134},\,-p_{124}).
\]

\begin{lemma}[Expansion]\label{lem:expand}
Let $q=q_i$ and $s_j=q_{i+j}-q$. If $i+j+3<N$ then
\[
\frac{b_{i+j}}a=\frac1q-\frac{1+s_j}{q^2}+\frac{s_j^2+2s_j+k_j+1}{q^3}-e_j,\qquad
0\le e_j\le\frac{\mu_{i+j}\bigl((x+s_j)^3\bigr)}{q^4}.
\]
\end{lemma}

\begin{proof}
Apply $\mu_{i+j}$ to $\frac1{q+y}=\frac1q-\frac y{q^2}+\frac{y^2}{q^3}-\frac{y^3}{q^3(q+y)}$ with
$y=s_j+x\ge0$, using $b_{i+j}/a=\nu_{i+j}(1)=\mu_{i+j}\bigl(1/(q+s_j+x)\bigr)$
and~\eqref{eq:lowmoments}.
\end{proof}

\begin{proposition}[All minors]\label{prop:allminors}
Let $i+8<N$, $b_i<2A$ and $k_0,\dots,k_6\le H$ at state $i$. Then every minor of $\widehat{\mathcal M}_i$
satisfies $|p|\le c_XA^4H^6/a^3$, for an absolute constant $c_X$.
\end{proposition}

\begin{proof}
Subtracting the expansions of Lemma~\ref{lem:expand} for $j$ and $j+1$ gives
$d_j/a=k_j/q^2-(r_j+k_j)/q^3+e_{j+1}-e_j$. So replacing the first row by
$d'_j=d_j-a(q^{-2}-q^{-3})k_j+aq^{-3}r_j=a(e_{j+1}-e_j)$ changes no minor. Now
$|d'_j|\ll aH^3/q^4$ by the error bound of Lemma~\ref{lem:expand}, while
$|k_j|\le H$ and $|r_j|\le10H^2$, so $|p|\ll aH^6/q^4$. Finally $a<4Aq$
(Lemma~\ref{lem:fourAq}).
\end{proof}

\subsection{Counting minor vectors}\label{sec:minorcount}

\begin{proposition}[Minor vectors]\label{prop:minorvec}
The number of vectors of minors of integer $3\times5$ matrices with all coordinates in $[-X,X]$ is
$\ll1+\sum_{\kappa\le X}\kappa^4\tau(\kappa)\ll X^{5+o(1)}$.
\end{proposition}

The minor vectors are the integral points of the affine cone over the
Grassmannian $\mathrm{Gr}(3,5)$ in its Pl\"ucker embedding, which we call the
\emph{Pl\"ucker cone}. This cone has
dimension seven, but its integral points of height at most $X$ number about
$X^5$, as for rational points of bounded height on $\mathrm{Gr}(3,5)$, whose
anticanonical class is five times the hyperplane class~\cite{FMT}. So the
exponent $5$ is optimal. The proof uses the standard affine chart of the Grassmannian.

\begin{proof}
For a nonzero vector $p$ choose a pivot $\pi$ with $\kappa=|p_\pi|$ maximal, and let $c_4,c_5$ be
the two columns outside $\pi$. Let $\mathbf u_r$ ($\mathbf v_r$) be the minor with the $r$-th pivot column
replaced by $c_4$ ($c_5$), with the sign of the column permutation. Write
$\mathcal M=\mathcal M_\pi(I\mid B)$ up to a permutation of columns, where $\mathcal M_\pi$ is the
pivot block; by Cramer's rule the two columns of $B$ are $\mathbf u/p_\pi$ and $\mathbf v/p_\pi$. The minors of
$(I\mid B)$ are $\pm1$, the entries of $B$, and the $2\times2$ minors of $B$; multiplying by
$\det\mathcal M_\pi=p_\pi$ shows that the three minors $p_{w_r}$ using both $c_4$ and $c_5$, where
$w_r$ consists of $c_4$, $c_5$ and the $r$th pivot column, satisfy
$p_\pi\,p_{w_r}=\pm(\mathbf u\times \mathbf v)_r$. So $p$ is determined by
$(\pi,\operatorname{sgn}p_\pi,\kappa,\mathbf u,\mathbf v)$, where $|\mathbf u_r|,|\mathbf v_r|\le\kappa$ and $\kappa$ divides
every coordinate of $\mathbf u\times \mathbf v$.

Fix $\kappa\ge1$ and $\mathbf u$, and let $g=\gcd(\kappa,\mathbf u_1,\mathbf u_2,\mathbf u_3)$, $\ell=\kappa/g$, $\mathbf u=g\mathbf u_0$. By
B\'ezout there are $z\in\mathbb Z^3$ and $t\in\mathbb Z$ with $z\cdot \mathbf u_0+t\ell=1$. If
$\kappa\mid \mathbf u\times \mathbf v$, then $\ell\mid \mathbf u_0\times \mathbf v$, and the identity
$\mathbf v-(z\cdot \mathbf v)\mathbf u_0=t\ell\,\mathbf v+z\times(\mathbf v\times \mathbf u_0)$ shows $\mathbf v=c\mathbf u_0+\ell y$ with $c=z\cdot \mathbf v$ and
$y\in\mathbb Z^3$. Hence $\mathbf v$ lies in one of $\ell$ classes modulo $\ell$, each meeting
$[-\kappa,\kappa]^3$ in at most $(3g)^3$ points: at most $27\kappa g^2$ choices. Summing over
$\mathbf u$, grouped by $g=e\mid\kappa$, gives
$\sum_{\mathbf u} g^2\le\sum_{e\mid\kappa}e^2(3\kappa/e)^3\le27\kappa^3\tau(\kappa)$, so at most
$729\kappa^4\tau(\kappa)$ pairs $(\mathbf u,\mathbf v)$. There are ten pivots and two signs.
\end{proof}

\section{Joint integer invariants and the optimised bound}\label{sec:joint}

We now prove Theorem~\ref{thm:joint}. Fix a window run in $[A,2A)$ with $T\ge36$
terms. Throughout this section constants may depend on fixed positive integers
$m,n$ (the powers of Section~\ref{sec:jointgraph}) and on a loss $\eta>0$, but
never on the dividend, the orbit or the pattern. The argument has six steps.
\begin{enumerate}
\item For a fixed pattern, the nodes of every good start lie on at most
$24$ curves (Section~\ref{sec:jointcurve}).
\item On these curves the quotient $q_i$ and the curvature $I_3(i)$ are
rational functions of the nodes. Components on which the reconstruction
identities fail contribute $O(1)$ starts (Section~\ref{sec:jointrecon}).
\item The directions at infinity of the curves are controlled by the
Newton coefficients of the pattern. This makes nonexceptional
components have even degree (Section~\ref{sec:jointinfinity}), so that the
orders $3e/2$ and $5e/2$ below are integers and every component left after the
conics has degree at least four; it also shows
that exceptional patterns are few, and excludes lines
(Section~\ref{sec:jointexc}).
\item The reconstructed functions, and hence two families of new integer
coordinates $W$ and $J$, have prescribed orders at infinity
(Sections~\ref{sec:jointorders} and~\ref{sec:jointcoords}).
\item Each nonexceptional conic contributes $a^{o(1)}$ starts
(Section~\ref{sec:jointconic}).
\item On the remaining components, adjoining $W_j^m$ and $J^n$ raises the
degree. The curve-counting theorem and a balancing of heights then give the
exponent (Sections~\ref{sec:jointgraph} and~\ref{sec:jointcount}).
\end{enumerate}

\emph{Notation.} In this section
\begin{itemize}
\item $\theta=(N_0,N_1,N_2,u_1,u_2,v_1,v_2)$ are the pattern integers of
Section~\ref{sec:patterns}, with $N_0=I_4(i)$, and $\rho,\varphi,\chi$ and
$\rho',\varphi',\chi'$ the normalised parameters~\eqref{eq:jointbar};
\item $s_j$ are the nodes, $k_j$ the increments, $P_4,R_4$ the two orthogonal
quartics, $\Psi$ the quartic norm and $f$ the pattern quadratic;
\item $D_{jl}=d_jk_l-d_lk_j$ are the $2\times2$ minors of drops and increments, so that
$D_{01}=I_3(i)$ is the curvature of Section~\ref{sec:curv};
\item $Q$ and $E_0$ are the reconstructed quotient and curvature,
$W=(W_0,W_1,W_2)$ and $J$ the joint coordinates, and $S$, $Y$ their height
bounds;
\item $\operatorname{ord}_P$ is the order at a point $P$, positive at zeros and
negative at poles, $e$ is a multiplicity at infinity, $d$ the degree of a component (not to be
confused with the drops $d_j$),
$\delta$ the degree of a graph and $\delta_{\min}=10n+7m$ its lower bound
(Lemma~\ref{lem:jointpowereddegree}).
\end{itemize}

\subsection{Good starts and the pattern curve}\label{sec:jointcurve}

Put
\[
 H=\left\lfloor\frac{28a}{AT}\right\rfloor+1,\qquad
 X=\left\lfloor\frac{c_XA^4H^6}{a^3}\right\rfloor,
\]
with $c_X$ the constant of Proposition~\ref{prop:allminors}. Let the good starts be as in Lemma~\ref{lem:good}; at least $T/2$ starts are good. By
Proposition~\ref{prop:allminors}, the minor vector of a good start has all
coordinates at most $X$, and $X\ge1$ because its coordinate $p_{012}=L_i$ is a
positive integer. The pattern of a good start (Section~\ref{sec:minors}) therefore has
height at most $X$, and it determines $\theta(i)$. By Proposition~\ref{prop:minorvec}, there are
$O_\eta(a^\eta X^5)$ patterns.

The \emph{nodes} of state $i$ are $s_j=q_{i+j}-q_i$. They are the support
points of the measure $\mu_i$ of Section~\ref{sec:hankel}, and the
interpolation nodes of the Newton polynomials of Section~\ref{sec:patterns}. The
three \emph{free nodes} $s_1,s_2,s_3$ of a good start are integers in $[0,3H]$.
Distinct good starts with the same pattern have distinct free-node triples:
the pattern and the free nodes determine $q_i$, which strictly increases along
the orbit. (That $q_i$ is determined is shown in Section~\ref{sec:jointrecon},
where it is the rational function $Q$; the argument there uses only
Lemma~\ref{lem:jointcurve}(1) below.)

We now write, as functions of the three free nodes, the objects that at a good
start are the orthogonal quartics of states $i$ and $i+1$ and the norm
$\nu_i(P_4^2)$; the pattern curve is cut out by the pattern quadratic and a
resultant identity among them. For the pattern integers $\theta$, put
\begin{equation}\label{eq:jointbar}
 (\rho,\varphi,\chi)=\frac{(N_1,u_1,u_2)}{N_0},\qquad
 (\rho',\varphi',\chi')=\frac{(N_2,v_1,v_2)}{N_1},
\end{equation}
the \emph{normalised parameters} of the pattern. Then $\rho>0$, and
$\varphi>0>\chi$ by Lemma~\ref{lem:signs}.
Treat the free nodes $s_1,s_2,s_3$ as coordinates and define
\[
 s_0=0,\quad
 s_4=(\rho+\varphi)s_1-(\varphi-\chi)s_2-(\chi-1)s_3,\quad
 k_j=s_{j+1}-s_j\quad(0\le j\le3),
\]
\[
 k_4=\rho' k_1-\varphi' k_2-\chi' k_3,\qquad s_5=s_4+k_4.
\]
At a good start these are the actual nodes and increments, by the two
$k$-relations of Lemma~\ref{lem:pattern}; the first is equivalent to
$\rho k_0=\varphi k_1+\chi k_2+k_3$. Let
\[
 \gamma=(\rho,\,-\rho-\varphi,\,\varphi-\chi,\,\chi-1,\,1),
\]
and let $\gamma'$ be the same expression in $\rho',\varphi',\chi'$. Put
\[
 P_4(x)=\sum_{j=0}^4\gamma_j\prod_{\ell=0}^{j-1}(x-s_\ell),
 \qquad
 R_4(x)=\sum_{j=0}^4\gamma'_j\prod_{\ell=1}^{j}(x-s_\ell).
\]
By Lemma~\ref{lem:orthquartic}, at a good start $P_4$ is the
monic orthogonal polynomial $\pi_i$ of degree four for $\nu_i$, and
$R_4(x)=\pi_{i+1}(x-k_0)$ is the corresponding polynomial of the next state,
written in the variable of the current state. Put $Z=xR_4$ and
$f=\sum_{j=0}^4\gamma_j(s_j^2+k_j)$, the \emph{pattern quadratic}, and let
\begin{align*}
 \Psi={}&\sum_{j=0}^4\gamma_j
 \bigl(s_j^3+3s_j^2+3s_j(1+k_j)+(1+k_j)^2\bigr)
 +\sum_{j=0}^3\gamma_jk_{j+1}\\
 &+\rho' r_1-\varphi' r_2-\chi' r_3-k_4(2s_4+k_4),
\end{align*}
where $r_j=k_j(s_j+s_{j+1})+k_{j+1}$ for $j=1,2,3$. This is a cubic
polynomial in the free nodes, the \emph{quartic norm}: at a good start it is
the squared norm $\nu_i(P_4^2)$ of the orthogonal quartic
(Lemma~\ref{lem:jointcurve}).

The degree of an affine curve is that of its projective closure, equivalently
the leading coefficient of its affine Hilbert function.
``Integral'' means that the defining ideal is prime over $\mathbb Q$; it does
not assert geometric irreducibility or a rational parametrisation.
For a monic polynomial $g$ we write
$\operatorname{Res}(g,h)=\prod_{g(\theta)=0}h(\theta)$, the product over the
roots of $g$ with multiplicity. For monic $g,h$ of degrees $m,n$ this is
$(-1)^{mn}\operatorname{Res}(h,g)$, and
$\operatorname{Res}(g,h)=\operatorname{Res}(g,h+ug)$ for every polynomial $u$.

\begin{lemma}[The pattern curve]\label{lem:jointcurve}
\begin{enumerate}
\item At a good start $i$, $\Psi=\nu_i(P_4^2)>0$, and the free nodes satisfy
\begin{equation}\label{eq:jointcurve}
 f=0,\qquad N_0\operatorname{Res}(P_4,Z)=a\Psi^4.
\end{equation}
We call the second equation the \emph{fixed-dividend equation}.
\item For every pattern with $\rho,\varphi>0>\chi$, the two
equations~\eqref{eq:jointcurve} have degrees two and at most twelve and no
common factor. Their common zero set in $\mathbb A^3$ is the union of at most
twenty-four integral affine curves over $\mathbb Q$, of total degree at most
twenty-four.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) Since $r_m=s_{m+1}^2-s_m^2+k_{m+1}$, summation by parts and the
$k$-relation at state $i$ show that $N_0f$ is the $r$-relation
$\sum_mU_m(i)r_m=0$ of Lemma~\ref{lem:pattern}. Next, the shift rule
$\mu_i\bigl(g\prod_{\ell<j}(x-s_\ell)\bigr)=\mu_{i+j}\bigl(g(x+s_j)\bigr)$ and
the low moments~\eqref{eq:lowmoments} give
$\mu_i(x^3P_4)=\sum_j\gamma_j\bigl(s_j^3+3s_j^2+3s_j(1+k_j)+(1+k_j)^2+k_{j+1}\bigr)$.
Eliminating $k_5$ with the $r$-relation at state $i+1$, together with the
$k$-relation at $i+1$ that defines $k_4$, turns this into $\Psi$.
Since $\nu_i(xg)=\mu_i(g)-q_i\nu_i(g)$ and $P_4$ is orthogonal to $x^3$,
$\nu_i(P_4^2)=\nu_i(x^4P_4)=\mu_i(x^3P_4)=\Psi$, and this is positive as in
the proof of Theorem~\ref{thm:hankel}.

For the resultant, Lemma~\ref{lem:moments}(3) gives
$\nu_i\bigl(xg(x)\bigr)=\nu_{i+1}\bigl(g(x+k_0)\bigr)$. Hence, for $j<4$,
$\nu_i(x^jZ)=\nu_{i+1}\bigl((x+k_0)^j\pi_{i+1}(x)\bigr)=0$. So the monic
quintic $Z$ differs from the monic orthogonal polynomial $\pi^{(5)}$ of degree
five for $\nu_i$ by a multiple of $P_4$, and $\operatorname{Res}(P_4,Z)=
\operatorname{Res}(P_4,\pi^{(5)})$. Let $\|\pi^{(j)}\|^2$ be the squared $\nu_i$-norm of
the monic orthogonal polynomial $\pi^{(j)}$ of degree $j$, so $\pi^{(4)}=P_4$. The
three-term recurrence
$\pi^{(j+1)}=(x-\mathfrak a_j)\pi^{(j)}-\mathfrak b_j\pi^{(j-1)}$, with
$\mathfrak b_j=\|\pi^{(j)}\|^2/\|\pi^{(j-1)}\|^2$, gives
$\pi^{(j+1)}\equiv-\mathfrak b_j\pi^{(j-1)}\pmod{\pi^{(j)}}$, so
$\operatorname{Res}(\pi^{(j)},\pi^{(j+1)})
=(-\mathfrak b_j)^j(-1)^{j(j-1)}\operatorname{Res}(\pi^{(j-1)},\pi^{(j)})$.
Reducing from $j=4$ down to $j=1$, the signs multiply to $(-1)^{4+3+2+1}=1$, and
\[
 \operatorname{Res}(P_4,\pi^{(5)})=\prod_{j=1}^4\mathfrak b_j^j
 =\frac{\|P_4\|^8}{\prod_{j<4}\|\pi^{(j)}\|^2}
 =\frac{\Psi^4}{\det\bigl(\nu_i(x^{j+l})\bigr)_{j,l<4}}
 =\frac{a\Psi^4}{I_4(i)}.
\]
Since $N_0=I_4(i)$, this is the second equation.

(2) The first equation has degree two. For the second, $\Psi$ is cubic, and we
show that the resultant has total degree at most ten in the nodes. Since
$Z=xR_4$ and the product of the roots of $P_4$ is $P_4(0)=\gamma_0$ (every term
with $j\ge1$ has the factor $x-s_0=x$),
$\operatorname{Res}(P_4,Z)=\gamma_0\operatorname{Res}(P_4,R_4)$.

We use the Newton basis $N_\alpha(x)=\prod_{1\le l\le\alpha}(x-s_l)$ at the
nodes $s_1,\dots,s_4$. From $F_{j+1}=xN_j=N_{j+1}+s_{j+1}N_j$,
\[
 P_4=\sum_{j\le4}a_jN_j,\qquad R_4-P_4=\sum_{\alpha\le3}c_\alpha N_\alpha,\qquad
 a_j=\gamma_j+\gamma_{j+1}s_{j+1},\quad c_\alpha=\gamma'_\alpha-a_\alpha,
\]
with $\gamma_5=0$, so $a_4=1$; every $a_j$ and $c_\alpha$ has degree at most one
in the nodes, with top-degree parts $[a_j]_1=\gamma_{j+1}s_{j+1}$ and
$[c_\alpha]_1=-\gamma_{\alpha+1}s_{\alpha+1}$. The \emph{B\'ezout matrix}
$\mathcal B=(\mathcal B_{\alpha\beta})_{\alpha,\beta<4}$ is defined by
\begin{equation}\label{eq:bezout}
 \frac{P_4(x)R_4(y)-P_4(y)R_4(x)}{x-y}
 =\sum_{\alpha,\beta<4}\mathcal B_{\alpha\beta}N_\alpha(x)N_\beta(y).
\end{equation}
Its determinant is $\operatorname{Res}(P_4,R_4)$: in the monomial basis this is
the classical B\'ezout formula for two monic quartics, and the change to the
Newton basis is unitriangular in both variables. Since
$R_4=P_4+\sum_\alpha c_\alpha N_\alpha$, the left side of~\eqref{eq:bezout} is
$\sum_{j>k}(a_jc_k-a_kc_j)K_{jk}$, where $c_4=0$ and
$K_{jk}=(N_j(x)N_k(y)-N_j(y)N_k(x))/(x-y)$. For $j>k$, telescoping
$\prod_{l=k+1}^j(x-s_l)-\prod_{l=k+1}^j(y-s_l)$ gives
\begin{equation}\label{eq:bezoutK}
 K_{jk}=\sum_{m=k+1}^{j}N_{m-1}(x)\,Y_{kmj}(y),\qquad
 Y_{kmj}(y)=N_k(y)\prod_{l=m+1}^{j}(y-s_l).
\end{equation}
The polynomial $Y_{kmj}$ is monic of degree $k+j-m$, homogeneous of that degree
in $y$ and the nodes, and divisible by $N_k$. So its coefficient of $N_\beta$
is homogeneous of degree $k+j-m-\beta$ in the nodes, and vanishes unless
$\beta\ge k$. Thus the triple $(j,k,m)$ contributes to $\mathcal B_{\alpha\beta}$
only when $\alpha=m-1$ and $k\le\min(\alpha,\beta)$, and its contribution has
degree at most $\deg(a_jc_k-a_kc_j)+k+j-1-\alpha-\beta$. For $j=4$ we have
$a_4c_k-a_kc_4=c_k$, of degree at most one, and otherwise $j\le3$; in both cases
the degree is at most $4+k-\alpha-\beta\le4-\max(\alpha,\beta)$. Hence
\begin{equation}\label{eq:bezoutdeg}
 \deg\mathcal B_{\alpha\beta}\le4-\max(\alpha,\beta),
\end{equation}
and each term $\prod_\alpha\mathcal B_{\alpha\sigma(\alpha)}$ of the determinant
has degree at most $\sum_\alpha(4-\max(\alpha,\sigma(\alpha)))\le10$, with
equality possible only for the identity permutation. Note that $\gamma_1<0<\gamma_2$, $\gamma_3=\chi-1<-1$ and
$\gamma_0,\gamma_4>0$. Since $s_0=0$ and $\sum_j\gamma_js_j=0$, the quadratic part of $f$ is
\[
 f_2=\gamma_1s_1^2+\gamma_2s_2^2+\gamma_3s_3^2+(\gamma_1s_1+\gamma_2s_2+\gamma_3s_3)^2,
\]
whose Hessian determinant is $8\gamma_1\gamma_2\gamma_3(1+\gamma_1+\gamma_2+\gamma_3)
=-8\gamma_0\gamma_1\gamma_2\gamma_3\ne0$. A ternary quadratic form of rank three is not a
product of linear forms, so $f_2$, and hence $f$, is irreducible. The degree-twelve part of
the second equation is $-ac^4$ with $c=\sum_j\gamma_js_j^3$. If $f$ divided the second
equation, the irreducible form $f_2$ would divide $c^4$, hence $c$. Here
$s_4=-(\gamma_1s_1+\gamma_2s_2+\gamma_3s_3)$. Divide $c$ by $f_2$ as
polynomials in $s_3$; the $s_3^2$ coefficient of $f_2$ is $\gamma_3(1+\gamma_3)\ne0$. The
remainder is $\mathfrak r_1s_3+\mathfrak r_0$, with $\mathfrak r_0,\mathfrak r_1$ forms in $s_1,s_2$ and
\[
 \mathfrak r_1=\frac{\gamma_3-1}{\gamma_3+1}\bigl(\gamma_1(\gamma_1+\gamma_3+1)s_1^2
 +2\gamma_1\gamma_2s_1s_2+\gamma_2(\gamma_2+\gamma_3+1)s_2^2\bigr),
\]
as one checks by multiplying out the quotient
$-\bigl((\gamma_3+2)(\gamma_1s_1+\gamma_2s_2)+(\gamma_3^2-1)s_3\bigr)/(\gamma_3+1)$.
The $s_1s_2$ coefficient $2\gamma_1\gamma_2(\gamma_3-1)/(\gamma_3+1)$ of $\mathfrak r_1$ is
nonzero by the signs above, so $f_2\nmid c$. So the two
equations have no common factor, their common zero set is a curve, and by B\'ezout's theorem
its components over $\mathbb Q$ have total degree at most $2\cdot12$.
\end{proof}

The following example shows why the fixed-dividend equation is needed. If $(\rho,\varphi,\chi)=(\rho',\varphi',\chi')=(\vartheta,2\vartheta+1,-\vartheta-2)$ with $\vartheta>0$, then
$\sum_l\gamma_lz^l=(z-1)^3(z-\vartheta)$, and constant increments $k_j=t$ satisfy the
linear and quadratic relations for every $t$: since $z=1$ is a triple root,
$\sum_l\gamma_ll^r=0$ for $r\le2$, and with $s_j=jt$ these sums are exactly the
$k$- and $r$-relations.


