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\title[Uniform irreducibility and Galois groups]{Uniform irreducibility and Galois groups of generalized Laguerre polynomials}
\author{Khai-Hoan Nguyen-Dang}
\thanks{ORCID: \href{https://orcid.org/0000-0002-6765-811X}{0000-0002-6765-811X}}
\date{}
\subjclass[2020]{Primary 11R09; Secondary 11C08, 11J86, 14H25, 20B15, 33C45}
\keywords{Generalized Laguerre polynomial, irreducibility, Galois group, Pad\'e approximation, Newton polygon, logarithmic energy}
\hypersetup{pdftitle={Uniform irreducibility and Galois groups of generalized Laguerre polynomials},pdfauthor={Khai-Hoan Nguyen-Dang}}
\raggedbottom

\newcommand{\Gal}{\operatorname{Gal}}
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\begin{document}

\begin{abstract}
We prove that the generalized Laguerre polynomial $H_{n,s}$ is irreducible over $\mathbb Q$ and has Galois group containing
$A_n$ for every positive integer $s$, once $n$ exceeds an effective
absolute constant.
Equivalently, every numerator and denominator in the exponential
Pad\'e table has these properties once its own degree is sufficiently
large.  The exceptional positive integral pairs form an effectively
computable finite set.
The proof combines integral traces at degenerate fibers, Pad\'e
resultants, and an energy inequality for Galois-invariant positive semidefinite forms on the roots. 
\end{abstract}

\maketitle

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

A classical theorem of Schur~\cite{Schur1929} states that the exponential Taylor polynomial
\[
 E_n(X)=\sum_{j=0}^n\frac{X^j}{j!}
\]
is irreducible over $\mathbb Q$ for every $n\ge1$. One also can determine its Galois group: it is $A_n$ when $4\mid n$ and $S_n$ otherwise \cite{Schur} (see also~\cite{Coleman}).

In fact, Schur family occurs among the generalized Laguerre polynomials. More precisely, we use the monic normalization
\begin{equation}\label{eq:definition}
 H_{n,s}(X)=\sum_{k=0}^n\binom nk(s)_kX^{n-k},
 \qquad (s)_0=1,\qquad
 (s)_k=s(s+1)\cdots(s+k-1).
\end{equation}
Here $n$ is a nonnegative integer and $s$ may be complex; in
particular, $H_{0,s}=1$. Our main arithmetic results concern
positive integers $n,s$, for which $H_{n,s}\in\mathbb Z[X]$. It is also a generalization of Bessel polynomials (see \cite{Grosswald1951}):
\[
 y_n(X)=\sum_{j=0}^n\frac{(n+j)!}{(n-j)!\,j!}
                         \left(\frac X2\right)^j.
\]

Filaseta proved $y_n$ is irreducible over $\Q$ for all sufficiently large $n$ in 1995
\cite{Filaseta1995}. After that, together with Trifonov, they proved the result for every $n$ \cite{FT}.

Note that the two classical families are the specializations
\begin{equation}\label{eq:classical-specializations}
 H_{n,1}(X)=n!E_n(X),\qquad
 H_{n,n+1}(X)=X^ny_n(2/X).
\end{equation}
The second identity uses the usual normalization of the Bessel
polynomial \cite[(18.34.1)]{DLMF}. Multiplication by a nonzero
rational constant and the reciprocal substitution in
\eqref{eq:classical-specializations} preserve irreducibility and
splitting fields.

Hajir proposed extending these arithmetic properties to every
positive integral pair $(n,s)$
\cite[Conjectures~1.1--1.2]{Hajir}. His first conjecture asserts
that every $H_{n,s}$ is irreducible over $\mathbb Q$. His second
asserts that, whenever $H_{n,s}$ is irreducible, its Galois group
contains $A_n$.  We prove both assertions in all sufficiently large degrees, with
an effective degree bound independent of $s$, and show that the remaining exceptional pairs form an effectively computable finite set.


\subsection{Uniform arithmetic}\label{intro:laguerre}

For positive integers $n,s$, write
$G_{n,s}=\operatorname{Gal}(H_{n,s}/\mathbb Q)$ for the Galois group
in its action on the roots. These roots are distinct, and the
action is transitive exactly when $H_{n,s}$ is
irreducible over $\mathbb Q$.

\begin{theorem}
\label{thm:uniform-combined}\label{thm:irreducibility}\label{thm:alternating}
There is an effectively computable absolute integer $N$ such that
\[
 n\ge N,\qquad s\in\mathbb Z_{\ge1}
 \quad\Longrightarrow\quad
 H_{n,s}\text{ is irreducible over }\mathbb Q
 \quad\text{and}\quad A_n\le G_{n,s}.
\]
\end{theorem}

The bound is uniform in the parameter: $s$ may grow arbitrarily
with $n$. Thus the theorem applies to every sequence $s=s(n)$ of
positive integers. Proposition~\ref{prop:qualitative-uniform-irreducibility}
proves irreducibility. Theorem~\ref{thm:conditional-alternating}
proves alternating containment for irreducible members, with a
second effective degree bound independent of $s$.

\begin{corollary}
\label{cor:intro-finite-exceptions}
The set of positive integral pairs $(n,s)$ for which $H_{n,s}$ is
reducible or $A_n\not\le G_{n,s}$ is finite and effectively computable.
\end{corollary}

For reducibility, the fixed-degree bound is explicit:
\begin{equation}\label{eq:intro-degree-cap}
 H_{n,s}\text{ reducible},\quad n\ge2
 \quad\Longrightarrow\quad
 s<\left\lfloor\frac{n^2}{4}\right\rfloor^n.
\end{equation}
For irreducible specializations with smaller Galois group, a
decomposition-group criterion at infinity permits an effective
application of Runge's theorem.  Combining these fixed-degree
results with Theorem~\ref{thm:uniform-combined} proves the corollary
in Section~\ref{sec:general-infinity}.
The exceptional set has not been enumerated, and we do not prove
that it is empty.  Thus the result gives uniform eventual forms
of Hajir's conjectures and an effective finite reduction of their
all-degree forms.

Earlier work controls complementary parts of the parameter plane.
Hajir proved both conclusions for each fixed $s$ and all sufficiently
large $n$ \cite[Theorem~1.3]{Hajir}.  Jindal--Laishram--Sarma gave
the explicit sufficient condition
\[
 r=s-1\ge2,\qquad
 n>\exp\!\left(r\left(1+\frac{1.2762}{\log r}\right)\right),
\]
and proved both conclusions in every degree for $1\le s\le61$
\cite[Theorems~1.1--1.2]{JLS}.  The published theorem of
Jindal--Nair--Shorey extends that interval to $1\le s\le93$
and proves the Galois assertion, conditional on irreducibility,
for $94\le s\le101$ \cite{JNS}.  Jindal--Laishram studied the neighboring Bessel lines
$n\le s\le n+4$ \cite{JindalLaishram2022}.
In the other direction, Hajir--Wong proved that for each fixed
$n\ge5$, all but finitely many rational parameters give
irreducibility and alternating containment
\cite[Theorem~1(a)]{HW}; the finiteness there is ineffective
\cite[Corollary~6.2 and the following remark]{Hajir}.
Finiteness on each fixed row and column does not exclude infinitely
many exceptional pairs with both coordinates unbounded.
Theorem~\ref{thm:uniform-combined} excludes such pairs.

The fixed-parameter results for $L_n^{(\alpha)}$, where
\begin{equation}\label{eq:laguerre}
 H_{n,s}(X)=(-1)^n n!L_n^{(-n-s)}(X),
\end{equation} 
with
$\alpha\in\mathbb Q\setminus\mathbb Z_{<0}$ have a different scope:
Filaseta--Lam proved eventual irreducibility, and Hajir proved
eventual alternating containment
\cite[Theorem~1]{FilasetaLam2002}\cite[Theorem~1.1]{Hajir2005}.
Here $\alpha=-n-s$ varies with the degree and is a negative integer.
For the Bessel specialization, the irreducibility theorem of
Filaseta--Trifonov and Grosswald's criterion give $S_n$ in every
degree \cite{FT}\cite[Chapter~12, Theorem~1$'$]{Grosswald1978}.

\subsection{Root geometry and the proof}\label{intro:root-geometry}

A root theorem valid for every $n\ge2$ supplies the complex
estimates in the arithmetic proof. Fix $n\ge2$, continue a root
or a conjugate pair of $H_{n,s}$ along $s>0$, and write $q(s)$
for its modulus.  Classical annular estimates for generalized Bessel
polynomials locate these roots \cite{DSV}.  The next theorem also
describes their motion.

\begin{theorem}
\label{thm:intro-root-moduli}
For every such branch there is a unique finite, absolutely continuous,
nonnegative measure $\mu$ on $[0,n-1]$ such that
\[
 q(s)=s+\int_0^{n-1}\frac{s}{s+t}\,d\mu(t)\qquad(s>0).
\]
If $\xi\ge0$ is the corresponding absolute Hermite zero, characterized
by $\alpha(s)=-s+i\xi\sqrt{s}+O(1)$ as $s\to+\infty$ for the
upper-half-plane root (and $\xi=0$ for the real root), then
\[
 \mu([0,n-1])=\frac{n-1}{3}+\frac{\xi^2}{6}.
\]
For every integer $r\ge1$ and every $s>0$,
\[
 (-1)^{r-1}\frac{d^r}{ds^r}(q(s)-s)>0.
\]
\end{theorem}

Here the Hermite polynomials have the probabilists' normalization,
specified in \eqref{geo:hermite-normalization}.
Theorem~\ref{geo:bernstein}, Corollary~\ref{geo:limiting-excess},
and Proposition~\ref{geo:exact-support} prove the statement and
identify the exact support of each measure.  In particular,
$q-s$ is strictly increasing and strictly concave, and is a
complete Bernstein function.  Castillo obtains related complete
Bernstein representations for scaled ultraspherical zeros
\cite{Castillo2026}; the present representation concerns complex
Laguerre roots in the range $\alpha<-n$.

\subsubsection*{Outline of the proof}\label{intro:proof}\label{intro:methods}

The proof separates three possibilities: a rational factor, an
imprimitive action on the roots, and a primitive Galois group
not containing $A_n$. Integral traces and one energy inequality
control factors and invariant partitions. Local permutations
handle the remaining primitive groups.

First, a tridiagonal realization and Fan's variational principle
\cite{Fan} give the strict trace estimate
\begin{equation}\label{eq:intro-trace}
 \left|\operatorname{Re}\sum_{i=1}^d(\alpha_i+s)\right|<d(n-d)
 \qquad(0<d<n).
\end{equation}
The centered trace of a rational factor is therefore an integer in
a bounded interval independent of $s$.  At $s=0,-1,-2,-3$, at most
three roots are nonzero.  Reduction at these four fibers restricts
the trace to finitely many subset sums.  For every $n\ge2$, a
reducible member consequently has
\[
 \operatorname{rad}\bigl((s+i)(s+j)\bigr)<n^6
 \quad\text{for some }0\le i<j\le3.
\]
Matveev's theorem \cite{Matveev} then gives
$\log s<n^\varepsilon$ for every fixed $\varepsilon>0$ and all
effectively sufficiently large $n$.
Theorem~\ref{new:finite-fiber-criterion} isolates this argument
for general monic families; Theorem~\ref{new:four-fiber} gives the
all-degree bound, including rational parameters.

Second, we attach to a factor or an invariant partition a
Galois-invariant positive semidefinite matrix $Q$ with
$Q\mathbf1=0$.  At a finite place, a cluster of roots contributes
a nonnegative squared difference between its actual and expected
occupation.  The product formula compares the sum of these local
contributions with complex logarithmic energy.  The root-modulus
representation supplies the complex bounds uniformly in $s$.
For factors, the energy estimate and a nonzero Pad\'e resultant
give incompatible restrictions on the factor degree.
The resulting contradiction proves uniform irreducibility.

For the Galois group, local permutations first exclude primitive
groups not containing $A_n$.  The partition energy and trace bounds
then reduce imprimitivity to a partition into two blocks. The
stabilizer of a block fixes a quadratic field.
The real quadratic case is controlled by traces.  In the imaginary
quadratic case, we construct integral functions whose block sums
are small and whose first nonzero Laurent terms distinguish the
blocks.  The underlying interpolation theorem identifies a cyclic
intermediate field as the obstruction to prescribing Laurent
coefficients.  Hermite expansions and Gaussian quadrature provide
the quantitative bounds needed after specialization.
The last contradiction uses a criterion for large prime valuations
at nearby integers.  The primitive-group step uses
Burness--Guralnick \cite{BG}; this is the only input to the main
theorem that uses the classification of finite simple groups.

\subsection{Consequences and related families}

\subsubsection*{The exponential Pad\'e table}\label{intro:pade-table}

The same question has a natural formulation in rational approximation.
For $a,b\ge0$, let $A_{a,b}/B_{a,b}$ be the Pad\'e approximant of
type $[a/b]$ to $e^X$, normalized by
$A_{a,b}(0)=B_{a,b}(0)=1$.  Thus
\[
 \deg A_{a,b}=a,\qquad \deg B_{a,b}=b,\qquad
 B_{a,b}(X)e^X-A_{a,b}(X)=O(X^{a+b+1}).
\]
Its numerator and denominator are
\begin{equation}\label{eq:intro-pade-table}
 A_{a,b}(X)=\frac{b!}{(a+b)!}H_{a,b+1}(X),\qquad
 B_{a,b}(X)=\frac{a!}{(a+b)!}H_{b,a+1}(-X).
\end{equation}
Cullinan--Scheel used these identities to study diagonal
approximants and special cases of irreducibility \cite{CS}.
The boundary $b=0$ gives the exponential Taylor polynomials,
and the diagonal $a=b$ gives the Bessel polynomials.

\begin{corollary}
\label{cor:pade-table}
For the absolute constant $N$ in Theorem~\ref{thm:uniform-combined},
$A_{a,b}$ is irreducible over $\mathbb Q$ with Galois group
containing $A_a$ whenever $a\ge N$, for every $b\ge0$.
Likewise, $B_{a,b}$ is irreducible with Galois group containing
$A_b$ whenever $b\ge N$, for every $a\ge0$.
\end{corollary}

\begin{proof}
Apply Theorem~\ref{thm:uniform-combined} to
\eqref{eq:intro-pade-table}.  Multiplication by a nonzero rational
constant and $X\mapsto-X$ preserve irreducibility and splitting fields.
\end{proof}

This is an equivalent form of the main theorem: the required bound
depends only on the degree of the numerator or denominator being
studied.  It also applies to the exponential family in
\cite[Question~1.3 and Theorem~1.4]{RabarisonPazukiMolin2024}.

\subsubsection*{The symmetric and alternating cases}\label{intro:discriminant}

Schur's discriminant formula \cite[p.~54, (7)]{Schur1931}, in the present normalization, is
\begin{equation}\label{eq:intro-discriminant}
 \Delta_{n,s}=(-1)^{n(n-1)/2}
       \prod_{j=2}^n j^j(s+n-j)^{j-1}.
\end{equation}
Once alternating containment is known, the square class of
$\Delta_{n,s}$ determines whether $G_{n,s}=A_n$ or $S_n$.
In particular, $G_{n,s}=S_n$ when $n\equiv2,3\pmod4$ and $n\ge N$.
The classification of square discriminants by
Banerjee--Filaseta--Finch--Leidy \cite{BFFL2013} gives an exact
description in sufficiently large degree: the alternating cases
lie on the ten parameter lines
\[
 s\in\{1,2,3,4,5,6,n-4,n,n+2,n+4\},
\]
with the conditions in Table~\ref{tab:negative-alternating-cases}.
The classification is due to the cited authors; the uniform theorem
supplies alternating containment throughout the parameter range.

\subsubsection*{Other families and geometric applications}
The methods also apply beyond this family.  The finite-fiber
criterion proves large-parameter irreducibility for the Appell
family generated by $e^{Xz}(1-z)^{-s}(1-2z)^{-s}$
(Theorem~\ref{new:two-scale-irreducibility}).  The prime-valuation
criterion gives deterministic symmetric-group results for trimmed
exponentials, under an irreducibility hypothesis or an explicit
endpoint condition (Theorem~\ref{thm:trimmed-deterministic} and
Corollary~\ref{cor:trimmed-endpoint}).  Their discriminants are
nonsquare in the range of Proposition~\ref{prop:trimmed-discriminant},
with no upper bound on the parameter.  The full conjecture of
Bary-Soroker--Ben-Porath remains open \cite{BarySorokerBenPorath2021}.

Finally, the uniform theorem permits an arbitrary positive integral
parameter in each degree in the root-set criterion of
Bary-Soroker--Kadets \cite{BarySorokerKadets2026} and in the
hyperelliptic-Jacobian theorems of Zarhin
\cite{Zarhin2002,Zarhin2003}.  Together with Lombardo's product
criterion \cite{Lombardo2016}, this gives the geometric consequences
in Section~\ref{sec:arithmetic-geometry}.  The general geometric
criteria are due to these authors; our contribution is their
uniform application to the present family.

\subsubsection*{Organization}\label{intro:organization}
Section~\ref{sec:geometry-traces} develops the root geometry and
finite-fiber arithmetic. Sections~\ref{sec:energy-irreducibility}
and~\ref{sec:galois} prove uniform irreducibility and alternating
containment, respectively. Section~\ref{sec:general-infinity}
proves the effective finite reduction. Section~\ref{sec:applications}
contains the applications and the remaining questions.
Appendix~\ref{app:modulus} gives the analytic-continuation argument
for the root-modulus representation; Appendix~\ref{app:appell-certificate}
provides the exact certificate for the Appell family.

\subsubsection*{Notation}
All logarithms are natural.  We normalize $v_p(p)=1$ and fix
embeddings in $\overline{\mathbb Q}_p$ when needed.
For $m\in\mathbb Z_{>0}$, put
$\operatorname{rad}(m)=\prod_{p\mid m}p$,
$\omega(m)=\#\{p:p\mid m\}$, and let $P(m)$ be its greatest
prime divisor; the values at $m=1$ are $1,0,1$, respectively.
An integer is $y$-smooth if all its prime divisors are at most $y$.
For a reduced rational number $a/b$ with $b>0$, put
$H(a/b)=\max\{|a|,b\}$ and $h(a/b)=\log H(a/b)$.
Factors of monic integral polynomials are taken monic.
A nontrivial block system has at least two blocks, each containing
at least two roots; in a transitive action their sizes are equal.
For a permutation $g$ on $\Omega$, write $\operatorname{supp}(g)$
for its support and
$\operatorname{fpr}(g)=|\operatorname{Fix}(g)|/|\Omega|$.
Write $\mu(G)$ for the least support size of a nonidentity element
of $G$.  A permutation is called homogeneous if all its nontrivial
cycles have the same length.  Every constant called effective is
computable, with its dependence on fixed parameters indicated.


\section{Geometry of the roots and integral traces}\label{sec:geometry-traces}

\subsection{Root geometry and a preliminary parameter bound}
\label{geo:section}

Two estimates on the roots will be used in the arithmetic proof.
After translation by $s$, the real part of every partial root
sum is bounded independently of $s$. This bounds the traces of
rational factors when $s$ is integral. A lower bound of order $n$
for $|\alpha|-s+1$, uniform for $s\ge1$, controls the Pad\'e norms.
We first work with real parameters and impose integrality when
it is needed. Put
\[
 C_{n,s}(Y)=H_{n,s}(Y-s).
\]

\subsubsection{Identities, root counts, and an annulus}

The exponential generating functions are
\begin{align}
 \sum_{n\ge0}H_{n,s}(X)\frac{t^n}{n!}
 &=e^{Xt}(1-t)^{-s},\label{geo:egf}\\
 \sum_{n\ge0}C_{n,s}(Y)\frac{t^n}{n!}
 &=\exp\left(Yt+s\sum_{j\ge2}\frac{t^j}{j}\right).
 \label{geo:centered-egf}
\end{align}
These are identities of formal power series. Coefficient comparison gives
\begin{align}
 H'_{n,s}(X)&=nH_{n-1,s}(X),\label{geo:derivative}\\
 H_{n+1,s}(X)&=(X+s+n)H_{n,s}(X)-nXH_{n-1,s}(X),
 \label{geo:recurrence}\\
 XH''_{n,s}(X)&-(X+n+s-1)H'_{n,s}(X)+nH_{n,s}(X)=0.
 \label{geo:ode}
\end{align}
For real $s>0$, expansion of the integrand also gives
\begin{equation}\label{geo:gamma}
 H_{n,s}(X)=\frac1{\Gamma(s)}
       \int_0^\infty(X+t)^n t^{s-1}e^{-t}\,dt.
\end{equation}
Thus $H_{n,s}$ has no real root in even degree. In odd degree its
derivative is positive on $\mathbb R$ by \eqref{geo:derivative}, so
it has exactly one real root, which is negative because
$H_{n,s}(0)=(s)_n>0$.

For any complex parameter, a multiple nonzero root is impossible.
Indeed, if $a\ne0$ were a root of multiplicity $r\ge2$, the term
$XH''_{n,s}$ in \eqref{geo:ode} would have a nonzero coefficient of
$(X-a)^{r-2}$, while the other terms would have higher order. A zero
root occurs precisely at $s=0,-1,\ldots,1-n$. Thus all roots are simple
when $s>0$.