\subsection{Reconstruction and retained components}\label{sec:jointrecon}

We need $q_i$ and $I_3(i)$ as rational functions on each component, with
computable orders at infinity. This works only on components where two further
identities hold identically; on any other component they cut out boundedly many
good starts, which we count directly. The idea is that the polynomials $P_4$ and $Z$, both orthogonal to
$1,x,x^2,x^3$ in $L^2(\nu_i)$, determine the
moment vector of $\nu_i$ up to scale, and the two normalised division
equations fix the scale; this recovers $q_i$ and $I_3(i)$ as rational
functions of the free nodes. Concretely, form the $8\times9$ matrix of
coefficients of $P_4,xP_4,x^2P_4,x^3P_4,Z,xZ,x^2Z,x^3Z$ in the basis
$1,x,\dots,x^8$. Let $c_j$ be $(-1)^j$ times the minor obtained by
deleting column $j$, and define
\[
 Q=\frac{c_1-c_2}{c_1-c_0},\qquad
 E_0=a\,\frac{\det(c_{r+s})_{0\le r,s<3}(c_1-c_0)^3}
                  {(c_1^2-c_0c_2)^3}.
\]
At a good start the matrix annihilates the moment vector
$(h_i(0),\dots,h_i(8))$ and has rank eight. Indeed, the resultant in
\eqref{eq:jointcurve} is positive, so $P_4$ and $Z$ are coprime.
So the kernel is spanned by $(c_j)$, and $h_i(j)=\lambda c_j$ for some $\lambda$.
The normalised division equations $q_ih_i(0)+h_i(1)=q_ih_i(1)+h_i(2)=1$ then
determine $q_i$ and $\lambda$, and $h_i(0)h_i(2)>h_i(1)^2$ shows that the
denominators below are nonzero. This gives
\[
 h_i(j)=\frac{c_j(c_1-c_0)}{c_1^2-c_0c_2}\quad(0\le j\le8),\qquad
 Q=q_i,\qquad E_0=I_3(i).
\]
In particular, all displayed denominators are nonzero at good starts.

To state the identities as equations on the curve, use the raw-moment
polynomials
\[
 M_0=1,\qquad
 M_{r+1}(k_0,k_1,\dots)
 =\sum_{t=0}^{r}\binom rt k_0^{r-t}M_t(k_1,k_2,\dots).
\]
Only $M_0,\dots,M_6$ are needed; they involve $k_0,\dots,k_4$.
At good starts, $M_r=\mu_i(x^r)$ for $0\le r\le6$, by the
weight recursion in Section~\ref{sec:hankel}. Set
\[
 H_3=\det(M_{r+s})_{0\le r,s<3},\qquad
 H_4=\det(M_{r+s})_{0\le r,s<4},\qquad
 H_{013}=\det(M_{r+s})_{\substack{r=0,1,2\\s=0,1,3}}.
\]
This $H_4$ is the polynomial of Lemma~\ref{lem:multid}. With
$p_3=\chi-1-s_1-s_2-s_3$, the coefficient of $x^3$ in $P_4$, consider the
\emph{normalisation} and \emph{resolvent identities}
\begin{equation}\label{eq:jointcuts}
 N_0P_4(-Q)=aH_4,\qquad
 E_0P_4(-Q)=a\bigl((Q-p_3)H_3-H_{013}\bigr).
\end{equation}
The first is Lemma~\ref{lem:multid} at good starts, where $Q=q_i$ and
$P_4=\pi_i$; the second is the following analogue for $I_3$.

\begin{lemma}[The resolvent identity]\label{lem:resolvent}
At a good start, $I_3(i)\,P_4(-q_i)=a\bigl((q_i-p_3)H_3-H_{013}\bigr)$.
\end{lemma}

\begin{proof}
This is the bordering argument of Lemma~\ref{lem:multid}, one size smaller.
Write $h_j=h_i(j)$, $M_j=\mu_i(x^j)=q_ih_j+h_{j+1}$ and
$P_4(x)=x^4+p_3x^3+p_2x^2+p_1x+p_0$. Since $P_4$ is orthogonal to $1,x,x^2$,
expanding along the last row gives
\[
 \det(h_{r+s})_{r,s<3}\,P_4(x)=
 \det\begin{pmatrix}
 h_j&h_{j+1}&h_{j+2}&h_{j+4}+p_3h_{j+3}\\
 1&x&x^2&x^4+p_3x^3
 \end{pmatrix}_{j=0,1,2}.
\]
At $x=-q_i$, add $q_i$ times column $1$ to column $2$, then $q_i$ times
column $0$ to column $1$; the first three columns of row $j$ become
$h_j,M_j,M_{j+1}$, and the last row becomes $(1,0,0,q_i^4-p_3q_i^3)$. Using
$h_{s+1}=M_s-q_ih_s$ repeatedly, the last column equals
$M_{j+3}+(p_3-q_i)M_{j+2}$ plus a combination of the columns $M_j$, $M_{j+1}$
and $q_i^3(q_i-p_3)$ times the column $(h_j;1)$; removing these leaves
$(M_{j+3}+(p_3-q_i)M_{j+2};0)$. Expanding along the last row gives
\[
 \det(h_{r+s})_{r,s<3}\,P_4(-q_i)=(q_i-p_3)H_3-H_{013},
\]
and $a\det(h_{r+s})_{r,s<3}=I_3(i)$.
\end{proof}

So both identities~\eqref{eq:jointcuts} hold at good starts. Clearing
denominators turns them into polynomial equations $\mathcal C_N=0$ and
$\mathcal C_R=0$ in the free nodes, the \emph{cuts} (written out in the proof
below). A component containing a good start is \emph{retained} if both cuts
vanish identically on it.

\begin{lemma}[Removing components by polynomial equations]\label{lem:jointcuts}
The cuts have degree at most an absolute constant $C_{\rm cut}$ in the free
nodes. On a degree-$d$ component, each cut either vanishes identically or
vanishes at no more than $C_{\rm cut}d$ good starts. Consequently each
component that is not retained contributes $O(1)$ good starts.
\end{lemma}
\begin{proof}
Each entry of the $8\times9$ coefficient matrix has degree at most four in
the nodes, so the cofactors $c_j$ have bounded degree. Write
$Q=Q_{\mathrm n}/Q_{\mathrm d}$ and $E_0=E_{\mathrm n}/E_{\mathrm d}$ using the
formulas above, and clear denominators:
\begin{align*}
 \mathcal C_N&=N_0\widehat P-aH_4Q_{\mathrm d}^4,\\
 \mathcal C_R&=E_{\mathrm n}\widehat P
 -a\bigl((Q_{\mathrm n}-p_3Q_{\mathrm d})H_3-H_{013}Q_{\mathrm d}\bigr)Q_{\mathrm d}^3E_{\mathrm d},
\end{align*}
where $\widehat P=Q_{\mathrm d}^4P_4(-Q_{\mathrm n}/Q_{\mathrm d})$. Every factor
here is a fixed polynomial expression in the nodes and the pattern
coordinates, so both have bounded degree in the nodes.

If a polynomial of degree $r$ does not vanish identically on an integral
curve of degree $d$, its zero set on that curve has at most $rd$ distinct
rational points, by B\'ezout's theorem. Every good start satisfies both
cleared equations, and distinct good starts of this pattern give distinct
free-node triples (Section~\ref{sec:jointcurve}). Applying the point bound to
whichever equation is not an identity proves the claim; the bound $d\le24$
makes it uniform.
\end{proof}

\subsection{Directions at infinity}\label{sec:jointinfinity}

At a point at infinity the leading nodes satisfy $\sum_j\gamma_jz_j^r=0$ for
$r\le3$, and if some of them coincide, the $\gamma_j$ in each group of equal
leading nodes sum to zero (Lemma~\ref{lem:jointleading}). Excluding vanishing
pair sums therefore makes the five leading nodes distinct, which fixes the
orders at infinity of everything below.
Call a pattern \emph{exceptional} if two of the five Newton coefficients
$\gamma_j$ sum to zero. Since $\rho>0$, $\varphi>0$ and $\chi<0$,
every $\gamma_j$ is nonzero, and a pair sum can vanish only in one of the
four cases
\begin{equation}\label{eq:jointexceptional}
 \rho+\chi=1,\qquad \rho+\chi=0,\qquad \varphi=1,\qquad \rho+\varphi=1.
\end{equation}
Equivalently, the integer pattern coordinates satisfy
$N_1+u_2=N_0$, $N_1+u_2=0$, $u_1=N_0$ or $N_1+u_1=N_0$.

\begin{lemma}[Leading nodes and determinant orders]\label{lem:jointleading}
Let $C$ be an integral component of positive degree of
\eqref{eq:jointcurve}. Let $P$ be a closed point above infinity on the
normalisation of the projective closure of $C$, write
$\operatorname{ord}=\operatorname{ord}_P$, and let $e>0$ be the
multiplicity of the hyperplane at infinity at $P$; equivalently,
$e=\max_{1\le j\le3}(-\operatorname{ord}s_j)$.
For a local parameter (uniformiser) $\varpi$ at $P$, let $z_j\in\mathbb Q(P)$ be the value
at $P$ of the regular function $\varpi^es_j$, $0\le j\le4$; we call $z_j$ the
\emph{leading node}.
Then the $z_j$ are not all equal and satisfy
\begin{equation}\label{eq:joint-leading-moments}
 \sum_{j=0}^4\gamma_jz_j^r=0\qquad(0\le r\le3).
\end{equation}
If two of these five leading nodes coincide, the pattern is exceptional.
For a nonexceptional pattern, therefore,
\begin{align}
 \operatorname{ord}(s_l-s_j)&=-e\quad(0\le j<l\le4),\label{eq:joint-node-orders}\\
 \operatorname{ord}\operatorname{Res}(P_4,Z)&=-10e,\qquad
 \operatorname{ord}H_3=-3e,\qquad \operatorname{ord}H_4=-6e.\label{eq:joint-determinant-orders}
\end{align}
\end{lemma}

\begin{proof}
All nonzero rational constants have order zero. By the description of $e$ in
the statement, each $\varpi^es_j$ is regular at $P$, and $s_4$ is a linear
combination of $s_1,s_2,s_3$. Thus a polynomial of total node degree at most $r$
has order at least $-re$. The values $z_j$ lie in the residue field $\mathbb Q(P)$; we do not
require $P$ to be rational.
We have $z_0=0$, and at least one of $z_1,z_2,z_3$ is nonzero.

\emph{The leading nodes are distinct.}
The definitions of the nodes give the moment equations of orders zero
and one. The highest-degree part of $f=0$ gives the equation of order two.
Finally, for the equation of order three, multiply the fixed-dividend
equation by $\varpi^{12e}$ and evaluate at $P$: the left side vanishes because
the resultant has total degree at most ten, and the right side is
$a(\sum_j\gamma_jz_j^3)^4$, since the cubic part of $\Psi$ is
$\sum_j\gamma_js_j^3$. This gives~\eqref{eq:joint-leading-moments}.

If two nodes coincided, there would be at most four distinct values among
the $z_j$. Grouping the sums in \eqref{eq:joint-leading-moments} by those
values and using the corresponding Vandermonde matrix shows that the
sum of the $\gamma_j$ in each group is zero.
Since there are five indices and at least two groups, some group has at
most two elements. It cannot have one, because every $\gamma_j$ is nonzero;
it must therefore give a vanishing pair sum. This proves the collision
assertion. For a nonexceptional pattern it is impossible, so all five
leading nodes are distinct, which is~\eqref{eq:joint-node-orders}.

\emph{The determinant orders.}
Write $\mathcal R=\operatorname{Res}(P_4,Z)$.
The leading homogeneous parts needed below are
\begin{align}
 [\mathcal R]_{10}
   &=\gamma_0\gamma_1\gamma_2\gamma_3
        \prod_{0\le j<l\le4}(s_l-s_j), \label{eq:joint-resultant-leading}\\
 [H_3]_3&=k_0k_1(k_0+k_1),\notag\\
 [H_4]_6&=k_0k_1k_2(k_0+k_1)(k_1+k_2)(k_0+k_1+k_2).
 \label{eq:joint-hankel-leading}
\end{align}
Here $[F]_r$ denotes the homogeneous part of total node degree $r$.

For the degree-ten formula we use the B\'ezout matrix of the proof of
Lemma~\ref{lem:jointcurve}(2). By~\eqref{eq:bezoutdeg}, only the diagonal
term of its determinant can have degree ten, so
$[\operatorname{Res}(P_4,R_4)]_{10}=\prod_{\alpha<4}[\mathcal B_{\alpha\alpha}]_{4-\alpha}$.
By the degree count there, a contribution of degree $4-\alpha$ to
$\mathcal B_{\alpha\alpha}$ needs $k=\alpha$, hence $m=\alpha+1$, and $j\in\{3,4\}$.
The coefficient of $N_\alpha$ in $Y_{\alpha,\alpha+1,j}=N_\alpha\prod_{l=\alpha+2}^{j}(y-s_l)$
is $\prod_{l=\alpha+2}^{j}(s_{\alpha+1}-s_l)$. For $j=3$ the degree-two part of
$a_3c_\alpha-a_\alpha c_3$ is
$s_4(-\gamma_{\alpha+1}s_{\alpha+1})-\gamma_{\alpha+1}s_{\alpha+1}(-s_4)=0$, so only
$j=4$ contributes, and
\[
 [\mathcal B_{\alpha\alpha}]_{4-\alpha}=[c_\alpha]_1\prod_{l=\alpha+2}^{4}(s_{\alpha+1}-s_l)
 =-\gamma_{\alpha+1}s_{\alpha+1}\prod_{l=\alpha+2}^{4}(s_{\alpha+1}-s_l).
\]
The product of these four entries is
$\gamma_1\gamma_2\gamma_3\prod_{0\le j<l\le4}(s_l-s_j)$, recalling $s_0=0$ and $\gamma_4=1$, and
$\operatorname{Res}(P_4,Z)=\gamma_0\operatorname{Res}(P_4,R_4)$
gives~\eqref{eq:joint-resultant-leading}.

For the Hankel polynomials, let $\mathcal L^{(l)}$ be the linear functional on
polynomials in $x$ with $\mathcal L^{(l)}(x^r)=M_r(k_l,k_{l+1},\dots)$. The
recurrence defining the $M_r$ says exactly that
$\mathcal L^{(l)}(xg(x))=\mathcal L^{(l+1)}(g(x+k_l))$, so, as for $\mu_i$ in
Section~\ref{sec:hankel},
$\mathcal L^{(0)}(F_jF_l)=\mathcal L^{(l)}\bigl(\prod_{m<j}(x+s_l-s_m)\bigr)$ for
$j\le l$. Since $M_r$ has degree $r-1$ for $r\ge1$ and $M_0=1$, the coefficient
of $x^t$ contributes degree at most $j-1$ when $t\ge1$, and
\begin{equation}\label{eq:gramtop}
 \deg\mathcal L^{(0)}(F_jF_l)\le\min(j,l),\qquad
 [\mathcal L^{(0)}(F_jF_l)]_j=\prod_{m<j}(s_l-s_m)\quad(j\le l).
\end{equation}
The unitriangular change from $1,x,x^2,x^3$ to $F_0,\dots,F_3$ shows that
$H_4=\det\bigl(\mathcal L^{(0)}(F_jF_l)\bigr)_{j,l<4}$. By~\eqref{eq:gramtop},
only the diagonal term can have degree six, so
$[H_4]_6=\prod_{l<4}\prod_{m<l}(s_l-s_m)$, which is the displayed product since
$s_1=k_0$, $s_2=k_0+k_1$ and $s_3=k_0+k_1+k_2$. For $H_3$ the same Gram matrix
with $j,l<3$ is explicit,
\[
 \begin{pmatrix}1&1&1\\1&1+s_1&1+s_2\\1&1+s_2&k_2+1+2s_2-s_1+s_2(s_2-s_1)\end{pmatrix},
 \qquad\text{so}\qquad
 H_3=k_0k_1(k_0+k_1)+k_0k_2-k_1^2.
\]
Finally, $x^3=F_3+(s_1+s_2)x^2-s_1s_2x$; in the minor $H_{013}$ the multiple of
the column of $x$ can be removed, so
$H_{013}=(s_1+s_2)H_3+\det\bigl(\mathcal L^{(0)}(F_jF_l)\bigr)_{j<3,\,l\in\{0,1,3\}}$,
and by~\eqref{eq:gramtop} the last determinant has degree at most three. Hence
\begin{equation}\label{eq:H013top}
 [H_{013}]_4=(s_1+s_2)[H_3]_3=(2k_0+k_1)k_0k_1(k_0+k_1).
\end{equation}

Every factor in these leading products has order $-e$ by
\eqref{eq:joint-node-orders}. The lower-degree terms have strictly larger
order, so they cannot cancel the leading terms, which
proves~\eqref{eq:joint-determinant-orders}.
\end{proof}

\begin{corollary}[Even component degrees]\label{cor:jointparity}
Every positive-degree nonexceptional component of \eqref{eq:jointcurve}
has even degree. At each point above infinity its hyperplane multiplicity
$e$ is even, and
$\operatorname{ord}\Psi=-5e/2$; so the function $J=N_0\Psi$ of
Section~\ref{sec:jointcoords} also has order $-5e/2$.
\end{corollary}
\begin{proof}
Taking orders in the fixed-dividend equation and using
Lemma~\ref{lem:jointleading} gives
$4\operatorname{ord}\Psi=-10e$.
Since orders are integers and $N_0\ne0$, this proves both local assertions.
The degree of the curve is the sum of the multiplicities $e$, weighted
by the residue-field degrees, so it is even. The retained identities
are not needed here.
\end{proof}

\subsection{Exceptional patterns and line components}\label{sec:jointexc}

Exceptional patterns may have components of odd degree, including lines, but
there are fewer of them. Lines must be excluded because the curve bound saves
nothing on them. We exclude them only in the range~\eqref{eq:jointscale}
below. Outside that range, $c_{\rm L}A^4H^5X\ge a^3$ together with
$X\asymp A^4H^6/a^3$ and $H\asymp a/(AT)$ already gives
$T^{17}\le a^{5+o(1)}$ (first case in the proof of
Corollary~\ref{cor:jointexponent}).

\begin{lemma}[Exceptional patterns and lines]\label{lem:jointcomponents}
There are $O(X^4\log(2X))=O_\eta(a^\eta X^4)$ exceptional patterns. In the range
\begin{equation}\label{eq:jointscale}
 c_{\rm L} A^4H^5X<a^3,
\end{equation}
where $c_{\rm L}$ is a suitable absolute constant, no line component contains a good start.
\end{lemma}
\begin{proof}
Each relation in \eqref{eq:jointexceptional} becomes a vanishing maximal
minor after a fixed invertible integer column change. The transformed
minor vector still has height $O(X)$. For example, if the original columns
are $C_0,\dots,C_4$, then
\[
 \det(C_1,C_2+C_0,C_3-C_0)=p_{123}-p_{013}-p_{012}
                              =N_1+u_2-N_0.
\]
The other three relations follow by the same column additions and
permutations.

A fixed zero minor reduces the Pl\"ucker relations to a simple shape by
Grassmann duality. Up to sign, the ten minors $p_{c_1c_2c_3}$ are the Pl\"ucker
coordinates $p^\perp_{c_4c_5}$, $\{c_1,\dots,c_5\}=\{0,\dots,4\}$, of the orthogonal
complement of the row space, a $2$-plane in $\mathbb Q^5$. After permuting
columns we may take the vanishing minor to be $p^\perp_{34}=0$. The Pl\"ucker
relations $p^\perp_{c_1c_2}p^\perp_{c_3c_4}-p^\perp_{c_1c_3}p^\perp_{c_2c_4}+p^\perp_{c_1c_4}p^\perp_{c_2c_3}=0$ then say exactly that
the integer vectors
\[
 \mathbf x=(p^\perp_{03},p^\perp_{13},p^\perp_{23}),\qquad \mathbf y=(p^\perp_{04},p^\perp_{14},p^\perp_{24}),\qquad
 \mathbf z=(p^\perp_{12},-p^\perp_{02},p^\perp_{01})
\]
satisfy $\mathbf x\times\mathbf y=0$ and
$\mathbf x\cdot\mathbf z=\mathbf y\cdot\mathbf z=0$. They have height $O(X)$
and are signed rearrangements of the nine remaining minors, so they determine
the minor vector.
The case $\mathbf x=\mathbf y=0$ has $O(X^3)$ choices. Otherwise write
$\mathbf x=\lambda_1\mathbf v$, $\mathbf y=\lambda_2\mathbf v$ with $\mathbf v$ primitive.
In a dyadic range $R\le|\mathbf v|_\infty<2R$, there are $O(R^3)$
choices for $\mathbf v$ and $O((X/R)^2)$ for $(\lambda_1,\lambda_2)$.
The equation $\mathbf v\cdot\mathbf z=0$ has $O(X^2/R)$ solutions
in the box. Indeed, take a coordinate of $\mathbf v$ of absolute value
$|\mathbf v|_\infty\in[R,2R)$. Primitivity gives $|\mathbf v|_\infty$ admissible pairs
of the other two coordinates of $\mathbf z$ modulo $|\mathbf v|_\infty$, and the
remaining coordinate is then determined. Since $|\mathbf v|_\infty\ll X$, this gives
$O(X^2/R)$ choices.
Each range therefore contributes $O(X^4)$, and summing over $O(\log(2X))$
ranges proves the pattern count. Since $X\le a^{O(1)}$, the logarithm is
absorbed into $a^\eta$.

For the second claim the strategy is as follows: a line through a good start
forces the pattern to be exceptional and the nodes to satisfy one of four
special relations; each relation makes some increment at most $2X$, and then
the label bound makes $q_i$ too small.
Now suppose that a line contained in the zero set of~\eqref{eq:jointcurve} passes
through the free-node triple $\mathbf s$ of a good start, with direction $w$. Extend $w$ to the five
nodes linearly, as in Section~\ref{sec:jointcurve}, and write $\mathbf s(t)=\mathbf s+tw$. Both equations
of~\eqref{eq:jointcurve} hold identically in $t$.
\begin{enumerate}
\item The resultant has degree at most ten in $t$, so the $t^{12}$ coefficient of the second
equation, $-a\bigl(\sum_j\gamma_jw_j^3\bigr)^4$, vanishes. Then $\Psi(\mathbf s(t))$ has degree at most
two, so $\Psi^4$ has degree at most eight, and the $t^{10}$ coefficient
$[\mathcal R]_{10}(w)$ of the resultant, given by~\eqref{eq:joint-resultant-leading}, vanishes
too: two coordinates of $w$ coincide.
\item The direction satisfies the four moment equations of Lemma~\ref{lem:jointleading}: those
of orders zero and one by the definition of the nodes, order two from the $t^2$ coefficient of
$f$, and order three from step~(1). The grouping argument in the proof of
Lemma~\ref{lem:jointleading} then shows that the pattern is exceptional, with exactly two
groups of equal coordinates, of sizes two and three. After scaling, and in the order
of~\eqref{eq:jointexceptional}, $w$ is one of
\[
 (0,1,1,0,1),\quad (0,1,1,0,0),\quad
 (0,0,1,1,0),\quad (0,1,0,0,1).
\]
\item For each of these four directions the resultant has degree at most six in $t$. Here the
B\'ezout matrix of~\eqref{eq:bezout} is not fine enough, because its degree bound ignores the
coinciding nodes. Instead write the B\'ezout matrix of $P_4$ and $R_4$ in a Newton basis
$\prod_{l<\alpha}(x-n_l)$ whose nodes $n_0,n_1,n_2$ are adapted to the direction. Along the
line, its $(\alpha,\beta)$ entry then has $t$-degree at most $\varrho_\alpha+\kappa_\beta$, with
weights summing to six, so every term of the determinant has degree at most six.
Appendix~\ref{app:linecertificates} lists the nodes and weights; the bounds hold for all values
of the $\gamma_j$ and $\gamma'_j$. Hence $\Psi^4$ has degree at most six, $\Psi(\mathbf s(t))$
is affine in $t$, and its second difference $\Psi(\mathbf s+w)+\Psi(\mathbf s-w)-2\Psi(\mathbf s)$
vanishes.
\item In each case, imposing the corresponding relation~\eqref{eq:jointexceptional} and expanding
the cubic $\Psi$ gives the polynomial identity
\[
 \Psi(\mathbf s+w)+\Psi(\mathbf s-w)-2\Psi(\mathbf s)=E_w(\mathbf s)+c_w\,\nabla f(\mathbf s)\cdot w,
\]
in the free nodes and the pattern parameters, with $c_w=4,2,2,4$ and respectively
\[
 E_w=-2\rho s_3,\qquad 2(\rho+\varphi)(k_1-1),\qquad 2(\chi-1)(k_2-1),\qquad
 2\bigl((\varphi-\chi)-(s_4-s_1)\bigr).
\]
The quadratic part of $f$ vanishes at $w$ (the moment equation of order two), so
$f(\mathbf s\pm w)=f(\mathbf s)\pm\nabla f(\mathbf s)\cdot w$, and $f=0$ along the line gives
$\nabla f(\mathbf s)\cdot w=0$. Hence $E_w(\mathbf s)=0$.
The factors $\rho$, $\rho+\varphi$ and $\chi-1$ are nonzero, so respectively
\[
 s_3=0,\qquad k_1=1,\qquad k_2=1,\qquad s_4-s_1=\varphi-\chi.
\]
\end{enumerate}
The first case is impossible at a good start, where $s_3\ge3$. Next bound the polynomial
$\mathcal G$ of Lemma~\ref{lem:labels}. At a good start $1\le k_j\le H$ and $s_j\le jH$, so its
three factors satisfy $k_0+1\le2H$,
\[
 k_1s_2+s_2+k_1+k_2+1\le k_1(2H+2)+2H+1\le7Hk_1\le7H^2,
\]
and $k_2(k_1+k_2)s_3+g_3\le6H^2k_2+28H^2\le34H^2k_2\le34H^3$, since the five groups of terms
of $g_3$ are at most $6H^2$, $3H^2$, $2H^2$, $12H^2$ and $5H^2$ in turn (the last for
$k_3^2+2k_3+k_4+1$). Hence
\[
 \mathcal G\le476\,H^5k_1,\qquad \mathcal G\le476\,H^5k_2.
\]
In the last obstruction, $k_1+k_2+k_3=\varphi-\chi=(u_1-u_2)/N_0\le2X$.
Thus each surviving case gives $k_1\le2X$ or $k_2\le2X$.
The label bound $N_0q_i^4\le a\mathcal G$, with $N_0\ge1$, now gives
\[
 q_i^4\le952\,aH^5X.
\]
But $q_i\ge a/(4A)$ (Lemma~\ref{lem:fourAq}), so $a^3\le4^4\cdot952\,A^4H^5X$.
Condition~\eqref{eq:jointscale} with $c_{\rm L}=4^4\cdot952$ gives the strict reverse
inequality. (The formalisation expands $\mathcal G$ more carefully, as
$\mathcal G\le H^5(12k_1+464)$, and uses $c_{\rm L}=4^4\cdot488$.)
\end{proof}

\subsection{Orders of the reconstructed functions}\label{sec:jointorders}

\begin{lemma}[Orders of the reconstructed functions]\label{lem:joint-infinity-orders}
Let $C$ be a retained nonexceptional component of positive degree
containing a good start.
At every point above infinity, with hyperplane multiplicity $e$, one has
\[
 \operatorname{ord}Q=-\frac{3e}{2},\qquad
 \operatorname{ord}E_0=\frac{3e}{2}.
\]
In particular, $E_0$ is a nonzero rational function vanishing at every
point above infinity.
\end{lemma}
\begin{proof}
The good start guarantees that the reconstruction denominators are
not identically zero on $C$. Thus $Q$ and $E_0$ are rational functions
on $C$, and retention makes both identities \eqref{eq:jointcuts}
identities in its function field.