The identity $H_{n,s}(X)=X^ny_n(2/X;s-n+1)$, in the notation of
\cite[(18.34.1)]{DLMF}, relates these roots to those of generalized
Bessel polynomials. Classical annular estimates for the latter were
given by de Bruin, Saff, and Varga \cite[Theorem~5.1]{DSV}.
The following matrix argument gives the bound needed here.

\begin{lemma}\label{geo:annulus}
For $n\ge2$ and real $s>0$, every root $\alpha$ of $H_{n,s}$ satisfies
\begin{equation}\label{geo:eq-annulus}
 s<|\alpha|<s+n-1.
\end{equation}
\end{lemma}
\begin{proof}
Let $T$ be the real symmetric tridiagonal matrix with zero diagonal
and adjacent entries $\sqrt1,\ldots,\sqrt{n-1}$, and put
$D_0=\diag(s,s+1,\ldots,s+n-1)$. The determinant recurrence and
\eqref{geo:recurrence} give
\[
 H_{n,s}(z^2)=\det(z^2I-zT+D_0).
\]
Choose $z^2=\alpha$ and a unit vector $v$ in the corresponding kernel.
Multiplication on the left by $v^*$ gives
\[
 z^2-(v^*Tv)z+v^*D_0v=0.
\]
Both coefficients are real. Since $\alpha$ is not nonnegative real,
$z$ is nonreal; the other root of this scalar quadratic is therefore
$\bar z$. Consequently
\[
 |\alpha|=|z|^2=v^*D_0v\in[s,s+n-1].
\]
Equality at either endpoint would force $v$ to be a multiple of the
first or last coordinate vector. The adjacent nonzero entry of $T$
then contradicts $(z^2I-zT+D_0)v=0$. Both bounds are strict.
\end{proof}

\subsubsection{Centered roots and integral factor traces}

After centering at $-s$, sums of roots satisfy bounds independent of
$s$. The estimate follows from Fan's maximum principle
\cite[Theorem~1]{Fan};
we include the matrix proof to obtain the strict inequality.

\begin{proposition}\label{geo:trace}
Let $n\ge2$ and let $s>0$ be real. For any $d$ distinct roots
$\alpha_1,\ldots,\alpha_d$ of $H_{n,s}$, with $0<d<n$,
\[
 \left|\operatorname{Re}\sum_{i=1}^d(\alpha_i+s)\right|<d(n-d).
\]
In particular, if $a$ is the root sum of a monic real factor of
$C_{n,s}$ of degree $d$, then
\begin{equation}\label{geo:eq-trace}
 |a|<d(n-d).
\end{equation}
For positive integral $s$, a monic rational factor has $a\in\Z$.
Moreover, every root $\alpha$ of $H_{n,s}$ satisfies
\begin{equation}\label{geo:eq-radius}
 |\alpha+s|\le n-1+2\sqrt{(n-1)(s+n-1)}.
\end{equation}
\end{proposition}
\begin{proof}
The centered polynomial is the characteristic polynomial of the
tridiagonal matrix $A$ with entries
\[
 \begin{aligned}
 A_{jj}&=2j-n-1 &&(1\le j\le n),\\
 A_{j,j+1}&=j,\qquad A_{j+1,j}=-(n+s-j) &&(1\le j<n).
 \end{aligned}
\]
To verify this, let $P_k(X)$ be the characteristic polynomial of the
leading $k$-by-$k$ block of $A-sI$. Its continuant recurrence is
\[
 P_k=(X+n+s+1-2k)P_{k-1}
       +(k-1)(n+s-k+1)P_{k-2},
\]
with $P_0=1$ and $P_1=X+n+s-1$. Substitution into this recurrence,
using Pascal's identity, verifies the explicit formula
\[
 P_k(X)=\sum_{j=0}^k\frac{k!}{j!}
                 \binom{n+s-1-j}{k-j}X^j.
\]
At $k=n$ its coefficient of $X^j$ is
$\binom nj(s)_{n-j}$, as required.

A positive diagonal similarity transforms $A$ into $D+K$, where
\[
 D=\diag(1-n,3-n,\ldots,n-1),\qquad K^*=-K,
\]
and
\[
 K_{j,j+1}=\sqrt{j(n+s-j)},\qquad K_{j+1,j}=-K_{j,j+1}.
\]
Eigenvectors belonging to the selected roots
span a $d$-dimensional invariant subspace $V$. If the columns of $U$ are
an orthonormal basis of $V$, the trace of $U^*(D+K)U$ is
$\sum_i(\alpha_i+s)$. Since $K$ is skew-Hermitian,
\[
 \operatorname{Re}\sum_{i=1}^d(\alpha_i+s)=\operatorname{tr}(U^*DU)
   =\sum_{j=1}^n(2j-n-1)w_j,\qquad
 0\le w_j\le1,\quad \sum_{j=1}^n w_j=d,
\]
where $w_j$ are the diagonal entries of the orthogonal projection
onto $V$. The minimum and maximum of this expression are the sums
of the first and last $d$ diagonal entries of $D$, namely
$-d(n-d)$ and $d(n-d)$. Because those entries are strictly increasing,
equality requires $V$ to be the corresponding coordinate subspace.
Neither subspace is invariant under $D+K$, as its boundary coupling
is nonzero. This proves the first assertion and hence
\eqref{geo:eq-trace}.

Finally, the row and column sum bounds for the operator norm give
\[
 \|D\|=n-1,\qquad
 \|K\|\le2\max_{1\le j<n}\sqrt{j(n+s-j)}
       \le2\sqrt{(n-1)(s+n-1)}.
\]
Each $\alpha+s$ is an eigenvalue of $D+K$, proving
\eqref{geo:eq-radius}. If $s$ is integral, Gauss's lemma makes every
monic rational factor integral, and hence its trace is an integer.
\end{proof}

To turn the trace bound into a divisibility condition, we first
consider a polynomial that reduces to a power of $X$ times a
linear factor. The two residue roots are allowed to coincide.

\begin{lemma}\label{geo:two-cluster}
Let $P\in\Z[X]$ be monic of degree $N\ge2$, and suppose that
\[
 P(X)\equiv X^{N-1}(X-c)\pmod M
\]
for integers $M\ge1$ and $c$. Suppose a submultiset of the roots of
$P$, counted with multiplicity, has integral sum $\tau$, with
$\tau\equiv A\pmod M$ and $A(A-c)\ne0$.
Then
\begin{equation}\label{geo:eq-two-cluster}
 M\mid\gcd(|A|,|A-c|)
          \operatorname{lcm}(|A|,|A-c|)^{N-1}.
\end{equation}
\end{lemma}
\begin{proof}
Fix $p^e\Vert M$ and set
\[
 t=\min\{v_p(A),v_p(A-c)\},\qquad
 b=\max\{v_p(A),v_p(A-c)\},\qquad u=v_p(c)\ge t,
\]
where $v_p(0)=+\infty$. Suppose, to obtain a contradiction, that
$e>t+(N-1)b$. In particular $e>Nt$. The polynomial
$p^{-Nt}P(p^tZ)$ is monic with $p$-integral coefficients.
If $u>t$, its reduction is $Z^N$, so every root of $P$ has valuation
greater than $t$. Their selected sum does too. But in this case
$v_p(A)=v_p(A-c)=t$, contrary to $\tau\equiv A\pmod{p^e}$.

If $u=t$, the reduction is $Z^{N-1}(Z-\gamma)$ with $\gamma\ne0$.
Hensel's lemma lifts the simple nonzero root to a unique root over
$\mathbb Z_p$. Division by this linear factor leaves a polynomial
reducing to $Z^{N-1}$. Thus exactly one
root $\beta$ of $P$ has valuation $t$, and all other roots have
valuation greater than $t$. Every root $\alpha$ is integral, and the coefficient
congruence implies
\[
 v_p\bigl(\alpha^{N-1}(\alpha-c)\bigr)\ge e.
\]
For $\alpha\ne\beta$ one has $v_p(\alpha-c)=t$, and therefore
\[
 v_p(\alpha)\ge\frac{e-t}{N-1}>b.
\]
For $\beta$ the same congruence gives
\[
 v_p(\beta-c)\ge e-(N-1)t>b.
\]
Thus, according as the selection excludes or includes $\beta$,
either $v_p(\tau)>b$ or $v_p(\tau-c)>b$. Since $e>b$, either conclusion
contradicts $\tau\equiv A\pmod{p^e}$. We have proved
$e\le t+(N-1)b$, which is exactly the $p$-part of
\eqref{geo:eq-two-cluster}.
\end{proof}

At $s=0$ and $s=-1$, all but at most one root of $H_{n,s}$ lie at
the origin. These two specializations, together with the trace bound,
give an effective restriction on the parameter of a reducible member.

\begin{proposition}\label{geo:parameter-cap}
Let $n\ge2$ and $s\ge1$ be integers. If a proper monic rational factor of
$C_{n,s}$ has degree $d$ and root sum $a$, then
\begin{equation}\label{geo:s-divides}
 s\mid a^n.
\end{equation}
If $a=0$, then
\begin{equation}\label{geo:splus-divides}
 s+1\mid\gcd(d,n-d)\operatorname{lcm}(d,n-d)^{n-1}.
\end{equation}
In particular, writing $B_d=d(n-d)$ and $g=\gcd(d,n-d)$, one has
\begin{equation}\label{geo:refined-cap}
 \begin{cases}
 s\le(B_d-1)^n,&a\ne0,\\
 s\le B_d^{\,n-1}g^{\,2-n}-1,&a=0.
 \end{cases}
\end{equation}
Consequently, reducibility of $H_{n,s}$ implies
\begin{equation}\label{geo:eq-parameter-cap}
 s<\left\lfloor\frac{n^2}{4}\right\rfloor^n,
 \qquad \log s<2n\log n.
\end{equation}
\end{proposition}
\begin{proof}
The coefficient formula gives $C_{n,s}(Y)\equiv Y^n\pmod s$.
For $p^e\Vert s$, every centered root has valuation at least $e/n$:
otherwise its leading term in the polynomial equation would have
strictly smaller valuation than all other terms. Its selected root
sum therefore satisfies $v_p(a)\ge e/n$, proving
\eqref{geo:s-divides}.

At the next integral fiber,
\[
 H_{n,s}(X)\equiv X^{n-1}(X-n)\pmod{s+1}.
\]
The uncentered root sum is $\tau=a-ds\equiv a+d\pmod{s+1}$.
When $a=0$, Lemma~\ref{geo:two-cluster}, with $A=d$ and $c=n$,
applies because $0<d<n$, and proves \eqref{geo:splus-divides}.

Since $a\in\mathbb Z$, \eqref{geo:eq-trace} gives
$|a|\le B_d-1$; also
$\gcd(d,n-d)\operatorname{lcm}(d,n-d)^{n-1}
=B_d^{n-1}g^{2-n}$. These observations prove
\eqref{geo:refined-cap}. Now $B_d\le B=\lfloor n^2/4\rfloor$.
If $a\ne0$, then $s\le(B-1)^n<B^n$; if $a=0$, then
$s+1\le B^{n-1}\le B^n$. The latter is impossible when $B=1$.
Thus \eqref{geo:eq-parameter-cap} follows in every case.
\end{proof}


\subsubsection{The motion of the root moduli}

The annulus bounds the excess $|\alpha|-s$ at each parameter.
Its variation with $s$ is more rigid: every root branch has a
positive integral representation. Monotonicity will then reduce
the required uniform gap to the single parameter $s=1$.

\begin{theorem}\label{geo:bernstein}
Fix $n\ge2$. Continue a conjugate pair of roots of $H_{n,s}$ along $s>0$, and let
$q(s)$ be their common modulus. When $n$ is odd, $q(s)$ may instead
be the modulus of the unique real root. There is a unique finite
nonnegative Borel measure $\mu$ on $[0,n-1]$ such that
\begin{equation}\label{geo:eq-bernstein}
 q(s)=s+\int_0^{n-1}\frac{s}{s+t}\,d\mu(t)\qquad(s>0).
\end{equation}
The measure is absolutely continuous. In particular, $q(s)-s$ is
nondecreasing.
\end{theorem}

Appendix~\ref{app:modulus} proves the representation by continuation
across the singular fibers and Cauchy's formula.
Proposition~\ref{geo:exact-support} identifies the exact support
of the measure for each branch.
In the standard terminology, $q$ and $q-s$ are complete Bernstein
functions. Related representations for scaled ultraspherical
zeros are established in \cite[Theorem~1.4]{Castillo2026}.

The expansion at infinity identifies the mass of the measure.
We fix the Hermite normalization used throughout the paper:
\begin{equation}\label{geo:hermite-normalization}
 h_m(T)=m![u^m]e^{Tu+u^2/2}
       =i^{-m}\operatorname{He}_m(iT),
\end{equation}
where $\operatorname{He}_m$ is the probabilists' Hermite polynomial.
The centered generating function gives the scaled polynomial
\begin{equation}\label{geo:hermite-scaling}
 F_n(T,v):=v^nC_{n,v^{-2}}(T/v)
 =n![u^n]\exp\left(Tu+\sum_{m\ge2}
                         \frac{v^{m-2}u^m}{m}\right).
\end{equation}
The right side extends polynomially to $v=0$, with $F_n(T,0)=h_n(T)$.
The zeros of $h_n$ are simple and purely imaginary. Indeed,
Rodrigues' formula
$\operatorname{He}_n(x)=(-1)^ne^{x^2/2}(d/dx)^ne^{-x^2/2}$
and integration by parts give orthogonality for the positive Gaussian
weight. Multiplication by the product of the sign-changing factors
then shows that $\operatorname{He}_n$ has $n$ distinct real zeros.
The scaled polynomial therefore has a separate holomorphic root
germ at each limiting zero.

\begin{corollary}
\label{geo:limiting-excess}
Let $q$ and $\mu$ be as in Theorem~\ref{geo:bernstein}. Let
$\xi\ge0$ be the absolute value of the zero of the probabilists'
Hermite polynomial $\operatorname{He}_n$ corresponding to this root
branch. For a conjugate pair, choose the root in the upper half-plane;
then the scaling is
\[
 \alpha(s)=-s+i\xi\sqrt{s}+O(1)\qquad(s\longrightarrow+\infty).
\]
For the real root in odd degree, take $\xi=0$. Then
\begin{equation}\label{geo:measure-mass}
 \mu([0,n-1])=\lim_{s\to\infty}(q(s)-s)
   =\frac{n-1}{3}+\frac{\xi^2}{6}.
\end{equation}
In particular, for every integer $r\ge1$ and every $s>0$,
\begin{equation}\label{geo:strict-derivatives}
 (-1)^{r-1}\frac{d^r}{ds^r}(q(s)-s)
   =r!\int_0^{n-1}\frac{t}{(s+t)^{r+1}}\,d\mu(t)>0.
\end{equation}
Thus $q'(s)>1$, $q''(s)<0$, and the root-modulus excess is strictly
increasing and strictly concave.
\end{corollary}

\begin{proof}
The polynomials in \eqref{geo:hermite-normalization} satisfy
$h_j'=jh_{j-1}$ and $h_j=Th_{j-1}+(j-1)h_{j-2}$.
For $n\ge3$, \eqref{geo:hermite-scaling} gives
\[
 F_n(w,z)=h_n(w)
   +\frac{n(n-1)(n-2)}3zh_{n-3}(w)+O(z^2).
\]
Let $w_0=i\xi$ and let
$w(z)=w_0+bz+O(z^2)$ be the corresponding root germ. Since $h_n(w_0)=0$
and $h_n'(w_0)\ne0$, implicit differentiation gives
\[
 b=-\frac{(n-1)(n-2)}3
       \frac{h_{n-3}(w_0)}{h_{n-1}(w_0)}.
\]
The recurrence, applied twice, yields
\[
 (n-1)(n-2)h_{n-3}(w_0)
    =(n-1+w_0^2)h_{n-1}(w_0).
\]
Consequently
\[
 b=-\frac{n-1+w_0^2}{3}
   =\frac{\xi^2-n+1}{3}.
\]
For $n=2$, one has $F_n(w,z)=w^2+1$ and $\xi^2=1$, so the
same formula holds with $b=0$. Substituting $z=s^{-1/2}$ gives
\[
 \alpha(s)=-s+i\xi\sqrt{s}
               +\frac{\xi^2-n+1}{3}+O(s^{-1/2}).
\]
Taking absolute values shows that
\[
 q(s)-s\longrightarrow
 -\frac{\xi^2-n+1}{3}+\frac{\xi^2}{2}
 =\frac{n-1}{3}+\frac{\xi^2}{6}.
\]
This also applies to $\xi=0$, when $q=-\alpha$. Dominated convergence
in \eqref{geo:eq-bernstein} identifies this limit with the mass of
$\mu$, proving \eqref{geo:measure-mass}.

Differentiation under the integral is valid on every compact
subinterval of $(0,\infty)$ and gives
\eqref{geo:strict-derivatives}. Its right-hand side is strictly
positive: the mass in \eqref{geo:measure-mass} is positive, and
absolute continuity excludes an atom at $0$.
\end{proof}

\paragraph{The quadratic example.}
For $n=2$ the roots are $-s\pm i\sqrt{s}$, so
$q(s)=\sqrt{s(s+1)}$.  The representing measure is
\[
 d\mu(t)=\frac1\pi\sqrt{\frac{1-t}{t}}\,
              \mathbf1_{(0,1)}(t)\,dt.
\]
Its mass is $1/2$, in agreement with \eqref{geo:measure-mass}.
This example exhibits both the full support and the strict derivative
signs directly.

\begin{proposition}\label{geo:hermite-order}
Let $n\ge2$. For every $s>0$, distinct roots of $H_{n,s}$ cannot lie
on the same nonreal ray. Label the roots in the upper half-plane,
together with the real root in odd degree, by
\[
 \alpha_j(s)\sim((n-1)!s)^{1/n}e^{(2j+1)\pi i/n},
 \qquad 0\le j\le\lfloor(n-1)/2\rfloor,
 \qquad s\downarrow0.
\]
Then the corresponding Hermite zeros in
$\alpha_j(s)=-s+i\xi_j\sqrt{s}+O(1)$ satisfy
\[
 \xi_j>\xi_{j+1}\quad(0\le j<\lfloor(n-1)/2\rfloor),
 \qquad \xi_{\lfloor(n-1)/2\rfloor}\ge0.
\]
Thus the exact support endpoint $n-1-j$ in
Proposition~\ref{geo:exact-support} corresponds to the $j$th
nonnegative Hermite zero in decreasing order, starting with $j=0$.
\end{proposition}
\begin{proof}
Put $A=n+s-1$ and $\kappa=n-A/2$.  On a ray $X=\rho e^{i\theta}$,
the substitution $H(X)=e^{e^{i\theta}\rho/2}\rho^{A/2}u(\rho)$ in
\eqref{geo:ode} gives
\begin{equation}\label{geo:ray-ode}
 u''+\left(-\frac{e^{2i\theta}}4+
 \frac{\kappa e^{i\theta}}\rho-\frac{A(A+2)}{4\rho^2}\right)u=0.
\end{equation}
Suppose two roots lie at radii $\rho_1<\rho_2$ on a nonreal ray.
Proposition~\ref{geo:trace} gives
$\rho\cos\theta<n-1-s=2\kappa-2$ at both endpoints and hence on
the whole radial segment.  Multiply the equation by $\overline u$,
integrate, and take imaginary parts.  The endpoint terms vanish, so
\[
 0=\sin\theta\int_{\rho_1}^{\rho_2}
   \left(\frac\kappa\rho-\frac{\cos\theta}{2}\right)|u|^2\,d\rho,
\]
although the parenthesized factor is greater than $1/\rho$.
This proves separation. A nonreal branch cannot cross the real axis
without meeting its conjugate, contrary to simplicity. Its initial
angular order is therefore preserved for all $s>0$.
The expansion $\arg\alpha_j(s)=\pi-\xi_j/\sqrt{s}+O(s^{-1})$
then orders the Hermite limits in the opposite direction.  These
limits are distinct by the simple-root perturbation of
\eqref{geo:hermite-scaling}, proving the strict inequalities.
\end{proof}

The arithmetic applications require a lower bound uniform in $s$.
Monotonicity reduces this estimate to the exponential Taylor
polynomial at $s=1$.

\begin{corollary}\label{geo:gap}
Let $\rho_0>0$ be the unique solution of
$\log\rho_0+\rho_0=-1$.  For $n\ge2$, real $s\ge1$, and every
root $\alpha$ of $H_{n,s}$,
\begin{equation}\label{geo:eq-gap}
 |\alpha|-s+1>\rho_0(n+1)>\frac{n+1}{4}.
\end{equation}
\end{corollary}
\begin{proof}
At $s=1$ one has $H_{n,1}(X)=n!\sum_{j=0}^nX^j/j!$.
Taylor's integral remainder at a root $\alpha$ gives
\[
 e^\alpha=\frac{\alpha^{n+1}}{n!}
             \int_0^1 e^{t\alpha}(1-t)^n\,dt.
\]
Dividing by $e^\alpha$ and taking absolute values, with $R=|\alpha|$
and $N=n+1$, yields
\[
 1\le\frac{R^{n+1}}{n!}
       \int_0^1e^{-(1-t)\Re\alpha}(1-t)^n\,dt
 \le\frac{R^N e^R}{N!}.
\]
Since $\log(N!)>N\log N-N$, this implies
$\log(R/N)+R/N>-1$.  The function $u\mapsto\log u+u$ is strictly
increasing from $-\infty$ to $+\infty$, so $R>\rho_0N$.
Moreover $\log4>5/4$, since $e<3$ and $3^5<4^4$; hence
$\rho_0>1/4$.

Continue the root or its conjugate pair from parameter $1$ to $s$.
Theorem~\ref{geo:bernstein} gives
$q(s)-s+1\ge q(1)>\rho_0(n+1)$.
The weaker constant $1/4$ will suffice in the quantitative arguments.
\end{proof}


\subsection{Finite fibers and large-parameter irreducibility}\label{sec:finite-fibers}

A monic rational factor of a monic integral polynomial has an
integral root sum. At a fiber where all but a bounded number of
roots vanish, reduction modulo a prime restricts this sum to a
subset sum of the residual roots. If the corresponding trace
relation is nonzero, its size bounds the prime divisors of the
parameter difference. Two such relations give a height bound by
linear forms in logarithms. We formulate this argument first, then
apply it to Laguerre polynomials. The Appell application is given
in Section~\ref{sec:appell}.

The arithmetic input comes from linear forms in logarithms.
We record the exact estimate before deriving the bound used here.

\subsubsection{Linear forms in logarithms}
We use the following rational specialization of Matveev's theorem
\cite[Corollary~2.3, p.~1219]{Matveev}. The number of logarithms
is allowed to vary; its dependence in the estimate will be needed
when the prime support grows with the degree.

\begin{lemma}\label{comp:matveev}
Let $a_1,\ldots,a_t$ be positive rational numbers different from $1$,
and let $b_1,\ldots,b_t$ be integers. Suppose that
\[
 \Lambda=b_1\log a_1+\cdots+b_t\log a_t\ne0.
\]
Put $B=\max\{1,|b_1|,\ldots,|b_t|\}$, and choose
\[
 A_j\ge\max\{h(a_j),|\log a_j|,0.16\}\qquad(1\le j\le t).
\]
Here $h(M/N)=\log\max\{M,N\}$ for coprime positive integers $M,N$.
Then
\begin{equation}\label{comp:matveev-bound}
 \log|\Lambda|>
 -C_t A_1\cdots A_t\log(eB),\qquad
 C_t=\min\left\{\frac e2\,30^{t+3}t^{9/2},\,2^{6t+20}\right\}.
\end{equation}
\end{lemma}
\begin{proof}
For $t\ge2$, reorder the terms so that $A_t=\max_j A_j$.
The coefficient parameter in \cite[(1.3)]{Matveev} then satisfies
\[
 B_{\mathrm M}=\max\{1,\max_j|b_j|A_j/A_t\}\le B.
\]
Corollary~2.3 of \cite[p.~1219]{Matveev}, with $D=\kappa=1$,
gives \eqref{comp:matveev-bound}. The corollary allows dependent
logarithms.
For $t=1$, write $a_1=M/N$ in lowest terms. Since $M\ne N$,
integration of $1/x$ between these positive integers gives
$|\log a_1|>e^{-h(a_1)}$; this implies the displayed bound.
\end{proof}

\subsubsection{Nearby integers with bounded prime support}

\begin{lemma}
\label{new:general-support-height}
Let $M,h$ be positive integers, and suppose
\[
 R\ge e^4,\qquad
 \log\operatorname{rad}(M(M+h))\le R.
\]
Put
\[
 T=\frac{2R}{\log R},\qquad
 C_R=\frac e2\,30^3 T^{9/2}(15\log R)^T.
\]
Then
\begin{equation}\label{new:support-height-bound}
 \log M<\max\{2\log(2h),\;4C_R\log(12eC_R)\}.
\end{equation}
\end{lemma}

\begin{proof}
The assertion is immediate if $M=1$ or $M\le h$.  Otherwise put
$L=\log M$ and $H=\log h$.  Factoring $(M+h)/M$ gives a nonzero
linear form
\[
 0<\Lambda=\log\frac{M+h}{M}
     =\sum_{\ell=1}^{t}b_\ell\log p_\ell<\frac hM,
 \qquad \sum_{\ell=1}^{t}\log p_\ell\le R,
\]
where the $p_\ell$ are distinct primes and all $b_\ell\ne0$.
Since the $\ell$th prime is at least $\ell+1$,
\[
 R\ge\log((t+1)!)\ge t\log t-t.
\]
We claim that $t\le T$.  Writing $w=\log R\ge4$, one has
\[
 T(\log T-1)-R
 =\frac Rw\bigl(w-2\log w+2\log2-2\bigr)>0.
\]
The expression in parentheses is positive at $w=4$ and increasing
thereafter.  Monotonicity of $z(\log z-1)$ for $z>1$ proves the claim.

Use the first constant in Lemma~\ref{comp:matveev}, the rational
case of \cite[Corollary~2.3, p.~1219]{Matveev}:
\[
 \log|\Lambda|>
 -\frac e2\,30^{t+3}t^{9/2}
       \left(\prod_{\ell=1}^{t}\log p_\ell\right)\log(eB),
 \qquad B=\max_\ell|b_\ell|.
\]
By the arithmetic--geometric mean inequality and $t\le T$,
\[
 \frac e2\,30^{t+3}t^{9/2}\prod\log p_\ell
 \le\frac e2\,30^3t^{9/2}(30R/t)^t\le C_R.
\]
The last inequality uses the increase of $z\log(30R/z)$ on
$0<z\le T$; indeed $T\le R/2$.  Also, since $M\ge2$ and $h<M$,
\[
 B\le\frac{\log(M+h)}{\log2}<3L.
\]
Consequently
\[
 L-H<C_R\log(3eL).
\]
If $L\le2H$, the claimed bound follows.  Otherwise
$L<2C_R\log(3eL)$.  For $a,b\ge1$ and $y>0$, the inequality
$y<a\log(by)$ implies $y<2a\log(2ab)$ whenever $2ab\ge e$:
at $y_0=2a\log(2ab)$ one has
$a\log(by_0)\le y_0$, and $y-a\log(by)$ increases for $y\ge y_0$.
Apply this with $a=2C_R$, $b=3e$.  It gives
$L<4C_R\log(12eC_R)$, as required.
\end{proof}

\subsubsection{The finite-fiber criterion}

If $R\in\mathbb Z[X]$ is monic of degree $r$, with roots
$\rho_1,\ldots,\rho_r$ counted with multiplicity, define its
\emph{subset-trace polynomial} by
\[
 \mathcal T_R(Z)=
 \prod_{J\subseteq\{1,\ldots,r\}}
        \left(Z-\sum_{j\in J}\rho_j\right).
\]
It is monic of degree $2^r$ and belongs to $\mathbb Z[Z]$: its
coefficients are algebraic integers and symmetric functions of the
roots.  The empty subset contributes the factor $Z$.

\begin{proposition}
\label{new:trace-specialization}
Let $P(T,X)\in\mathbb Z[T,X]$ be monic of degree $n$ in $X$, and
let $c(T)\in\mathbb Z[T]$.  Suppose that, at distinct integers
$t_1,\ldots,t_q$,
\[
 P(t_i,X)=X^{n-r_i}R_i(X),\qquad 0\le r_i\le n,
 \qquad R_i\in\mathbb Z[X]\text{ monic of degree }r_i.
\]
Let $t\in\mathbb Z\setminus\{t_1,\ldots,t_q\}$, and let a monic
integral factor of $P(t,X)$ have
degree $d$, root sum $\tau$, and centered trace $A=\tau-dc(t)$.
Then
\begin{equation}\label{new:trace-specialization-divisibility}
 \operatorname{rad}(|t-t_i|)
 \mid \mathcal T_{R_i}\bigl(A+dc(t_i)\bigr).
\end{equation}
Whenever two of the integers on the right are nonzero, the radical
of the product of the corresponding two parameter differences
divides the absolute value of their product.
\end{proposition}

\begin{proof}
For a prime $p\mid t-t_i$, reduction modulo $p$ takes the given
factor to a monic divisor of $X^{n-r_i}R_i(X)$.  Its roots, counted
with multiplicity, select a submultiset of the roots of $R_i$
together with some zero roots.  Its trace is therefore a root of
$\mathcal T_{R_i}$ modulo $p$.  This trace is congruent to
$A+dc(t_i)$, proving the divisibility prime by prime.  Reduction of
the symmetric definition of $\mathcal T_{R_i}$ is valid even when
roots coalesce modulo $p$.
\end{proof}

The next theorem reduces large-parameter irreducibility to a strip
bound and a condition on the subset traces of finitely many fibers.

\begin{theorem}
\label{new:finite-fiber-criterion}
Fix an integer $n_0\ge2$.  For each integer $n\ge n_0$, let
$P_n(T,X)\in\mathbb Z[T,X]$ be monic of degree $n$ in $X$, and let
$c_n(T)\in\mathbb Z[T]$.  Suppose that there are fixed constants
$C\ge1$, $\kappa\ge1$, and integers $q\ge3$, $r_0\ge0$, with the
following properties.
\begin{enumerate}
\item For every positive integer $t$ and every complex root $\alpha$
of $P_n(t,X)$,
\[
 |\operatorname{Re}(\alpha-c_n(t))|\le Cn^\kappa.
\]
\item There are distinct integers $t_{n,1},\ldots,t_{n,q}$ satisfying
\[
 |t_{n,i}|+|c_n(t_{n,i})|\le Cn^\kappa
\]
and specializations
\[
 P_n(t_{n,i},X)=X^{n-r_{n,i}}R_{n,i}(X),\qquad
 0\le r_{n,i}\le\min\{n,r_0\},
\]
where $R_{n,i}$ is monic of degree $r_{n,i}$ and every coefficient has absolute value
at most $Cn^\kappa$.
\item Put $U_{n,i}=\mathcal T_{R_{n,i}}$. For every integer
$1\le d\le n/2$ and every $A\in\mathbb Z$, at least two of
\[
 U_{n,i}\bigl(A+dc_n(t_{n,i})\bigr),\qquad 1\le i\le q,
\]
are nonzero.
\end{enumerate}
Then, for every $\varepsilon>0$, there is an effective
$N=N(C,\kappa,q,r_0,\varepsilon,n_0)$ such that
$P_n(t,X)$ is irreducible over $\mathbb Q$ whenever
\[
 n\ge N,\qquad t\in\mathbb Z_{>0},\qquad
 \log t\ge n^\varepsilon.
\]
\end{theorem}

The last condition says equivalently that, after omitting any one
index $j$, the remaining shifted trace polynomials have no common
integer root. A sufficient algebraic condition is
\begin{equation}\label{new:trace-gcd-condition}
 \gcd_{\mathbb Q[Z]}
 \{U_{n,i}(Z+dc_n(t_{n,i})):i\ne j\}=1
 \quad\text{for every $d,j$ as above}.
\end{equation}
The distinction is useful: the theorem only needs exclusion of
integer roots, whereas a resultant can verify the stronger condition.

\begin{proof}
Suppose a monic factor has degree $1\le d\le n/2$, and let $\tau$
be its root sum.  Gauss's lemma gives $\tau\in\mathbb Z$, so the
centered trace $A=\tau-dc_n(t)$ is integral.  Because it is real,
the first hypothesis bounds it by
\[
 |A|\le dCn^\kappa\le Cn^{\kappa+1}.
\]
By the third hypothesis, at least two of the integers
\[
 U_{n,i}\bigl(A+dc_n(t_{n,i})\bigr)
\]
are nonzero.

There are constants $C_1,K_1$, depending only on $C,\kappa,r_0$,
such that the absolute value of every nonzero displayed integer
is at most $C_1n^{K_1}$.  To see this directly, every root of
$R_{n,i}$ has absolute value at most $1+Cn^\kappa$ by Cauchy's
bound.  Every subset sum has absolute value at most
$r_0(1+Cn^\kappa)$, and there are at most $2^{r_0}$ factors in
$U_{n,i}$.  Also
\[
 |A+dc_n(t_{n,i})|\le2Cn^{\kappa+1}.
\]
These estimates give, for example, a uniform bound
\[
 |U_{n,i}(A+dc_n(t_{n,i}))|
 \le\bigl(2C+r_0(1+C)\bigr)^{2^{r_0}}
          n^{(\kappa+1)2^{r_0}}.
\]
Choose indices $i\ne j$ with nonzero evaluations.  For all large
$n$, the inequality $\log t\ge n^\varepsilon$ gives
$t>2Cn^\kappa$.  Therefore the two integers $t-t_{n,i}$ and
$t-t_{n,j}$ are positive.  Proposition~\ref{new:trace-specialization}
gives
\[
 \log\operatorname{rad}
 \bigl((t-t_{n,i})(t-t_{n,j})\bigr)
 \le2\log C_1+2K_1\log n.
\]
Their difference $h$ lies between $1$ and $2Cn^\kappa$, while
their smaller member $M$ satisfies $M\ge t/2$.
Apply Lemma~\ref{new:general-support-height} with
$R=\max\{e^4,2\log C_1+2K_1\log n\}$.
Its explicit bound gives
\[
 \log M\le
 \exp\!\left(O\!\left(
 \frac{\log n\,\log\log\log n}{\log\log n}\right)\right),
\]
for all sufficiently large $n$, with effective constants depending
only on the stated data. The bound is $n^{o(1)}$. In particular,
$\log M<n^{\varepsilon/2}$ for all sufficiently large $n$.
But $M\ge t/2$ gives
$\log M\ge n^\varepsilon-\log2>n^{\varepsilon/2}$, a contradiction.
\end{proof}

\begin{corollary}
\label{new:finite-fiber-bootstrap}
Under the hypotheses of Theorem~\ref{new:finite-fiber-criterion},
suppose that, for some fixed $\delta>0$, $P_n(t,X)$ is irreducible
for every integer $1\le t\le\exp(n^\delta)$ once $n$ is sufficiently
large.  Then $P_n(t,X)$ is irreducible for every positive integer
$t$ once $n$ is sufficiently large.
\end{corollary}

\begin{proof}
Apply the theorem with $\varepsilon=\delta$ to cover the
complementary range.
\end{proof}

\paragraph{Scope.}
The geometric hypothesis is stronger than necessary: a uniform
bound for the centered trace of every rational factor suffices.
Likewise, one may replace $U_{n,i}$ by any integral polynomial
annihilating the relevant subset traces, and may remove repeated
prime factors after evaluation.  The four-fiber Laguerre argument
uses both refinements, giving the much smaller exponent $6$ in its
prime-support bound.

The same proof also permits fibers of the form
\[
 P_n(t_{n,i},X)=(X-b_{n,i})^{n-r_{n,i}}
                   R_{n,i}(X-b_{n,i}),\qquad b_{n,i}\in\mathbb Z.
\]
One adds the bound $|b_{n,i}|\le Cn^\kappa$ to the hypotheses and
replaces $c_n(t_{n,i})$ in every trace evaluation by
$c_n(t_{n,i})-b_{n,i}$.  A degree-$d$ factor then has trace
$db_{n,i}$ plus a subset sum of the residual roots after
specialization.  All height and prime-support estimates are unchanged
apart from their constants.

The trace condition excludes persistent factors that the other
hypotheses do not detect.  For example, the Appell family
$n![z^n]e^{Xz}(1-z^2)^{-t}$ has a factor $X$ for every odd $n$,
because its coefficients have fixed parity. Its negative-integer
fibers have bounded residual degree. The convolution argument in
Section~\ref{sec:appell}, applied to $H_{n,t}(X)$ and
$(-1)^nH_{n,t}(-X)$, places every root in
$|\operatorname{Re}X|\le2(n-1)$, independently of $t$.
All the trace polynomials nevertheless vanish at the trace of
the factor $X$.

\begin{corollary}
\label{new:rational-finite-fiber-criterion}
Assume the hypotheses of Theorem~\ref{new:finite-fiber-criterion},
with its strip bound valid for every real parameter $t>0$.
Suppose also that $c_n(T)$ has degree at most one and that
\[
 \deg_T\bigl([X^{n-k}]P_n(T,X)\bigr)\le k
 \qquad(0\le k\le n).
\]
For every fixed $B\ge0$ and $\varepsilon>0$, there is an effective
$N=N(C,\kappa,q,r_0,n_0,B,\varepsilon)$ such that
$P_n(a/b,X)$ is irreducible over $\mathbb Q$ whenever
\[
 n\ge N,\qquad a,b\in\mathbb Z_{>0},\qquad (a,b)=1,
 \qquad b\le n^B,\qquad \log a\ge n^\varepsilon.
\]
\end{corollary}

\begin{proof}
The coefficient-degree assumption makes
\[
 F(X)=b^nP_n(a/b,X/b)
\]
monic and integral.  If it has a monic integral factor of degree
$1\le d\le n/2$ and trace $\tau$, put
$A=\tau-db c_n(a/b)$.  The affine condition gives $A\in\mathbb Z$,
and the real-parameter strip bound gives
$|A/b|\le Cn^{\kappa+1}$.
Write $e_i=\deg U_{n,i}\le 2^{r_0}$.  If
$p\mid a-bt_{n,i}$, then $p\nmid b$, and reduction to the fiber
$t_{n,i}$ gives
\[
 p\mid b^{e_i}U_{n,i}\bigl(A/b+d c_n(t_{n,i})\bigr).
\]
These evaluations are integers.  At least two are nonzero: otherwise
$A/b$ would be a common rational root of all but one of the shifted
trace polynomials.  Each of those polynomials is monic and integral,
so this rational root would be an integer, contrary to the trace
condition.

For all sufficiently large $n$, the hypotheses give
$a>2Cb n^\kappa$, so every $a-bt_{n,i}$ is positive and at least
$a/2$.  The coefficient and trace estimates in the proof of
Theorem~\ref{new:finite-fiber-criterion} bound each nonzero evaluation
by $b^{2^{r_0}}C_1n^{K_1}$, with the same effective constants
$C_1,K_1$.  For two nonzero evaluations, the corresponding positive
integers $a-bt_{n,i}$ and $a-bt_{n,j}$ therefore satisfy
\[
 \log\operatorname{rad}\bigl((a-bt_{n,i})(a-bt_{n,j})\bigr)
 \le 2\log C_1+2K_1\log n+2^{r_0+1}\log b.
\]
Their positive difference is at most $2Cb n^\kappa$.
Since $b\le n^B$, the radical bound is $O_B(\log n)$ and the
logarithm of the difference is $O_B(\log n)$.
Lemma~\ref{new:general-support-height} gives
$\log a=n^{o_B(1)}$, effectively and uniformly in $b$, contradicting
$\log a\ge n^\varepsilon$.
The rational change of variable defining $F$ preserves irreducibility.
\end{proof}

\subsubsection{Four degenerate Laguerre fibers}

For relatively prime positive integers $a,b$, put
\[
 F_{n,a,b}(X)=b^nH_{n,a/b}(X/b)
 =\sum_{k=0}^{n}\binom nk
      \left(\prod_{j=0}^{k-1}(a+jb)\right)X^{n-k}.
\]
This polynomial is monic and integral.  Its irreducibility over
$\mathbb Q$ is equivalent to that of $H_{n,a/b}$.

\begin{theorem}
\label{new:four-fiber}
Let $n\ge2$, and let $a,b$ be relatively prime positive integers.
If $F_{n,a,b}$ is reducible over $\mathbb Q$, then there are
$0\le i<j\le3$ such that
\[
 \operatorname{rad}\bigl((a+ib)(a+jb)\bigr)<b^3n^6.
\]
\end{theorem}

\begin{proof}
For $n=2$, the polynomial $H_{2,a/b}$ has discriminant $-4a/b<0$,
so it is irreducible and there is nothing to prove.  Assume $n\ge3$.
Choose a monic integral factor of degree $1\le d\le n/2$, with
roots $\beta_1,\ldots,\beta_d$.  Its centered trace
\[
 A=\sum_{\nu=1}^{d}(\beta_\nu+a)
\]
is an integer.  Since $\beta_\nu/b$ are roots of $H_{n,a/b}$, the
real-parameter trace estimate in Proposition~\ref{geo:trace} gives
\begin{equation}\label{new:centered-trace}
 |A|<bd(n-d).
\end{equation}
At parameters $s=0,-1,-2,-3$, the nonzero parts of $H_{n,s}$ are
\[
 1,\qquad X-n,\qquad X^2-2nX+n(n-1),\qquad R(X),
\]
where
\[
 R(X)=(X-n)^3-3n(X-n)-2n.
\]
Polynomials annihilating all subset sums of the nonzero roots are
\begin{align*}
 T_0(Z)&=Z,\\
 T_1(Z)&=Z(Z-n),\\
 T_2(Z)&=Z(Z-2n)\bigl((Z-n)^2-n\bigr),\\
 T_3(Z)&=-Z(Z-3n)R(Z)R(3n-Z).
\end{align*}
For the last identity, the subset sums are $0$, $3n$, the three
roots of $R$, and their complements in $3n$.