\emph{The quotient order.}
Write
$P_4(x)=x^4+p_3x^3+p_2x^2+p_1x+p_0$.
Its Newton expression gives $\deg p_j\le4-j$, and hence
$\operatorname{ord}p_j\ge-(4-j)e$.
The first retained identity gives
\[
 \operatorname{ord}P_4(-Q)=\operatorname{ord}(aH_4/N_0)=-6e.
\]
If $\operatorname{ord}Q\ge-e$, every term of $P_4(-Q)$ would have order
at least $-4e$, a contradiction. Put $r=\operatorname{ord}Q<-e$.
For $0\le j<4$,
\[
 \operatorname{ord}\bigl(p_j(-Q)^j\bigr)-4r
 \ \ge\ (4-j)(-e-r)>0.
\]
Thus $Q^4$ is the unique term of lowest order. It follows that
$4r=-6e$, or $\operatorname{ord}Q=-3e/2$.

\emph{The reconstructed curvature order.}
The coefficient $p_3=\chi-1-s_1-s_2-s_3$ has order at least $-e$.
By the proof of Lemma~\ref{lem:jointleading}, $H_{013}$ has degree at most
four, with degree-four part~\eqref{eq:H013top}.
Consequently
\[
 \operatorname{ord}(Q-p_3)=-3e/2,\qquad
 \operatorname{ord}\bigl((Q-p_3)H_3\bigr)=-9e/2
       <-4e\le\operatorname{ord}H_{013}.
\]
There is again a unique term of lowest order. Dividing the second
retained identity by $P_4(-Q)$ therefore gives
\[
 \operatorname{ord}E_0=-9e/2-(-6e)=3e/2.
\]
The argument uses the discrete valuation on the normalisation, so it
also covers branches above singular points of the original curve.
\end{proof}

\subsection{The joint coordinates}\label{sec:jointcoords}

Put $d_j=b_{i+j}-b_{i+j+1}$ and
\[
 D_{01}=d_0k_1-d_1k_0,\quad D_{02}=d_0k_2-d_2k_0,\quad
 D_{12}=d_1k_2-d_2k_1.
\]
The first of these, $D_{01}$, is the curvature $I_3(i)$. With $s_j=q_{i+j}-q_i$ and
$r_j=k_j(s_j+s_{j+1})+k_{j+1}$, the Hankel polynomial $H_3$ of Section~\ref{sec:jointrecon}
and its shift are
\begin{align*}
 H_3&=k_0r_1-k_1r_0=\sigma(k_0,k_1,k_2),\\
 H_3'&=k_1r_2-k_2r_1=\sigma(k_1,k_2,k_3),
\end{align*}
with $\sigma$ as in Section~\ref{sec:six}. They are positive at every state at which they are
defined. At a good start, $W=(D_{12},-D_{02},D_{01})=\mathbf d\times\mathbf k$ satisfies
$W\cdot\mathbf k=0$ and $W\cdot\mathbf r=\det(\mathbf d,\mathbf k,\mathbf r)=p_{012}=N_0$;
these two linear equations determine $W$ from its last coordinate $D_{01}=I_3(i)$.
Solving them with $E_0$ in place of $I_3(i)$, on a retained component, gives
\[
 E_{12}=\frac{H_3'E_0-N_0k_1}{H_3},\qquad
 E_{02}=\frac{k_0E_{12}+k_2E_0}{k_1},\qquad
 W=(E_{12},-E_{02},E_0),\qquad J=N_0\Psi.
\]
The determinant and cross-product identities give, identically,
\begin{equation}\label{eq:jointdot}
 W\cdot(k_0,k_1,k_2)=0,\qquad W\cdot(r_0,r_1,r_2)=N_0.
\end{equation}
At a good start these rational functions take the integer values
$W=(D_{12},-D_{02},D_{01})$. Changing
the rank-five Gram basis to $1,x,x^2,x^3,P_4$ gives
\[
 \det\bigl(\nu_i(x^{r+s})\bigr)_{r,s<5}=\det\bigl(\nu_i(x^{r+s})\bigr)_{r,s<4}\,\nu_i(P_4^2).
\]
Consequently $J=I_5(i)$ is a positive integer at good starts. The heights of these
integers are controlled by
\[
 S=\left\lfloor\frac{c_S A^3H^3}{a^2}\right\rfloor,\qquad
 Y=\left\lfloor\frac{c_Y A^5H^{10}}{a^4}\right\rfloor,
\]
for suitable absolute constants $c_S,c_Y$.

\begin{lemma}[Heights of the joint coordinates]\label{lem:jointheights}
At good starts, all ten $2\times2$ minors $D_{jl}=d_jk_l-d_lk_j$ with $0\le j<l\le4$ satisfy
$|D_{jl}|\le S$; in particular $|W_j|\le S$. Moreover $1\le J\le Y$.
\end{lemma}
\begin{proof}
Write $q=q_i$, and recall $q\ge a/(4A)$ (Lemma~\ref{lem:fourAq}). The proof of
Proposition~\ref{prop:allminors} gives $d_j/a=k_j/q^2+O(H^2/q^3)$ for $j\le4$,
the error terms of Lemma~\ref{lem:expand} being absorbed because
$H\le1+28a/(AT)\le29a/A\le116q$.
In $D_{jl}=d_jk_l-d_lk_j$ the leading terms cancel, so $|D_{jl}|q^3\ll aH^3$, and
$q\ge a/(4A)$ gives $|D_{jl}|\le S$ for suitable $c_S$, including the floor because the
minors are integers.

For $J=I_5(i)$, argue as in the proof of Lemma~\ref{lem:labels} with five Newton polynomials
instead of four: Hadamard's inequality and $\nu_i\le\mu_i/q$ give
$Jq^5\le a\prod_{j<5}\mu_i(F_j^2)$. By the shift rule,
$\mu_i(F_j^2)=\mu_{i+j}\bigl(F_j(x+s_j)\bigr)$ is a polynomial of degree $j$ in the increments, and
the increment bound gives $\mu_i(F_j^2)\ll H^j$. Hence $Jq^5\ll aH^{10}$, and $q\ge a/(4A)$
gives $J\le Y$ for suitable $c_Y$.
\end{proof}

\begin{corollary}[Orders of the joint coordinates]\label{cor:joint-curvature-orders}
In the setting of Lemma~\ref{lem:joint-infinity-orders}, the joint
coordinates satisfy
\begin{equation}\label{eq:jointorders}
 \operatorname{ord}W_0=\operatorname{ord}W_2=3e/2,\qquad
 \operatorname{ord}W_1\ge3e/2,\qquad
 \operatorname{ord}J=-5e/2.
\end{equation}
\end{corollary}
\begin{proof}
The cubic leading part of $H_3'$ is $k_1k_2(k_1+k_2)$.
By \eqref{eq:joint-node-orders}, it has order $-3e$, whereas the
quadratic remainder $k_1k_3-k_2^2$ has order at least $-2e$.
Thus $\operatorname{ord}H_3=\operatorname{ord}H_3'=-3e$ and
$\operatorname{ord}k_1=-e$.
In the numerator defining $E_{12}$ the two terms have different orders:
\[
 \operatorname{ord}(H_3'E_0)=-3e/2
       <-e=\operatorname{ord}(N_0k_1).
\]
Therefore $\operatorname{ord}E_{12}=3e/2$.
Each term in $k_0E_{12}+k_2E_0$ has order $e/2$, so their sum has
order at least $e/2$. Dividing by $k_1$ gives
$\operatorname{ord}E_{02}\ge3e/2$. Cancellation here is harmless:
equality for the first and third coordinates already gives
$\min_j\operatorname{ord}W_j=3e/2$.
The formula for $J=N_0\Psi$ is Corollary~\ref{cor:jointparity}.
\end{proof}

\subsection{Counting the nonexceptional conics}\label{sec:jointconic}

\begin{proposition}[Conics]\label{prop:jointconic}
A nonexceptional component of degree two contains the free-node triples of $O_\eta(a^\eta)$
good starts, with an implied constant independent of the pattern and the component.
Consequently all nonexceptional conics together contribute $O_\eta(a^\eta X^5)$ good starts.
\end{proposition}

In outline: a component that is not retained contributes $O(1)$ good starts
(Lemma~\ref{lem:jointcuts}). A retained conic through three free-node triples of good starts
lies in a plane with integer coefficients of size $a^{O(1)}$, and Corollary~\ref{cor:jointparity} leaves it a single rational branch at
infinity, of multiplicity two. So it is a parabola, the image of a quadratic polynomial map
$t\mapsto(\xi_0(t),\xi_1(t),\xi_2(t))/\Delta_c$ whose parameter $t$ is an integer at every
integer point. By Lemma~\ref{lem:joint-infinity-orders}, the reconstructed curvature $E_0$ is
a rational function of $t$ with a zero of order three at infinity, and it takes the integer
value $I_3(i)$ at each good start. Its nonconstant denominator must then divide a fixed nonzero
resultant at each such $t$, which leaves $a^{o(1)}$ possibilities. Appendix~\ref{app:conic}
gives the details, including the uniform bounds on the coefficients.

\subsection{The extra degree and the powered graph}\label{sec:jointgraph}

The conic count leaves retained nonexceptional components of degree at
least four. Fix such a component $C$ of degree $d$. The additional
integer coordinates will increase its degree without increasing their
heights too much. The key is one extra zero over the affine part of $C$.

All divisors in this argument are on the normalisation $\widetilde C$
of the projective closure of $C$, over $\mathbb Q$.
For a closed point $P$ of $\widetilde C$, write
$\deg P=[\mathbb Q(P):\mathbb Q]$, where $\mathbb Q(P)$ is the residue field; thus
$\deg(\sum_P a_P[P])=\sum_P a_P\deg P$.
Let $E_\infty=\sum_{P\mid\infty}e_P[P]$ be the pullback of the original
hyperplane at infinity. Then
$\sum_{P\mid\infty}e_P\deg P=d$.  We do not require the closed points, or the
branches over a rational point of $C$, to be rational.

Let $\mathcal F$ be the pole envelope of the curvature coordinates:
\[
 \mathcal F=\sum_P \max\{0,-\operatorname{ord}_P(W_0),
                   -\operatorname{ord}_P(W_1),
                   -\operatorname{ord}_P(W_2)\}[P],
\]
with $\operatorname{ord}_P(0)=+\infty$.
The orders in \eqref{eq:jointorders} show that $\mathcal F$ is supported over the
affine part of $C$, and hence is disjoint from $E_\infty$.

\begin{lemma}[An additional affine zero]\label{lem:jointaffinezero}
There is a rational linear combination $\Phi$ of $W_0,W_1,W_2$ which is
not the zero function and has a positive-order zero at a closed point
$P_0$ of $\widetilde C$ outside $E_\infty$.
\end{lemma}

\begin{proof}
First, the $W_j$ cannot all be scalar multiples, in $\mathbb Q(C)$, of a
fixed nonzero vector $v\in\mathbb Q^3$.  Such a relation, together with
$W\cdot(k_0,k_1,k_2)=0$, would put $C$ in the plane
$v_0k_0+v_1k_1+v_2k_2=0$.  This is a genuine plane: the first three increments
are an invertible linear change of the three node coordinates.
The pattern quadratic is irreducible, so its restriction to this plane
is nonzero and has degree at most two.  Every curve contained in this
plane section therefore has degree at most two, contrary to $d\ge4$.

Choose the free-node triple $\mathbf s_*\in C(\mathbb Q)$ of a good start.  All the $W_j$
are regular there, because their common reconstruction denominator
does not vanish at a good start.  Write $w_j=W_j(\mathbf s_*)$.
The identity $W\cdot(r_0,r_1,r_2)=N_0\ne0$ gives $w\ne0$.
Choose $l$ with $w_l\ne0$.  At least one function
$w_lW_j-w_jW_l$ is nonzero: otherwise
$W=(W_l/w_l)w$, contradicting the preceding paragraph.
Take such a function for $\Phi$.  It has rational coefficients and
satisfies $\Phi(\mathbf s_*)=0$.

This zero survives normalisation even if $\mathbf s_*$ is singular.
Indeed, write $\Phi=g/g_0$ near $\mathbf s_*$, with $g$ not identically zero on $C$,
$g(\mathbf s_*)=0$ and $g_0(\mathbf s_*)\ne0$.
The finite normalisation has a closed point $P_0$ above $\mathbf s_*$.
In its discrete valuation ring the numerator is in the maximal ideal
and the denominator is a unit, so $\operatorname{ord}_{P_0}(\Phi)>0$.
Since $\mathbf s_*$ is affine, $P_0$ does not lie in $E_\infty$.
\end{proof}

\begin{lemma}[The curvature pole degree]\label{lem:jointpoledegree}
The pole envelope satisfies
\begin{equation}\label{eq:jointextra}
 \deg \mathcal F\ge \frac{3d}{2}+1.
\end{equation}
\end{lemma}

\begin{proof}
Use the function $\Phi$ from Lemma~\ref{lem:jointaffinezero}.
At every point over infinity, each $W_j$ has order at least $3e_P/2$.
The same is true of a rational linear combination; cancellation can
only increase its zero order.  Consequently
\[
 (\Phi)_0\ \ge\ \frac32 E_\infty+[P_0],
 \qquad
 (\Phi)_\infty\ \le\ \mathcal F .
\]
The first divisor is integral because every $e_P$ is even.
The function $\Phi$ is nonzero and has a zero, so its principal divisor
has equal zero and pole degrees.  Hence
\[
 \deg \mathcal F\ \ge\ \deg(\Phi)_\infty
 =\deg(\Phi)_0
 \ \ge\ \frac32\sum_{P\mid\infty}e_P\deg P+\deg P_0
 \ \ge\ \frac{3d}{2}+1.
\]
The last inequality uses only $\deg P_0\ge1$, not that $P_0$ is rational.
\end{proof}

For positive integers $m,n$, let $\Gamma_{m,n}\subseteq\mathbb A^7$ be
the closure of the graph obtained by adjoining
$W_0^m,W_1^m,W_2^m,J^n$ to the three original node coordinates.
The graph is taken on the common domain of these rational functions.
It is an integral curve, and projection to the first three coordinates
is birational: on that domain the four remaining coordinates are
uniquely determined.  Thus $\Gamma_{m,n}$ and $C$ have the same function
field.  Taking powers does not lose this property, because the original
coordinates remain in the graph.

\begin{lemma}[Degree of the powered graph]\label{lem:jointpowereddegree}
If $\delta=\deg\Gamma_{m,n}$, then
\begin{equation}\label{eq:jointdegree}
 2\delta\ge(5n+3m)d+2m,
 \qquad
 \delta\ge10n+7m=:\delta_{\min}.
\end{equation}
Moreover $\delta\le C_{\rm gr}(m+n)d$ for an absolute constant $C_{\rm gr}$.
For $m=n=1$ the lower bound reads $\delta\ge4d+1\ge17$.
\end{lemma}

\begin{proof}
\emph{Lower bound.}
At a point $P$ over the old infinity hyperplane, the original node
coordinates have pole orders at most $e_P$, while
$J^n$ has pole order $5ne_P/2$ and all $W_j^m$ vanish.
Since $n\ge1$, the pole envelope of the seven graph coordinates at
$P$ is exactly $5ne_P/2$.
Over the affine part, the node coordinates and $J$ are regular:
$J=N_0\Psi$ is polynomial in the nodes.  There the envelope is exactly
$m$ times the curvature envelope.  The complete envelope of the graph
coordinates is therefore
\[
 \mathcal F_{m,n}=\frac{5n}{2}E_\infty+m\mathcal F.
\]

There is a single rational linear combination $\psi$ of the seven
coordinates whose pole divisor is $\mathcal F_{m,n}$.  To see this, fix a point $P$
in the finite support of $\mathcal F_{m,n}$ and multiply all seven coordinates by a
local uniformiser to the power $\mathcal F_{m,n}(P)$.  Their values in $\mathbb Q(P)$
are not all zero.  Cancellation of the leading pole is therefore a
proper $\mathbb Q$-linear condition on the seven coefficients.
The finitely many such proper subspaces do not cover $\mathbb Q^7$.
Choosing coefficients outside their union gives the asserted
simultaneous absence of cancellation, also when $\mathbb Q(P)\ne\mathbb Q$.

As a linear function on $\Gamma_{m,n}$, $\psi$ has degree at most
$\delta$.  Indeed, homogenise it on the projective closure of the
graph: its pencil with the homogenising coordinate consists of
hyperplane sections of degree $\delta$.  Removing common zeros can
only decrease the degree of the resulting map to $\mathbb P^1$.
Birational projection identifies its function field with $\mathbb Q(C)$,
so this degree is $\deg(\psi)_\infty=\deg \mathcal F_{m,n}$.  It follows that
\[
 \delta\ \ge\ \frac{5n}{2}d+m\deg \mathcal F
 \ \ge\ \frac{5n+3m}{2}d+m.
\]
This is the first inequality in \eqref{eq:jointdegree}; $d\ge4$ gives
the second.  The auxiliary linear combination is used only to measure
degree: no bound on its coefficients enters the integer-point box.

\emph{Upper bound.} This is a routine count of polynomial degrees, given in
Appendix~\ref{app:degree}.
\end{proof}

Every good start whose free-node triple lies on $C$ gives an integer point on this graph.
By Lemma~\ref{lem:jointheights}, its coordinates have absolute values
at most
\[
 \max(3H,S^m,Y^n).
\]
The original node coordinates also show that distinct such node
triples remain distinct after adjoining the powers.

\subsection{The window bound and the final exponents}\label{sec:jointcount}

We use the coefficient-uniform affine-curve theorem of
Castryck--Cluckers--Dittmann--Nguyen~\cite[Theorem~3]{CCDN}, also stated as
\cite[Corollary~2.2]{CDHNV}: an integral affine curve $C$ over $\mathbb Q$ of degree $\delta$
in $\mathbb A^r$, with $r\ge2$ fixed, has
\begin{equation}\label{eq:jointexternal}
 N_{\rm aff}(C,B)\ll_r \delta^3 B^{1/\delta}(\log B+\delta),\qquad B\ge1,
\end{equation}
integer points in the box $[-B,B]^r$, with an implied constant
independent of the coefficients of $C$. We use it in the form proved in the
library cited in Section~\ref{sec:lean}: ``integral'' means prime over
$\mathbb Q$, and $B\ge1$. These two points need no new input: a curve that is
integral over $\mathbb Q$ but not geometrically integral has at most $\delta^2$
rational points, which lie in the intersection of its conjugate components, and
the range $1\le B<2$ follows from $B=2$.

\begin{proposition}[Window bound]\label{prop:jointwindow}
Fix positive integers $m,n$ and $\eta>0$, and put $\mathcal B=\max(H,S^m,Y^n)$.
For a window run in $[A,2A)$ with $T\ge36$ satisfying~\eqref{eq:jointscale},
\begin{equation}\label{eq:jointcount}
 T\ll_{m,n,\eta}a^\eta\left(
 X^5+X^4H^{1/2}+X^5\mathcal B^{1/\delta_{\min}}\right).
\end{equation}
\end{proposition}

\begin{proof}
At least $T/2$ starts are good, and for each pattern the free-node triples of
its good starts are distinct (Section~\ref{sec:jointcurve}). By
Lemma~\ref{lem:jointcurve}, these triples lie on at most twenty-four integral
curves of degree at most twenty-four, and we bound the good starts on each
curve, allowing overlaps. All the heights $H$, $S$, $Y$ and $X$ are at most a fixed
power of $a$, so the logarithm in~\eqref{eq:jointexternal} is $O_\eta(a^\eta)$.
\begin{itemize}
\item \emph{Exceptional patterns.} There are $O_\eta(a^\eta X^4)$ of them
(Lemma~\ref{lem:jointcomponents}). Under~\eqref{eq:jointscale} no line
contains a good start, so every relevant curve has degree $2\le d\le24$, and
\eqref{eq:jointexternal} in $\mathbb A^3$ with $B=3H$ gives $O_\eta(a^\eta H^{1/2})$
good starts per pattern. This gives the term $X^4H^{1/2}$.
\item \emph{Nonexceptional patterns: components that are not retained, and
conics.} By Lemma~\ref{lem:jointcuts} and Section~\ref{sec:jointconic},
these contribute $O_\eta(a^\eta)$ per pattern, hence $O_\eta(a^\eta X^5)$ in all.
\item \emph{Remaining nonexceptional components.} Lines do not
occur, by Lemma~\ref{lem:jointcomponents}, and odd degrees do not occur, by
Corollary~\ref{cor:jointparity}, so these components have degree $d\ge4$.
Their good starts give distinct integer points of $\Gamma_{m,n}\subseteq\mathbb A^7$ in a box of
side $\max(3H,S^m,Y^n)\le3\mathcal B$. By Lemma~\ref{lem:jointpowereddegree}, the degree
$\delta$ of $\Gamma_{m,n}$ satisfies $\delta_{\min}\le\delta\le24C_{\rm gr}(m+n)$, so
\eqref{eq:jointexternal} gives $O_{m,n,\eta}(a^\eta\mathcal B^{1/\delta_{\min}})$
good starts per component. This gives the term $X^5\mathcal B^{1/\delta_{\min}}$.
\end{itemize}
The added coordinates $W_j$ and $J$ are functions of the free nodes on each
component. They therefore introduce no further freely counted coordinates, and
the pattern count is not enlarged.
\end{proof}

\begin{corollary}\label{cor:jointexponent}
For fixed positive integers $m,n$ and every $\varepsilon>0$,
\[
 P(a,b)\ll_{m,n,\varepsilon}a^{E(m,n)+\varepsilon},\qquad
 E(m,n)=\max\left(\frac{15\delta_{\min}+m}{51\delta_{\min}+3m},\
           \frac{15\delta_{\min}+6n}{51\delta_{\min}+20n}\right),
\]
where $\delta_{\min}=10n+7m$. In particular $E(1,1)=128/435$ and
$E(50,51)=2201/7480$.
\end{corollary}

\begin{proof}
We bound the number $T$ of terms in each dyadic window, eliminating the
window scale $A$. If $T<36$ there is nothing to prove. If $AT>28a$, then
Proposition~\ref{prop:six} gives $T^4\ll a$. Otherwise $HAT\le56a$, and
Proposition~\ref{prop:drop} gives $T^2\le8\Delta A$. Substituting $H\le56a/(AT)$ and then
$A\ge T^2/(8\Delta)$ into the definitions of $X$, $S$ and $Y$ gives
\begin{equation}\label{eq:jointscales}
 H\ll\frac{\Delta a}{T^3},\qquad X\ll\frac{\Delta^2a^3}{T^{10}},\qquad
 S\ll\frac{a}{T^3},\qquad Y\ll\frac{\Delta^5a^6}{T^{20}}.
\end{equation}
Every power of $\Delta$ is $O(a^\eta)$, and we absorb such factors into $a^{O(\eta)}$.
\begin{itemize}
\item If~\eqref{eq:jointscale} fails, then $c_{\rm L}A^4H^5X\ge a^3$. With
$Xa^3\le c_XA^4H^6$ this gives $A^8H^{11}\gg a^6$, and $H\le56a/(AT)$ turns it into
$A^3T^{11}\ll a^5$, so $T^{17}\ll\Delta^3a^5$.
\item Otherwise Proposition~\ref{prop:jointwindow} applies. By~\eqref{eq:jointscales}, its
first term gives $T\ll a^{O(\eta)}a^{15}/T^{50}$, so $T^{51}\ll a^{15+O(\eta)}$. Its second term
gives $T\ll a^{O(\eta)}a^{12+1/2}/T^{40+3/2}$, so $T^{85}\ll a^{25+O(\eta)}$.
\item For the graph term, $T\ll a^{O(\eta)}X^5\mathcal B^{1/\delta_{\min}}$ gives
$T^{51\delta_{\min}}\ll a^{O(\eta)}a^{15\delta_{\min}}\mathcal B$. Since $H\ge1$,
$\max(H,S^m)\le\max(H,S)^m\ll a^{O(\eta)}(a/T^3)^m$, so if $\mathcal B=\max(H,S^m)$ then
$T^{51\delta_{\min}+3m}\ll a^{15\delta_{\min}+m+O(\eta)}$. If $\mathcal B=Y^n$ then
$T^{51\delta_{\min}+20n}\ll a^{15\delta_{\min}+6n+O(\eta)}$.
\end{itemize}
The first three exponents all equal $5/17$, both fractions in $E(m,n)$ exceed $5/17$, and
$T^4\ll a$ is stronger still. Choosing $\eta$ small in terms of $\varepsilon$, $m$ and $n$, every
window has $T\ll a^{E(m,n)+\varepsilon/2}$, and summing over the $O(\log a)$ windows proves
the bound. For $m=n=1$ the two fractions are $256/870=128/435$ and $261/887$, and the first is
larger. For $(m,n)=(50,51)$ the larger fraction is $2201/7480$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:joint}]
Set $r=n/m$. The two fractions in $E(m,n)$ become
\[
 E_S(r)=\frac{150r+106}{510r+360},\qquad
 E_Y(r)=\frac{156r+105}{530r+357}.
\]
The first decreases and the second increases on $r>0$. Their crossing satisfies
$30r^2-10r-21=0$, so $r_*=(5+\sqrt{655})/30$ and their common value is
$\alpha_*=(1310-\sqrt{655})/4365$.
Positive rational ratios approximate $r_*$; choose the fixed positive integers
$m,n$ to make $E(m,n)<\alpha_*+\varepsilon/2$, and apply
Corollary~\ref{cor:jointexponent} with loss $\varepsilon/2$.
The powers and all constants depend only on $\varepsilon$, not on $a,b$.
\end{proof}

\begin{remark}\label{rem:jointscales}
In terms of the scales, at the critical point $T\approx a^{\alpha_*}$ one has
$A\approx a^{2\alpha_*}$, $H\approx S\approx a^{1-3\alpha_*}\approx a^{0.117}$,
$X\approx a^{3-10\alpha_*}\approx a^{0.057}$ and
$Y\approx a^{6-20\alpha_*}\approx a^{0.115}$. The pattern count $X^5\approx a^{0.287}$
alone would give the exponent $5/17$. The graph term multiplies it by
$\mathcal B^{1/\delta_{\min}}$, and at $n/m\approx r_*$ the two box radii $S^m$ and $Y^n$
balance, at $\mathcal B\approx a^{0.117m}$. This gives the extra factor
$a^{0.117/(10r_*+7)}\approx a^{0.0068}$. Scaling $m$ and $n$ together does not
change it, since $\log\mathcal B$ and $\delta_{\min}$ grow in proportion.
\end{remark}