If $p\mid a+ib$, then $p\nmid b$, because $\gcd(a,b)=1$.
The chosen factor reduces to a monic divisor of the fiber at
$a=-ib$.  Its uncentered trace is congruent to $A+ibd$ modulo $p$.
Consequently
\begin{equation}\label{new:fiber-divisibility}
 p\mid b^{\deg T_i}T_i(A/b+id).
\end{equation}
The quantity on the right is an integer.  The assertion remains
valid in every characteristic, since roots are counted with
multiplicity.  Whenever $A/b$ is integral, powers of $b$ can be
discarded from this divisibility.

First suppose $A\notin\{0,-bd,b(n-d)\}$.  Taking $(i,j)=(0,1)$
in \eqref{new:fiber-divisibility} shows that the required radical
divides the nonzero integer
\[
 N=A(A+bd)(A+bd-bn).
\]
By \eqref{new:centered-trace},
\[
 |A|<bdn,\qquad |A+bd|<bdn,\qquad
 |A+bd-bn|<b(d+1)n,
\]
and hence $|N|<b^3n^3d^2(d+1)<b^3n^6$.

For the three exceptional trace values, the following integers
contain the required prime supports whenever they are nonzero:
\begin{equation}\label{new:exceptional-support-table}
\begin{array}{c|c|c}
 A &(i,j)&N\\ \hline
 -bd &(0,2)&d(d-2n)((d-n)^2-n)\\
 0 &(1,2)&2d(d-n)((2d-n)^2-n)\\
 b(n-d)&(0,2)&(n-d)(n+d)(d^2-n).
\end{array}
\end{equation}
This follows by substitution in $T_0,T_1,T_2$; replacing repeated
factors by one copy preserves their prime support.  Every nonzero
entry has absolute value less than $n^4$.
The first entry can vanish only when $(n,d)=(4,2)$: writing
$n=m^2$ and $n-d=m$ gives $m^2-m\le m^2/2$, hence $m\le2$.
At this exceptional trace $A=-2b$, use fibers $0$ and $3$.
Their evaluations are $T_0(-2)=-2$ and $T_3(4)=-2048$, so all
relevant primes divide $2$.  The required bound follows.

The other two entries leave only the following cases:
\[
 n=m^2,\quad (A,d)=\left(0,\frac{m^2-m}{2}\right),
 \qquad\text{or}\qquad
 n=m^2,\quad (A,d)=\bigl(b(m^2-m),m\bigr).
\]
Here $m\ge2$.

In the first case use fibers $1$ and $3$.  Writing
$d=m(m-1)/2$, direct substitution gives
\[
 T_1(d)=-\frac{m^2(m^2-1)}4
\]
and
\[
 T_3(3d)=\frac{9m^6(m^2-1)^3
 (m^2-9m+16)(m^2+9m+16)}{256}.
\]
Every prime divisor of either integer divides
\[
 N=3m(m^2-1)(m^2-9m+16)(m^2+9m+16).
\]
Neither quadratic factor vanishes at an integer: the first has
discriminant $17$, and the second is positive.  For $m\ge3$,
\[
 |m^4-49m^2+256|
 \le m^4+49m^2+256<10m^4,
\]
so $0<|N|<30m^7<m^{12}=n^6$.
For $m=2$, the same integer has absolute value
$1368<4096=m^{12}$.

In the second case use fibers $0$ and $3$.  Put
\[
 J(m)=m(m-1)(m-4)+2.
\]
Then
\[
 T_3(n+2m)=4m^6(m-1)^3(m+2)J(m).
\]
The prime supports of $n-m=m(m-1)$ and this value are contained
in that of
\[
 N=2m(m-1)(m+2)J(m).
\]
Here $J(2)=-2$ and $J(3)=-4$, while $0<J(m)<m^3$ for $m\ge4$.
Thus $0<|J(m)|<m^3$ for every $m\ge2$, and
$0<|N|<4m^6<m^{12}=n^6$.
Powers of $b$ introduce no relevant primes.  This completes the proof.
\end{proof}

\begin{corollary}
\label{new:rational-large-parameter}
For every fixed $B\ge0$ and $\varepsilon>0$, there is an effective
integer $N(B,\varepsilon)$ such that $F_{n,a,b}$ is irreducible over
$\mathbb Q$ whenever
\[
 n\ge N(B,\varepsilon),\qquad
 \gcd(a,b)=1,\qquad 1\le b\le n^B,
 \qquad \log a\ge n^\varepsilon.
\]
More explicitly, a reducible member with $n\ge2$ satisfies
\[
 \log a<\max\{2\log(6b),\;4C_R\log(12eC_R)\},\qquad
 R=\max\{e^4,\,3\log b+6\log n\},
\]
where $C_R$ is defined in Lemma~\ref{new:general-support-height}.
\end{corollary}

\begin{proof}
Theorem~\ref{new:four-fiber} supplies $M=a+ib\ge a$ and
$h=(j-i)b\le3b$ to which Lemma~\ref{new:general-support-height}
applies.  This proves the explicit estimate.
For $b\le n^B$, write $u=\log n$.  Then $R=O_B(u)$ and
\[
 \log C_R
 =\log((e/2)30^3)+\frac92\log\frac{2R}{\log R}
   +\frac{2R\log(15\log R)}{\log R}=o_B(u).
\]
It follows that
$4C_R\log(12eC_R)=\exp(o_B(u))=n^{o_B(1)}$, whereas
$2\log(6b)=O_B(u)$.  Both are eventually smaller than
$n^\varepsilon$.  All quantities in the explicit bound are
elementary and effectively computable, so the threshold is effective.
\end{proof}

Positive rational parameters can give reducible polynomials.  For example,
\[
 H_{3,1/4}(X)=\left(X+\frac34\right)
                  \left(X^2+\frac{15}{16}\right),
 \qquad F_{3,1,4}(X)=(X+3)(X^2+15).
\]
Thus the numerator and denominator conditions in the rational theorem
have genuine content; positivity alone is not sufficient.

\section{Uniform irreducibility}\label{sec:energy-irreducibility}

\subsection{Local arithmetic and energy}\label{sec:common-local}

A rational factor selects a Galois-invariant set of roots; a block
system gives an invariant partition. Both restrict how roots can
occupy a local cluster. We express these restrictions by positive
semidefinite matrices and compare the resulting local contributions
with a bound from complex root separation. The same inequality will
therefore exclude factors and block systems.

We first determine the tame clusters of $H_{n,s}$ and prove the
separation estimates needed at all places. The energy theorem is
then stated for a general monic integral polynomial. Its applications
depend on finding clusters with enough imbalance; the differential
equation alone supplies only the separation bounds.

\subsubsection{Tame root clusters and inertia}\label{sec:inertia}

The local calculation rests on the following form of Hensel's lemma.
No irreducibility assumption is needed.

\begin{lemma}
\label{lem:general-tame-packet}
Let $F\in\mathbb Z_p[X]$ be monic, and suppose
$\overline F=X^N\overline R$ with $1\le N\le\deg F$,
$\overline R$ squarefree, and $\overline R(0)\ne0$.
Let $A$ be the degree-$N$ Hensel factor lifting $X^N$.
Suppose $p\nmid N$ and, for an integer $e>0$, every nonleading
coefficient of $A$ has valuation at least $e$, with $v_p(A(0))=e$.
Put $g=\gcd(N,e)$ and $q=N/g$.
Then every root of $A$ has valuation $e/N$, as does every difference
between two distinct roots of $A$. All other roots are units, and
every difference of distinct roots involving one of those roots has
valuation zero.
The inertia group of the splitting field over $\mathbb Q_p$ is
cyclic of order $q$, with a generator of cycle type
$q^g1^{\deg F-N}$.
\end{lemma}
\begin{proof}
Write $N=gq$ and $e=ga$, where $\gcd(a,q)=1$.
Choose a finite unramified extension $K_0/\mathbb Q_p$ whose residue
field splits $\overline R$ and $Y^N+c$, where $c$ is the reduction
of $p^{-e}A(0)$, and which contains the $q$th roots of unity.
This is possible since $p\nmid q$. If $\pi^q=p$, then
$p^{-e}A(\pi^aY)$ is integral and reduces to $Y^N+c$.
That polynomial is separable because $p\nmid N$, so its roots lift
to units of $K_0(\pi)$ with distinct residues. Rescaling gives the
asserted valuations of the roots of $A$ and their differences.
The complementary Hensel factor splits over $K_0$, with distinct
nonzero residues.

The splitting field is therefore contained in the tame extension
$K_0(\pi)$, whose ramification index is $q$. Conversely, the valuation
$a/q$ of every root of $A$ forces its root field to have ramification
index divisible by $q$. Thus the splitting field and every such root
field have ramification index exactly $q$. Tame inertia is cyclic,
and its orbit on a root has size equal to the ramification index of
that root field. Its generator consequently has $g$ cycles of length
$q$, while the complementary roots are fixed.
\end{proof}

For the Laguerre family, a prime above the degree can divide only
one of the consecutive parameter nodes $s,\ldots,s+n-1$.
The preceding lemma then determines the full inertia action.

\begin{proposition}\label{prop:exact-inertia}
Let $n\ge2$ and $s\ge1$ be integers. Let $p>n$ be prime and
suppose $p\mid s+j$, where $0\le j<n$.
Set
\[
 e=v_p(s+j),\qquad N=n-j,\qquad g=\gcd(N,e),\qquad q=N/g.
\]
In the splitting field of $H_{n,s}$, inertia at $p$ is cyclic of order
$q$ and has a generator of cycle type $q^g1^j$. In particular, inertia
is trivial exactly when $N\mid e$.
\end{proposition}

\begin{proof}
There is at most one node divisible by $p$, since the nodes lie in an
interval of length $n$. Modulo $p$ we have
\[
 H_{n,s}(X)\equiv X^NR_{n,j}(X)\pmod p,\qquad
 R_{n,j}(X)=\sum_{k=0}^j(-1)^k\binom nk\frac{j!}{(j-k)!}X^{j-k}.
\]
The constant coefficient of $R_{n,j}$ is nonzero. Also
$R_{n,j}=(-1)^j j!L_j^{(N)}$ and hence
\begin{equation}\label{eq:residual-ODE}
 XR_{n,j}''+(N+1-X)R_{n,j}'+jR_{n,j}=0.
\end{equation}
The residual polynomial is separable. Indeed, a repeated nonzero root,
followed through successive derivatives of \eqref{eq:residual-ODE},
would make all derivatives through order $j$ vanish there, whereas
$R_{n,j}^{(j)}=j!\ne0$ in characteristic $p$. The case $j=0$ is immediate.

Hensel factorization gives $H_{n,s}=FU$ over $\Z_p$, with $F,U$ monic,
$\deg F=N$, $\overline F=X^N$, and $\overline U=R_{n,j}$.
Every coefficient of $H_{n,s}$ below degree $N$ has valuation exactly
$e$. Since $U(0)$ is a unit, induction on the coefficient index in
$H_{n,s}=FU$ gives
\[
 v_p(F(0))=e,\qquad v_p([X^h]F)\ge e\quad(0<h<N).
\]
Lemma~\ref{lem:general-tame-packet} now applies and gives the
stated inertia group and cycle type. Its order is one exactly when
$N\mid e$.
\end{proof}


\subsubsection{Root separation}
A \emph{valuation packet} is the full set of roots with a specified
valuation at a fixed prime. For the Laguerre family put
\[
 \epsilon_p=\begin{cases}
 1/(p-1),&p\le n,\\
 0,&p>n.
 \end{cases}
\]

The differential equation bounds the distance between distinct roots
at every place. The $p$-adic bound determines how far a cluster can
persist; the complex bound supplies disjoint disks around the roots.

\begin{lemma}\label{ene:separation}
Let $n\ge2$ and $s\ge1$ be integers. Any two distinct roots of
$H_{n,s}$ satisfy
\begin{equation}\label{ene:padic-separation}
 v_p(\alpha-\beta)\le v_p(\alpha)+\epsilon_p.
\end{equation}
If $s\ge n$, they also satisfy
\begin{equation}\label{ene:complex-separation}
 |\alpha-\beta|>\frac1{12}\sqrt{\frac{s}{n}}.
\end{equation}
\end{lemma}
\begin{proof}
The roots are simple by Section~\ref{geo:section}.
Fix a root $\alpha$ and set
\[
 D_k=\frac{H_{n,s}^{(k)}(\alpha)}{H'_{n,s}(\alpha)},
 \qquad D_0=0,\quad D_1=1.
\]
Differentiating \eqref{geo:ode} $k$ times gives, for $0\le k\le n-2$,
\begin{equation}\label{ene:derivative-recursion}
 \alpha D_{k+2}-(\alpha+n+s-1-k)D_{k+1}+(n-k)D_k=0.
\end{equation}
The roots are algebraic integers. If $a=v_p(\alpha)\ge0$, induction
in this recurrence gives $v_p(D_k)\ge-(k-1)a$ for $k\ge1$.
For $v_p(z)>a+\epsilon_p$, every term after the constant term in
\[
 \frac{H_{n,s}(\alpha+z)}{H'_{n,s}(\alpha)z}
 =1+\sum_{k=2}^n\frac{D_k}{k!}z^{k-1}
\]
has positive valuation. Here we use
$v_p(k!)\le(k-1)/(p-1)$ when $p\le n$, and $v_p(k!)=0$ when $p>n$.
The displayed quotient cannot vanish, proving
\eqref{ene:padic-separation}.

For the complex estimate put $t=n/s\le1$.
The annulus \eqref{geo:eq-annulus} and centered-radius estimate
\eqref{geo:eq-radius} imply, for $0\le k\le n-2$,
\[
 \left|\frac{\alpha+n+s-1-k}{\alpha}\right|
 \le2t+2\sqrt{2t}<5\sqrt t,
 \qquad \left|\frac{n-k}{\alpha}\right|\le t.
\]
With $M=6\sqrt t$ we have $5\sqrt t\,M+t<M^2$.
The recurrence therefore gives $|D_k|\le M^{k-1}$ by induction.
If $0<|z|\le1/(2M)$, the sum after the constant term in the same
Taylor quotient has absolute value at most
\[
 \sum_{k\ge2}\frac{2^{1-k}}{k!}=2e^{1/2}-3<1.
\]
This proves \eqref{ene:complex-separation}.
\end{proof}
\subsubsection{An adelic energy inequality}

The local positivity and the complex estimate below are elementary
logarithmic-energy arguments. Galois invariance allows their
comparison by the product formula, even when the matrix entries
are arbitrary real numbers. The condition $Q\mathbf1=0$ means
zero total charge at the complex place, and positive semidefiniteness
makes every local cluster contribution nonnegative. Encoding a
factor or a block system in $Q$ gives a common form of the two
arithmetic obstructions. A comparison of finite-place and complex
bounds for a factor resultant is also used in Deng's preprint
\cite[Sections~4--5]{Deng2026}.

\begin{theorem}
\label{thm:adelic-matrix-energy}
Let $F\in\mathbb Z[X]$ be monic and separable of degree $n\ge2$,
with nonzero roots $\alpha_1,\ldots,\alpha_n$. For every prime $p$,
fix an extension of $v_p$ to $\overline{\mathbb Q}$, normalized by
$v_p(p)=1$.  Suppose that numbers
$\epsilon_p\geq0$, zero for all but finitely many $p$, satisfy
\begin{equation}
\label{eq:general-local-separation}
 v_p(\alpha_i-\alpha_j)\leq v_p(\alpha_i)+\epsilon_p
 \qquad(i\neq j).
\end{equation}
Let $Q=(q_{ij})$ be a real symmetric positive semidefinite matrix,
invariant under the Galois action on the roots, and suppose
$Q\mathbf1=0$.  Put $\tau=\operatorname{tr}Q$ and
\[
 E_p(Q)=\sum_{i\neq j}q_{ij}v_p(\alpha_i-\alpha_j)
       +\sum_iq_{ii}\bigl(v_p(\alpha_i)+\epsilon_p\bigr).
\]
Then only finitely many $E_p(Q)$ are nonzero, and
\begin{equation}
\label{eq:general-local-energy}
 E_p(Q)\geq\int_0^\infty
       \mathbf1_{C_p(t)}^{\mathsf T}Q\mathbf1_{C_p(t)}\,dt\geq0,
 \qquad C_p(t)=\{i:v_p(\alpha_i)\geq t\}.
\end{equation}
Furthermore,
\begin{align}
\label{eq:general-product-energy}
 \sum_pE_p(Q)\log p
 &=\sum_{i\neq j}q_{ij}\log|\alpha_i-\alpha_j|
   +\sum_iq_{ii}\log|\alpha_i|
   +\tau\sum_p\epsilon_p\log p.
\end{align}
If the closed disks of radii $\rho_i>0$ about the complex roots are
pairwise disjoint, then
\begin{equation}
\label{eq:general-energy-bound}
 0\leq\sum_pE_p(Q)\log p
 \leq\sum_iq_{ii}\log\frac{|\alpha_i|}{\rho_i}
       +\tau\sum_p\epsilon_p\log p.
\end{equation}
In particular, if $|\alpha_i|\leq R$ and $\rho_i=\rho$, the upper
bound is $\tau(\log(R/\rho)+\sum_p\epsilon_p\log p)$.

If, at some prime with $\epsilon_p=0$, a set $C$ of $N\geq1$ roots
has valuation $e/N$ for some $e>0$, all differences between its distinct roots have
valuation $e/N$, and every other root and every other difference has
valuation zero, call $C$ an \emph{exact packet} at $p$. Then
\begin{equation}
\label{eq:general-packet-energy}
 E_p(Q)=\frac eN\mathbf1_C^{\mathsf T}Q\mathbf1_C.
\end{equation}
The theorem can also be applied after replacing each $\alpha_i$ by
$\alpha_i-a$, for a fixed integer $a$ with $F(a)\neq0$, provided
\eqref{eq:general-local-separation} is verified for the shifted roots;
the correction terms may change.
\end{theorem}

\begin{proof}
Write $h_i=v_p(\alpha_i)+\epsilon_p$.  At height $t>0$ retain the
indices with $h_i\geq t$ and partition them by the equivalence
relation $v_p(\alpha_i-\alpha_j)\geq t$.  The ultrametric inequality
makes this relation transitive.  Hypothesis
\eqref{eq:general-local-separation} ensures that every distinct pair
with difference valuation at least $t$ is retained.  Consequently
\[
 E_p(Q)=\int_0^\infty\sum_C
        \mathbf1_C^{\mathsf T}Q\mathbf1_C\,dt.
\]
Every term is nonnegative.  The indices with
$v_p(\alpha_i)\geq t$ form one entire retained class when this set is
nonempty: their pairwise differences have valuation at least $t$,
whereas a root of smaller valuation cannot belong to that class.
This proves \eqref{eq:general-local-energy}.  Finiteness follows
because, outside the supports of $\epsilon_p$, $F(0)$, and
$\Disc F$, all valuations in the definition are zero.
The packet formula is immediate from the same definition.

To apply the product formula with real matrix entries, first group
the terms by Galois orbits. The entries $q_{ii}$ are constant on each
root orbit $I$, and the entries $q_{ij}$ are constant on each orbit
$\mathcal O$ of unordered pairs of distinct roots. The orbit products
\[
 \prod_{i\in I}\alpha_i
 \quad\text{and}\quad
 \prod_{\{i,j\}\in\mathcal O}(\alpha_i-\alpha_j)^2
\]
are nonzero rational numbers.  Apply the product formula to each
one, weight the resulting identity by the corresponding real matrix
entry, and sum over the orbits.  The squares in the pair products
account for the two ordered terms $q_{ij}$ and $q_{ji}$.  Adding
$\tau\sum_p\epsilon_p\log p$ gives
\eqref{eq:general-product-energy}.

Decompose $Q=\sum_\nu c_\nu c_\nu^{\mathsf T}$ with real vectors
$c_\nu$ orthogonal to $\mathbf1$. For one such vector $c$, let
$\sigma_i$ be normalized arclength on the circle of radius $\rho_i$
about $\alpha_i$, and put $\eta=\sum_i c_i\sigma_i$. Its logarithmic
potential is the continuous, piecewise harmonic function
\[
 U(z)=\sum_i c_i\log\max\{\rho_i,|z-\alpha_i|\}.
\]
Because $\sum_i c_i=0$, we have $U(z)=O(|z|^{-1})$ and
$\nabla U(z)=O(|z|^{-2})$ at infinity. Integration by parts on the
disks and their exterior, using $\Delta U=2\pi\eta$, gives
\[
 \iint\log|z-w|\,d\eta(z)\,d\eta(w)
 =-\frac1{2\pi}\int_{\mathbb C}|\nabla U|^2\,dA\le0.
\]
The radial derivative jumps by $c_i/\rho_i$ across the $i$th circle,
and the outer boundary term tends to zero. The mean-value property
identifies the interaction between distinct circles with
$\log|\alpha_i-\alpha_j|$ and the self-interaction of the $i$th circle
with $\log\rho_i$. Summing over $\nu$ yields
\[
 \sum_{i\neq j}q_{ij}\log|\alpha_i-\alpha_j|
       +\sum_iq_{ii}\log\rho_i\leq0.
\]
Together with \eqref{eq:general-product-energy}, this proves
\eqref{eq:general-energy-bound}.  Translating by an integer preserves
monicity, integrality, the differences, and the Galois action.
\end{proof}


\begin{lemma}
\label{lem:matrix-energy}
Let $n\geq2$, $s\geq1$ be integers and let $Q$ satisfy the hypotheses
of Theorem~\ref{thm:adelic-matrix-energy} for the roots of $H_{n,s}$.
Use
\[
 \epsilon_p=\begin{cases}1/(p-1),&p\leq n,\\0,&p>n.\end{cases}
\]
Then \eqref{eq:general-local-energy}--\eqref{eq:general-energy-bound}
hold.  If $p>n$ and $p^e\Vert s+j$, $0\leq j<n$, put $N=n-j$ and
let $C$ be the set of roots of positive $p$-adic valuation.  Then
$|C|=N$ and
\begin{equation}
\label{eq:matrix-packet}
 E_p(Q)=\frac eN\mathbf1_C^{\mathsf T}Q\mathbf1_C.
\end{equation}
If $s\geq n$, $\log n\geq10000$, and $Q\neq0$, then
\[
 \sum_pE_p(Q)\log p
 <\operatorname{tr}(Q)
   \left(\frac12\log s+\frac32\log n+\log48\right).
\]
\end{lemma}

\begin{proof}
Lemma~\ref{ene:separation} supplies the local-separation hypothesis.
The proof of Proposition~\ref{prop:exact-inertia} identifies the $N$
positive-valuation roots as the tame packet of
Lemma~\ref{lem:general-tame-packet}, so
\eqref{eq:general-packet-energy} gives \eqref{eq:matrix-packet}.  Finally, use
$|\alpha_i|<s+n\leq2s$ from Lemma~\ref{geo:annulus}, the disjoint
disk radius $\rho=\sqrt{s/n}/24$ from Lemma~\ref{ene:separation}, and
$\sum_{p\leq n}\log p/(p-1)<\log n$ from
\eqref{pade:dusart-sum}.
\end{proof}

\begin{proposition}
\label{prop:general-projectors}
In the notation of Theorem~\ref{thm:adelic-matrix-energy}, a
Galois-invariant subset $A$ of $k$ roots, $0<k<n$, gives the admissible
rank-one matrix $Q=cc^{\mathsf T}$, where
\[
 c_i=\begin{cases}1-k/n,&i\in A,\\-k/n,&i\notin A,\end{cases}
 \qquad \operatorname{tr}Q=\frac{k(n-k)}n.
\]
For an exact packet $C$ of size $N$ as in
\eqref{eq:general-packet-energy}, putting $m=|A\cap C|$ gives
\[
 E_p(Q)=\frac eN\left(m-\frac{kN}{n}\right)^2.
\]

A Galois-invariant partition into $a\geq2$ blocks $B_1,\ldots,B_a$,
each of size $b\geq2$, with $n=ab$, gives the admissible projector
\[
 q_{ij}=\frac{\mathbf1_{\{i,j\text{ in the same block}\}}}{b}
          -\frac1n,
 \qquad \operatorname{tr}Q=a-1.
\]
For an exact packet and $k_\ell=|B_\ell\cap C|$,
\begin{equation}
\label{eq:general-block-energy}
 E_p(Q)=\frac eN
       \left(\frac1b\sum_{\ell=1}^a k_\ell^2-\frac{N^2}{n}\right).
\end{equation}
If an inertia element of prime order $P>a$ has support exactly $C$,
write $N=P\ell$ and $\ell=au+r$, $0\leq r<a$.  Then
\begin{equation}
\label{eq:general-block-imbalance}
 E_p(Q)\geq\frac{eP^2r(a-r)}{nN}.
\end{equation}
When the uniform disk and modulus bounds $\rho,R$ hold, take such
packets at distinct primes. If the sum of the displayed lower bounds,
each multiplied by $\log p$, exceeds
$(a-1)(\log(R/\rho)+\sum_p\epsilon_p\log p)$, then that block
system is impossible.  Independently, if an element of prime order $P>b$ has
support $C$, then $bP$ must divide $|C|$.
\end{proposition}

\begin{proof}
The rank-one formula is immediate.  The block matrix is the
orthogonal projector onto the vectors constant on each block and
orthogonal to $\mathbf1$, giving positive semidefiniteness, its trace,
and \eqref{eq:general-block-energy}.  An element of order $P>a$ acts
trivially on the $a$ blocks.  Its nontrivial cycles have length $P$,
so every $k_\ell$ is divisible by $P$.  Among nonnegative integers
with sum $\ell$, the sum of squares is minimized by $a-r$ entries
equal to $u$ and $r$ entries equal to $u+1$.  Hence
\[
 \sum k_\ell^2\geq
  P^2\left(\frac{\ell^2}{a}+\frac{r(a-r)}a\right).
\]
Substitution proves \eqref{eq:general-block-imbalance}; summing it and
using nonnegativity at all other primes proves the energy criterion.
For the final divisibility assertion, any fixed block has fewer than
$P$ elements and is therefore fixed pointwise.  Every moved block lies
in an orbit of $P$ blocks and contributes all its $b$ elements to the
support.
\end{proof}

\begin{corollary}
\label{cor:general-energy-irreducibility}
Under the hypotheses of Theorem~\ref{thm:adelic-matrix-energy},
suppose $|\alpha_i|\leq R$ and the closed disks of radius $\rho>0$
about the roots are pairwise disjoint.  Put
\[
 \mathcal B=\log(R/\rho)+\sum_p\epsilon_p\log p.
\]
Let $\mathcal P$ be a finite set of exact packet primes, with
$\epsilon_p=0$, packet size $N_p$, height $e_p/N_p$, and all inertia
orbits in the packet of a common length $q_p$.  For
$1\leq k\leq\lfloor n/2\rfloor$, define
\[
 \mathcal M_{p,k}=\{m\in\mathbb Z:
  \max(0,k-n+N_p)\leq m\leq\min(k,N_p),\quad q_p\mid m\}.
\]
If, for every such $k$, either some $\mathcal M_{p,k}$ is empty or
\begin{equation}
\label{eq:general-numerical-irreducibility}
 \sum_{p\in\mathcal P}
  \frac{e_p\log p}{nN_pk(n-k)}
  \min_{m\in\mathcal M_{p,k}}(nm-kN_p)^2
  >\mathcal B,
\end{equation}
then $F$ is irreducible over $\mathbb Q$.
For the tame packets of Lemma~\ref{lem:general-tame-packet}, one may
take $q_p=N_p/\gcd(N_p,e_p)$.
\end{corollary}

\begin{proof}
A proper rational factor has degree $k\leq n/2$ after choosing the
smaller complementary factor.  Its root set is a union of inertia
orbits, so its packet intersection has size in $\mathcal M_{p,k}$.
Apply the factor matrix of Proposition~\ref{prop:general-projectors}.
Dividing its local energy by
$\operatorname{tr}Q=k(n-k)/n$ gives the summand on the left of
\eqref{eq:general-numerical-irreducibility}.  The global upper bound
is $\mathcal B$, a contradiction.
\end{proof}
\subsubsection{Differential equations as a source of energy bounds}

The proof of Lemma~\ref{ene:separation} applies more generally to
a second-order differential equation. Its leading coefficient
replaces the root in the diagonal term of the energy. Since that
coefficient need only be nonzero at the roots, the polynomial itself
may have a zero root.

\begin{proposition}\label{new:ode-energy}
Let $P\in\mathbb Z[X]$ be monic and separable of degree $n\ge2$,
and suppose
\[
 A P''+B P'+CP=0,
 \qquad A,B\in\mathbb Z[X],\ C\in\mathbb Z,
 \quad\deg A\le2,\quad\deg B\le1,\quad\gcd(A,P)=1.
\]
Write its roots as $\alpha_i$, and put
$\epsilon_p=1/(p-1)$ for $p\le n$ and $\epsilon_p=0$ otherwise.
Then
\[
 v_p(\alpha_i-\alpha_j)
 \le\min\{v_p(A(\alpha_i)),v_p(A(\alpha_j))\}+\epsilon_p
 \quad(i\ne j).
\]
For any real symmetric Galois-invariant positive semidefinite matrix $Q$
with $Q\mathbf1=0$, define
\[
 E^A_p(Q)=\sum_{i\ne j}q_{ij}v_p(\alpha_i-\alpha_j)
       +\sum_iq_{ii}\bigl(v_p(A(\alpha_i))+\epsilon_p\bigr).
\]
These energies are nonnegative, and with $\tau=\operatorname{tr}Q$,
\begin{align*}
 \sum_pE^A_p(Q)\log p
 &=\sum_{i\ne j}q_{ij}\log|\alpha_i-\alpha_j|
       +\sum_iq_{ii}\log|A(\alpha_i)|+\tau S(n)\\
 &\le\tau\bigl(\log(4MR)+S(n)\bigr),
 \qquad S(n)=\sum_{p\le n}\frac{\log p}{p-1},
\end{align*}
where $R=\max\{1,\max_i|A(\alpha_i)|\}$,
$M=\max\{1,U+\sqrt V\}$, and
\begin{align*}
 U&=\max_{i,\,0\le k\le n-2}
   \left|\frac{kA'(\alpha_i)+B(\alpha_i)}{A(\alpha_i)}\right|,\\
 V&=\max_{i,\,0\le k\le n-2}
   \left|\frac{\binom{k}{2}A''(\alpha_i)+kB'(\alpha_i)+C}
                  {A(\alpha_i)}\right|.
\end{align*}
\end{proposition}
\begin{proof}
At a root $\alpha$ put $D_k=P^{(k)}(\alpha)/P'(\alpha)$.
Differentiating the equation $k$ times gives
\[
 A(\alpha)D_{k+2}+(kA'(\alpha)+B(\alpha))D_{k+1}
 +\left(\binom{k}{2}A''(\alpha)+kB'(\alpha)+C\right)D_k=0,
\]
with $D_0=0$, $D_1=1$.
Integrality implies
$v_p(D_k)\ge-(k-1)v_p(A(\alpha))$.
Since $v_p(k!)\le(k-1)\epsilon_p$ for $2\le k\le n$, the quotient
\[
 \frac{P(\alpha+z)}{zP'(\alpha)}
   =1+\sum_{k=2}^n\frac{D_k}{k!}z^{k-1}
\]
cannot vanish if $v_p(z)>v_p(A(\alpha))+\epsilon_p$.
Interchanging the two roots proves the local bound.

Use the cluster-integral proof of
Theorem~\ref{thm:adelic-matrix-energy}, now with diagonal heights
$h_i=v_p(A(\alpha_i))+\epsilon_p$.  Each pair height is at most both
diagonal heights, so the same integral gives $E^A_p(Q)\ge0$.
For each Galois orbit of roots, the product of the corresponding
$A(\alpha_i)$ is a nonzero rational number. The orbit-product argument
in Theorem~\ref{thm:adelic-matrix-energy} therefore gives the displayed
global identity. Only primes dividing
$\Disc(P)\operatorname{Res}(P,A)$ or at most $n$ can
contribute.

For complex separation, $M^2\ge UM+V$ and the recurrence imply
$|D_k|\le M^{k-1}$.  If $0<|z|\le1/(2M)$, the nonconstant part of
the Taylor quotient has absolute value at most
\mbox{$\sum_{k\ge2}2^{1-k}/k!=2e^{1/2}-3<1$}.
Thus the closed disks of radius $1/(4M)$ about the roots are
disjoint.  The circle-energy estimate in the proof of
Theorem~\ref{thm:adelic-matrix-energy} gives
$\sum_{i\ne j}q_{ij}\log|\alpha_i-\alpha_j|\le\tau\log(4M)$.
This proves the global bound.
\end{proof}

For Laguerre polynomials, $A(X)=X$ recovers the earlier diagonal
term. The monic probabilists' Hermite polynomials satisfy
$P''-XP'+nP=0$, so the proposition applies with $A=1$, including in
odd degree when $P$ has the factor $X$. To deduce irreducibility one
must still find local clusters whose energy exceeds the global bound.

\subsection{Pad\'e norms and the prime-square obstruction}
\label{pade:section}

The root-modulus gap also has an arithmetic consequence. We use
the exponential Pad\'e pair to construct a nonzero integral
resultant. Its local valuations and its complex absolute value give
opposite bounds, which constrain the prime divisors of a factor's
constant term. A Newton-polygon obstruction then excludes factors
throughout a polynomial range of parameters. The Pad\'e connection
for this family also appears in Cullinan and Scheel~\cite{CS}.
For positive integers $K$ and $d$, put
\begin{equation}\label{pade:Td}
 T_d(K)=\sum_{p\mid K}\min\left\{v_p(K),\frac d{p-1}\right\}\log p.
\end{equation}
\subsubsection{The exponential Pad\'e pair}

\begin{proposition}\label{pade:prop-budget}
Let $n\ge3$ and $s\ge1$ be integers. If $G$ is a monic rational
factor of $H_{n,s}$ of degree $d\ge1$, and $K=|G(0)|$, then,
with $\rho_0$ as in Corollary~\ref{geo:gap},
\begin{equation}\label{pade:budget}
 \log\bigl(\rho_0(n+1)\bigr)
 <\frac{T_d(K)}d+\frac{\omega(K)\log s}{n+1}.
\end{equation}
If $s\ge n+2$, one also has
\begin{equation}\label{pade:tangent-budget}
 \frac{\log s+\log n}{2}-2
 <\frac{T_d(K)}d+\frac{\omega(K)\log s}{n+1}.
\end{equation}
In the latter inequality, the constant $2$ can be replaced by
$\log2+1/2$.
\end{proposition}
\begin{proof}
Gauss's lemma gives $G\in\mathbb Z[X]$. Set
$r=s-1$, $P(Z)=H_{n,s}(-Z)$, and $Q(Z)=H_{r,n+1}(Z)$,
with the convention $H_{0,n+1}=1$.
Coefficient extraction from $(1-t)^{j-r-1}$ gives
\[
 \binom{n+r-j}{n}
 =\sum_{a=0}^n(-1)^a\binom{n+r-a}{n-a}\binom ja.
\]
Consequently, as formal power series,
\begin{equation}\label{pade:formal}
 e^Z\frac{P(Z)}{n!}
 =\sum_{j\ge0}\binom{n+r-j}{n}\frac{Z^j}{j!}
 =\frac{Q(Z)}{r!}+O(Z^{n+r+1}).
\end{equation}
The generalized binomial coefficients in the infinite sum are
integers. Those with $r+1\le j\le n+r$ vanish.
If $E=e^ZP/n!-Q/r!$, then
\[
 PE'-(P'+P)E=-\frac{PQ'-P'Q-PQ}{r!}=O(Z^{n+r}).
\]
The numerator is a polynomial of degree at most $n+r$ with leading
coefficient $(-1)^{n+1}$. Thus
\begin{equation}\label{pade:wronskian}
 PQ'-P'Q-PQ=(-1)^{n+1}Z^{n+r}.
\end{equation}
Since $P(0)\ne0$, $P$ and $Q$ have no common root.
Write $-M_1,\ldots,-M_d$ for the roots of $G$.

Fix a prime $p$ and a root with $e=v_p(M_i)>0$; the valuation $e$
need not be an integer.  Let $s_p(r)$ denote
the sum of the base-$p$ digits of $r$. We claim that
\begin{equation}\label{pade:local}
 v_pQ(M_i)\ge(n+s)e-(n+1)\min\{e,(p-1)^{-1}\}
                         -\frac{s_p(r)}{p-1}.
\end{equation}
If $e>1/(p-1)$, the series in \eqref{pade:formal} converge at $M_i$.
Because $P(M_i)=0$, $Q(M_i)/r!$ is the negative of the tail beginning
at $j=n+s$. Legendre's formula gives, for $j\ge1$,
\[
 je-v_p(j!)\ge j\left(e-\frac1{p-1}\right)+\frac1{p-1}.
\]
It follows that
\[
 v_pQ(M_i)\ge(n+s)e-\frac{n+s_p(r)}{p-1},
\]
which is stronger than \eqref{pade:local}.
If $0<e\le1/(p-1)$, use the finite polynomial identity
\[
 Q(M_i)=r!\sum_{j=0}^r\binom{n+r-j}{n}\frac{M_i^j}{j!}.
\]
For $0\le j\le r$,
$je-v_p(j!)\ge r(e-1/(p-1))$. Hence
$v_pQ(M_i)\ge re-s_p(r)/(p-1)$, which is exactly the required bound.
This argument includes $r=0$ and the convergence boundary.

We shall use the elementary digit estimate
\begin{equation}\label{pade:digit}
 p^{s_p(r)/(p-1)}\le r+1=s.
\end{equation}
Indeed, concavity of $\log(a+1)$ on $0\le a\le p-1$ gives
$p^{a/(p-1)}\le a+1$. If $a_i$ are the digits of $r$, then
$\prod_i(a_i+1)\le r+1$: the product counts the integers whose digits
are bounded by the corresponding $a_i$, and all of them lie in
$[0,r]$.

The resultant
\[
 \mathcal R=\Res\bigl((-1)^dG(-Z),Q(Z)\bigr)
             =\prod_{i=1}^dQ(M_i)
\]
is a nonzero integer. Both $M_i$ and $Q(M_i)$ are algebraic integers,
so their finite valuations are nonnegative. For a fixed $p$,
$\sum_i v_p(M_i)=v_p(K)$, and
\[
 \sum_{i:v_p(M_i)>0}\min\{v_p(M_i),(p-1)^{-1}\}
 \le\min\{v_p(K),d/(p-1)\}.
\]
For roots with zero valuation, use only $v_pQ(M_i)\ge0$.
Summing \eqref{pade:local} over the remaining roots, then over
$p\mid K$, and applying \eqref{pade:digit}, yields
\begin{equation}\label{pade:norm-lower}
 \log|\mathcal R|\ge(n+s)\log K-(n+1)T_d(K)
                              -d\,\omega(K)\log s.
\end{equation}
Primes not dividing $K$ make nonnegative contributions to the left
side, so omitting them preserves this inequality.

For $R=|M_i|>s$, the gamma integral gives
\begin{align*}
 |Q(M_i)|
 &\le\frac{R^r}{n!}\int_0^\infty
                     t^ne^{-t}(1+t/R)^r\,dt\\
 &\le\frac{R^{n+s}}{(R-r)^{n+1}}.
\end{align*}
The root gap \eqref{geo:eq-gap} gives $R-r>\rho_0(n+1)$. Therefore
\[
 \log|\mathcal R|
 <(n+s)\log K-d(n+1)\log\bigl(\rho_0(n+1)\bigr).
\]
This is strict also when $r=0$. Comparison with
\eqref{pade:norm-lower} proves \eqref{pade:budget}.

For the second inequality, the tangent line to $\log(1+x)$ at $v>0$
gives
\[
 \log(1+x)\le\phi(v)+\frac{x}{1+v},\qquad
 \phi(v)=\log(1+v)-\frac{v}{1+v}\le\frac{v^2}{2}
 \quad(x\ge0).
\]
The last bound follows by integrating
$\phi'(v)=v/(1+v)^2\le v$ from zero. Since $R>s>r$, the same gamma
integral now gives
\begin{equation}\label{pade:tangent-denominator}
 |Q(M_i)|<\frac{R^{n+s}}{L_v^{n+1}},\qquad
 L_v=e^{-r\phi(v)/(n+1)}\left(R-\frac r{1+v}\right).
\end{equation}
The inequality is strict because $r>0$ and the tangent-line inequality
is strict except at one integration point. Take
$v=\sqrt{(n+1)/r}\le1$, using $s\ge n+2$. Then
\[
 L_v>e^{-1/2}\left(1+\frac{rv}{1+v}\right)
 >\frac{\sqrt{r(n+1)}}{2\sqrt e}
 \ge\frac{\sqrt{sn}}{2\sqrt e}.
\]
The last step uses $(s-1)(n+1)\ge sn$. Multiplying
\eqref{pade:tangent-denominator} over the roots and comparing with
\eqref{pade:norm-lower} proves the assertion with constant
$\log2+1/2<2$.
\end{proof}

In the estimates below we use the weaker bound $\rho_0>1/4$
to keep the numerical constants simple.

\subsubsection{Prime estimates}

Write
\[
 \pi(x)=\#\{p\le x:p\text{ prime}\},\qquad
 \vartheta(x)=\sum_{p\le x}\log p,
 \qquad S(x)=\sum_{p\le x}\frac{\log p}{p-1}.
\]
We use the following explicit estimates:
\begin{align}
 |\vartheta(x)-x|&\le\frac{x}{100(\log x)^2}
 &&(x>7\,713\,133\,853),\label{pade:dusart-theta}\\
 \pi(x)&\le\frac{x}{\log x-1.1}
 &&(x>60\,184),\label{pade:dusart-pi}\\
 S(x)&<\log x&&(x\ge319).\label{pade:dusart-sum}
\end{align}
The first two estimates are \cite[Theorem~5.2, p.~4, and
Theorem~6.9, (6.6)]{Dusart2010}.
For the third, Rosser and Schoenfeld \cite[Theorem~6, (3.22)]{RS}
prove
\[
 \sum_{p\le x}\frac{\log p}{p}
 <\log x+E+\frac1{2\log x}\qquad(x\ge319),
\]
where \cite[(2.8)]{RS} defines
\[
 E=-\gamma-\sum_p\frac{\log p}{p(p-1)}.
\]
Adding the sum over $p\le x$ of $\log p/(p(p-1))$ gives
\[
 S(x)<\log x-\gamma+\frac1{2\log x}<\log x.
\]
The last inequality follows, for example, from
$\gamma>1-\log2>3/10$ and $\log x>2$.
Here $\gamma>1-\log2$ follows by writing $\gamma$ as the increasing
limit of $\sum_{k=1}^N(1/k-\log(1+1/k))$.