\section{Exterior lattice coordinates}\label{sec:lattice}

We now prove Theorem~\ref{thm:lattice}. The idea is that the pivot parts of the three rows of
the pattern matrix satisfy congruences modulo the pivot minor fixed by the pattern, so they
lie in a lattice $\Lambda_p$. In a reduced basis the increment row has coordinates
$O(H/\lambda)$, and the curvature vector, a cross product, has coordinates
$O(S/\lambda^2)$, even though the drop row itself is large; here $\lambda$ is the sup norm of
a shortest nonzero vector of $\Lambda_p$. We keep the notation and the conventions on constants
of Section~\ref{sec:joint}, and write $\delta_{\min}=10n+7m$ as there. Only the count of
good starts on retained nonexceptional components of degree at least four changes.
Proposition~\ref{prop:latticewindow} replaces the term $X^5\mathcal B^{1/\delta_{\min}}$ of
Proposition~\ref{prop:jointwindow} by
\[
 X^{5-1/(3\delta_{\min})}H^{1/\delta_{\min}}+X^{5-2m/(3\delta_{\min})}S^{m/\delta_{\min}}+X^5Y^{n/\delta_{\min}}.
\]
In effect $H$ is
replaced by $H/X^{1/3}$ and $S$ by $S/X^{2/3}$, while $Y$ is unchanged. The node term $X^{5-1/(3\delta_{\min})}H^{1/\delta_{\min}}$ gives the exponent
$45\delta_{\min}/(153\delta_{\min}-1)$, which is below $\gamma_*$ once
$\delta_{\min}\ge18$, so of the two fractions balanced in the proof of
Theorem~\ref{thm:joint} only the curvature fraction improves, and the crossing moves from
$\alpha_*=0.2942513\ldots$ to $\gamma_*=0.2942294\ldots$. Every pattern has
$\lambda(p)\ll X^{1/3}$ (Lemma~\ref{lem:index}), and the weighted count of
Lemma~\ref{lem:weights} shows that this saving is available on average: the patterns with an
unusually short vector are few enough to pay for their larger boxes.

Section~\ref{sec:latticepattern} attaches to each pattern $p$ a rank-three lattice
$\Lambda_p\subseteq\mathbb Z^3$ that contains the restrictions of the rows of
$\widehat{\mathcal M}_i$ to three pivot columns, and gives it a reduced basis. Section~\ref{sec:latticeshort} shows
that few patterns have a short vector in $\Lambda_p$. Section~\ref{sec:latticeexterior} writes
the nodes and the curvature coordinates in the reduced basis.
Section~\ref{sec:latticegraph} checks that this linear change of coordinates keeps the degree
bounds of Lemma~\ref{lem:jointpowereddegree}, and Section~\ref{sec:latticecount} sums over
patterns and optimises the exponent.

\subsection{The lattice of a pattern}\label{sec:latticepattern}
Let $p\ne0$ be the minor vector of a $3\times5$ integer matrix $\mathcal M$, with pivot $\pi$
as in the proof of Proposition~\ref{prop:minorvec}; so $|p_\pi|$ is maximal among the
coordinates of $p$. Put $\kappa=|p_\pi|\ge1$, let $j_1<j_2<j_3$ be the pivot columns and
$j_4<j_5$ the other two, and call $x_\pi=(x_{j_1},x_{j_2},x_{j_3})$ the \emph{pivot part} of a
row $x$ of $\mathcal M$. Let $\mathbf u,\mathbf v\in\mathbb Z^3$ be as in that proof: $\mathbf u_r$ ($\mathbf v_r$) is the minor
with the $r$-th pivot column replaced by column $j_4$ ($j_5$), with the sign of the column
permutation. So the coordinates of $\mathbf u$ and $\mathbf v$ are, up to sign, coordinates of $p$, and
$p_\pi x_{j_4}=\mathbf u\cdot x_\pi$ and $p_\pi x_{j_5}=\mathbf v\cdot x_\pi$ for every row $x$
(Lemma~\ref{lem:lift}). Let $\Lambda_p\subseteq\mathbb Z^3$ be the subgroup of $x$ with
$\kappa\mid \mathbf u\cdot x$ and $\kappa\mid \mathbf v\cdot x$. It contains $\kappa\mathbb Z^3$. Write $|x|$
for the Euclidean norm and $|x|_\infty$ for the sup norm, and let the \emph{minimum}
$\lambda(p)$ be the least sup norm of a nonzero vector of $\Lambda_p$. Then
$|x|\ge\lambda(p)$ for every nonzero $x\in\Lambda_p$.

\begin{lemma}[Lifting]\label{lem:lift}
For a column $l$, let $\mathbf L_l\in\mathbb Z^3$ be $p_\pi\mathbf e_r$ if $l=j_r$ is a pivot column,
$\mathbf u$ if $l=j_4$ and $\mathbf v$ if $l=j_5$, where $\mathbf e_1,\mathbf e_2,\mathbf e_3$ is the standard
basis. For every row $x$ of $\mathcal M$ and every column $l$, $p_\pi x_l=\mathbf L_l\cdot x_\pi$.
Moreover $\det(\mathbf L_0,\mathbf L_1,\mathbf L_2)=p_\pi^2p_{012}$,
and $|\mathbf L_l|_\infty\le\kappa$.
\end{lemma}

\begin{proof}
The vectors $\mathbf L_l$ are the rows of $\mathcal M^{\mathsf T}\operatorname{adj}(\mathcal M_\pi^{\mathsf T})$,
where $\mathcal M_\pi$ is the pivot $3\times3$ block, and the first identity is Cramer's rule
for the system $\mathcal M_\pi c=(\text{column }l)$, whose solution is
$c=\mathbf L_l/p_\pi$. Taking the rows $l=0,1,2$ gives
$\det(\mathbf L_0,\mathbf L_1,\mathbf L_2)=p_{012}\det\operatorname{adj}(\mathcal M_\pi^{\mathsf T})=p_{012}p_\pi^2$. The
bound holds because the coordinates of $\mathbf u$ and $\mathbf v$ are, up to sign, coordinates
of $p$, and $|p_\pi|$ is maximal.
\end{proof}

So the pivot parts $x_\pi$ of the rows of $\mathcal M$ lie in $\Lambda_p$. They are the rows of
$\mathcal M_\pi$, which span a sublattice of index $|\det\mathcal M_\pi|=\kappa$, so
$[\mathbb Z^3:\Lambda_p]\le\kappa$.

\begin{lemma}[Index]\label{lem:index}
If $K$ is a nonnegative integer and $(K+1)^3>\kappa$, then $\lambda(p)\le K$.
\end{lemma}

\begin{proof}
The box $[0,K]^3$ has more than $\kappa\ge[\mathbb Z^3:\Lambda_p]$ points, so two of them lie in
the same coset of $\Lambda_p$, and their difference is a nonzero vector of $\Lambda_p$ with sup
norm at most $K$.
\end{proof}

The next lemma is the classical reduction of a three-dimensional lattice: any
basis with $|\mathbf f_1||\mathbf f_2||\mathbf f_3|\ll\det\Lambda$ would do, and the constant does
not affect the exponents. For completeness we give the elementary proof in
Appendix~\ref{app:reduction}.

\begin{lemma}[Reduction]\label{lem:reduce}
Let $\Lambda\subseteq\mathbb Z^3$ be a subgroup containing $\kappa\mathbb Z^3$, $\kappa\ge1$.
There is a basis $\mathbf f_1,\mathbf f_2,\mathbf f_3$ of $\Lambda$ with $\det\Lambda=\det(\mathbf f_1,\mathbf f_2,\mathbf f_3)$,
$0<\det\Lambda\le\kappa^3$ and
\[
27\,|\mathbf f_1|^2|\mathbf f_2|^2|\mathbf f_3|^2\le64\,(\det\Lambda)^2 .
\]
Hence every $x\in\Lambda$ is $\sum\xi_i\mathbf f_i$ with $\xi\in\mathbb Z^3$ and
$27|\mathbf f_i|^2\xi_i^2\le64|x|^2$.
\end{lemma}

\subsection{Few patterns have a short vector}\label{sec:latticeshort}

\begin{lemma}[Divisibility count]\label{lem:dotcount}
For $\ell\ge1$ and $x\in\mathbb Z^3$, at most $27\ell^2\gcd(\ell,x)$ vectors
$y\in[-\ell,\ell]^3$ have $\ell\mid y\cdot x$. Here $\gcd(\ell,x)=\gcd(\ell,x_1,x_2,x_3)$.
\end{lemma}

\begin{proof}
Each residue class modulo $\ell$ meets $[-\ell,\ell]^3$ in at most $3^3$ points, so it suffices
to show that the set $\mathcal K$ of solutions in $[0,\ell)^3$ has at most $\ell^2e$ elements, where
$e=\gcd(\ell,x)$. By B\'ezout choose $z\in\mathbb Z^3$ with $z\cdot x\equiv e\pmod\ell$. The map
$(y,j)\mapsto(y+jz)\bmod\ell$ is injective on $\mathcal K\times[0,\ell/e)$: equal images give
$\ell\mid(j-j')e$, so $j=j'$, and then $y=y'$. Hence $|\mathcal K|\,\ell/e\le\ell^3$.
\end{proof}

\begin{proposition}[Short vectors]\label{prop:short}
Let $\mathcal S$ be a set of nonzero minor vectors with coordinates in $[-X,X]$.
Then for every $R\ge1$,
\[
\#\{p\in\mathcal S:\lambda(p)\le R\}\ll R^3X^4\tau_{\max}(X)^2 .
\]
\end{proposition}

\begin{proof}
If $\lambda(p)\le R$, some nonzero $x\in[-R,R]^3$ lies in $\Lambda_p$, so
$\kappa\mid \mathbf u\cdot x$, where $\kappa=|p_\pi|\le X$; we drop the condition $\kappa\mid \mathbf v\cdot x$. For fixed $x$, the encoding $p\mapsto(\pi,\operatorname{sgn}p_\pi,\kappa,\mathbf u,\mathbf v)$
of the proof of Proposition~\ref{prop:minorvec} gives at most $20\sum_{\kappa\le X}N(\kappa,x)$
such $p$, where $N(\kappa,x)$ counts pairs $(\mathbf u,\mathbf v)$ in $[-\kappa,\kappa]^3$ with
$\kappa\mid \mathbf u\times \mathbf v$ and $\kappa\mid \mathbf u\cdot x$. By the proof of
Proposition~\ref{prop:minorvec}, for each $\mathbf u$ there are at most $27\kappa\gcd(\kappa,\mathbf u)^2$
vectors $\mathbf v$. Group the $\mathbf u$ by $g=\gcd(\kappa,\mathbf u)$ and write $\mathbf u=g\mathbf u_0$ with $(\kappa/g)\mid \mathbf u_0\cdot x$.
Lemma~\ref{lem:dotcount} with $\ell=\kappa/g$, and $\gcd(\kappa/g,x)\le\gcd(\kappa,x)$, give
\[
 \sum_{\mathbf u}g^2\le\sum_{g\mid\kappa}g^2\cdot27\Bigl(\frac\kappa g\Bigr)^2\gcd(\kappa,x)
 =27\kappa^2\tau(\kappa)\gcd(\kappa,x),
\]
so $N(\kappa,x)\le729\kappa^3\tau(\kappa)\gcd(\kappa,x)$. Next,
$\gcd(\kappa,x)\le\sum_{e\mid\kappa,\,e\mid x}e$, and
swapping the sums,
\[
 \sum_{0<|x|_\infty\le R}\gcd(\kappa,x)
 \le\sum_{e\mid\kappa}e\,\#\{x\ne0:|x|_\infty\le R,\ e\mid x\}
 \le27R^3\tau(\kappa).
\]
Indeed, a divisor $e>R$ of $\kappa$ divides no nonzero $x$ in the box, and for $e\le R$
the multiples of $e$ number at most $(2R/e+1)^3$, with
$e(2R/e+1)^3\le27R^3$. Sum over $x$ and then over $\kappa\le X$,
using $\sum_{\kappa\le X}\kappa^3\le X^4$.
\end{proof}

\begin{lemma}[Fractional weights]\label{lem:weights}
Let $\mathcal S$ be a set of nonzero minor vectors with coordinates in $[-X,X]$, $X\ge1$.
Then for $0\le t\le1$,
\[
 \sum_{p\in\mathcal S}\lambda(p)^{-t}\ll\tau_{\max}(X)^2X^{5-t/3}.
\]
\end{lemma}

\begin{proof}
Choose an integer $K$ with $X\le K^3\le8X$. By Lemma~\ref{lem:index}, $1\le\lambda(p)\le K$
for $p\in\mathcal S$. Writing $1/\lambda=1/K+\sum_{\lambda\le j<K}1/(j(j+1))$ and using
Proposition~\ref{prop:short} with $R=j$,
\[
 K\sum_{p\in\mathcal S}\frac1{\lambda(p)}
 \le\#\mathcal S+O\Bigl(KX^4\tau_{\max}(X)^2\sum_{1\le j<K}\frac{j^2}{j+1}\Bigr)
 \le\#\mathcal S+O(X^4\tau_{\max}(X)^2K^3) .
\]
The proof of Proposition~\ref{prop:minorvec} gives
$\#\mathcal S\ll\tau_{\max}(X)X^5$. Since $K^3\le8X$ and $\tau_{\max}(X)\ge1$,
the right side is $O(\tau_{\max}(X)^2X^5)$. Dividing by $K\ge X^{1/3}$
proves the case $t=1$. For $0<t<1$, H\"older's inequality
with exponents $1/(1-t)$ and $1/t$ gives
$\sum\lambda^{-t}\le(\#\mathcal S)^{1-t}\bigl(\sum\lambda^{-1}\bigr)^t$, which is the claim;
the cases $t=0$ and $t=1$ are the two bounds above.
\end{proof}

\subsection{Exterior coordinates}\label{sec:latticeexterior}

\begin{lemma}[Exterior coordinates]\label{lem:exterior}
Let $\mathbf f_1,\mathbf f_2,\mathbf f_3$ be a basis of $\Lambda\subseteq\mathbb Z^3$ as in
Lemma~\ref{lem:reduce}, with $|\mathbf f_i|\ge\lambda>0$. Let $x,x'\in\Lambda$ have coordinates
$\xi,\xi'\in\mathbb Z^3$, and put $w=x\times x'$ and $c=\xi\times\xi'\in\mathbb Z^3$. Then
\[
 w=c_1\,\mathbf f_2\times\mathbf f_3+c_2\,\mathbf f_3\times\mathbf f_1+c_3\,\mathbf f_1\times\mathbf f_2,\qquad
 \mathbf f_i\cdot w=(\det\Lambda)\,c_i,\qquad 27\lambda^4c_i^2\le64|w|^2 .
\]
\end{lemma}

\begin{proof}
The first identity follows by expanding $(\sum\xi_i\mathbf f_i)\times(\sum\xi'_i\mathbf f_i)$
bilinearly, and taking the dot product with $\mathbf f_i$ gives the second. By Cauchy--Schwarz and
Lemma~\ref{lem:reduce},
\[
 27|\mathbf f_1|^2|\mathbf f_2|^2|\mathbf f_3|^2c_i^2\le64(\det\Lambda)^2c_i^2=64(\mathbf f_i\cdot w)^2
 \le64|\mathbf f_i|^2|w|^2 .
\]
Cancelling $|\mathbf f_i|^2$ and using $|\mathbf f_j|\ge\lambda$ for the other two vectors gives the
claim.
\end{proof}

The lemma replaces the bounds $|\xi|\ll|x|/\lambda$ and $|\xi'|\ll|x'|/\lambda$, which would give
$|c|\ll|x||x'|/\lambda^2$, by $|c|\ll|x\times x'|/\lambda^2$. This matters because the drop row
is large while the curvature minors are small.

Now fix a pattern $p$ that occurs at a good start, and let $\mathbf f_1,\mathbf f_2,\mathbf f_3$ be a
basis of $\Lambda_p$ as in Lemma~\ref{lem:reduce}. Write $\lambda=\lambda(p)$. At a good start
$i$ with pattern $p$, the pivot parts $d_\pi$ and $k_\pi$ of the drop and increment rows of
$\widehat{\mathcal M}_i$ lie in $\Lambda_p$, with coordinates $\xi$ and $\xi'$. Put
\[
 W'=\xi\times\xi',\qquad w=d_\pi\times k_\pi.
\]
The entries of $w$ are, up to sign, the minors $D_{jl}$ with $j,l$ pivot columns. The next lemma
shows that $W'=\Upsilon_pW$ for a constant matrix $\Upsilon_p$. We use $W'$ rather than $\Upsilon_pW$ because $W'$ is
visibly integral, while $\Upsilon_p$ has rational entries.

\begin{lemma}[Coordinates in the reduced basis]\label{lem:latticecoords}
There are matrices $\mathcal T_p,\Upsilon_p\in\mathrm{GL}_3(\mathbb Q)$, depending only on $p$, such that at
every good start with pattern $p$
\[
 (s_1,s_2,s_3)=\mathcal T_p\xi',\qquad W'=\Upsilon_pW,\qquad
 |\xi'|_\infty\le\frac{3H}{\lambda},\qquad |W'|_\infty\le\frac{8S}{3\lambda^2}.
\]
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:lift}, $p_\pi(k_0,k_1,k_2)=\mathcal Lk_\pi$, where $\mathcal L$ has rows
$\mathbf L_0,\mathbf L_1,\mathbf L_2$ and $\det\mathcal L=p_\pi^2p_{012}\ne0$, since $p_{012}=N_0=I_4(i)\ge1$ at a good
start. With $k_\pi=(\mathbf f_1\,\mathbf f_2\,\mathbf f_3)\xi'$ and the cumulative sums $s_j=k_0+\dots+k_{j-1}$,
this gives $\mathcal T_p$. Since $|k_\pi|^2\le3H^2$ and $|\mathbf f_i|\ge\lambda$, Lemma~\ref{lem:reduce} gives
$|\xi'_i|\le(8/3)H/\lambda\le3H/\lambda$.

For the curvature, $W=(D_{12},-D_{02},D_{01})=(d_0,d_1,d_2)\times(k_0,k_1,k_2)$, and
$p_\pi(d_0,d_1,d_2)=\mathcal Ld_\pi$. Let $\operatorname{Cof}(\mathcal L)$ have rows
$\mathbf L_1\times \mathbf L_2$, $\mathbf L_2\times \mathbf L_0$, $\mathbf L_0\times \mathbf L_1$; its determinant is
$(\det\mathcal L)^2\ne0$. The Binet--Cauchy identity
$(\mathbf L\cdot x)(\mathbf L'\cdot y)-(\mathbf L'\cdot x)(\mathbf L\cdot y)=(\mathbf L\times\mathbf L')\cdot(x\times y)$, applied to pairs of
rows of $\mathcal L$, gives $(\mathcal Lx)\times(\mathcal Ly)=\operatorname{Cof}(\mathcal L)(x\times
y)$, so $p_\pi^2W=\operatorname{Cof}(\mathcal L)\,w$. Lemma~\ref{lem:exterior} gives
$(\det\Lambda_p)W'=(\mathbf f_1\,\mathbf f_2\,\mathbf f_3)^{\mathsf T}w$. Together,
$W'=p_\pi^2(\det\Lambda_p)^{-1}(\mathbf f_1\,\mathbf f_2\,\mathbf f_3)^{\mathsf T}\operatorname{Cof}(\mathcal L)^{-1}W$,
which defines $\Upsilon_p$. Finally $|w|^2\le3S^2$ by Lemma~\ref{lem:jointheights}, and
Lemma~\ref{lem:exterior} gives $3\lambda^2|W'_i|\le8S$.
\end{proof}

\subsection{The lattice graph}\label{sec:latticegraph}
On a retained nonexceptional component $C$ of degree $d\ge4$ of a pattern $p$, regard
$W'=\Upsilon_pW$ as a vector of rational functions on $C$, and let $\Gamma'_{m,n}$ be the image of the
graph of
\[
 \bigl(s_1,s_2,s_3,(W'_0)^m,(W'_1)^m,(W'_2)^m,J^n\bigr)
\]
under the linear map that replaces the nodes $s$ by $\xi'=\mathcal T_p^{-1}s$.

\begin{lemma}[Degree of the lattice graph]\label{lem:latticedegree}
The degree $\delta'$ of $\Gamma'_{m,n}$ satisfies
$\delta_{\min}\le\delta'\le C_{\rm gr}(m+n)d$.
\end{lemma}

\begin{proof}
An invertible linear change of the node coordinates is a linear automorphism of
$\mathbb A^7$, so it preserves the affine Hilbert function, and we may keep the nodes $s$. The
proof of Lemma~\ref{lem:jointpowereddegree}, with Appendix~\ref{app:degree}, uses the curvature
coordinates only through the following facts.
\begin{enumerate}
\item At every point $P$ over infinity, $\operatorname{ord}_PW_j\ge3e_P/2>0$ for each $j$
(Corollary~\ref{cor:joint-curvature-orders}).
\item The pole envelope $\mathcal F$ is supported over the affine part of $C$, and
$\deg\mathcal F\ge3d/2+1$ (Lemma~\ref{lem:jointpoledegree}).
\item Some rational linear combination of the seven graph coordinates has pole divisor
$\frac{5n}2E_\infty+m\mathcal F$; its construction uses only this envelope.
\item The $W_j$ have a common denominator $\Omega$ and numerators of degree at most $C_W$
(Appendix~\ref{app:degree}).
\end{enumerate}
Each of these is unchanged when $W$ is replaced by $W'=\Upsilon_pW$ for a constant invertible matrix
$\Upsilon_p$. Each $W'_j$ is a constant combination of the $W_l$, and each $W_l$ is a constant
combination of the $W'_j$, so for every point $P$
\[
 \min_j\operatorname{ord}_PW'_j=\min_l\operatorname{ord}_P W_l .
\]
This gives (1) for $W'$, and it shows that the pole envelope of $W'$ is $\mathcal F$, which gives
(2) and (3). The numerators of $W'$ over $\Omega$ are constant combinations of those of $W$,
which gives (4). The proof of Lemma~\ref{lem:jointpowereddegree} therefore applies unchanged.
\end{proof}

\subsection{The window bound and the optimised exponent}\label{sec:latticecount}

\begin{proposition}[Lattice window bound]\label{prop:latticewindow}
Fix positive integers $m,n$ and $\eta>0$. For a window run in $[A,2A)$ with $T\ge36$
satisfying~\eqref{eq:jointscale},
\[
 T\ll_{m,n,\eta}a^\eta\left(X^5+X^4H^{1/2}+X^{5-1/(3\delta_{\min})}H^{1/\delta_{\min}}
 +X^{5-2m/(3\delta_{\min})}S^{m/\delta_{\min}}+X^5Y^{n/\delta_{\min}}\right).
\]
\end{proposition}

\begin{proof}
The proof of Proposition~\ref{prop:jointwindow} applies, except for the remaining components of
the nonexceptional patterns. For such a component of a pattern $p$, the good starts give
integer points $(\xi',(W'_0)^m,(W'_1)^m,(W'_2)^m,J^n)$ of $\Gamma'_{m,n}$, and distinct node triples
give distinct points because $\xi'$ determines $s$. By Lemma~\ref{lem:latticecoords} they lie
in the box of side
\[
 \mathcal B_p=\max\Bigl(1,\frac{3H}{\lambda(p)},\Bigl(\frac{8S}{3\lambda(p)^2}\Bigr)^m,Y^n\Bigr).
\]
By Lemma~\ref{lem:latticedegree} and~\eqref{eq:jointexternal}, there are
$O_{m,n,\eta}(a^\eta\mathcal B_p^{1/\delta_{\min}})$ of them, and
\[
 \mathcal B_p^{1/\delta_{\min}}\le1+3H^{1/\delta_{\min}}\lambda(p)^{-1/\delta_{\min}}+8S^{m/\delta_{\min}}\lambda(p)^{-2m/\delta_{\min}}+Y^{n/\delta_{\min}}.
\]
Sum over the $O_\eta(a^\eta X^5)$ patterns. Since $\delta_{\min}\ge2m$, both $1/\delta_{\min}$ and $2m/\delta_{\min}$ lie in
$[0,1]$, and Lemma~\ref{lem:weights}, with $\tau_{\max}(X)=a^{o(1)}$, gives
$\sum_p\lambda(p)^{-1/\delta_{\min}}\ll a^\eta X^{5-1/(3\delta_{\min})}$ and
$\sum_p\lambda(p)^{-2m/\delta_{\min}}\ll a^\eta X^{5-2m/(3\delta_{\min})}$.
\end{proof}

\begin{corollary}\label{cor:latticeexponent}
For fixed positive integers $m,n$ and every $\varepsilon>0$,
\begin{gather*}
 P(a,b)\ll_{m,n,\varepsilon}a^{E'(m,n)+\varepsilon},\\
 E'(m,n)=\max\left(\frac5{17},\ \frac{45\delta_{\min}}{153\delta_{\min}-1},\ \frac{45\delta_{\min}-3m}{153\delta_{\min}-11m},\
 \frac{15\delta_{\min}+6n}{51\delta_{\min}+20n}\right),
\end{gather*}
where $\delta_{\min}=10n+7m$. In particular $E'(3,2)=153/520$.
\end{corollary}

\begin{proof}
Argue as in the proof of Corollary~\ref{cor:jointexponent}. The first two terms and the windows in which~\eqref{eq:jointscale}
fails give the exponent $5/17$. For the third term, $T^{3\delta_{\min}}\ll a^{O(\eta)}X^{15\delta_{\min}-1}H^3$, and
\eqref{eq:jointscales} gives $T^{153\delta_{\min}-1}\ll a^{45\delta_{\min}+O(\eta)}$. For the fourth,
$T^{3\delta_{\min}}\ll a^{O(\eta)}X^{15\delta_{\min}-2m}S^{3m}$ gives $T^{153\delta_{\min}-11m}\ll a^{45\delta_{\min}-3m+O(\eta)}$. For the
fifth, $T^{\delta_{\min}}\ll a^{O(\eta)}X^{5\delta_{\min}}Y^n$ gives $T^{51\delta_{\min}+20n}\ll a^{15\delta_{\min}+6n+O(\eta)}$. For
$(m,n)=(3,2)$, $\delta_{\min}=41$ and the last three fractions are $1845/6272$, $153/520$ and $627/2131$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:lattice}]
Set $r=n/m$. The last two fractions in $E'(m,n)$ become
\[
 E'_S(r)=\frac{450r+312}{1530r+1060},\qquad
 E_Y(r)=\frac{156r+105}{530r+357},
\]
where $E_Y$ is the same function as in the proof of Theorem~\ref{thm:joint}. The first decreases
and the second increases on $r>0$. Their crossing satisfies $15r^2=7$, so $r_*=\sqrt{105}/15$, and
their common value is $\gamma_*=471/1559-6\sqrt{105}/7795$. The fraction $45\delta_{\min}/(153\delta_{\min}-1)$ decreases
in $\delta_{\min}$, and it is below $\gamma_*$ once $\delta_{\min}>\gamma_*/(153\gamma_*-45)\approx17.2$; also
$5/17<\gamma_*$. Choose a positive rational $r$ close to $r_*$ with
$\max(E'_S(r),E_Y(r))<\gamma_*+\varepsilon/2$, and write $r=n/m$ with
$\delta_{\min}=10n+7m\ge18$, scaling $m$ and $n$ by a common factor if necessary.
Then $45\delta_{\min}/(153\delta_{\min}-1)<\gamma_*$, so
$E'(m,n)<\gamma_*+\varepsilon/2$, and Corollary~\ref{cor:latticeexponent} with loss
$\varepsilon/2$ gives the theorem.
\end{proof}

\section{Exact polar projection and the Hilbert sieve}\label{sec:polar}

The curvature vector $W$ is a multiple of an affine-linear vector
$G\mathbf k+\ell$ in the increments (Proposition~\ref{prop:polaridentity}). Away
from the single common zero of $G\mathbf k+\ell$, every zero of $J$ is
therefore a pole of $W$, and this raises the graph degree provided that
(i) little cancellation occurs at that base point and (ii) the pattern curve is
irreducible, of the full degree $24$. A local intersection calculation gives~(i), and a sieve
for branches at infinity gives~(ii). We then count the branch-bad and
degenerate patterns that these steps exclude, and obtain Theorem~\ref{thm:polar}.

\subsection{An exact identity at orbit states}

Recall the normalised parameters~\eqref{eq:jointbar}, with $\rho,\varphi>0>\chi$ and
$\gamma=(\rho,-\rho-\varphi,\varphi-\chi,\chi-1,1)$; primes indicate the next state.
We retain the notation $N_0$, $\Psi$, $P_4$, $R_4$ and $D_{jl}$ from
Section~\ref{sec:joint}. Write $\mathbf k=(k_0,k_1,k_2)^{\mathsf T}$ and
$\mathbf d=(d_0,d_1,d_2)^{\mathsf T}$, so that
$W=(D_{12},-D_{02},D_{01})^{\mathsf T}=\mathbf d\times\mathbf k$.

The increment and drop rows of the pattern matrix satisfy the same linear relations
(Lemma~\ref{lem:pattern}): the cofactor relations at $i$ and $i+1$ give
\[
 k_3=\rho k_0-\varphi k_1-\chi k_2,\qquad k_4=\rho'k_1-\varphi'k_2-\chi'k_3,
\]
and the same formulas express $d_3,d_4$ in terms of $\mathbf d$. For
$\mathbf y\in\mathbb Q^3$ extend $\mathbf y$ to $(y_0,\dots,y_4)$ by these formulas,
and let $\langle\mathbf y\rangle_j=y_0+\dots+y_{j-1}$ be its partial sums, so that
$\langle\mathbf k\rangle_j=s_j$ and $\langle\mathbf d\rangle_j=b_i-b_{i+j}$. Define $G$ and $\ell$ by
\[
 \sum_{j=0}^4\gamma_j\langle\mathbf y\rangle_j^2=\mathbf y^{\mathsf T}G\mathbf y,\qquad
 \sum_{j=0}^4\gamma_jy_j=\ell^{\mathsf T}\mathbf y.
\]
Explicitly,
\[
 G=\begin{pmatrix}
 \rho(\rho+1)&-\rho(\varphi-1)&-\rho(\chi-1)\\
 -\rho(\varphi-1)&\varphi(\varphi-1)&\varphi(\chi-1)\\
 -\rho(\chi-1)&\varphi(\chi-1)&\chi(\chi-1)
 \end{pmatrix},\qquad
 \ell=\begin{pmatrix}
 \rho(\chi-\chi')\\ \rho'-\rho-\varphi(\chi-\chi')\\ \varphi-\varphi'-\chi(\chi-\chi')
 \end{pmatrix},
\]
and $\det G=\rho(\rho+\varphi)(\varphi-\chi)(\chi-1)\ne0$. The pattern
quadratic is $f=\sum_j\gamma_j(s_j^2+k_j)=\mathbf k^{\mathsf T}G\mathbf k+\ell^{\mathsf T}\mathbf k$.
Put
\[
 V=G\mathbf k+\ell,\qquad\text{so that}\qquad f=\mathbf k\cdot V.
\]
The linear form $\mathbf y\mapsto\mathbf y\cdot V$ is the polarisation of $f$
at $\mathbf k$: it is $\mathbf y^{\mathsf T}G\mathbf k+\ell^{\mathsf T}\mathbf y$,
and it recovers $f$ at $\mathbf y=\mathbf k$. We call $V$ the \emph{polar vector}.

\begin{proposition}[Polar identity]\label{prop:polaridentity}
At every state $i$ with $i+7<N$,
\[
 \Psi W=N_0V,\qquad JW=N_0^2V.
\]
In particular, $W$ is a positive multiple of $V$.
\end{proposition}
\begin{proof}
The vector $V$ is orthogonal to $\mathbf k$, since $\mathbf k\cdot V=f=0$. It is
also orthogonal to $\mathbf d$; this is where the division equations enter.
By polarisation,
\[
 \mathbf d\cdot V=\sum_{j=0}^4\gamma_js_j\langle\mathbf d\rangle_j+\sum_{j=0}^4\gamma_jd_j.
\]
By the definition of $\gamma$, $\sum_j\gamma_j=0$. The relations of
Lemma~\ref{lem:pattern} give
$\sum_j\gamma_j\langle\mathbf y\rangle_j=\sum_ly_l(\gamma_{l+1}+\dots+\gamma_4)=0$ for
$\mathbf y=\mathbf k,\mathbf d$ (Lemma~\ref{lem:orthquartic}), that is,
$\sum_j\gamma_js_j=\sum_j\gamma_jb_{i+j}=0$. Multiply the division equations
$(q_i+s_j)b_{i+j}+b_{i+j+1}=a$ by $\gamma_j$ and sum over $j\le4$. The terms in $a$
and in $q_i$ cancel, and writing $b_{i+j}=b_i-\langle\mathbf d\rangle_j$ and
$b_{i+j+1}=b_{i+j}-d_j$ leaves
$-\sum_j\gamma_js_j\langle\mathbf d\rangle_j-\sum_j\gamma_jd_j=0$. Hence $\mathbf d\cdot V=0$.

Since $W\cdot(r_0,r_1,r_2)=\det(\mathbf d,\mathbf k,\mathbf r)=p_{012}=N_0>0$, $W\ne0$, so $\mathbf d$
and $\mathbf k$ are independent and $V$ is a multiple of $W=\mathbf d\times\mathbf k$.
To find the multiple, put $r_j=k_j(s_j+s_{j+1})+k_{j+1}$. Substituting the
definitions of $s_4$, $k_0,\dots,k_4$, $r_0,r_1,r_2$, $G$, $\ell$ and $\Psi$ and
expanding gives the polynomial identity
\[
 V\cdot(r_0,r_1,r_2)-\Psi=-(s_4+1+\chi-\chi')f
\]
in $s_1,s_2,s_3$ and the six normalised parameters; both sides are cubic in the
nodes. At an orbit state $f=0$, so $V\cdot r=\Psi$ while $W\cdot r=N_0$. Thus $\Psi W=N_0V$. Now use $J=N_0\Psi$
and $\Psi>0$.
\end{proof}

There is one point where this fails: at the common zero of the three forms,
a zero of $J$ need not be a pole of $W$, and every pattern curve passes
through that point. In projective coordinates $[\mathbf k:k_h]$, the three homogeneous forms
$G\mathbf k+\ell k_h$ have a unique common zero, the finite point
$\mathbf k_*=-G^{-1}\ell$. Since $V(\mathbf k_*)=0$, the identities $f=\mathbf k\cdot V$
and $V\cdot r-\Psi=-(s_4+1+\chi-\chi')f$ from the proof of
Proposition~\ref{prop:polaridentity} give $f(\mathbf k_*)=\Psi(\mathbf k_*)=0$. The
point also lies on the fixed-dividend curve for every $a$. Indeed, its nodes are
\[
 s_0^*=0,\qquad
 s_j^*=\frac{\gamma'_{j-1}-\gamma_{j-1}}{\gamma_j}\ (1\le j\le3),
 \qquad s_4^*=\chi'-\chi,
\]
as one checks by substituting them in $V$. Divide $Z(x)=x R_4(x)$ by $P_4(x)$
in the Newton basis $F_0,\dots,F_4$ at $s_0,\dots,s_3$, with $F_5=F_4(x-s_4)$.
Since $s_0=0$, we have $xN_j=F_{j+1}$ for the Newton polynomials $N_j$ of $R_4$,
so $Z=\sum_j\gamma'_jF_{j+1}$, while $xF_j=F_{j+1}+s_jF_j$. Hence, with
$c_Z=\chi'-\chi-s_4$, the quotient is $x+c_Z$ and the remainder
$Z-(x+c_Z)P_4$ has the Newton coefficients
\[
 -c_Z\gamma_0,\qquad
 \gamma'_{j-1}-\gamma_{j-1}-\gamma_j(s_j+c_Z)\quad(1\le j\le3);
\]
the coefficient of $F_4$ is $\gamma'_3-\gamma_3-s_4-c_Z=0$.
They all vanish at $\mathbf k_*$, so $Z_*=x P_{4,*}$ and
$N_0\operatorname{Res}(P_{4,*},Z_*)-a\Psi(\mathbf k_*)^4=0$. Hence $\mathbf k_*$ lies
on every pattern curve, and Lemma~\ref{lem:polarbase} below bounds the loss there.

We need $JW=N_0^2V$ as an identity in the function field, not only at good
starts, so we add the three cleared polar identities to the cuts
$\mathcal C_N,\mathcal C_R$ of Section~\ref{sec:jointrecon}. They have bounded degree, so, as in
Lemma~\ref{lem:jointcuts}, a component on which one of them does not vanish
identically contains $O(1)$ good starts. Throughout this section, \emph{retained}
means that all five cuts vanish identically; on a retained component,
\[
 JW=N_0^2V
\]
holds in the function field.

\subsection{The base point and the graph degree}

For an integral component $C$ of a pattern curve, the \emph{base divisor} of
$V$ is
\[
 \sum_P\Bigl(\min_j\operatorname{ord}_PV_j\Bigr)P,
\]
summed over the affine points $P$ of the normalisation of $C$; its degree
weights $P$ by $[\mathbb Q(P):\mathbb Q]$. Since
$V=G(\mathbf k-\mathbf k_*)$, it is supported over $\mathbf k_*$,
the only common zero of the homogenised forms.

\begin{lemma}[The base divisor]\label{lem:polarbase}
There is a fixed rational polynomial $\Theta$ in the ten minor coordinates,
independent of $a$ and not identically zero on the Pl\"ucker cone, with the
following property. For every pattern with $\Theta\ne0$ and every integral
component $C$ of the pattern curve, the base divisor of $V$ on
$C$ has degree at most four.
\end{lemma}
\begin{proof}
\emph{Step 1: the quartic initial form.} We work on the chart where $N_0N_1\ne0$, so that the normalised parameters are
defined, and
\[
 \gamma_1\gamma_2\gamma_3\cdot\det(G)\cdot G_{00}\cdot\ell_0\ne0.
\]
Here $\det G\ne0$ defines the base point $\mathbf k_*$, and
$\gamma_1\gamma_2\gamma_3\ne0$ defines its nodes below. Since
$\nabla f(\mathbf k_*)=-\ell$ and $\ell_0\ne0$, the quadric is smooth at
$\mathbf k_*$. Take $\xi_1=k_1-k_{*,1}$ and $\xi_2=k_2-k_{*,2}$ as local
coordinates. Since $G_{00}\ne0$, $f$ is a genuine quadratic in $k_0$, and
$\ell_0\ne0$ makes $\mathbf k_*$ a simple root of it; so we can solve $f=0$
for $k_0$ by formal power series.
Let $Z_1(x;\xi_1,\xi_2)$ be the linear part of the remainder of $Z$ modulo $P_4$
computed after Proposition~\ref{prop:polaridentity},
and let $\Psi_1(\xi_1,\xi_2)$ be the linear part of $\Psi$ on this tangent plane.
Let $F=N_0\operatorname{Res}(P_4,Z)-a\Psi^4$ be the fixed-dividend equation,
restricted to the quadric $f=0$. It has order at least four at $\mathbf k_*$, with
quartic initial form
\begin{equation}\label{eq:polarinitial}
 N_0\mathcal R_*(\xi_1,\xi_2)-a\Psi_1(\xi_1,\xi_2)^4,\qquad
 \mathcal R_*(\xi_1,\xi_2)=\operatorname{Res}_{x}(P_{4,*},Z_1).
\end{equation}
To see that only first derivatives enter, compute the resultant as a
$4\times4$ B\'ezout determinant. Each row vanishes at $\mathbf k_*$, so its term of
degree four uses exactly the linear part of every row. Higher terms in the
formal solution of $f=0$ cannot change~\eqref{eq:polarinitial}.

\emph{Step 2: the polynomial $\Theta$.} Consider the rational function of the
pattern parameters
\[
 M=\mathcal R_*(1,0)\Psi_1(0,1)^4-\mathcal R_*(0,1)\Psi_1(1,0)^4.
\]
If $M\ne0$ and $N_0\ne0$, the two axis coefficients in
\eqref{eq:polarinitial} cannot both vanish, for any $a$: if they did,
multiplying the two equations crosswise would give $N_0M=0$. Now write $M$, the
chart conditions and $N_0N_1$ as rational functions of the ten minor
coordinates, by~\eqref{eq:jointbar}, each in lowest terms, and let $\Theta$ be
the product of all their numerators and denominators. Then $\Theta\ne0$ implies
that each of these functions is defined and nonzero. To see that
$\Theta$ is not identically zero on the Pl\"ucker cone, take the maximal
minors of
\[
 \begin{pmatrix}1&0&0&2&10\\0&1&0&-5&-22\\0&0&1&3&9\end{pmatrix},
\]
which give $(\rho,\varphi,\chi,\rho',\varphi',\chi')=(2,5,-3,3,6,-5)$ and
$\mathbf k_*=(-1/7,-3/28,-1/2)$. There the chart conditions hold, and
Appendix~\ref{app:polarcertificates} computes the tangent frame, the quartic and
the first derivatives, giving
\[
 M=\frac{26291112152499}{60236288}\ne0.
\]

\emph{Step 3: from the initial form to the base divisor.}
Let $\mathcal O$ be the local ring of the smooth quadric at $\mathbf k_*$, with $F$ as in
Step~1. Choose $g=\xi_1$ or $g=\xi_2$ so that the complementary axis in
\eqref{eq:polarinitial} has nonzero coefficient. The completion of
$\mathcal O/(g)$ is a power-series ring in one variable, in which $F$ has order
exactly four. Consequently
$\dim_{\mathbb Q}\mathcal O/(F,g)=4$.
If $C$ passes through $\mathbf k_*$, its local ring $\mathcal O_C=\mathcal O/I_C$ is a
quotient with $F\in I_C$, so
\[
 \dim_{\mathbb Q}\mathcal O_C/g\mathcal O_C\le4.
\]
In particular, $g$ is nonzero on this one-dimensional integral local ring.

Let $\overline{\mathcal O}_C$ be its finite normalisation. The quotient
$\overline{\mathcal O}_C/\mathcal O_C$ is finite-dimensional over $\mathbb Q$.
Apply multiplication by $g$ to
$0\to\mathcal O_C\to\overline{\mathcal O}_C\to\overline{\mathcal O}_C/\mathcal O_C\to0$.
Multiplication is injective on the first two rings, and its kernel and cokernel
on the finite-dimensional quotient have equal dimensions. The resulting exact
sequence therefore gives
\[
 \dim_{\mathbb Q}\overline{\mathcal O}_C/g\overline{\mathcal O}_C
 =\dim_{\mathbb Q}\mathcal O_C/g\mathcal O_C
 =\sum_{P\mid\mathbf k_*}[\mathbb Q(P):\mathbb Q]\operatorname{ord}_P g.
\]
The final equality follows by factoring the ideal $(g)$ in the semilocal
Dedekind ring $\overline{\mathcal O}_C$. Since $G$ is invertible, the coordinates of
$V=G(\mathbf k-\mathbf k_*)$ generate the centred maximal ideal. Their minimum order
at each $P$ is at most $\operatorname{ord}_P g$. The weighted sum of these
minimum orders, which is the degree of the base divisor, is therefore at most
four. A component missing $\mathbf k_*$ has zero base divisor.
\end{proof}

Call a pattern \emph{degenerate} if $\Theta=0$ at its minor vector.

\begin{lemma}[The polar graph degree]\label{lem:polargraphdegree}
Let $\Theta\ne0$, and let $C$ be a retained nonexceptional component of degree
$24$. Then the powered graph $\Gamma_{m,n}$ of $C$, and the lattice graph
$\Gamma'_{m,n}$ of Section~\ref{sec:lattice}, have degree at least
$\delta_{\rm pol}:=60n+56m$.
\end{lemma}

\begin{proof}
At each point at infinity,
$\operatorname{ord}_P J=-5e_P/2$, by Corollary~\ref{cor:jointparity}, and
$\sum e_P[\mathbb Q(P):\mathbb Q]=24$. Thus $J$ has pole divisor, and hence
zero divisor, of degree $60$; all its zeros are affine. At an affine point $P$
the coordinates of $V$ are regular, and $W=N_0^2V/J$ gives
\[
 \max_j\bigl(-\operatorname{ord}_PW_j\bigr)
 =\operatorname{ord}_PJ-\min_j\operatorname{ord}_PV_j,
\]
where the subtracted term is the multiplicity of $P$ in the base divisor. Summing over
affine points, Lemma~\ref{lem:polarbase} gives an affine curvature pole envelope
of degree at least $60-4=56$, in place of the bound $3\cdot24/2+1=37$ of
Lemma~\ref{lem:jointpoledegree}. These poles are disjoint from the poles of $J$
at infinity, so the pole formula of Lemma~\ref{lem:jointpowereddegree} gives
\[
 \deg\Gamma_{m,n}\ge\frac{5n}2\cdot24+56m=60n+56m.
\]
An invertible constant change of curvature coordinates preserves the minimum
valuation, so the same bound holds for $\Gamma'_{m,n}$.
\end{proof}

It remains to show that components of degree less than $24$ occur for
relatively few patterns. Since the pattern curve has total degree at most $24$
(Lemma~\ref{lem:jointcurve}), a component of degree $24$ is the whole curve:
we must show that the pattern curve is usually irreducible over $\mathbb Q$.

\subsection{An explicit label for a branch at infinity}

The plan is as follows. At a branch at infinity of a nonexceptional
component, the residue field contains the ratio $t$ of two leading nodes,
which is a root of a fixed sextic, and a square root $\beta$ coming from the
fixed-dividend equation. If the minimal polynomial of $t+c\beta$ has degree
twelve for some slope $c$, every branch at infinity has residue degree at
least twelve; since its multiplicity $e$ is even, the component then has the
maximal degree $24$. The \emph{label} below is a polynomial whose
irreducibility certifies this, and which can be sieved over patterns.