We shall repeatedly use the elementary comparisons
\begin{equation}\label{pade:comparisons}
 \frac{\log u}{u}<\frac1{1000},\qquad
 e^u>10^{30}u^5,\qquad e^{u/2}>10^{25}u^4
 \qquad(u\ge10000).
\end{equation}
At $u=10000$, the logarithmic comparisons follow from
$\log10000<10$ and $\log10<7/3$. The function $(\log u)/u$
decreases, while the logarithms of the last two ratios have derivatives
$1-5/u>0$ and $1/2-4/u>0$.
We also use $2/3<\log2<7/10$, $\log13<3$, and $\log52<4$.
These elementary bounds follow from the exponential series and
\[
 \log2=2\sum_{j\ge0}\frac{1}{(2j+1)3^{2j+1}}.
\]

The Newton-polygon arguments require lower bounds for the number
of primes in intervals proportional to $n$. The same estimates
also control the total number of primes up to $n$.

\begin{lemma}\label{pade:lem-prime-counts}
If $u=\log n\ge10000$, then
\begin{equation}\label{pade:prime-counts}
 \pi(n)<\frac{1003n}{1000u},\qquad
 \pi(n/\ell)-\pi(n/(4\ell))>
              \frac{749n}{1000\ell u}\quad(1\le\ell\le13).
\end{equation}
In particular, $\pi(n)<n/100$.
\end{lemma}
\begin{proof}
The estimate \eqref{pade:dusart-pi} gives directly
\[
 \pi(n)\le\frac{n}{u-11/10}<\frac{1003n}{1000u}
 \qquad(u\ge10000).
\]
For $t=n/\ell$, all arguments exceed the threshold in
\eqref{pade:dusart-theta}, by \eqref{pade:comparisons}, and
$\log(t/4)>u-4$. Hence
\[
 \vartheta(t)-\vartheta(t/4)
 \ge\frac{3t}{4}-\frac{5t}{400(u-4)^2}
 >\frac{749t}{1000}.
\]
Division by $\log t\le u$ gives the second estimate.
\end{proof}

The Pad\'e inequalities involve $\omega(K)$ and $T_d(K)$.
Both can be estimated from an upper bound for $\log K$, which
the root annulus supplies for a factor constant term.

\begin{lemma}\label{pade:lem-support}
Let $K,d$ be positive integers, and suppose $\log K<X$, where
$X\ge10000$. Then
\begin{equation}\label{pade:support}
 \omega(K)<\frac{2X}{\log X},\qquad
 \frac{T_d(K)}d<\log X+1.001.
\end{equation}
\end{lemma}
\begin{proof}
Put $w=\omega(K)$. Since the $i$th prime is at least $i+1$,
$K\ge(w+1)!$. If $w\ge1$, integration of $\log t$ gives
$\log K>w\log w-w$.
At $w=2X/\log X$, this last expression exceeds $X$: writing
$y=\log X\ge4$, the required comparison is
$y/2-\log y+\log2-1>0$, which holds at $y=4$ and increases thereafter.
The function $w\log w-w$ increases for $w>1$, proving the first bound;
the case $w=0$ is immediate.

For the second bound, use
$T_d(K)/d\le\sum_{p\mid K}\log p/(p-1)$ and split at $X$.
The smaller primes contribute at most $S(X)<\log X$ by
\eqref{pade:dusart-sum}. The remaining contribution is at most
$\log K/(X-1)<X/(X-1)<1.001$.
\end{proof}

When $s$ has polynomial size in $n$, the first Pad\'e inequality
already excludes small factors.

\begin{lemma}\label{pade:lem-small-degree-lower}
If $\log n\ge10000$, $1\le s\le n^3$, and $H_{n,s}$ has a proper
rational factor of degree $d$, then
\begin{equation}\label{pade:small-degree-lower}
 d>\frac n{100\log n}.
\end{equation}
\end{lemma}
\begin{proof}
Let $G$ be a monic factor of degree $d$, and set
$u=\log n$, $t=\log(s+n)$, $X=dt$, and $K=|G(0)|$.
The annulus \eqref{geo:eq-annulus} gives $\log K<X$, and
$u<t<3.01u$. If $d\le n/(100u)$, then
$10000\le u<X<n/32$. Lemma~\ref{pade:lem-support} implies
\[
 \frac{T_d(K)}d<u-5\log2+1.001<u-2.3.
\]
As $X/\log X$ increases in this range,
\[
 \frac{\omega(K)\log s}{n+1}
 <\frac{3u}{16(u-\log32)}<\frac15.
\]
The right side of \eqref{pade:budget} is less than $u-2.1$,
whereas its left side exceeds $u-\log4>u-2$, a contradiction.
\end{proof}

\subsubsection{The prime-square obstruction}

For primes $p$ with $\sqrt n<p\le n$, the absence of a multiple of $p^2$ in the
parameter interval forces a Newton-polygon side of slope $-1/p$.
Its denominator restricts the degrees of rational factors.

\begin{proposition}\label{pade:prop-square}
Suppose $H_{n,s}$ has a rational factor of degree $d$. If a prime $p$ satisfies $p^2>n$
and $p\mid\binom nd$, then some integer in $[s,s+n-1]$ is divisible
by $p^2$. Moreover, if $n\ge5$ and $1\le d\le n/2$, then
\begin{equation}\label{pade:square-count}
 \pi(n)-\pi(n-d)
 \le\left\lfloor\frac{s+n-1}{(n-d+1)^2}\right\rfloor
      -\left\lceil\frac{s}{n^2}\right\rceil+1.
\end{equation}
\end{proposition}
\begin{proof}
Suppose that the interval contains no multiple of $p^2$.
The asserted divisibility of $\binom nd$ requires $p\le n$.
Write $n=ap+b$ and $s-1=cp+t$, with $0\le b,t<p$ and $a<p$.
For $j=\ell p+v$, $0\le v<p$, the coefficient of $X^j$ has valuation
\begin{equation}\label{pade:coefficient-valuation}
 a-\ell+\left\lfloor\frac{t+b-v}{p}\right\rfloor
                   +\mathbf1_{v>b}.
\end{equation}
Indeed, the rising factorial $(s)_{n-j}$ counts multiples of $p$,
each of valuation one, and, since $n<p^2$,
$v_p\binom nj=\mathbf1_{v>b}$.

If $t+b<p$, the lower Newton polygon has vertices
\[
 (0,a),\qquad(ap,0),\qquad(n,0).
\]
If $t+b\ge p$, put $v_0=t+b-p<b$. Its vertices are
\[
 (0,a+1),\qquad(v_0+1,a),\qquad(ap+v_0+1,0),\qquad(n,0).
\]
These assertions follow directly from \eqref{pade:coefficient-valuation};
coincident vertices are omitted. In the second case the initial slope
is $-1/(v_0+1)$, where $1\le v_0+1\le b<p$, so it is distinct
from $-1/p$. The Newton polygon theorem therefore shows that exactly
$ap$ roots have valuation $1/p$, with only $b$ other roots.

The roots selected by the rational factor form a set invariant under
the local Galois group. Intersecting it with the set of roots of
valuation $1/p$ again gives an invariant set. Its product belongs to
$\mathbb Q_p^\times$, and hence has integer valuation. The number of
selected roots of valuation $1/p$ must consequently be a multiple
of $p$. Thus $d=kp+e$ with $0\le k\le a$ and $0\le e\le b$.
The same base-$p$ factorial valuation formula now gives
$v_p\binom nd=0$, a contradiction.

For the second assertion, every prime $p\in(n-d,n]$ satisfies
$p>n/2>\sqrt n$ and divides $\binom nd$. Choose an integer $k_p\ge1$
with $s\le k_pp^2\le s+n-1$. Distinct primes have distinct
multipliers: if $p<q$ had the same multiplier $k$, then
$k(q^2-p^2)\ge2p+1>n-1$, impossible in an interval of length $n-1$.
Every multiplier lies between
$\lceil s/n^2\rceil$ and
$\lfloor(s+n-1)/(n-d+1)^2\rfloor$, proving \eqref{pade:square-count}.
\end{proof}

The degree bound and the prime estimates ensure that $(n-d,n]$
contains many primes. The square-divisibility obstruction then
requires more distinct multipliers than a small value of $s$ permits.

\begin{proposition}\label{pade:prop-cubic-region}
If $\log n\ge10000$ and
\begin{equation}\label{pade:cubic-region}
 1\le s\le\frac{n^3}{10\log n},
\end{equation}
then $H_{n,s}$ is irreducible over $\mathbb Q$.
\end{proposition}
\begin{proof}
If $H_{n,s}$ is reducible, choose a factor of degree
$1\le d\le n/2$ and put $u=\log n$.
Lemma~\ref{pade:lem-small-degree-lower} gives $d>n/(100u)$.
Both $n$ and $n-d$ exceed the threshold of \eqref{pade:dusart-theta},
and $\log(n-d)\ge u-\log2$. Therefore
\[
 \vartheta(n)-\vartheta(n-d)
 >d-\frac{3n}{100u^2}>\frac{4d}{5},
 \qquad \pi(n)-\pi(n-d)>\frac{4d}{5u}.
\]
On the other hand, the right side of \eqref{pade:square-count} is less
than
\[
 \frac{6sd}{n^3}+1+\frac4n
 \le\frac{3d}{5u}+1+\frac4n.
\]
Here we used $n-d\ge n/2$ and
\[
 \frac1{(n-d)^2}-\frac1{n^2}
 =\frac{d(2n-d)}{n^2(n-d)^2}\le\frac{6d}{n^3}.
\]
But $d/(5u)>n/(500u^2)>1+4/n$ by
\eqref{pade:comparisons}. This is a contradiction.
\end{proof}

For larger parameters, the tangent estimate in
Proposition~\ref{pade:prop-budget} gives a stronger restriction.
The resulting degree bound depends on $s$ only through $\log(s+n)$.

\begin{lemma}\label{pade:lem-tangent-degree}
Suppose $\log n\ge10000$, $s\ge n^{5/2}$, and $H_{n,s}$ has a
rational factor of degree $d\ge1$. Put $A=\log(s+n)/\log n$. Then
\begin{equation}\label{pade:tangent-degree}
 d>\frac n{16A}.
\end{equation}
\end{lemma}
\begin{proof}
Write $u=\log n$, suppose $d\le n/(16A)$, and put
$X=dAu$ and $K=|G(0)|$ for a monic factor $G$ of degree $d$.
The root annulus gives $\log K<X$, while
$10000\le Au\le X\le nu/16$.
Lemma~\ref{pade:lem-support} yields
\[
 \frac{T_d(K)}d<u+\log u,\qquad \omega(K)<\frac n4.
\]
The estimate for $T_d(K)$ uses $\log16>1.001$. To bound $\omega(K)$,
apply monotonicity of $X/\log X$ and the inequality $\log(nu/16)>u/2$.
Moreover $Au-1<\log s<Au$. Thus \eqref{pade:tangent-budget}
would imply
\[
 \frac{Au+u}{2}-3<u+\log u+\frac{Au}{4}.
\]
This is impossible: $A\ge5/2$, and
\[
 u\left(\frac A4-\frac12\right)-\log u-3
 \ge\frac u8-\log u-3>0
\]
by \eqref{pade:comparisons}.
\end{proof}


\subsection{Root energy and uniform eventual irreducibility}
\label{ene:section}

We apply the matrix-energy inequality to a rational factor.
Its proportion of roots in a local cluster need not equal its
proportion in the full root set; the square of this discrepancy
contributes to the energy. The resulting degree restrictions will
combine with the trace criterion to cover all parameters.
Throughout this section, $n,s$ are positive integers, $u=\log n$,
and all factors are defined over $\mathbb Q$.

\subsubsection{The energy of a rational factor}

Suppose $H_{n,s}=GJ$, where $G,J$ are monic rational factors and
$1\le d=\deg G\le n/2$. They are integral by Gauss's lemma.
Put $x=d/n$, and use the rank-one matrix $Q=cc^{\mathsf T}$ with
$c_i=1-x$ on the roots of $G$ and $c_i=-x$ on the roots of $J$.
Thus
\begin{equation}\label{ene:charges}
 \sum_i c_i=0,\qquad \sum_i c_i^2=nx(1-x).
\end{equation}
At a prime $p$, define stopping heights and local energy by
\begin{equation}\label{ene:local-energy}
 h_{p,i}=v_p(\alpha_i)+\epsilon_p,\qquad
 E_p=\sum_{i\ne j}c_ic_jv_p(\alpha_i-\alpha_j)
                         +\sum_i c_i^2h_{p,i}.
\end{equation}
There are only finitely many nonzero contributions: outside the primes
at most $n$ and those dividing the constant term or discriminant,
all the valuations occurring here vanish.
\begin{lemma}\label{ene:energy}
Every $E_p$ is nonnegative. If $N_p(t)$ counts all roots of valuation
at least $t$, and $K_p(t)$ counts the selected roots of valuation at
least $t$, then
\begin{equation}\label{ene:origin-energy}
 E_p\ge\int_0^\infty(K_p(t)-xN_p(t))^2\,dt.
\end{equation}
Write
\[
 \begin{gathered}
 K=|G(0)|,\qquad P=(s)_n,\qquad
 S(n)=\sum_{p\le n}\frac{\log p}{p-1},\\
 C_\infty=\sum_{i\ne j}c_ic_j\log|\alpha_i-\alpha_j|.
 \end{gathered}
\]
Then the global energy $\mathcal E=\sum_pE_p\log p$ satisfies
\begin{equation}\label{ene:global-energy}
 \mathcal E=C_\infty+(1-x)^2\log K+x^2\log(P/K)
                                      +nx(1-x)S(n).
\end{equation}
If the closed complex disks of radius $\rho$ around the roots are
pairwise disjoint, then
\begin{equation}\label{ene:arch-energy}
 C_\infty\le-nx(1-x)\log\rho.
\end{equation}
\end{lemma}


\begin{proof}
The root sets of the rational factors are Galois invariant, so
$Q=cc^{\mathsf T}$ is invariant and has trace $nx(1-x)$.
Lemma~\ref{lem:matrix-energy} therefore applies. On the roots
of valuation at least $t$, the sum of the charges is
$K_p(t)-xN_p(t)$, giving \eqref{ene:origin-energy}. The diagonal
contribution to the product formula is
$(1-x)^2\log K+x^2\log(P/K)$, which gives
\eqref{ene:global-energy}. The complex bound follows from the same
lemma.
\end{proof}
The annulus and separation estimates now give an upper bound.

\begin{lemma}\label{ene:upper-bounds}
Suppose $u=\log n\ge10000$ and a factor of degree
$1\le d\le n/2$ exists. If $s\ge n^{5/2}$, then
\begin{equation}\label{ene:rough-energy}
 \mathcal E<3nx(1-x)\log s.
\end{equation}
\end{lemma}
\begin{proof}
For $s\ge n$, apply Lemma~\ref{lem:matrix-energy} to
$Q=cc^{\mathsf T}$, whose trace is $nx(1-x)$.  It gives
\begin{equation}\label{ene:energy-common-bound}
 \mathcal E<nx(1-x)
       \left(\frac12\log s+\frac32u+\log48\right).
\end{equation}
For $s\ge n^{5/2}$ this is less than \eqref{ene:rough-energy}.
\end{proof}

\subsubsection{Packets and deletion of small-prime contributions}

For $p>n$, the local calculation in
Proposition~\ref{prop:exact-inertia} concerns a single node $s+j$.
The following specialization of the packet-energy formula records
both its energy and the restriction on the number of roots selected
by a factor.

\begin{lemma}\label{ene:packet}
Suppose $p>n$ and $p^e\Vert s+j$, where $0\le j<n$, and put $N=n-j$.
Exactly $N$ roots have valuation $e/N$; the other $j$ roots are units.
If a rational factor selects $k$ of the former roots, then
\begin{equation}\label{ene:packet-occupation}
 N\mid ke,\qquad
 E_p\log p= N\left(\frac{k}{N}-x\right)^2\log(p^e).
\end{equation}
\end{lemma}
\begin{proof}
The tame-packet construction in Proposition~\ref{prop:exact-inertia}
and Lemma~\ref{lem:general-tame-packet} gives the stated valuations
and inertia orbit length $q=N/\gcd(N,e)$.  A rational factor selects
a union of these orbits, so $q\mid k$, equivalently $N\mid ke$.
Formula~\eqref{eq:matrix-packet}, applied to $Q=cc^{\mathsf T}$,
gives
\[
 E_p=\frac eN(k-xN)^2,
\]
which proves the energy identity, including when $N=1$.
\end{proof}

Before summing over large primes, we bound the parts of the nodes
supported on small primes. For each small prime, remove its full
contribution from one node of maximal valuation. The remaining
product divides a factorial. Deleting those nodes altogether gives
the form used below. This is the allocation argument of Erd\H{o}s
and Selfridge \cite[Lemma~2]{ES}.

\begin{lemma}\label{ene:deletion}
Let $\mathcal I\subseteq\{0,\ldots,n-1\}$ and $Y\ge2$.
Let $\mathcal P$ be the set of primes $p\le Y$ dividing at least
one node $s+j$ with $j\in\mathcal I$.
For each $p\in\mathcal P$, choose an index $j_p\in\mathcal I$
of maximal $p$-valuation.
Remove the $p$-part only from the node assigned to $p$, and put
\[
 R_j=\prod_{\substack{p\in\mathcal P\\j\ne j_p}}p^{v_p(s+j)}.
\]
Then $\prod_{j\in\mathcal I}R_j\mid(n-1)!$.
In particular, deleting the union of the chosen indices removes at
most $\pi(Y)$ nodes. If $\mathcal J$ is the remaining set and
$S_Y(m)$ denotes the part of $m$ supported on primes at most $Y$,
then
\begin{equation}\label{ene:smooth-product}
 \prod_{j\in\mathcal J}S_Y(s+j)\mid(n-1)!.
\end{equation}
\end{lemma}
\begin{proof}
For $j\ne j_p$, maximality gives
$v_p(s+j)\le v_p(j-j_p)$.
The product of the distances on either side of $j_p$ divides
$j_p!(n-1-j_p)!$, which divides $(n-1)!$.
This proves the first assertion prime by prime. Every undeleted
node retains its entire $Y$-smooth part, giving the second.
\end{proof}

From now on take $Y=n$, write $S_j=S_n(s+j)$, and put
$B_j=(s+j)/S_j$. Thus all primes dividing $B_j$ are greater than $n$.
A prime greater than $n$ occurs at only one parameter node.
Moreover, for $p^e\Vert B_j$,
\begin{equation}\label{ene:exponent-bound}
 e< A:=\frac{\log(s+n)}{\log n}.
\end{equation}
The reduced denominator of a selected fraction $q=k/(n-j)$ divides
both $n-j$ and $e$, by \eqref{ene:packet-occupation}.

\subsubsection{A restriction on the factor degree}

Since $e<A$, every selected fraction has reduced denominator less
than $A$. Distinct such fractions are separated by at least
$A^{-2}$. We choose packet widths that exclude the one possible
fraction close to $x=d/n$ and then sum the resulting energy bounds.

\begin{lemma}\label{ene:farey}
Suppose $u=\log n\ge10000$, $s\ge n^{5/2}$, and a rational factor
of degree $1\le d\le n/2$ exists. Then
\begin{equation}\label{ene:farey-dichotomy}
 d<4800\qquad\hbox{or}\qquad A>(n/4800)^{1/4}.
\end{equation}
\end{lemma}
\begin{proof}
Put $x=d/n$ and $\eta=\min\{x/2,1/(4A^2)\}$.
Distinct reduced fractions with denominators at most $A$ differ by at
least $A^{-2}$. Hence at most one lies in $(x-\eta,x+\eta)$.
If present, its denominator $b$ is at least two, since
$0<x-\eta<x+\eta<1$. Choose the widths
$\lceil n/8\rceil\le N\le n$, omitting the multiples of $b$ if
that close fraction exists. There are at least $7n/16-2$ choices.
Every reduced fraction whose denominator is at most $A$ and divides
one of the chosen widths $N$ is therefore at distance at least
$\eta$ from $x$.