Let $P$ be a point above infinity on the normalisation of a nonexceptional component.
The five leading nodes are distinct by Lemma~\ref{lem:jointleading}.
Scale them to $(0,1,t,y,z)$. The moment equations~\eqref{eq:joint-leading-moments}
of orders zero and one give $\gamma_0+\gamma_1+\gamma_2+\gamma_3+1=0$ and
\[
 z=-\gamma_1-\gamma_2t-\gamma_3y.
\]
Those of orders two and three become
\[
 Q_2(t,y)=\gamma_1+\gamma_2t^2+\gamma_3y^2+z^2=0,\qquad
 Q_3(t,y)=\gamma_1+\gamma_2t^3+\gamma_3y^3+z^3=0.
\]
Define the universal sextic $E_\gamma(t)$ by the polynomial identity
\begin{equation}\label{eq:branchsextic}
 \operatorname{Res}_y(Q_2,Q_3)=\gamma_3^2(\gamma_3+1)E_\gamma(t).
\end{equation}
The quotient on the right is a polynomial over
$\mathbb Z[\gamma_1,\gamma_2,\gamma_3]$, of degree six in $t$.
At an orbit pattern $\gamma_3=\chi-1<-1$, so
$\gamma_3^2(\gamma_3+1)\ne0$. Cancelling this factor in the resultant
and linear subresultant identities gives $E_\gamma(t)=0$ and
$\mathfrak d(t)y+\mathfrak n(t)=0$, where
\begin{align*}
 \mathfrak d(t)&=(\gamma_3-1)\bigl(\gamma_2(\gamma_2+\gamma_3+1)t^2
            +2\gamma_1\gamma_2t+\gamma_1(\gamma_1+\gamma_3+1)\bigr),\\
 \mathfrak n(t)&=\gamma_2(\gamma_2+1)(\gamma_2+\gamma_3+1)t^3
       +\gamma_1\gamma_2(3\gamma_2+\gamma_3+2)t^2\\
     &\quad+\gamma_1\gamma_2(3\gamma_1+\gamma_3+2)t
       +\gamma_1(\gamma_1+1)(\gamma_1+\gamma_3+1).
\end{align*}
These are polynomial identities, so the cleared relation remains valid when
a specialised denominator vanishes.

Write
\[
 \mathcal V_4=t y(t-1)(y-1)(y-t),\qquad \mathcal V_5=z(z-1)(z-t)(z-y).
\]
By~\eqref{eq:joint-resultant-leading}, the value of the leading part
$[\mathcal R]_{10}$ at these nodes is
$\mathcal R_0=\gamma_0\gamma_1\gamma_2\gamma_3\mathcal V_4\mathcal V_5$.
The four moment equations give the barycentric identity
\begin{equation}\label{eq:branchbarycentric}
 \mathcal R_0\mathcal V_4=\mathcal V_5^4.
\end{equation}
Indeed, write the nodes as $z_0,\dots,z_3,z_4=z$, with $\gamma_4=1$. Taking
$g=\prod_{l<4,\,l\ne j}(x-z_l)$ in $\sum_k\gamma_kg(z_k)=0$ (valid since $\deg g\le3$)
gives $\gamma_j\prod_{l<4,\,l\ne j}(z_j-z_l)=-\prod_{l<4,\,l\ne j}(z-z_l)$, and multiplying
by $z_j-z$ gives $\gamma_j\prod_{l\le4,\,l\ne j}(z_j-z_l)=\prod_{l<4}(z-z_l)=\mathcal V_5$ for
$0\le j<4$. In the product of these four identities each pair $l<m<4$ contributes
$-(z_m-z_l)^2$, and $\prod_{j<4}(z_j-z)=\mathcal V_5$, so the product is
$\gamma_0\gamma_1\gamma_2\gamma_3\mathcal V_4^2\mathcal V_5=\mathcal V_5^4$,
which is~\eqref{eq:branchbarycentric}.
The value
$\psi_P=(\Psi^2/s_1^5)(P)\in\mathbb Q(P)$ is defined and nonzero, because
$\operatorname{ord}_P\Psi=-5e_P/2$.
Taking leading terms in the curve equation gives
$N_0\mathcal R_0=a\psi_P^2$. Thus the residue field $\mathbb Q(P)$ contains
\begin{equation}\label{eq:branchsign}
 \beta=\mathcal V_5^2/\psi_P,\qquad N_0\beta^2=a\mathcal V_4.
\end{equation}
So $\mathbb Q(P)$ contains $\mathbb Q(t,\beta)$, which generically has degree
twelve, not just six.

For a slope $c$ define
\[
 D_\gamma(t)=-t(t-1)\,\mathfrak n(t)\bigl(\mathfrak n(t)+\mathfrak d(t)\bigr)\bigl(\mathfrak n(t)+t\mathfrak d(t)\bigr)
\]
and the label polynomial
\begin{equation}\label{eq:branchlabel}
 \mathfrak L_{a,N_0,c,\gamma}(\zeta)=
 \operatorname{Res}_t\bigl(E_\gamma(t),
       N_0\mathfrak d(t)^3(\zeta-t)^2-c^2aD_\gamma(t)\bigr).
\end{equation}
Here the resultant means the Sylvester determinant with degree bounds $6$
and $11$, padding missing coefficients with zero. This convention makes
\eqref{eq:branchlabel} a polynomial family under every specialisation.
The relation $\mathfrak d(t)y+\mathfrak n(t)=0$ gives $D_\gamma(t)=\mathfrak d(t)^3\mathcal V_4$, so
\eqref{eq:branchsign} implies
\[
 \mathfrak L_{a,N_0,c,\gamma}(t+c\beta)=0
\]
in every such residue field, including when $\mathfrak d(t)=0$.

\begin{lemma}[The degree-twelve label]\label{lem:branchlabel}
Fix $a\in\mathbb Q^\times$, and regard $N_0,\gamma_1,\gamma_2,\gamma_3$
as independent parameters. For some integer $c$ with $0\le c\le144$,
$\mathfrak L_{a,N_0,c,\gamma}$ is irreducible of degree twelve over
$\mathbb Q(N_0,\gamma_1,\gamma_2,\gamma_3)$.
For a specialised nonexceptional pattern, if one such label polynomial is
irreducible of degree twelve over $\mathbb Q$, every integral component of
its pattern curve has degree twenty-four; that is, the pattern curve is
irreducible over $\mathbb Q$.
\end{lemma}
\begin{proof}
At $(\gamma_1,\gamma_2,\gamma_3)=(-7,8,-4)$, the sextic in
\eqref{eq:branchsextic} equals $24E_\star(t)$, where
\[
 E_\star(t)=2400t^6-20160t^5+69776t^4-126448t^3
                  +125636t^2-64680t+13475.
\]
Appendix~\ref{app:polarcertificates} proves $E_\star$ irreducible modulo $71$.
This specialisation preserves degree six. Gauss's lemma therefore shows
that $E_\gamma$ is irreducible over
$K=\mathbb Q(\gamma_1,\gamma_2,\gamma_3)$: a factorisation into two factors of positive $t$-degree over the
parameter ring would specialise to two factors of positive degree.
Choose a root $t$ of $E_\gamma$, so $[K(t):K]=6$.
In this field $\mathfrak d(t)\ne0$ since its degree is less than six, and
$y=-\mathfrak n(t)/\mathfrak d(t)$. Each factor $t$, $t-1$, $\mathfrak n$,
$\mathfrak n+\mathfrak d$, $\mathfrak n+t\mathfrak d$ of
$D_\gamma=\mathfrak d(t)^3\mathcal V_4$ is a nonzero polynomial of degree at most
three, so none vanishes at $t$, and $\mathcal V_4\ne0$.

Adjoin the independent parameter $N_0$. The element $a\mathcal V_4/N_0$ has
odd valuation at $N_0=0$, so it is not a square in $K(t)(N_0)$.
Thus $(t,\beta)$ generates an extension of degree twelve over $K(N_0)$.
Among its twelve embeddings, equality of two values $t+c\beta$ excludes
at most one slope per pair, so at most $12^2=144$ slopes are excluded and one of
$0,\dots,144$ gives a primitive element. (Counting only unordered pairs of distinct
embeddings would give the smaller range $0,\dots,66$; we keep the cruder bound, which is the
one formalised, since only the finiteness of the range matters.) Its minimal polynomial divides~\eqref{eq:branchlabel},
whose degree is at most twelve; hence that label polynomial is irreducible
of degree twelve. This argument holds for each fixed nonzero $a$; the
chosen slope may depend on $a$.

For an actual pattern, the residue field at any branch at infinity contains
the root $t+c\beta$. Irreducibility of the specialised degree-twelve label
forces $[\mathbb Q(P):\mathbb Q]\ge12$.
Corollary~\ref{cor:jointparity} gives $e_P\ge2$, so
\[
 \deg C=\sum_{P\text{ at infinity}}e_P[\mathbb Q(P):\mathbb Q]\ge24.
\]
Lemma~\ref{lem:jointcurve} gives the reverse inequality.
\end{proof}

For each $a$, fix a slope $c(a)\in\{0,\dots,144\}$ given by
Lemma~\ref{lem:branchlabel}. Call a pattern \emph{branch-good} if
$\mathfrak L_{a,N_0,c(a),\gamma}$ is irreducible of degree twelve over $\mathbb Q$, and
\emph{branch-bad} otherwise.

\subsection{The Hilbert sieve}

The following monic variant of~\cite[Theorem~1.10]{CDHNV}, with explicit
logarithmic powers, is proved along the same lines in Appendix~\ref{app:hilbert}.

\begin{theorem}[Monic specialisation estimate]\label{thm:monichilbert}
For fixed integers $r,d\ge1$ there is a constant $C_{r,d}$ with the following
property. Let $F\in\mathbb Z[u_1,\dots,u_r][\zeta]$ be monic in $\zeta$ and
irreducible over $\mathbb Q(u_1,\dots,u_r)$. Suppose
$\deg_\zeta F\le d$, each coefficient of $F$ has total parameter degree at most
$d$, and every integer coefficient has absolute value at most $\mathcal H\ge1$.
For every integer $B\ge1$,
\[
 \#\{v\in[-B,B]^r\cap\mathbb Z^r:
              F(v,\zeta)\text{ is reducible over }\mathbb Q\}
 \le C_{r,d}(1+\log(\mathcal H+1))^4 B^{r-1/2}(1+\log B)^{4r+1}.
\]
\end{theorem}

The label polynomial~\eqref{eq:branchlabel} depends on the pattern only through
$N_0$ and $\gamma_1,\gamma_2,\gamma_3$, which are functions of the four minors on the
first four columns. We therefore sieve over these minors, and then count completions
to full patterns (Lemma~\ref{lem:branchpatterns}). To apply the theorem with $a$
fixed, call the four minors on the first four columns,
\[
 \mathbf p_\square=(p_{012},p_{013},p_{023},p_{123}),
\]
the \emph{vertex} of a minor vector $p$. For $p_{012}=N_0\ne0$ the vertex
determines the independent parameters above:
\begin{equation}\label{eq:branchvertex}
 \gamma_1=-\frac{p_{123}+p_{023}}{p_{012}},\qquad
 \gamma_2=\frac{p_{023}+p_{013}}{p_{012}},\qquad
 \gamma_3=-\frac{p_{013}+p_{012}}{p_{012}}.
\end{equation}
This change is birational. Substitute it in~\eqref{eq:branchlabel} and
clear denominators by a fixed power of $p_{012}$. To make the resulting
degree-twelve polynomial monic without changing its generic irreducibility,
use the following polynomial operation. For $F(\zeta)=\sum_{i=0}^e F_i\zeta^i$ put
\[
 \operatorname{mon}_e(F)(\zeta)=\zeta^e+\sum_{i<e}F_i F_e^{e-1-i}\zeta^i.
\]
When $F_e\ne0$, this is $F_e^{e-1}F(\zeta/F_e)$.
When $F_e=0$ and $e\ge2$, it is $\zeta^{e-1}(\zeta+F_{e-1})$, hence reducible.
Thus irreducibility of a specialised monic normalisation implies both
preservation of degree and irreducibility of the original label.

For each fixed positive integer $a$, use the slope $c(a)$ and set $e=12$.
We obtain a generically irreducible monic integer family in the vertex,
with uniformly bounded degree and coefficient height at most $C(a+1)^{C'}$,
for absolute constants $C,C'$. The same is true after dilation
$\mathbf p_\square=g\mathbf z$, with height at most $C(a+1)^{C'}g^{C'}$.
The slope may depend on $a$, but the bounds are uniform because there are
only $145$ choices. Counting vertices with $p_{012}=0$ as bad, which adds the
$O(B^3)$ points of that slice, Theorem~\ref{thm:monichilbert} gives
\begin{equation}\label{eq:badvertices}
 \#\{\mathbf z\in[-B,B]^4\cap\mathbb Z^4:
                  g\mathbf z\text{ is branch-bad}\}
 \ll_\eta a^\eta B^{7/2}
\end{equation}
whenever $1\le g,B\le a^7$. Any fixed power of $a$ would do here: the
range only has to contain the dilations and boxes used for orbit patterns
below, which are at most the pattern height $X\ll a^3$
(by~\eqref{eq:jointscales}, or directly from $X\ll A^4H^6/a^3$ and
$H\ll a/A$).

\begin{lemma}[Branch-bad full patterns]\label{lem:branchpatterns}
Among the minor vectors of coordinate height at most $X$ with $p_{012}\ne0$,
the number that are branch-bad is $O_\eta(a^\eta X^{9/2})$.
\end{lemma}
\begin{proof}
Since $p_{012}\ne0$, the vertex is nonzero. Partition the vertices into dyadic
bands with $|\mathbf p_\square|_\infty\asymp R$, $1\le R\le X$, and choose a maximal coordinate as signed pivot $\kappa$.
The other three vertex coordinates form $\mathbf u$; put
$g=\gcd(\kappa,\mathbf u)$. Completing the fifth column amounts to choosing
$\mathbf v\in[-X,X]^3$ with $\kappa\mid\mathbf u\times\mathbf v$, as in the proof of
Proposition~\ref{prop:minorvec}. Since $\kappa$ is maximal only among the
first four columns, $\mathbf v$ ranges over the full $X$-box rather than
$[-\kappa,\kappa]^3$, and the congruence count is
\[
 O(X^3g^2/R^2).
\]
Apply~\eqref{eq:badvertices} after dilation by $g$.
There are $O_\eta(a^\eta(R/g)^{7/2})$ bad vertices in this common-factor
class; imposing the exact gcd only reduces this count. Their total
contribution is at most
\[
 O_\eta\left(a^\eta X^3R^{3/2}\sum_{g\ge1}g^{-3/2}\right).
\]
The series converges, and summing the bands $R\le X$ proves the claim.
The finitely many pivot choices only change the constant.
\end{proof}

\subsection{Counting degenerate patterns}

The degenerate patterns, on the hypersurface $\Theta=0$, must be counted with a saving over
the $X^{5+o(1)}$ patterns in total. The following estimate saves $X^{1/3}$.

\begin{proposition}[Hypersurface pattern count]\label{prop:polarpatterns}
Let $\Theta$ be a rational polynomial in the ten minor coordinates, with
$\Theta\ne0$ at the minor vector of some rational $3\times5$ matrix $\mathcal M_0$.
For every $\eta>0$, the number of integral points of the Pl\"ucker cone of
coordinate height at most $X\ge1$ satisfying $\Theta=0$ is
\[
 O_{\Theta,\eta}(X^{14/3+\eta}).
\]
The count includes nonprimitive vectors and the zero vector.
\end{proposition}
\begin{proof}
Choose a minor of maximal absolute value $\kappa>0$ and record its sign.
The six one-column replacements form $\mathbf u,\mathbf v\in[-\kappa,\kappa]^3$, and the other
coordinates are signed entries of $(\mathbf u\times \mathbf v)/\kappa$.
There are twenty signed pivot charts. In each chart, clearing denominators
in $\Theta$ gives a nonzero polynomial $\Theta_\pi(\kappa,\mathbf u,\mathbf v)$.
Indeed, the product of the pivot minor and $\Theta$ at the minor vector of $\mathcal M_0$
is a nonzero polynomial in the matrix entries, so it is nonzero at some
rational matrix; every pivot chart meets the nonvanishing locus.

Put $g=\gcd(\kappa,\mathbf u)$, ${\bar\kappa}=\kappa/g$ and $\mathbf u_0=\mathbf u/g$.
Since $\gcd({\bar\kappa},\mathbf u_0)=1$, the congruence $\mathbf u_0\times \mathbf v\equiv0\pmod {\bar\kappa}$
is equivalent to
\[
 \mathbf v\in\Lambda:=\mathbb Z\mathbf u_0+{\bar\kappa}\mathbb Z^3.
\]
This lattice contains ${\bar\kappa}\mathbb Z^3$, so Lemma~\ref{lem:reduce} gives a basis
$\mathbf f_1,\mathbf f_2,\mathbf f_3$ of $\Lambda$. We need two bounds on their lengths.
\begin{enumerate}
\item[(a)] $|\mathbf f_i|\ll {\bar\kappa}$. The vectors ${\bar\kappa}\mathbf e_1,{\bar\kappa}\mathbf e_2,{\bar\kappa}\mathbf e_3$ lie
in $\Lambda$ and span a sublattice of full rank, so for each $i$ one of them
has a nonzero $i$th coordinate in the basis, and the coordinate bound of
Lemma~\ref{lem:reduce} gives $27|\mathbf f_i|^2\le64{\bar\kappa}^2$.
\item[(b)] $|\mathbf f_i||\mathbf f_j|\ge {\bar\kappa}\lambda^\perp(\mathbf u_0)$ for $i\ne j$, where
$\lambda^\perp(\mathbf u_0)$ is the least sup norm of a nonzero vector of the
normal lattice $\Lambda_{\bar\kappa}^\perp=\{z\in\mathbb Z^3:{\bar\kappa}\mid z\cdot\mathbf u_0\}$.
Indeed, if $x,y\in\Lambda$ are independent, then $x\times y={\bar\kappa}z$ for a nonzero
$z\in\Lambda_{\bar\kappa}^\perp$, so $|x||y|\ge|x\times y|\ge {\bar\kappa}\lambda^\perp(\mathbf u_0)$.
\end{enumerate}
By Lemma~\ref{lem:reduce}, the vectors $\mathbf v\in\Lambda$ with
$|\mathbf v|_\infty\le\kappa$ have basis coordinates in a box with sides
$B_i=O(\kappa/|\mathbf f_i|)$, and by~(a) and ${\bar\kappa}\le\kappa$ these sides are $\gg1$. A
nonzero polynomial of bounded degree vanishes at
$\ll B_1B_2B_3/\min_iB_i$ integer points of such a box (the polynomial
grid-zero bound). Applied to a nonzero fibre $\Theta_\pi(\kappa,\mathbf u,\cdot)$
in the basis coordinates, this and~(b) give
\begin{equation}\label{eq:polarfiber}
 \#\{\mathbf v\in\Lambda\cap[-\kappa,\kappa]^3:\Theta_\pi(\kappa,\mathbf u,\mathbf v)=0\}
 \ll_\Theta\sum_{i<j}\frac{\kappa^2}{|\mathbf f_i||\mathbf f_j|}
 \ll_\Theta\frac{\kappa^2}{{\bar\kappa}\lambda^\perp(\mathbf u_0)}.
\end{equation}

Here is the reciprocal estimate that produces the exponent $14/3$:
\begin{equation}\label{eq:normalreciprocal}
 \sum_{\mathbf u_0\in[-{\bar\kappa},{\bar\kappa}]^3\cap\mathbb Z^3}\frac1{\lambda^\perp(\mathbf u_0)}
 \ll\tau({\bar\kappa}){\bar\kappa}^{8/3}.
\end{equation}
It holds for all $\mathbf u_0$, without a primitivity restriction.
Pigeonholing $z\mapsto \mathbf u_0\cdot z\pmod {\bar\kappa}$ gives $\lambda^\perp(\mathbf u_0)\ll {\bar\kappa}^{1/3}$.
For $R\ge1$, Lemma~\ref{lem:dotcount} and a divisor sum give
\begin{align*}
 \#\{\mathbf u_0:\lambda^\perp(\mathbf u_0)\le R\}
 &\le\sum_{0<|z|_\infty\le R}\#\{\mathbf u_0\in[-{\bar\kappa},{\bar\kappa}]^3:{\bar\kappa}\mid \mathbf u_0\cdot z\}\\
 &\ll {\bar\kappa}^2\sum_{0<|z|_\infty\le R}\gcd({\bar\kappa},z_1,z_2,z_3)
 \ll\tau({\bar\kappa}){\bar\kappa}^2R^3.
\end{align*}
For the last bound, expand the gcd as a divisor sum and note that a
nonzero $z$ of size at most $R$ has no common divisor greater than $R$.
Partial summation up to $K\asymp {\bar\kappa}^{1/3}$ now bounds the reciprocal sum by
$O({\bar\kappa}^3/K+\tau({\bar\kappa}){\bar\kappa}^2K^2)$, proving~\eqref{eq:normalreciprocal}.

For fixed $\kappa,g$, equations~\eqref{eq:polarfiber} and
\eqref{eq:normalreciprocal} bound the nonzero fibres by
$O_\Theta(\tau({\bar\kappa})\kappa^2{\bar\kappa}^{5/3})$. Sum over $g\mid \kappa$ and $\kappa\le X$ to obtain
\[
 O_\Theta\bigl(\tau_{\max}(X)^2X^{14/3}\bigr).
\]

Finally, an identically zero fibre forces one fixed nonzero coefficient
polynomial $\Theta_0(\kappa,\mathbf u)$ to vanish. For fixed $g$, the polynomial
$\Theta_0(g{\bar\kappa},g\mathbf u_0)$ is still nonzero. The four-variable grid-zero bound supplies
$O_\Theta((X/g)^3)$ possible pairs $({\bar\kappa},\mathbf u_0)$. Each has
$O(\kappa g^2)\le O(Xg^2)$ completions in its $\kappa$-box, so these fibres total
\[
 O_\Theta\left(X^4\sum_{g\le X}\frac1g\right)
 =O_\Theta(X^4\log(2X)).
\]
The divisor estimate absorbs $\tau_{\max}(X)^2$ into $X^\eta$.
Summing the charts and adding the zero vector completes the proof.
\end{proof}

\subsection{The window estimate and the polar exponent}

Retain the window notation $[A,2A)$, $T\ge36$, and the heights $H$, $X$, $S$
and $Y$ of Sections~\ref{sec:joint} and~\ref{sec:lattice}.
Only windows satisfying~\eqref{eq:jointscale} need the new estimate.

\begin{proposition}[Polar window estimate]\label{prop:polarwindow}
Fix positive integers $m,n$ and $\eta>0$, and let $\delta_{\rm pol}=60n+56m$.
Every window above satisfying~\eqref{eq:jointscale} obeys
\begin{equation}\label{eq:polarsix}
\begin{split}
 T\ll_{m,n,\eta}a^\eta\bigl(&X^5+X^4H^{1/2}
 +X^{14/3}\max(H,S,Y)^{1/17}\\
 &+X^{5-1/(3\delta_{\rm pol})}H^{1/\delta_{\rm pol}}
 +X^{5-2m/(3\delta_{\rm pol})}S^{m/\delta_{\rm pol}}+X^5Y^{n/\delta_{\rm pol}}\bigr).
\end{split}
\end{equation}
\end{proposition}
\begin{proof}
The earlier exceptional patterns, conics and non-retained components contribute
$X^5+X^4H^{1/2}$, as in Proposition~\ref{prop:latticewindow}.
Among the remaining patterns, Lemma~\ref{lem:branchpatterns} and
Proposition~\ref{prop:polarpatterns} bound those that are branch-bad or
degenerate by $O_\eta(a^\eta X^{14/3})$; orbit patterns have $p_{012}=N_0\ge1$, as
Lemma~\ref{lem:branchpatterns} requires.
Here $X\ll a^3$, so the extra $X^\eta$ loss is absorbed into $a^\eta$.
For these patterns use the unpowered graph of Section~\ref{sec:joint}
($m=n=1$, so $\delta_{\min}=17$), whose box has size $\max(H,S,Y)$; this gives
the third term of~\eqref{eq:polarsix}.

For each remaining retained component, Lemma~\ref{lem:branchlabel}
gives degree $24$, and Lemma~\ref{lem:polargraphdegree} gives graph degree
at least $\delta_{\rm pol}$. Its upper degree is bounded in terms of $m,n$ by
Lemma~\ref{lem:latticedegree}. The coefficient-uniform curve bound
and Lemma~\ref{lem:latticecoords} therefore give, for a pattern with shortest
lattice length $\lambda$, a contribution
\[
 \ll_{m,n,\eta}a^\eta\bigl((H/\lambda)^{1/\delta_{\rm pol}}+(S/\lambda^2)^{m/\delta_{\rm pol}}+Y^{n/\delta_{\rm pol}}\bigr).
\]
Summing with Lemma~\ref{lem:weights}, at exponents $1/\delta_{\rm pol}$ and $2m/\delta_{\rm pol}$,
produces the last three terms.
\end{proof}

\begin{theorem}[Polar bound]\label{thm:polarconditional}
For every fixed pair of positive integers $m,n$ and every $\varepsilon>0$,
\[
 P(a,b)\ll_{m,n,\varepsilon} a^{E_{\rm pol}(m,n)+\varepsilon}
 \quad(0<b<a),
\]
where, with $\delta_{\rm pol}=60n+56m$,
\[
 E_{\rm pol}(m,n)=\max\left\{\frac5{17},\frac{45\delta_{\rm pol}}{153\delta_{\rm pol}-1},
 \frac{45\delta_{\rm pol}-3m}{153\delta_{\rm pol}-11m},\frac{15\delta_{\rm pol}+6n}{51\delta_{\rm pol}+20n}\right\}.
\]
For $(m,n)=(3,2)$ this exponent is $4317/14677$.
Optimising over the fixed powers gives
\[
 P(a,b)\ll_\varepsilon a^{\beta_*+\varepsilon},\qquad
 \beta_*=\frac{41301-3\sqrt{82849}}{137480}
        =0.2941336553\ldots.
\]
\end{theorem}
\begin{proof}[Proof of Theorems~\ref{thm:polarconditional} and~\ref{thm:polar}]
Apply the window argument of Section~\ref{sec:lattice} to
\eqref{eq:polarsix}. The three terms from the contribution
$X^{14/3}\max(H,S,Y)^{1/17}$ of branch-bad and degenerate patterns give exponents
$717/2440$ (twice, since $H$ and $S$ satisfy the same bound $\ll a^{1+o(1)}/T^3$) and $732/2491$, all below $5/17$.
The other terms give the displayed maximum. Windows with $AT>28a$, or
failing~\eqref{eq:jointscale}, are handled as in the proof of
Corollary~\ref{cor:jointexponent} and give at most $5/17$; dyadic summation and
the factor $a^\eta$ absorb the remaining losses.

For $(m,n)=(3,2)$, $\delta_{\rm pol}=288$ and the last three fractions are
$12960/44063$, $4317/14677$ and $1083/3682$.
For optimisation put $r=n/m$. The curvature and rank-five fractions become
\[
 E_S^{\rm pol}(r)=\frac{2700r+2517}{9180r+8557},\qquad
 E_Y^{\rm pol}(r)=\frac{906r+840}{3080r+2856}.
\]
The first decreases and the second increases. They cross at
$r_*=(\sqrt{82849}-47)/360$, with common value $\beta_*$.
Approximate $r_*$ by positive rationals and scale the corresponding
numerator and denominator until $45\delta_{\rm pol}/(153\delta_{\rm pol}-1)<\beta_*$.
The arbitrarily small exponent loss absorbs the approximation, proving
Theorem~\ref{thm:polar}.
\end{proof}

\section{Rises in the quotient increments}\label{sec:rises}

Throughout this section $k_i=q_{i+1}-q_i\ge1$, $d_i=b_i-b_{i+1}\ge1$, and a
\emph{rise} is an index $0\le i\le N-3$ with $k_i<k_{i+1}$.

\begin{lemma}[Six quotients in arithmetic progression]\label{lem:ap}
Let $q,h\ge1$ and $x_0>x_1>\dots>x_6\ge0$ satisfy $x_j(q+jh)+x_{j+1}=a$ for $0\le j\le5$. Then
$q^4\le 324\,h^3a$.
\end{lemma}

\begin{proof}
Let $f_j=x_j-x_{j+1}$, and let $E=(f_1-f_2)h$ and $F=(f_2-f_3)h$ be the curvatures of the blocks
starting at $x_1$ and $x_2$ (all their increments equal $h$). With $U=f_0-2f_1+f_2$, the
division equations $\mathcal E_j\colon x_j(q+jh)+x_{j+1}-a=0$ give the exact identity
\[
qhU=(3h-1)E+F,\qquad\text{namely}\qquad qhU-(3h-1)E-F=h(\mathcal E_0-3\mathcal E_1+3\mathcal E_2-\mathcal E_3).
\]
This is the case $k_0=k_1=k_2=k_3=h$ of Lemma~\ref{lem:six}, where $L=2h^3U$. By
Lemma~\ref{lem:curv}, $E,F\ge1$, so the integer $U$ is at least $1$.

Now bound $E$ and $F$. Subtracting consecutive division equations gives
$q_jf_j\le hx_{j+1}$, where $q_j=q+jh$. Since $x_{j+1}q_{j+1}\le a$, this yields
$q^2f_j\le ha$ for $1\le j\le4$. By Lemma~\ref{lem:size},
$Ex_3\le(h+1)(f_1+f_2+f_3)^2\le 9h^2(h+1)a^2/q^4$. Also $a<x_3(q+3h+1)$, so
$Eq^4\le 9h^2(h+1)(q+3h+1)a$, and likewise $Fq^4\le 9h^2(h+1)(q+4h+1)a$. Hence
\[
q^5h\le q^4\bigl((3h-1)E+F\bigr)\le 27h^3(h+1)(q+4h+1)a .
\]
If $q\ge h$ this gives $q^4\le 324h^3a$. If $q<h$ then $q^4<h^4\le h^3a$, since $h\le q_1\le a$.
\end{proof}

\begin{lemma}[Drops decrease between rises]\label{lem:norise}
Whenever $b_{i+2}>0$, $d_{i+1}k_i<d_ik_{i+1}$. In particular, if $k_{i+1}\le k_i$ then
$d_{i+1}<d_i$.
\end{lemma}

\begin{proof}
This is $D=d_ik_{i+1}-d_{i+1}k_i\ge1$ (Lemma~\ref{lem:curv}). If $k_{i+1}\le k_i$ and
$d_{i+1}\ge d_i$, then $d_ik_{i+1}\le d_{i+1}k_i$, which is a contradiction.
\end{proof}

\begin{lemma}[Bounded multiplicity]\label{lem:mult}
Let $c_j\ge1$ be integers, $j\in J$, such that no value occurs more than $M$ times. Then
$|J|^2\le 2M\sum_{j\in J}c_j$.
\end{lemma}

\begin{proof}
Induct on $|J|$. Let $c_{j_0}$ be the largest value. The values lie in $[1,c_{j_0}]$, so
$|J|\le Mc_{j_0}$. Removing $j_0$ and using $|J|^2=(|J|-1)^2+2|J|-1$ completes the induction.
\end{proof}

\begin{proposition}[Rises in a window]\label{prop:rises}
Let $x_0,\dots,x_{T-1}$ be a window run in $[A,2A)$ with $T\ge2$, and let $r$ be the number
of rises among its $T-1$ increments. Then:
\begin{enumerate}
\item $(T-1)^2\le 2(r+1)(A-1)$;
\item $T-1\le\lfloor a/A\rfloor$;
\item if $6A\le a$, then $T-1=\ell+s$ with $\ell^2\le 8(r+1)\lfloor a/A\rfloor$ and
$s^3\le c_{\rm low}A$, where $c_{\rm low}=324\cdot3^4=26244$.
\end{enumerate}
\end{proposition}

\begin{proof}
Split the increments into \emph{stretches}: maximal blocks containing no rise. There are at most
$r+1$ stretches.

(1) By Lemma~\ref{lem:norise}, the drops strictly decrease along a stretch, so each value of $d_j$
occurs at most once per stretch, hence at most $r+1$ times. Lemma~\ref{lem:mult} and
$\sum_jd_j=x_0-x_{T-1}\le A-1$ give the claim.

(2) The increments are $\ge1$, and they sum to $q_{T-1}-q_0\le\lfloor a/A\rfloor$.

(3) In $c_{\rm low}=324\cdot3^4$, the factor $324$ comes from Lemma~\ref{lem:ap}
and $3^4$ from the bound $a<3Aq$ below. Call an increment $h$ \emph{low} if
$c_{\rm low}h^3A^4<a^3$. Five consecutive equal increments
$h$ inside a stretch give six quotients in arithmetic progression. Lemma~\ref{lem:ap} then
gives $q^4\le324h^3a$. On the other hand, $a<3Aq$, which follows from $a=qx_j+x_{j+1}$, where $x_j$ is the
first term of the progression, $x_{j+1}<x_j<2A$ and $6A\le a$. Together these give $a^3\le c_{\rm low}h^3A^4$, so $h$ is not low.
Within a stretch the increments are nonincreasing, so equal values are consecutive. Hence each
low value occurs at most $4$ times per stretch, and at most $4(r+1)$ times in all.
Lemma~\ref{lem:mult} bounds the number $\ell$ of low increments by
$\ell^2\le8(r+1)\sum_jk_j\le8(r+1)\lfloor a/A\rfloor$. Let $s$ be the number of high
increments. If $s=0$, its bound is immediate. Otherwise let $h_*$ be their
smallest value. Then $sh_*\le\lfloor a/A\rfloor$ and
$a^3\le c_{\rm low}h_*^3A^4$, so $s^3a^3\le c_{\rm low}A^4(a/A)^3=c_{\rm low}Aa^3$.
\end{proof}

\begin{lemma}[Geometric sums]\label{lem:geom}
Let $\alpha,\beta,s,t>0$ and $K\ge0$. Then
\[
\sum_{k<K}\min\bigl(\alpha2^{-sk},\ \beta2^{tk}\bigr)\le
\Bigl(\frac1{1-2^{-s}}+\frac{2^t}{2^t-1}\Bigr)\alpha^{t/(s+t)}\beta^{s/(s+t)} .
\]
\end{lemma}

\begin{proof}
Let $M=\alpha^{t/(s+t)}\beta^{s/(s+t)}$. Then $(\alpha2^{-sk})^t(\beta2^{tk})^s=M^{s+t}$, so for each
$k$ either $\beta2^{tk}\le M$ or $\alpha2^{-sk}<M$. The terms $\beta2^{tk}\le M$ grow with ratio
$2^t$, so they sum to at most $M2^t/(2^t-1)$. The terms $\alpha2^{-sk}<M$ decay with ratio $2^{-s}$,
so they sum to at most $M/(1-2^{-s})$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:rises}]
The rises inside a window are rises of the orbit, so $r+1\le R$. Combining the three parts
of Proposition~\ref{prop:rises} gives, for the window $[2^k,2^{k+1})$,
\[
T_k\le 1+\sqrt R\,\min\bigl(\sqrt{8a}\,2^{-k/2},\ \sqrt2\,2^{k/2}\bigr)
+\min\bigl(a\,2^{-k},\ 30\cdot2^{k/3}\bigr);
\]
when $6A>a$, part~(2) gives $T-1\le a/A<6\le30A^{1/3}$ instead of part~(3).
Lemma~\ref{lem:geom} gives
\[
 \sum_k\min(\sqrt{8a}\,2^{-k/2},\sqrt2\,2^{k/2})\ll(\sqrt{8a}\sqrt2)^{1/2}\ll a^{1/4}.
\]
It also gives
$\sum_k\min(a2^{-k},30\cdot2^{k/3})\ll a^{1/4}$. Since $\log_2a+1\ll a^{1/4}$, summing over $k$
gives the theorem.
\end{proof}

\begin{remark}
By Theorem~\ref{thm:rises}, an orbit with $N\ge a^\theta$, $\theta>1/4$, must have
$R\gg a^{2\theta-1/2}$; for $\theta$ near the exponent of Theorem~\ref{thm:polar} this is
about $a^{0.088}$. Proposition~\ref{prop:rises}(3) also localises the rises. There is an
absolute constant $c>0$ such that if $6A\le a$, $A\le ca^{3\theta}$ and a window run in
$[A,2A)$ has $T\ge a^\theta$ terms, then $s\le T/4$, so $\ell\ge T/4$ and the number $r$ of
rises in the run satisfies $r+1\gg Aa^{2\theta-1}$.
\end{remark}

\section{The Lean formalisation}\label{sec:lean}

The source repository is
\url{https://github.com/jasondavies/pierce-length-bound}.
The proof baseline is commit
\lean{6da6511cea9f8526054dd5ad60d77164f45de52e},
which is registered in the Palomar registry as
\href{https://palomar-registry.org/entry?id=PALOMAR-2026-10-27-000017&version=1}{\lean{PALOMAR-2026-10-27-000017}},
version~1.
The development uses Lean~4 \lean{v4.35.0} and Mathlib \lean{v4.35.0}.
Every theorem stated in this paper is proved there in full; the one
conditional result in the repository is described below.

The strongest bound is exported as
\begin{quote}\ttfamily\small
theorem P\_le\_polar\_optimal :\\
\hspace*{2em}$\forall$ $\varepsilon$ > (0 : $\mathbb R$), $\exists$ C > 0, $\forall$ a b : $\mathbb N$,\\
\hspace*{2em}0 < b $\to$ b < a $\to$ (P a b : $\mathbb R$) $\le$\\
\hspace*{2em}C * (a : $\mathbb R$) \^{} (((41301 : $\mathbb R$) - 3 * Real.sqrt 82849) / 137480 + $\varepsilon$)
\end{quote}
The quantifier order makes uniformity explicit: $C$ may depend on
$\varepsilon$, but not on $a$ or $b$. The function \texttt{P a b} is defined
by well-founded recursion,
\[
 P(a,0)=0,\qquad P(a,b)=1+P(a,a\bmod b)\quad(b>0),
\]
and counts the positive terms of the zero-indexed orbit, as in Section~\ref{sec:intro}.
The rational bound $4317/14677$ is exported separately as
\lean{P_le_4317_14677}. Both results are assembled in
\lean{PolarProjectionMain} from proved arithmetic, geometric and
counting estimates. The repository README gives the correspondence between
the other results and their Lean declarations.

The affine-curve theorem~\eqref{eq:jointexternal} is imported from the separate
\href{https://github.com/jasondavies/ccdn-dimension-growth/tree/bb025e7207f3a35a21b898629f5678b72aa7e44f}{\lean{ccdn-dimension-growth} library at revision \lean{bb025e7}},
which is registered in the Palomar registry as \lean{PALOMAR-2026-10-22-000015}, version~1.
That development proves the plane estimate, degree-preserving projection
and box rescaling, using the same affine Hilbert degree as this paper.
The present repository proves the monic Hilbert irreducibility estimate
(Theorem~\ref{thm:monichilbert}) and its application to the branch-label
family, so neither counting input is assumed.

The repository also contains a conditional refinement of
Theorem~\ref{thm:main}, \lean{P_le_loglog}, which proves
$P(a,b)\ll a^{3/10}\sqrt{\log\log3a}$ from two explicit hypotheses: the
short-sum estimate for the Erd\H{o}s--Hooley function from~\cite{NT}, over
intervals $(x,x+y]$ with $y\ge x^{1/4}$, and the mean-value estimate
from~\cite{BT}. Their statements include the constants and thresholds; their
proofs are not formalised here. This result is weaker than
Theorem~\ref{thm:polar} and is not used elsewhere.

The proof library contains no \lean{sorry} and uses no \lean{native_decide}.
Its axiom audit permits only \lean{propext}, \lean{Classical.choice} and
\lean{Quot.sound}, the standard axioms used by Mathlib. Sixteen public
statements are compared with independent Mathlib-only specifications, and
their proofs are replayed by Lean, NanoDa and con-ron. The specification
file \lean{PierceLengthBoundPalomar/Challenge.lean} deliberately contains theorem holes; it is
separate from the proved solution. Build instructions, the full theorem
map, and the verification commands are recorded in the README.

The divisor estimate uses \lean{DivisorPowerBound.lean} from
\lean{openai/math}~\cite{OAIMath} under Apache-2.0, with unchanged proof
bodies and adapted imports and module visibility. The repository records
the source revision and the provenance of its other dependencies.

\section{Limits of the method}\label{sec:limits}

The polar refinement brings the exponent close to $5/17$, but does not
remove that threshold. As explained in Section~\ref{sec:intro}, $5/17$ is the
exponent obtained if every pattern carries $O(1)$ starts, and the estimates
here never do better than that: the non-retained components and conics contribute
$a^{o(1)}$ starts for each of the $X^{5+o(1)}$ patterns, and the exceptional patterns and
the windows where the line-exclusion condition~\eqref{eq:jointscale} fails
also lead to exponent $5/17$, independently of the coordinate powers $m,n$.
The branch-bad and degenerate patterns give only the exponents
$717/2440$ and $732/2491$, below this threshold.

The improved graph degree gives
\[
 \beta_*-\frac5{17}=0.00001600826\ldots.
\]
Increasing the graph degree further can reduce this gap. Going below
$5/17$ requires an improvement in the other counting contributions, or a
different way to combine the information from successive orbit states.
This is a limitation of the present estimates, not a lower bound for the
Pierce iteration.

The elementary $3/10$ argument has a similar barrier: even with no divisor
loss, its two window bounds balance at $A\asymp a^{3/5}$
(Section~\ref{sec:proof310}), so better constants or short-interval divisor
estimates alone do not change that exponent.

Finally, there is no obvious monotone quantity along an orbit that could
replace the window-by-window counting; in particular the curvatures need not
decrease. For $a=13242$, the orbit of
$255$ begins
\[
 255,\ 237,\ 207,\ 201,\ 177,\ 144,\ 138,\ 132,\ldots
\]
with quotients $51,55,63,65,74,91,95,100,\ldots$ and first five curvatures
$24,12,6,111,30$. All these remainders lie in the single window $[128,256)$.
The rise bound gives one constraint on the order of the increments, but
the present argument does not capture all compatibility conditions along
an entire orbit. Shallit's conjectured $O((\log a)^2)$ bound remains open.

\appendix

\section{Conic components}\label{app:conic}

In this appendix we prove Proposition~\ref{prop:jointconic}. Let $C$ be a nonexceptional
component of degree two. A component that is not retained
already contributes $O(1)$ starts, by Lemma~\ref{lem:jointcuts}. We may therefore assume that the component
$C$ is retained. If it contains at most two free-node triples of good starts,
there is nothing to prove. In the remaining case we first construct a
parameter whose value is an integer at every such triple.

\begin{lemma}[A bounded integral parameter]\label{lem:jointconicparameter}
Let $C$ be a retained nonexceptional degree-two component containing the
free-node triples of at least three good starts. After permuting the three
free-node coordinates to $(x,y,z)$, there are
an integer $\Delta_c\ne0$ and polynomials $\xi_0,\xi_1,\xi_2\in\mathbb Z[t]$ such
that
\[
 \deg \xi_j\le2,\qquad
 |\Delta_c|,\ \max_{j,k}|[t^k]\xi_j|\le(aX)^{O(1)}.
\]
The map
\[
 t\longmapsto
 \bigl(\xi_0(t),\xi_1(t),\xi_2(t)\bigr)/\Delta_c
\]
parametrises $C$ over $\mathbb Q$. Its inverse takes an integer value at
every integer point of $C$.
\end{lemma}
\begin{proof}
Every free node of a good start has absolute value at most $a$. Three distinct
such triples lie on an integer plane
\[
 \lambda_zz+\lambda_xx+\lambda_yy+\lambda_1=0
\]
with coefficients of size $a^{O(1)}$: take a cross product of two differences
as the normal, or, if it vanishes, a nonzero perpendicular vector of the same
size. A plane not containing an integral curve
of degree two meets it in at most two distinct points. Thus this plane
contains $C$. Choose the eliminated coordinate $z$ with $\lambda_z\ne0$.

The six rational pattern coordinates~\eqref{eq:jointbar} have common denominator
$N_0N_1\ne0$ and integer numerators of size at most $X^2$, so the pattern
quadratic $f$ has rational coefficients with numerators and denominators of size
$X^{O(1)}$. Substitute $z=-(\lambda_xx+\lambda_yy+\lambda_1)/\lambda_z$ and clear
denominators. The resulting plane equation is
\begin{equation}\label{eq:jointconicplane}
 \alpha_c x^2+\beta_c xy+\gamma_c y^2+\ell_xx+\ell_yy+\ell_0=0,
\end{equation}
with integer coefficients of size $(aX)^{O(1)}$. The restriction of $f$ to the plane is nonzero: otherwise
the plane would be a component of the irreducible pattern quadric.
It has degree two because it contains $C$, and hence defines $C$ in
that plane.

The pullback of the hyperplane at infinity to the normalisation of $C$
has degree two. Corollary~\ref{cor:jointparity} says
that each of its positive multiplicities is even. Counting residue
degrees, there is therefore just one branch over infinity, of
multiplicity two and residue degree one. The normalisation map is
finite and surjective, so the projective conic meets the line at
infinity only at the image of this rational point. Hence the leading
binary quadratic in \eqref{eq:jointconicplane} has a single rational
zero, of multiplicity two: it is a nonzero rational scalar times the
square of a rational linear form. Interchanging $x$ and $y$ if necessary gives
\[
 \alpha_c\ne0,\qquad \beta_c^2-4\alpha_c\gamma_c=0.
\]
Set
\[
 \kappa_c=2\beta_c\ell_x-4\alpha_c\ell_y,\qquad
 \epsilon_c=\ell_x^2-4\alpha_c\ell_0,\qquad
 t=2\alpha_c x+\beta_c y+\ell_x.
\]
Completing the square in \eqref{eq:jointconicplane} gives
\begin{equation}\label{eq:jointconicparameter}
 t^2=\kappa_c y+\epsilon_c.
\end{equation}
Here $\kappa_c\ne0$. If it were zero, a rational point of $C$ with
parameter $t_0$ would give $\epsilon_c=t_0^2$, and the equation would
factor as $(t-t_0)(t+t_0)=0$. Since $C$ is irreducible, it would lie in
one of these two lines, contradicting its degree.

We can therefore solve explicitly:
\[
 y=\frac{t^2-\epsilon_c}{\kappa_c},\qquad
 x=\frac{t-\beta_c y-\ell_x}{2\alpha_c},\qquad
 z=-\frac{\lambda_xx+\lambda_yy+\lambda_1}{\lambda_z}.
\]
Substitution verifies the conic and plane equations for every rational
$t$, and the displayed linear expression in $x,y$ recovers $t$.
In particular, $t$ is an integer whenever $(x,y,z)$ is an integer point.
Distinct points have distinct parameters.
Taking $\Delta_c=2\alpha_c\kappa_c\lambda_z$ clears all three denominators, and all
the quantities involved are polynomials of bounded degree in coefficients of
size $(aX)^{O(1)}$, which gives the stated bounds.
\end{proof}

\begin{lemma}[Properness and coefficient height]\label{lem:jointconicproper}
In the parameter of Lemma~\ref{lem:jointconicparameter}, write
\[
 E_0(t)=\frac{F_E(t)}{G_E(t)},\qquad F_E,G_E\in\mathbb Z[t],
 \qquad \gcd_{\mathbb Q[t]}(F_E,G_E)=1.
\]
The polynomials may be chosen so that
\[
 \deg G_E-\deg F_E=3,\qquad
 \max(\deg F_E,\deg G_E)\le C_{\rm deg},\qquad
 \max_k\bigl(|[t^k]F_E|,|[t^k]G_E|\bigr)\le C_{\rm ht}a^{C_{\rm exp}},
\]
for absolute constants $C_{\rm deg},C_{\rm ht},C_{\rm exp}$. In particular, $G_E$ is
nonconstant. At the parameter $t$ of every good start, $G_E(t)\ne0$ and
$F_E(t)/G_E(t)=I_3(i)$ is an integer.
\end{lemma}
\begin{proof}
The explicit parametrisation identifies $C$ with the affine line and
its smooth projective model with $\mathbb P^1$. Its point at infinity
has hyperplane multiplicity two. Lemma~\ref{lem:joint-infinity-orders} therefore gives
\[
 \operatorname{ord}_{t=\infty}E_0=3.
\]
The function is nonzero, since it is positive at a good start.
For a reduced rational function on $\mathbb P^1$, its order at infinity
is $\deg G_E-\deg F_E$, proving properness.

We next bound the coefficients uniformly in the component, including
after cancellation. The six rational pattern coordinates, the three coordinate
polynomials $\xi_j/\Delta_c$, and the integer $a$ can be written over the common
nonzero denominator $N_0N_1\Delta_c$, and their numerators are polynomials in $t$ of
degree at most two with coefficients of size $(aX)^{O(1)}$. The cofactor
formula defining $E_0$ is a quotient of two fixed polynomials in these inputs.
Substituting and clearing a fixed power of the common denominator gives
nonzero integer polynomials $\tilde n,\tilde d$ of bounded degree with
coefficients of size $(aX)^{O(1)}$.