Apply Lemma~\ref{ene:deletion} to the corresponding indices $j=n-N$.
The bound $\pi(n)<n/100$ from \eqref{pade:prime-counts} leaves more
than $0.42n$ indices. Consequently
\[
 \sum_{j\in\mathcal J}\log B_j
 >0.42n\log s-\log((n-1)!)
 >\frac{n}{50}\log s,
\]
since $\log s\ge(5/2)u$ and $\log((n-1)!)<nu$.
For every prime power in these $B_j$, \eqref{ene:exponent-bound}
and the preceding denominator exclusion give
$|k/N-x|\ge\eta$. Since $N\ge n/8$, summing the packet energies
without repetition of any large prime yields
\[
 \mathcal E>\frac{n^2\eta^2}{400}\log s.
\]
Comparison with \eqref{ene:rough-energy} gives
$n\eta^2<1200x(1-x)$.
If $\eta=x/2$, then $d<4800$.
Otherwise $n/(16A^4)<1200x(1-x)\le300$, giving the other alternative.
\end{proof}

\subsubsection{Uniform eventual irreducibility}

For a reducible member, the four-fiber bound makes $\log s$ smaller
than every fixed positive power of $n$ once $n$ is sufficiently large.
Within that range, the Pad\'e lower bound for the factor degree
contradicts the energy dichotomy. Together these estimates give uniform eventual irreducibility.

\begin{proposition}\label{prop:qualitative-uniform-irreducibility}
There is an effective absolute integer $N_0$ such that $H_{n,s}$ is
irreducible for every $n\ge N_0$ and every positive integer $s$.
\end{proposition}
\begin{proof}
Suppose a factor of degree $1\le d\le n/2$ exists.  Corollary~
\ref{new:rational-large-parameter}, with $B=0$ and $\varepsilon=1/8$,
gives $\log s<n^{1/8}$ for all sufficiently large $n$.
For sufficiently large $n$, one has
$n^{5/2}\le n^3/(10\log n)$, so
Proposition~\ref{pade:prop-cubic-region} excludes $s\le n^{5/2}$.
In the remaining range put $u=\log n$ and $A=\log(s+n)/u$.
For all sufficiently large $n$,
\[
 A<\frac{2n^{1/8}}u<(n/4800)^{1/4},\qquad
 d>\frac n{16A}>4800,
\]
where the middle degree bound is Lemma~\ref{pade:lem-tangent-degree}.
These inequalities contradict Lemma~\ref{ene:farey}.  Every threshold
used here is effective, so their maximum gives the asserted $N_0$.
\end{proof}

\section{Galois groups}\label{sec:galois}
\label{sec:valuations-groups}\label{sec:partitions-galois}

Assuming irreducibility, we prove that the Galois group contains
$A_n$ once $n$ exceeds an effective absolute constant. A local
criterion first excludes the primitive groups that omit $A_n$.
For an imprimitive group, the energy estimate restricts its
invariant partitions to two blocks. Integral traces handle a
real quadratic block field; Laurent interpolation handles an
imaginary quadratic block field. The estimates are uniform in $s$.

\subsection{Prime valuations in short intervals}\label{sec:nearby-integers}

We begin with an arithmetic criterion for alternating containment.
A short interval cannot contain many integers whose large-prime
valuations are all divisible by exponents comparable to its length.
Newton polygons turn this statement into a criterion for alternating
containment. We apply it first to Laguerre polynomials; the
trimmed-exponential application is given in
Section~\ref{sec:trimmed-exponentials}.
Sections~\ref{sec:groups}--\ref{sec:final-galois}
will place every remaining Laguerre obstruction in its parameter range.

\subsubsection{Independent nodes and simultaneous logarithms}
\label{sec:simultaneous-logarithms}

For a rational number $a/b$ in lowest terms, put
$H(a/b)=\max\{|a|,|b|\}$ and $h(a/b)=\log H(a/b)$.
We use Loxton's estimate for simultaneous linear forms
\cite[Theorem~4]{Loxton1986}, in the form stated by
Stewart \cite[Lemma~1]{Stewart2008}.
Let $r\ge2$, $1\le m\le r$, and let
$\beta_1,\ldots,\beta_r$ be multiplicatively independent positive
rational numbers. Suppose the rational matrix $(b_{ij})$ has rank $m$,
and put
\[
 \Lambda_i=\sum_{j=1}^r b_{ij}\log\beta_j\qquad(1\le i\le m).
\]
For $A_j,B\ge4$ with $H(\beta_j)\le A_j$ and $H(b_{ij})\le B$,
write $\Omega=\prod_{j=1}^r\log A_j$. Then
\begin{equation}\label{fel:loxton}
 \max_i|\Lambda_i|>
 \exp\!\left(-(16r)^{200r}(\Omega\log\Omega)^{1/m}
                                      \log(B\Omega)\right).
\end{equation}

The forms must retain full rank after dependent bases are removed.
We ensure this by first choosing multiplicatively independent
integers in the parameter interval. The following elementary
version of Stewart's argument is convenient
\cite[Theorem~1 and Corollary~3]{Stewart2008}.

\begin{lemma}\label{fea:independent-nearby}
Fix an integer $t\ge2$, and put
\[
 K_t=t\bigl((4(t-1))^{t-1}+1\bigr).
\]
Let $M,s$ be positive integers with $s>2M^2$.
Any set of at least $K_t$ distinct integers in $[s,s+M]$
contains $t$ multiplicatively independent integers.
\end{lemma}
\begin{proof}
Suppose every $t$-element subset is dependent. Choose an integral
multiplicative relation for each such subset and omit an index
whose coefficient has largest absolute value. If there are $k$
numbers, some fixed $(t-1)$-element subset is associated with at least
\[
 \left\lceil\frac{\binom kt}{\binom k{t-1}}\right\rceil
 =\left\lceil\frac{k-t+1}{t}\right\rceil
\]
subsets. Write its elements as $a_1,\ldots,a_{t-1}$. For every
associated extra element $z$, the chosen relation has the form
\[
 z^\ell=\prod_{i=1}^{t-1}a_i^{c_i},
 \qquad \ell>0,\qquad |c_i|\le\ell.
\]
Divide each coordinate interval of $[-1,1]^{t-1}$ into
$4(t-1)$ intervals. Since $k\ge K_t$, two distinct extra elements
$z,z'$ have associated vectors satisfying
\[
 \left|\frac{c_i}{\ell}-\frac{c'_i}{\ell'}\right|
 \le\frac1{2(t-1)}\qquad(1\le i<t).
\]
Raise the quotient of the two relations to clear denominators.
The identities $h(x^j)=|j|h(x)$ and
$h(xy)\le h(x)+h(y)$ then give
\[
 h(z/z')\le
 \sum_{i=1}^{t-1}
 \left|\frac{c_i}{\ell}-\frac{c'_i}{\ell'}\right|h(a_i)
 \le\frac12\log(2s).
\]
Here $s+M<2s$. But $\gcd(z,z')\mid z-z'$ and
$0<|z-z'|\le M$, so
\[
 h(z/z')\ge\log(s/M)>\tfrac12\log(2s),
\]
a contradiction.
\end{proof}

The next bound allows a fixed number of small primes to have large
exponents. It will be used after assigning each small prime to one
node where its valuation is maximal.

\begin{lemma}\label{fel:weighted-powers}
Fix $C\ge1$ and integers $m\ge2$, $R\ge0$.
There is an effectively computable constant $C_{m,R,C}>0$
with the following property. Let $n\ge3$ and $s$ be positive
integers, put $L=\log s$, and suppose
\[
 L\ge2\log(Cn),\qquad 2\le H\le L.
\]
Let $z_0,\ldots,z_m$ be multiplicatively independent integers in
$[s,s+Cn]$. Suppose that, for $1\le i\le m$,
\[
 z_i=d_i x_i^{k_i}
          \prod_{\ell=1}^{R_i}p_{i\ell}^{e_{i\ell}},
\]
where $d_i,x_i,k_i$ are positive integers, the $p_{i\ell}$ are
primes, and
\[
 \log d_i\le H,\quad n/2\le k_i\le n,\quad
 R_i\le R,\quad p_{i\ell}\le Cn,\quad 0\le e_{i\ell}\le n.
\]
Then
\begin{equation}\label{fel:height-inequality}
 L^{1-1/m}\le C_{m,R,C}\,
 H(\log n)^R\max\{1,L/n\}
                     \bigl(\log(nL)\bigr)^{1+1/m}.
\end{equation}
\end{lemma}
\begin{proof}
Use the $m$ real forms
\[
 \Lambda_i=k_i\log x_i+\log d_i+
       \sum_{\ell=1}^{R_i}e_{i\ell}\log p_{i\ell}-\log z_0
       =\log(z_i/z_0).
\]
Their values are linearly independent over $\mathbb Q$, since a
relation among them would give a multiplicative relation among
$z_0,\ldots,z_m$. Moreover,
\begin{equation}\label{fel:small-forms}
 0<\max_i|\Lambda_i|\le Cn/s.
\end{equation}

List the positive integers occurring as bases, omitting $1$ and
repetitions, and choose a maximal independent subset
$\beta_1,\ldots,\beta_r$. There are at most
$r_0=m(R+2)+1$ original bases, and $m\le r\le r_0$.
Every original logarithm is a rational linear combination of this
basis. All original bases have logarithmic height at most $2L$:
indeed $s\ge(Cn)^2$, and
\[
 \log x_i\le\frac{\log(s+Cn)}{k_i}\le\frac{4L}{n}.
\]
Choose $r$ primes for which the valuation matrix of the
$\beta_j$ is nonsingular. Its entries, and the entries of the
valuation vector of each original base, are at most $4L$.
Cramer's rule gives a common denominator $\det V$ and numerators
of absolute value at most
\[
 D_0=r!(4L)^r.
\]
These are relations at every prime: the selected valuation vectors
form a basis of the span of all original valuation vectors, and
restriction to the chosen $r$ primes is injective on that span.

After substitution, the coefficient matrix of the $\Lambda_i$
still has rank $m$, by the independence of their values. Each
entry has rational height at most
$B_0=(R+3)nD_0$, since a row has at most $R+3$ original terms
and their integer coefficients have absolute value at most $n$.
Thus
\begin{equation}\label{fel:coefficient-bound}
 \log\max\{4,B_0\}\ll_{m,R}\log(nL).
\end{equation}

Put $U=\max\{1,L/n\}$. We may use logarithmic height bounds
$H$, $4U$, $2\log(Cn)$, and $2L$ for the bases
$d_i$, $x_i$, $p_{i\ell}$, and $z_0$, respectively.
Every bound exceeds $\log4$ and hence also $1$.
Because the independent basis is a subset of the original list,
\begin{equation}\label{fel:omega-bound}
 \Omega\le2L\left(4HU(2\log(Cn))^R\right)^m,
 \qquad \log\Omega\ll_{m,R,C}\log(nL).
\end{equation}
Repeated bases can be assigned any one of their corresponding
bounds; removing repetitions or dependent bases decreases the
upper bound for the product.
Applying \eqref{fel:loxton} and using
\eqref{fel:coefficient-bound}--\eqref{fel:omega-bound}, we obtain
\[
 -\log\max_i|\Lambda_i|
 \ll_{m,R,C}L^{1/m}HU(\log n)^R
                         \bigl(\log(nL)\bigr)^{1+1/m}.
\]
By \eqref{fel:small-forms}, the left side is at least
$L-\log(Cn)\ge L/2$. This proves
\eqref{fel:height-inequality}.
\end{proof}

\subsubsection{Large-prime valuations at nearby integers}
\label{sec:prime-width-allocation}

The exponent in the next theorem may vary from one integer to
another and need not be prime. This flexibility is what permits
one to use different Newton-polygon widths simultaneously.

\begin{theorem}
\label{thm:dense-nearby-nodes}
Fix real numbers $C\ge1$, $c>0$, $D>0$, and
\[
 \kappa>\lfloor C/c\rfloor+3.
\]
There is an effectively computable integer
$N=N(C,c,D,\kappa)$ such that the following holds.
Let $n\ge N$ and $s\in\mathbb Z_{>0}$ satisfy
\[
 (\log n)^\kappa\le\log s\le n^D.
\]
If
\[
 \mathcal Z\subseteq[s,s+\lfloor Cn\rfloor]\cap\mathbb Z,
 \qquad |\mathcal Z|\ge c n/\log n,
\]
then, for every assignment of integers $k_z$ with
$n/2\le k_z\le n$, there are $z\in\mathcal Z$ and a prime $p>Cn$
for which
\[
 k_z\nmid v_p(z).
\]
\end{theorem}
\begin{proof}
Put $M=\lfloor Cn\rfloor$, $L=\log s$, and suppose instead that
$k_z\mid v_p(z)$ for all $z\in\mathcal Z$ and all primes $p>Cn$.
For every prime $p\le M$ dividing a node, assign $p$ to one node
$z_p$ where $v_p(z_p)$ is maximal. Reduce every valuation at a
node $z$ modulo $k_z$. Put the remainders at its assigned primes
into separate factors and all other remainders into $d_z$. This gives
\[
 z=d_zx_z^{k_z}\prod_{p\text{ assigned to }z}p^{r_{z,p}},
 \qquad 0\le r_{z,p}<k_z.
\]
There are no remainders at primes exceeding $M$, since an integer
prime exceeds $M$ exactly when it exceeds $Cn$.

The unassigned parts satisfy
\begin{equation}\label{eq:nearby-factorial-allocation}
 \prod_{z\in\mathcal Z}d_z\mid M!.
\end{equation}
To check this at a prime $p\le M$, consider any level $h\ge1$.
At most $\lfloor M/p^h\rfloor+1$ integers in an interval of
length $M$ are divisible by $p^h$. If there is one, the chosen
maximal node $z_p$ is among them. The total valuation of the
unassigned parts is therefore at most
$\sum_{h\ge1}\lfloor M/p^h\rfloor=v_p(M!)$.
Reducing valuations modulo the individual $k_z$ can only decrease
this total, proving \eqref{eq:nearby-factorial-allocation}.

Set
\[
 R=\lfloor C/c\rfloor,\qquad
 \gamma=c-\frac{C}{R+1}>0,\qquad
 A=\max\left\{2,\frac{4(C+1)}{\gamma}\right\},\qquad
 H=A(\log n)^2.
\]
For all effectively sufficiently large $n$, the prime estimate
\eqref{pade:dusart-pi} and the elementary factorial bound give
\[
 \pi(M)\le\left(C+\frac{\gamma(R+1)}4\right)
                    \frac n{\log n},\qquad
 \sum_z\log d_z\le\log(M!)\le(C+1)n\log n.
\]
The number of nodes with more than $R$ assigned primes is at most
\[
 \frac{\pi(M)}{R+1}
 \le\left(c-\frac{3\gamma}4\right)\frac n{\log n}.
\]
At most $\gamma n/(4\log n)$ nodes have $\log d_z>H$.
Consequently at least $\gamma n/(2\log n)$ nodes remain, each
with at most $R$ separate prime factors and $\log d_z\le H$.

Choose an integer
\[
 m>\max\left\{2,D,\frac{\kappa+1}{\kappa-R-3}\right\}.
\]
Then $D/m<1$ and
\[
 \kappa(1-1/m)>R+3+1/m.
\]
For sufficiently large $n$, there are at least $K_{m+1}$ remaining
nodes, $s>2M^2$, $L\ge2\log(Cn)$, and $2\le H\le L$.
Lemma~\ref{fea:independent-nearby} gives independent nodes
$z_0,\ldots,z_m$. Retain the displayed factorization for
$z_1,\ldots,z_m$ and use $z_0$ as a single base.
Lemma~\ref{fel:weighted-powers} yields
\[
 L^{1-1/m}\ll_{C,c,D,\kappa}
 (\log n)^{R+3+1/m}\max\{1,L/n\},
\]
where we used $L\le n^D$ to bound $\log(nL)$.
If $L\ge n$, this implies
\[
 n\ll_{C,c,D,\kappa}
 n^{D/m}(\log n)^{R+3+1/m},
\]
which is impossible for large $n$ because $D/m<1$.
If $L<n$, it instead implies
\[
 (\log n)^{\kappa(1-1/m)}
 \ll_{C,c,D,\kappa}(\log n)^{R+3+1/m},
\]
again a contradiction by the choice of $m$.
All constants depend only on the displayed fixed parameters.
The prime estimates and logarithmic-form bound are effective,
so the required threshold is effective as well.
\end{proof}

For every fixed $0<\varepsilon\le D$, the same conclusion holds
in the range $n^\varepsilon\le\log s\le n^D$, with a threshold
depending on $C,c,\varepsilon,D$. Indeed, choose
$\kappa=\lfloor C/c\rfloor+4$ and use
$(\log n)^\kappa\le n^\varepsilon$ for sufficiently large $n$.

\subsubsection{A Newton-polygon criterion}
\label{sec:newton-node-criterion}

A single edge whose slope has a sufficiently large prime denominator
forces alternating containment in a transitive group
\cite[Theorems~2.1 and~2.2]{Hajir2005}. The preceding theorem
supplies such a denominator from a collection of nearby integers.

\begin{theorem}
\label{thm:newton-node-criterion}
Fix $C\ge1$, $c>0$, $D>0$, and
$\kappa>\lfloor C/c\rfloor+3$.
For all effectively sufficiently large $n$, the following holds.
Let $s\in\mathbb Z_{>0}$ satisfy
$(\log n)^\kappa\le\log s\le n^D$, and let $F\in\mathbb Z[X]$
be separable of degree $n$, with $F(0)\ne0$.
Suppose there are a set of primes
\[
 \mathcal P\subseteq(n/2,n-3],\qquad
 |\mathcal P|\ge c n/\log n,
\]
and distinct integers $j_P\in\{0,\ldots,\lfloor Cn\rfloor\}$
with the following property: whenever $p>Cn$ is prime and
$e=v_p(s+j_P)>0$ is not divisible by $P$, the $p$-adic Newton
polygon of $F$ has an edge of slope $-e/P$.
For each pair $(p,P)$, this edge condition may instead hold for
$F(X+a_{p,P})$ with some integer $a_{p,P}$.

Then $\operatorname{Gal}(F/\mathbb Q)$ contains a $P$-cycle for
some $P\in\mathcal P$ and preserves no nontrivial partition into
equal blocks. If $F$ is irreducible over $\mathbb Q$, then
\[
 A_n\le\operatorname{Gal}(F/\mathbb Q).
\]
The threshold depends only on $C,c,D,\kappa$.
\end{theorem}
\begin{proof}
Apply Theorem~\ref{thm:dense-nearby-nodes} to the distinct integers
$z_P=s+j_P$, with exponent $k_{z_P}=P$.
It gives a prime $p>Cn$ and a width $P\in\mathcal P$ for which
$P\nmid e=v_p(z_P)$. In particular $e>0$.
The corresponding Newton-polygon edge supplies a root of the
relevant translate with $p$-adic valuation $e/P$. The translate
has the same splitting field as $F$. Since this fraction has denominator $P$,
the ramification index of the local splitting field, and hence the
order of the global Galois group, is divisible by $P$.
Cauchy's theorem gives an element of order $P$. It is a single
$P$-cycle because $2P>n$.

A group preserving $a$ blocks of size $b$, where $a,b\ge2$ and
$ab=n$, embeds in $S_b\wr S_a$. Its order divides $(b!)^a a!$,
which is not divisible by $P>n/2\ge\max\{a,b\}$.
Thus no such partition exists. If $F$ is irreducible, its root
action is transitive and therefore primitive. Jordan's theorem,
applied to the $P$-cycle with $P\le n-3$, gives $A_n$.
\end{proof}

\begin{corollary}\label{fea:polynomial-height-exclusion}
Fix $D>0$ and $\kappa>5$. For all effectively sufficiently large
$n$, the Galois group of $H_{n,s}$ preserves no nontrivial partition
into equal blocks whenever
\[
 (\log n)^\kappa\le\log s\le n^D.
\]
If $H_{n,s}$ is irreducible in this range, then $A_n\le G_{n,s}$.
These conclusions do not use the classification of finite simple groups.
In particular, they hold in every range
$n^\varepsilon\le\log s\le n^D$ with fixed $0<\varepsilon\le D$,
after increasing the effective threshold in terms of $\varepsilon,D$.
\end{corollary}
\begin{proof}
Use all primes $n/2<P\le n-3$ and put $j_P=n-P$.
Their number is at least $2n/(5\log n)$ for all effectively
sufficiently large $n$. For $p>n$ and $e=v_p(s+j_P)>0$,
write $H_{n,s}(X)=\sum_{k=0}^n A_kX^k$. Then
\[
 A_k=\binom nk(s)_{n-k},\qquad
 v_p(A_k)=
 \begin{cases}
 e,&0\le k<P,\\
 0,&P\le k\le n.
 \end{cases}
\]
Indeed $s+j_P$ is the only multiple of $p$ among
$s,\ldots,s+n-1$, and $p\nmid\binom nk$.
The Newton polygon contains the edge from $(0,e)$ to $(P,0)$.
Theorem~\ref{thm:newton-node-criterion} applies with $C=1$ and
$c=2/5$, since $\lfloor C/c\rfloor+3=5$;
separability follows from \eqref{eq:intro-discriminant}.
The final assertion follows by taking $\kappa=6$ and then
$(\log n)^6\le n^\varepsilon$.
\end{proof}


\subsubsection{Powers in dense subsets of a progression}\label{sec:dense}

The preceding criterion requires an upper bound on $\log s$.
We also need an estimate with no upper bound on the parameter.
The following theorem, independent of the polynomial family,
supplies it. The initial term may be as small as
$k^{2+\varepsilon}$, and the excluded density is explicit. The argument combines Erd\H{o}s's maximal-valuation deletion
\cite{Erdos} with the determinant count of
Filaseta--Laishram--Saradha \cite[Lemma~3.1]{FLS}. For earlier work on
products after terms have been omitted, see
Erd\H{o}s--Turk \cite[Theorem~6.1]{ErdosTurk}.