Choose a primitive $\Pi\in\mathbb Z[t]$ representing
$\gcd_{\mathbb Q[t]}(\tilde n,\tilde d)$. By Gauss's lemma, $\Pi$ divides both
$\tilde n$ and $\tilde d$ in $\mathbb Z[t]$, so $F_E=\tilde n/\Pi$ and
$G_E=\tilde d/\Pi$ are coprime integer polynomials with the same ratio. Every
integer factor of a nonzero integer polynomial of degree at most $C_{\rm deg}$ and
height $\mathcal H$ has height at most $2^{C_{\rm deg}}(C_{\rm deg}+1)\mathcal H$, by
comparing Mahler measures; so cancellation preserves the bound
$(aX)^{O(1)}$. Since $X\le a^{O(1)}$, this is the asserted $C_{\rm ht}a^{C_{\rm exp}}$.
Finally, the reconstruction denominator is nonzero at every good start, hence
so are $\tilde d$ and its factor $G_E$ at the corresponding parameter, and the
reconstruction identity gives $F_E(t)/G_E(t)=I_3(i)\in\mathbb Z$.
\end{proof}

\begin{lemma}[Integer values of a rational function]\label{lem:jointconicdivisors}
Let $F_E,G_E\in\mathbb Z[t]$ be coprime over $\mathbb Q$, of degrees at most
$C_{\rm deg}$, with $G_E$ nonconstant and coefficient heights at most
$C_{\rm ht}a^{C_{\rm exp}}$. For every $\eta>0$, the number of integers $t$ such that
$G_E(t)\ne0$ and $F_E(t)/G_E(t)\in\mathbb Z$ is $O_\eta(a^\eta)$.
The implied constant depends only on $\eta,C_{\rm deg},C_{\rm ht},C_{\rm exp}$.
\end{lemma}
\begin{proof}
The integer resultant $R=\operatorname{Res}(F_E,G_E)$ is nonzero.
The adjugate of the Sylvester matrix gives $U,V\in\mathbb Z[t]$ with
$UF_E+VG_E=R$. At any parameter in question, $G_E(t)\mid F_E(t)$, and hence
$G_E(t)\mid R$. There are $2\tau(|R|)$ possible signed values of $G_E(t)$.
For each such value $v$, the nonzero polynomial $G_E-v$ has at most
$\deg G_E\le C_{\rm deg}$ roots. Thus the number of parameters is at most
\[
 2C_{\rm deg}\tau(|R|).
\]
The Sylvester matrix has size at most $2C_{\rm deg}$, and its entries are
bounded by $C_{\rm ht}a^{C_{\rm exp}}$. Expanding its determinant gives
\[
 1\le |R|\le(2C_{\rm deg})!\bigl(C_{\rm ht}a^{C_{\rm exp}}\bigr)^{2C_{\rm deg}}.
\]
The divisor estimate $\tau(n)\ll_\varepsilon n^\varepsilon$, with
$\varepsilon$ chosen in terms of $\eta$ and the fixed degree and height
exponents, proves the claim.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:jointconic}]
Apply the last lemma to the reconstructed invariant $E_0$. The
parameter is injective on free-node triples, and reconstruction
of the strictly increasing $q_i$ makes those triples injective in the
good starts. Each retained nonexceptional conic therefore
contributes $O_\eta(a^\eta)$ starts. The components with at most two
free-node triples of good starts and the nonretained components were already bounded
absolutely. Apply the conic estimate with $\eta/2$ and the pattern
estimate $O_{\eta/2}(a^{\eta/2}X^5)$. Since there are at most twenty-four
components per pattern, all nonexceptional conics together contribute
$O_\eta(a^\eta X^5)$.
\end{proof}

\section{The upper bound for graph degrees}\label{app:degree}

\begin{proof}[Proof of the upper bound in Lemma~\ref{lem:jointpowereddegree}]
Only total polynomial degrees in the nodes matter here, not the heights of the
pattern-dependent coefficients. The cofactors $c_j$ of
Section~\ref{sec:jointrecon} have bounded degree, hence so do the numerator
and denominator of $E_0=E_{\mathrm n}/E_{\mathrm d}$. Since $H_3,H_3'$ have
degree at most three and the increments have degree one, the three curvature
coordinates have the common denominator $\Omega=E_{\mathrm d}H_3k_1$, and
$\Omega$ and the numerators $\mathcal N_0,\mathcal N_1,\mathcal N_2$ have
degree at most an absolute constant $C_W$. The numerator for $J$ with this
denominator is $\mathcal N_3=J\Omega$, of degree at most $C_W+3$. The
denominator is nonzero on $C$, since it is nonzero at a good start.

Put $\bar m=m+n$, and represent the powered coordinates with denominator
$\Omega^{\bar m}$ and numerators
\[
 \mathcal N_j^m\Omega^{\bar m-m}\quad(0\le j\le2),
 \qquad
 \mathcal N_3^n\Omega^{\bar m-n}.
\]
All these numerators, and those of the node coordinates written with the same
denominator, have degree at most $(C_W+3)\bar m+1$. So multiplying a
polynomial of degree at most $t$ in the graph coordinates by $\Omega^{\bar mt}$
gives a polynomial in the original nodes of degree at most
$\bigl((C_W+3)\bar m+1\bigr)t$. This operation is injective on the
corresponding spaces of functions on $C$, because $\Omega$ is nonzero in
$\mathbb Q(C)$. Comparing affine Hilbert functions gives
\[
 \delta t+O(1)\le d\bigl((C_W+3)\bar m+1\bigr)t+O(1),
\]
and hence the asserted bound with $C_{\rm gr}=C_W+4$.
\end{proof}

\section{A reduced basis in dimension three}\label{app:reduction}

The constant $64/27=(4/3)(16/9)$ in Lemma~\ref{lem:reduce} comes from two
successive size reductions. We give the elementary argument for completeness.

\begin{proof}[Proof of Lemma~\ref{lem:reduce}]
Write $\det\Lambda$ for the least nonzero $|\det|$ of a triple in $\Lambda$; the triple
$(\kappa\mathbf e_1,\kappa\mathbf e_2,\kappa\mathbf e_3)$ gives $\det\Lambda\le\kappa^3$. A triple
$(\mathbf f_1,\mathbf f_2,\mathbf f_3)$ with $|\det|=\det\Lambda$ is a basis: if not, some $x\in\Lambda$ has a Cramer
coordinate that is not an integer, say the first, $\det(x,\mathbf f_2,\mathbf f_3)/\det(\mathbf f_1,\mathbf f_2,
\mathbf f_3)$, and subtracting the integer part of that coordinate times $\mathbf f_1$ from $x$ gives a
triple with smaller nonzero determinant. Among all bases (all have $|\det|=\det\Lambda$) minimise
$g_1=|\mathbf f_1|^2$, and then the Gram determinant
$g_2=g_1|\mathbf f_2|^2-\langle\mathbf f_1,\mathbf f_2\rangle^2$. Put $g_3=(\det\Lambda)^2$.

Choose an integer $z_1$ with $|2(\langle\mathbf f_1,\mathbf f_2\rangle+z_1g_1)|\le g_1$. Then
$(\mathbf f_2+z_1\mathbf f_1,\mathbf f_1,\mathbf f_3)$ is a basis, so $g_1\le|\mathbf f_2+z_1\mathbf f_1|^2$, and
$g_1|\mathbf f_2+z_1\mathbf f_1|^2=g_2+(\langle\mathbf f_1,\mathbf f_2\rangle+z_1g_1)^2$. Hence
$3g_1^2\le4g_2$.

Let $\langle x,x'\rangle_1=g_1\langle x,x'\rangle-\langle\mathbf f_1,x\rangle\langle\mathbf f_1,x'\rangle$,
so that $\langle\mathbf f_2,\mathbf f_2\rangle_1=g_2$, and Sylvester's identity gives
$\langle\mathbf f_2,\mathbf f_2\rangle_1\langle\mathbf f_3,\mathbf f_3\rangle_1-\langle\mathbf f_2,\mathbf f_3\rangle_1^2
=g_1g_3$. Choose an integer $z_2$ with $|2(\langle\mathbf f_2,\mathbf f_3\rangle_1+z_2g_2)|\le g_2$ and
put $y=\mathbf f_3+z_2\mathbf f_2$. Then $(\mathbf f_1,y,\mathbf f_2)$ is a basis with first vector $\mathbf f_1$,
so $g_2\le g_1|y|^2-\langle\mathbf f_1,y\rangle^2=\langle y,y\rangle_1$, while
$g_2\langle y,y\rangle_1=(\langle\mathbf f_2,\mathbf f_3\rangle_1+z_2g_2)^2+g_1g_3$. Hence
$3g_2^2\le4g_1g_3$.

Finally choose an integer $z_3$ with $|2(\langle\mathbf f_1,y\rangle+z_3g_1)|\le g_1$, and use the
basis $\mathbf f_1$, $\mathbf f_2'=\mathbf f_2+z_1\mathbf f_1$, $\mathbf f_3'=y+z_3\mathbf f_1$. The identities above
give $3g_1|\mathbf f_2'|^2\le4g_2$ and
$4g_1g_2|\mathbf f_3'|^2\le g_2^2+4g_1g_3+g_1^2g_2$. With the two inequalities this gives
$9g_2|\mathbf f_3'|^2\le16g_3$. Since $|\mathbf f_1|^2=g_1$, the product of the bounds
$|\mathbf f_2'|^2\le\frac43g_2/g_1$ and $|\mathbf f_3'|^2\le\frac{16}9g_3/g_2$ is the claim for the
basis $\mathbf f_1,\mathbf f_2',\mathbf f_3'$; rename $\mathbf f_2',\mathbf f_3'$ as $\mathbf f_2,\mathbf f_3$.
Replacing $\mathbf f_1$ by $-\mathbf f_1$ if necessary makes the determinant positive. For the
coordinates, $x_1\det\Lambda=\det(x,\mathbf f_2,\mathbf f_3)$, and Hadamard's inequality gives
$\det(x,\mathbf f_2,\mathbf f_3)^2\le|x|^2|\mathbf f_2|^2|\mathbf f_3|^2$, so
$27|\mathbf f_1|^2x_1^2(\det\Lambda)^2\le64(\det\Lambda)^2|x|^2$; likewise for $x_2$ and $x_3$.
\end{proof}

\section{Effective Hilbert irreducibility for monic families}\label{app:hilbert}

We prove Theorem~\ref{thm:monichilbert}. The letters in this appendix are local to it.

\begin{proof}[Proof of Theorem~\ref{thm:monichilbert}]
Write $\operatorname{lg}(x)=1+\log(x+1)$. Implied constants depend only on $r,d$.

\emph{Step 1: integer roots in one parameter.} Let $\mathcal A\in\mathbb Z[u][Z]$ be
monic in $Z$, of $Z$-degree $e\ge2$, irreducible over $\mathbb Q(u)$, of total
degree at most $D$ and height at most $H_0$. We claim that
\[
 \#\{v\in[-B,B]\cap\mathbb Z:\ \mathcal A(v,Z)\text{ has an integer root}\}
 \ll_D\operatorname{lg}(H_0)^4B^{1/e}(1+\log B).
\]
An integer root $z$ of $\mathcal A(v,Z)$ satisfies $|z|\le1+(D+1)H_0B^D$. Choose an integer
$k>D$ with $k\asymp_D\operatorname{lg}(H_0)$, and apply the shear $(u,Z)\mapsto(u,Z+u^k)$.
The polynomial $\mathcal A_k(u,Z)=\mathcal A(u,Z-u^k)$ is irreducible in $\mathbb Q[u,Z]$, and its
total degree is exactly $ke$: the term $(Z-u^k)^e$ contributes $\pm u^{ke}$,
while every other term has degree at most $D+k(e-1)<ke$. The points
$(v,z+v^k)$ are integer points of the plane curve $\mathcal A_k=0$ in a box of size
$B'\ll_DH_0B^k$. By~\eqref{eq:jointexternal} with $\delta=ke$ there are
$\ll(ke)^3B'^{1/(ke)}(\log B'+ke)$ of them. Here
$B'^{1/(ke)}\ll_DH_0^{1/(ke)}B^{1/e}\ll B^{1/e}$ by the choice of $k$, while
$ke\ll\operatorname{lg}(H_0)$ and $\log B'\ll\operatorname{lg}(H_0)(1+\log B)$. This proves
the claim.

\emph{Step 2: one parameter.} Let $r=1$, and let $y_1,\dots,y_e$ be the roots of
$F$ in an algebraic closure of $\mathbb Q(u)$; they are integral over
$\mathbb Z[u]$. For a nonempty proper subset $I\subset\{1,\dots,e\}$ and
$1\le j\le|I|$, let $\sigma_{I,j}$ be the $j$th elementary symmetric function of
$(y_i)_{i\in I}$. Since $F$ is irreducible, $\prod_{i\in I}(\zeta-y_i)\notin\mathbb Q(u)[\zeta]$,
so some $\sigma_I:=\sigma_{I,j(I)}$ lies outside $\mathbb Q(u)$. The resolvent
$\prod_{|I'|=|I|}(Z-\sigma_{I',j(I)})$ is monic in $Z$, and its coefficients are
fixed integer polynomials in the coefficients of $F$; so it lies in
$\mathbb Z[u][Z]$ with degree $O(1)$ and height $O(\mathcal H^{O(1)})$. Let $\mathcal A_I$ be its
monic irreducible factor over $\mathbb Q(u)$ that vanishes at $\sigma_I$. By
Gauss's lemma $\mathcal A_I\in\mathbb Z[u][Z]$; by Gelfond's factor-height inequality its
height is $O(\mathcal H^{O(1)})$; and its $Z$-degree is at least two, because
$\sigma_I\notin\mathbb Q(u)$.

Now suppose that $F(v,\zeta)$ is reducible over $\mathbb Q$. By Gauss's lemma it has
a monic factor $g\in\mathbb Z[\zeta]$ of degree $e'$ with $1\le e'<e$. Extend the specialisation
$u\mapsto v$ to a ring homomorphism $\phi$ from $\mathbb Z[u][y_1,\dots,y_e]$ to
$\overline{\mathbb Q}$, which is possible because this ring is integral over
$\mathbb Z[u]$. Then $F(v,\zeta)=\prod_i(\zeta-\phi(y_i))$, so the roots of $g$, with
multiplicity, are $\phi(y_i)$ for $i$ in some $I$ with $|I|=e'$, and the
coefficients of $g$ are $\pm\phi(\sigma_{I,j})$. In particular $\phi(\sigma_I)$ is
an integer root of $\mathcal A_I(v,Z)$. Step~1, applied to the $O(1)$ polynomials $\mathcal A_I$,
bounds the number of such $v$ by $O(\operatorname{lg}(\mathcal H)^4B^{1/2}(1+\log B))$.

\emph{Step 3: one parameter, the others free.} Let $r\ge2$. We claim that for
all but $O(\operatorname{lg}(\mathcal H)^4B^{1/2}(1+\log B))$ integers $v_1\in[-B,B]$, the
polynomial $F(v_1,u_2,\dots,u_r,\zeta)$ is irreducible over $\mathbb Q(u_2,\dots,u_r)$.
Fix an integer $b>d$ and let $\mathcal K$ be the Kronecker map
\[
 u_2^{a_2}\cdots u_r^{a_r}\zeta^{a_{r+1}}\longmapsto\omega^{a_2+a_3b+\dots+a_{r+1}b^{r-1}}.
\]
It is a
linear bijection from polynomials with all partial degrees less than $b$ onto
polynomials in $\omega$ of degree less than $b^r$, it commutes with specialising
$u_1$, and it is multiplicative on products whose partial degrees stay below $b$.
Since $\zeta$ occupies the highest position, $\mathcal K(F)\in\mathbb Z[u_1][\omega]$ is monic.
Factor $\mathcal K(F)=\prod_\nu \mathcal A_\nu$ into monic irreducible factors over
$\mathbb Q(u_1)$. Again $\mathcal A_\nu$ lies in $\mathbb Z[u_1][\omega]$, with bounded
degree and height $O(\mathcal H^{O(1)})$. By Step~2, outside an exceptional set of the stated size, every
$\mathcal A_\nu(v_1,\omega)$ is irreducible over $\mathbb Q$.

Let $v_1$ be outside this set, and suppose
$F(v_1,\cdot)=gh$ with $g,h\in\mathbb Q[u_2,\dots,u_r][\zeta]$ monic in $\zeta$ of positive
$\zeta$-degree. All partial degrees of $g$ and $h$ are at most those of $F$, hence
less than $b$, so $\mathcal K(g)\mathcal K(h)=\prod_\nu \mathcal A_\nu(v_1,\omega)$, and by unique
factorisation $\mathcal K(g)=\prod_{\nu\in S}\mathcal A_\nu(v_1,\omega)$ for some set $S$ of
indices. Put $g_S=\mathcal K^{-1}\bigl(\prod_{\nu\in S}\mathcal A_\nu\bigr)$ and define
$g_{S^c}$ likewise, so that $g_S(v_1)=g$ and $g_{S^c}(v_1)=h$. If
$g_Sg_{S^c}=F$ identically, this is a factorisation of $F$ over
$\mathbb Q(u_1,\dots,u_r)$ into factors of positive $\zeta$-degree, which is impossible.
Otherwise some coefficient of $g_Sg_{S^c}-F$ is a nonzero polynomial in $u_1$ of
bounded degree vanishing at $v_1$. There are $O(1)$ sets $S$, so this adds
$O(1)$ exceptional values.

\emph{Step 4: induction on $r$.} The case $r=1$ is Step~2. For $r\ge2$, the
exceptional values $v_1$ of Step~3 contribute at most their number times
$(2B+1)^{r-1}$. For every other $v_1\in[-B,B]$, the family
$F(v_1,u_2,\dots,u_r,\zeta)$ is monic, irreducible over $\mathbb Q(u_2,\dots,u_r)$, of
the same degree bounds, and of height at most $(d+1)\mathcal HB^d$. Since
$\operatorname{lg}((d+1)\mathcal HB^d)\ll\operatorname{lg}(\mathcal H)(1+\log B)$, the induction hypothesis
bounds its reducible specialisations by
$O(\operatorname{lg}(\mathcal H)^4B^{r-3/2}(1+\log B)^{4r+1})$. Summing over the $2B+1$
values of $v_1$ gives the theorem.
\end{proof}

\section{Degree certificates along the line directions}\label{app:linecertificates}

This appendix records the certificates used in step~3 of the proof of
Lemma~\ref{lem:jointcomponents}. Fix one of the four directions $w$, write the nodes along
the line as $s_j+tw_j$, and expand
\[
 \frac{P_4(x)R_4(y)-P_4(y)R_4(x)}{x-y}
 =\sum_{\alpha,\beta<4}\mathcal B'_{\alpha\beta}M_\alpha(x)M_\beta(y),\qquad
 M_\alpha(x)=\prod_{l<\alpha}(x-n_l),
\]
with the nodes $n_0,n_1,n_2$ below. As for~\eqref{eq:bezout},
$\det\mathcal B'=\operatorname{Res}(P_4,R_4)$, and
$\operatorname{Res}(P_4,Z)=\gamma_0\operatorname{Res}(P_4,R_4)$ has the same degree in $t$.
In each case the entry $\mathcal B'_{\alpha\beta}$ has $t$-degree at most
$\varrho_\alpha+\kappa_\beta$, for all values of the $\gamma_j$, the $\gamma'_j$ and the
nodes $s_j$; since $\sum_\alpha\varrho_\alpha+\sum_\beta\kappa_\beta=6$, every term of the
determinant has $t$-degree at most six.
\begin{center}\small
\begin{tabular}{@{}llll@{}}
\toprule
Direction $w$ & Nodes $(n_0,n_1,n_2)$ & Row weights $\varrho$ & Column weights $\kappa$\\
\midrule
$(0,1,1,0,1)$ & $(s_1,s_2,0)$ & $(1,1,0,0)$ & $(1,1,1,1)$\\
$(0,1,1,0,0)$ & $(s_1,0,s_2)$ & $(2,1,1,0)$ & $(1,1,0,0)$\\
$(0,0,1,1,0)$ & $(s_1,s_2,0)$ & $(1,1,1,0)$ & $(1,1,1,0)$\\
$(0,1,0,0,1)$ & $(0,s_1,s_2)$ & $(2,1,1,1)$ & $(1,0,0,0)$\\
\bottomrule
\end{tabular}
\end{center}
Each entry is an explicit polynomial of degree at most four in $t$, so each bound is a finite
expansion. We checked all sixteen entries in each case with a computer algebra system, and
the bounds are formalised as \lean{line_resultant_natDegree_le_six}. For generic
coefficients the determinant has degree exactly six in each case.

\section{Certificates for the polar bound}\label{app:polarcertificates}

\subsection{The quartic initial form at the base point}

At $(\rho,\varphi,\chi,\rho',\varphi',\chi')=(2,5,-3,3,6,-5)$ the polar base point is
$\mathbf k_*=(-1/7,-3/28,-1/2)$ and $\nabla f(\mathbf k_*)=(-4,9,-5)$.
Use tangent coordinates $(\xi_1,\xi_2)$ with first-order displacement
\[
 \delta\mathbf k=((9\xi_1-5\xi_2)/4,\xi_1,\xi_2).
\]
Thus the two axes used in Lemma~\ref{lem:polarbase} have tangent vectors
$(9/4,1,0)$ and $(-5/4,0,1)$. Direct substitution in the Newton quartic,
its remainder and the norm polynomial gives
\begin{align*}
 P_{4,*}(x)&=x^4-\frac{20}{7}x^3+\frac{757}{112}x^2
                         -\frac{669}{112}x+2,\\
 Z_1(x;\xi_1,\xi_2)&=\frac{77\xi_1+7\xi_2}{14}
       +\frac{-56\xi_1-124\xi_2}{14}x
       +\frac{-45\xi_1+157\xi_2}{14}x^2
       +\frac{28\xi_1-28\xi_2}{14}x^3,\\
 \Psi_1(\xi_1,\xi_2)&=\frac{3\xi_1+85\xi_2}{28}.
\end{align*}
The resultant is
\[
 \mathcal R_*(\xi_1,\xi_2)=\frac{
 32233977\xi_1^4-95047956\xi_1^3\xi_2+147873750\xi_1^2\xi_2^2
             -111199572\xi_1\xi_2^3+31702169\xi_2^4}{6272}.
\]
In particular,
\[
 \mathcal R_*(1,0)\Psi_1(0,1)^4-\mathcal R_*(0,1)\Psi_1(1,0)^4
 =\frac{26291112152499}{60236288}.
\]
The realising matrix in Lemma~\ref{lem:polarbase} shows that $\Theta$ is
nonzero on the Pl\"ucker cone; no orbit with these parameters is required.

\subsection{The irreducible sextic}

The specialisation $E_\gamma=24E_\star$ at
$(\gamma_1,\gamma_2,\gamma_3)=(-7,8,-4)$ reduces modulo $71$ to a nonzero
scalar multiple of
\[
 q(T)=T^6+20T^5-14T^4+15T^3-28T^2+5T-4.
\]
More precisely, $E_\star=57q$ in $\mathbb F_{71}[T]$.
The following certificates prove that $q$ is irreducible. Repeated squaring
modulo $q$ gives
\begin{align*}
 T^{71^2}&\equiv r_2(T):=36T^5+48T^4+49T^3+53T^2+36T+42,\\
 T^{71^3}&\equiv r_3(T):=37T^5+22T^4+60T^3+13T^2+66T+35.
\end{align*}
For checking the iteration one may first use
\[
 T^{71}\equiv14T^5+33T^4+31T^3+34T^2+48T\pmod q
\]
and raise each remainder to the $71$st power. Define
\begin{align*}
 U_2&=20T^4+34T^3+31T^2+13T+24,\\
 V_2&=31T^5+61T^4+42T^3+61T^2+39T+4,\\
 U_3&=25T^4+11T^3+53T^2+59T+48,\\
 V_3&=7T^5+30T^4+65T^3+64T^2+25T+40.
\end{align*}
Direct multiplication in $\mathbb F_{71}[T]$ verifies
\[
 U_2q+V_2(r_2-T)=1,\qquad U_3q+V_3(r_3-T)=1.
\]
Thus $q$ has no irreducible factor of degree one or two, since such a factor
would divide $T^{71^2}-T$, and none of degree three, since it would divide
$T^{71^3}-T$. Every reducible sextic has an irreducible factor of degree at
most three, so $q$ is irreducible. Reduction modulo $71$ preserves degree,
and Gauss's lemma proves $E_\star$ irreducible over $\mathbb Q$.

\section*{Declaration of AI-generated research and authorship}
This manuscript is the product of an AI-driven research process. The human
author initiated the investigation and directed AI systems throughout, and
wrote no part of the manuscript or of the Lean code. GPT-6-Astra (OpenAI) found
the results and the key steps of their proofs. These include the orthogonal
polynomials and adjacent labels behind Section~\ref{sec:patterns}, the first
proof of a bound $a^{3/10-\delta}$ through resultants, the use of
Lemma~\ref{lem:multid} to bound the number of states per pattern, the sign
identities of Lemma~\ref{lem:signs} and the slice count behind the $11/37$
bound, the observation that the pattern integers form a Pl\"ucker vector, the
weighting of minor vectors by the shortest vectors of their lattices behind
the $45/152$ bound, the joint-invariant graphs with their normalisation and
valuation arguments, the exterior-lattice coordinates, and the polar identity
and Hilbert sieve. The AI systems also developed the presentation, the
formalisation and the writing. The systems were GPT-6-Astra (OpenAI), used
through Codex, and Claude Opus~5.5 (Anthropic), used through Claude Code. Both
contributed to the Lean formalisation and to the review and revision of the
manuscript.

The Lean development gives a machine-checkable version of the results. Lean's
kernel checks the proofs, including the Mathlib results and the separate
curve-counting library they use, relative to the axioms \lean{propext},
\lean{Classical.choice} and \lean{Quot.sound} (Section~\ref{sec:lean}); the
sixteen selected statements are also replayed by two independent kernels. The
conditional refinement described in Section~\ref{sec:lean} assumes two
published theorems as explicit hypotheses. These automated checks are not
independent human peer review.

This declaration distinguishes the provenance of the work from conventional
AI-assisted writing. The human author accepts responsibility for the
submission and for the accuracy of this disclosure.

\section*{Acknowledgements}
We thank the Mathlib community, and the authors of \lean{openai/math} for the
formalised divisor bound.

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\end{document}