For $\varepsilon>0$, define
\[
 a_\varepsilon=\min\left\{\varepsilon,
                   \frac{3}{\sqrt7}-1\right\},\qquad
 \delta_\varepsilon=
 \frac{a_\varepsilon(1-\tfrac72a_\varepsilon)}
      {6(1+a_\varepsilon)}.
\]

\begin{theorem}
\label{thm:dense-power-general}
Fix positive integers $d,C$ with $C\ge d$, and real numbers
$\varepsilon>0$ and $0<\delta<\delta_\varepsilon$.
There is an effectively computable $k_0(d,C,\varepsilon,\delta)$ such
that, whenever positive integers $k,A$ satisfy
\[
 k\ge k_0,\qquad A\ge k^{2+\varepsilon},\qquad \gcd(A,d)=1,
\]
the following assertions hold simultaneously for every integer $r\ge2$.
\begin{enumerate}[label=\textup{(\roman*)}]
\item Fewer than $(1-\delta)k$ indices $0\le j<k$ admit a representation
\[
 A+jd=a_jx_j^r,\qquad a_j,x_j\in\Z_{>0},\qquad P(a_j)\le Ck.
\]
\item More than $\delta k$ distinct primes $p>Ck$ satisfy
\[
 r\nmid v_p\left(\prod_{j=0}^{k-1}(A+jd)\right).
\]
\item If $I\subseteq\{0,\ldots,k-1\}$ has at least $(1-\delta)k$
elements, then
\[
 \prod_{j\in I}(A+jd)\ne c y^r
 \quad\text{for all }c,y\in\Z_{>0}\text{ with }P(c)\le Ck.
\]
\end{enumerate}
In particular, for $\varepsilon\ge1/10$ one may take $\delta=1/102$.
The threshold is independent of $A$ and $r$.
\end{theorem}

\begin{proof}
Choose fixed real numbers $a,b>0$ such that
\begin{equation}\label{eq:dense-parameters}
 a\le\varepsilon,\qquad \frac72a+6b<1,\qquad
 \delta<\frac{ab}{1+a}.
\end{equation}
Such a choice exists by the definition of $\delta_\varepsilon$:
the maximum of $a(1-\tfrac72a)/(6(1+a))$ on $0<a\le\varepsilon$
occurs at $a=a_\varepsilon$. All estimates below are uniform in $A$
and in the integer $r\ge2$.

Suppose that a set $I$ of at least $(1-\delta)k$ indices satisfies
the representation in~\textup{(i)}. Replace each $a_j$ by the
$r$th-power-free part of $A+jd$. For every prime $p\le Ck$ dividing
one of these terms, remove one index in $I$ maximizing
$v_p(A+jd)$. Let $J$ be the remaining set and $m=|J|$. Then
\begin{equation}\label{eq:dense-power-deletion}
 m\ge(1-\delta)k-\pi(Ck)=(1-\delta-o(1))k,
 \qquad \prod_{j\in J}a_j\mid(k-1)!.
\end{equation}
Indeed, if $j_p$ is a removed maximizing index, then $p\nmid d$ and
$v_p(A+jd)\le v_p(j-j_p)$ for every remaining $j$.
The product of the nonzero distances from $j_p$, on both sides,
divides $j_p!(k-1-j_p)!$, which divides $(k-1)!$.
In particular,
\[
 v_p(a_j)\le\left\lfloor\frac{\log(k-1)}{\log p}\right\rfloor
 \qquad(j\in J).
\]

Put $\alpha=k^a$, $\beta=k^{-b}$, and
$U=\{j\in J:a_j\le\alpha k\}$.
We claim that at most
\begin{equation}\label{eq:dense-power-exceptional}
 O_d(\alpha^{7/2}\beta^{-5})
\end{equation}
primes in $[\beta k,k)$ divide two of the cofactors indexed by $U$.
For each such prime $p$, select an ordered pair of distinct indices
$(i,j)$. There are at most $(\alpha/\beta)^2$ possible ordered pairs
$(a_i/p,a_j/p)$. Fix such a pair $(h,h')$ and one prime $p$ with
its associated $(x_i,x_j)$. For any other prime $q$ with the same
cofactor pair, write its associated variables as $(x_u,x_v)$.
For sufficiently large $k$, all four variables lie in
\[
 L=(A/(\alpha k))^{1/r}\le x\le
 R=(2A/(\beta k))^{1/r},
\]
and
\[
 |hx_i^r-h'x_j^r|\le d/\beta,\qquad
 |hx_u^r-h'x_v^r|\le d/\beta.
\]
Multiplying by $x_v^r$ and $x_j^r$, respectively, and subtracting gives
\[
 h\bigl|(x_ix_v)^r-(x_jx_u)^r\bigr|
 \le \frac{4dA}{\beta^2 k}.
\]
Consequently
\[
 |x_ix_v-x_jx_u|
 \le\frac{4d}{r}\alpha^{2-2/r}\beta^{-2}(A/k)^{-1+2/r}
 \le2d\alpha\beta^{-2}.
\]
The last inequality uses $A/k\ge\alpha$ and $-1+2/r\le0$.
If $e=\gcd(x_i,x_j)$, then $pe^r$ divides the difference of the
two corresponding progression terms and is coprime to $d$.
Thus $pe^r\le k-1$ and $e\le\beta^{-1/r}$.
For a fixed integer determinant $x_ix_v-x_jx_u$, its solutions
$(x_u,x_v)$ lie on an arithmetic line whose coordinate steps are
$x_i/e$ and $x_j/e$. The number lying in $[L,R]^2$ is at most
\[
 2Re/L+1\le2(2\alpha)^{1/r}\beta^{-2/r}+1
 \le2\sqrt{2\alpha}\,\beta^{-1}+1.
\]
Since $\gcd(A,d)=1$, each $x_j$ is coprime to $d$.
The variables $x_j$, $j\in U$, are therefore pairwise distinct:
equality of two would give $x_j^r\mid i-j$, although
$x_j^r\ge A/(\alpha k)\ge k> |i-j|$.
Thus a pair $(x_u,x_v)$ cannot be assigned to two different primes.
Counting the cofactor pairs, determinants, and points on each line
proves~\eqref{eq:dense-power-exceptional}.

Since $b<1/6$, every prime $p\ge\beta k$ has exponent at most one
in each retained $a_j$. An exceptional prime from
\eqref{eq:dense-power-exceptional} divides at most $2/\beta$ terms;
every other prime in $[\beta k,k)$ occurs at most once in the product
over $U$. No prime at least $k$ divides a retained cofactor. Hence
\begin{align*}
 \sum_{j\in U}\log a_j
 &\le k\sum_{p\le\beta k}\frac{\log p}{p-1}
       +\sum_{p<k}\log p
       +O_d(\alpha^{7/2}\beta^{-6}\log k)\\
 &\le (1-b)k\log k+o(k\log k).
\end{align*}
Here \eqref{eq:dense-parameters} makes the last error $o(k\log k)$;
the effective prime estimates are those of~\cite[Theorems~4 and~6]{RS}.

All cofactors $a_j$, $j\in J$, are distinct. Otherwise, for $a_i=a_j=c$,
\[
 d|i-j|=c|x_i^r-x_j^r|
 \ge r c^{1/r}A^{1-1/r}\ge2\sqrt A>dk,
\]
a contradiction. The uniform Stirling bound
\[
 \log(u!)\ge u\log k-O(k)\qquad(0\le u\le k)
\]
therefore shows that $|U|\le(1-b+o(1))k$.
Writing $W=J\setminus U$, we obtain
$|W|\ge(b-\delta-o(1))k$, and then
\begin{align*}
 \sum_{j\in J}\log a_j
 &\ge\log(|U|!)+(1+a)|W|\log k\\
 &\ge \bigl(1+ab-(1+a)\delta-o(1)\bigr)k\log k.
\end{align*}
This contradicts~\eqref{eq:dense-power-deletion}, since
$ab>(1+a)\delta$. This proves~\textup{(i)}.

Every index excluded from~\textup{(i)} has a prime $p>Ck$ whose
valuation in that term is not divisible by $r$.
Such a prime divides no other progression term, because
$p>Ck\ge dk>d|i-j|$. Choosing one for each excluded index proves
\textup{(ii)}. The same uniqueness shows that a product as in
\textup{(iii)} would force every selected index to satisfy
\textup{(i)}, a contradiction. Finally, if $\varepsilon\ge1/10$,
the choices $a=1/10$, $b=27/250$, and $\delta=1/102$ satisfy
\eqref{eq:dense-parameters}: indeed,
$(7/2)a+6b=499/500<1$ and $ab/(1+a)=27/2750>1/102$.
Every limiting estimate used above
has an effective error bound, proving effectivity.
\end{proof}

Theorem~\ref{thm:dense-power-general}\textup{(iii)} allows a fixed
positive proportion of arbitrary omissions. It does not settle the
corresponding problem for unions of widely separated intervals,
or the problem of powerful products: a valuation not divisible by
$r$ need not equal one.

\subsection{Local cycles and primitive groups}\label{sec:groups}

We first obtain a large prime cycle when the parameter is at most cubic
in the degree.  For larger parameters, the local packets give elements
with small support.  We combine these with a progression estimate and
the primitive-group theorem of Burness and Guralnick.  Irreducibility is
assumed only in the statements that require transitivity.

\subsubsection{A first parameter range}\label{sec:cubic}
For a prime close to the degree, the absence of a square multiple in the
parameter interval produces the required prime cycle.

\begin{lemma}\label{lem:cubic-polygon}
Let $n/2<p<n$ be prime.  If no integer in $[s,s+n-1]$ is divisible
by $p^2$, the $p$-adic Newton polygon of $H_{n,s}$ has an edge of
width $p$ and slope $-1/p$.  In particular, $p$ divides $|G_{n,s}|$.
\end{lemma}

\begin{proof}
Write $n=p+b$, with $0\le b<p$. The coefficient calculation
\eqref{pade:coefficient-valuation}, and the two polygons in the proof
of Proposition~\ref{pade:prop-square}, show that exactly $p$ roots
have valuation $1/p$. A corresponding local ramification index is
divisible by $p$, hence so is the order of the global Galois group.
This is the Newton-index argument of
\cite[Theorems~2.1 and~2.2]{Hajir2005}.
\end{proof}


The obstruction in the lemma is a square multiple in a short interval.
Such multiples cannot occur simultaneously for all primes close to
$n$ unless $s$ is sufficiently large.

\begin{proposition}\label{prop:cubic-range}
There is an absolute effective integer $n_{\mathrm{cub}}$ such that,
if $n\ge n_{\mathrm{cub}}$, $H_{n,s}$ is irreducible over $\mathbb Q$,
and
\[
 1\le s\le\frac{n^3}{10\log n},
\]
then $G_{n,s}$ contains $A_n$.
\end{proposition}
\begin{proof}
Put $u=\log n$ and
\[
 \mathcal P=\{p\text{ prime}:9n/10<p<n-2\}.
\]
The effective prime number theorem gives
\begin{equation}\label{eq:cubic-prime-count}
 |\mathcal P|>\frac{n}{20u}
\end{equation}
for all sufficiently large $n$.  For example, this follows directly
from the effective bounds for $\vartheta$ in
\cite[Theorem~4.2]{Dusart}, after removing the at most three primes
in the upper endpoint interval.

Suppose every $p\in\mathcal P$ has a square multiple
$k_pp^2\in[s,s+n-1]$.  Distinct primes have distinct multipliers:
if $p<q$ had the same multiplier $k$, then
\[
 k(q^2-p^2)\ge q^2-p^2>2p>n-1,
\]
which is impossible for two integers in this interval.  The multipliers
all lie in
\[
 \left[\frac{s}{n^2},\frac{100(s+n-1)}{81n^2}\right].
\]
The number of integers in this real interval is at most its length
plus one.  Consequently
\[
 |\mathcal P|
 \le\frac{19s}{81n^2}+\frac{100}{81n}+1
 \le\frac{19n}{810u}+\frac{100}{81n}+1
 <\frac{n}{20u}
\]
for sufficiently large $n$, contradicting
\eqref{eq:cubic-prime-count}.

Choose $p\in\mathcal P$ for which there is no square multiple.
Lemma~\ref{lem:cubic-polygon} gives $p\mid|G_{n,s}|$.  Cauchy's
theorem supplies an element of order $p$, necessarily a single
$p$-cycle because $2p>n$.  Since irreducibility makes the root action
transitive and $p>n/2$, this action is primitive.  Indeed, in a
nontrivial block system an order-$p$ element either moves at least
$p$ whole blocks, requiring at least $2p$ points, or fixes every
block and has its $p$-cycle inside one block, requiring a block of
size at least $p>n/2$; both alternatives are impossible.
Jordan's theorem now gives $G_{n,s}\supseteq A_n$, since
$p<n-2$; equivalently, apply \cite[Theorem~2.2]{Hajir2005}.
Every numerical threshold used above is effective.
\end{proof}


\subsubsection{Small supports and Cartesian products}\label{sec:products}

The progression estimate first gives small nonidentity permutations
uniformly in the parameter. No rate of decay is needed for the
subsequent argument.

\begin{proposition}\label{prop:small-support}
For every fixed $\delta\in(0,1)$ there is an effective $n_\delta$ such
that
\[
 \mu(G_{n,s})\le\delta n
\]
whenever $n\ge n_\delta$, $s\ge1$, and $H_{n,s}$ is irreducible.
\end{proposition}

\begin{proof}
For $s\le n^3/(10\log n)$, Proposition~\ref{prop:cubic-range} gives
$G_{n,s}\supseteq A_n$ in sufficiently large degree. A $3$-cycle has
support $3\le\delta n$ once $n$ is large enough.

Suppose therefore that $s>n^3/(10\log n)$, put
$k=\lfloor\delta n/2\rfloor$, and assume $\mu(G_{n,s})>\delta n$.
For each width $N=2,4,\ldots,2k$ and each prime $p>n$ dividing
$s+n-N$, Proposition~\ref{prop:exact-inertia} would give a
nonidentity element of support $N\le\delta n$ unless
$N\mid v_p(s+n-N)$. Thus all these valuations are even.

The resulting nodes form a progression of length $k$ and step $2$.
Divide its terms by $h=\gcd(s+n,2)$. Its starting point and step are
\[
 A=\frac{s+n-2k}{h},\qquad d=\frac2h\in\{1,2\},
 \qquad\gcd(A,d)=1.
\]
Every term is a square times an $n$-smooth integer. Choose a fixed
integer $C\ge4/\delta$; then $n\le Ck$ for large $n$. Moreover
$A\ge s/2>n^3/(20\log n)>k^{5/2}$ for large $n$.
Theorem~\ref{thm:dense-power-general}\textup{(i)}, with
$\varepsilon=1/2$, $\delta=1/102$, and $r=2$, contradicts the
fact that all $k$ indices qualify.
\end{proof}

The next observation relates fixed-point counts to divisors. Its
strength is that it applies to the full symmetric group on each
coordinate, irrespective of the smaller group acting there.

\begin{lemma}\label{lem:cartesian-fixed}
Let $m\ge2$, $a\ge2$, and $n=m^a$. In the product action of
$S_m\wr S_a$ on $\{1,\ldots,m\}^a$, an element fixing at least
$15n/16$ points has trivial coordinate permutation. Its number of
fixed points is a product $f_1\cdots f_a$ with
$15m/16\le f_i\le m$. In particular it has a divisor in $(m/2,m]$.
\end{lemma}

\begin{proof}
A cycle of $r\ge2$ coordinates permits at most $m$ choices for those
entries of a fixed tuple, rather than $m^r$. A nontrivial coordinate
permutation therefore has fixed-point ratio at most $1/m\le1/2$.
In the base group the fixed-point ratio is
$\prod_i(f_i/m)$. Since each factor lies in $[0,1]$, a product at
least $15/16$ forces each factor to be at least $15/16$.
\end{proof}

Ford's divisor estimate makes the set of possible fixed-point counts
in this lemma sparse when $m$ grows. This excludes the whole product
action, including its odd-base cases.

\begin{proposition}\label{prop:no-cartesian}
For all sufficiently large $n$, effectively and uniformly in $s\ge1$
with $H_{n,s}$ irreducible, the root action of $G_{n,s}$ is not
permutation-isomorphic to a subgroup of $S_m\wr S_a$ in its product
action, for any $m\ge2$, $a\ge2$, $n=m^a$.
\end{proposition}

\begin{proof}
In this action a nontrivial coordinate permutation moves at least
$(1-1/m)n$ points, and a nontrivial base element moves at least
$2n/m$ points. Thus the minimal degree is at least
$n\min\{1-1/m,2/m\}$. Proposition~\ref{prop:small-support} implies
that $m\to\infty$ along any sequence of such exceptional actions
with $n\to\infty$.

For large $n$, the displayed lower bound for the support exceeds $3$
for every $m\ge2$, $a\ge2$. Thus the product action cannot contain
$A_n$. Proposition~\ref{prop:cubic-range} again permits us to assume
$s>n^3/(10\log n)$. Put $k=\lfloor n/32\rfloor$.
For a width $N\in\{2,4,\ldots,2k\}$ set $j=n-N$.
If $p>n$ divides $s+j$ and its inertia is nontrivial, that inertia
has support $N\le n/16$. Lemma~\ref{lem:cartesian-fixed} implies
that $j$ has a divisor in $(m/2,m]$.

Let $H(x,y,z)$ count the positive integers at most $x$ with a divisor
in $(y,z]$. Ford proved the uniform estimate
\begin{equation}\label{eq:Ford}
 H(x,y,2y)\ll
 \frac{x}{(\log y)^\kappa(\log\log y)^{3/2}},\qquad
 3\le y\le\sqrt x,
 \quad\kappa=1-\frac{1+\log\log2}{\log2}>0;
\end{equation}
see \cite[Theorem~1(v) and Corollary~2]{Ford}. Its upper bound is effective. Since $a\ge2$,
we can apply it with $x=n$ and $y=m/2$. As $m\to\infty$, the number
of exceptional labels $j$ is $o(n)=o(k)$.

At every other label all inertia at primes above $n$ is trivial.
Proposition~\ref{prop:exact-inertia} then shows that every such prime
has valuation divisible by the even integer $N$. Consequently all
but $o(k)$ terms of
\[
 s+n-2k,\ s+n-2k+2,\ldots,s+n-2
\]
are squares times $n$-smooth integers.
Divide by $h=\gcd(s+n,2)$ to obtain a progression with step $d=1$ or
$2$, coprime starting point $A\ge s/2>k^{5/2}$, and length $k$.
For large $n$, $n\le33k$. Theorem~\ref{thm:dense-power-general}\textup{(i)},
with $C=33$, $r=2$, $\varepsilon=1/2$, and $\delta=1/102$,
contradicts this density.
For effectivity, first use \eqref{eq:Ford} to choose a fixed $m_0$ for
which the exceptional proportion is smaller than $1/102$. Then apply
Proposition~\ref{prop:small-support} with a fixed positive support
proportion smaller than
$\min_{2\leq m<m_0}\min\{1-1/m,2/m\}$ to exclude the remaining
values of $m$.  This gives a single effective threshold.
\end{proof}

\subsubsection{Primitive groups}\label{sec:primitive}

After Cartesian products have been excluded, the primitive-group
classification leaves actions of symmetric or alternating groups on
subsets. Their homogeneous permutations have a simple restriction.

\begin{lemma}\label{lem:subset-cycles}
Suppose $2\le\ell<m/2$. In the action of $S_m$ on the $\ell$-element
subsets of $\{1,\ldots,m\}$, a homogeneous permutation with fixed-point
ratio at least $15/16$ cannot have even nontrivial cycle length greater
than two.
\end{lemma}

\begin{proof}
Let $\sigma\in S_m$ move $u$ letters, and let $I$ be a uniformly chosen
$\ell$-element subset. Counting the contributions of the moved letters
to the symmetric difference gives
\[
 \mathbb E|I\mathbin\triangle\sigma(I)|
 =\frac{2u\ell(m-\ell)}{m(m-1)}.
\]
Since the symmetric difference is at most $2\ell$ and is zero on
fixed subsets,
\[
 1-\fpr(\sigma)\ge\frac{u(m-\ell)}{m(m-1)}>\frac{u}{2m}.
\]
Thus $u<m/8$. In particular there are at least $\ell$ fixed letters.
Selecting one letter from a nontrivial letter cycle of length $r$,
together with $\ell-1$ fixed letters, gives a subset orbit of length
$r$. If the subset action is homogeneous of length $q$, every
nontrivial letter cycle therefore has length $q$.
If $q>2$ is even, select two opposite letters from a $q$-cycle and
$\ell-2$ fixed letters. Their subset orbit has length $q/2>1$,
contradicting homogeneity.
\end{proof}

The lemma converts odd valuations into forbidden permutations on a
full progression of widths.

\begin{proposition}\label{prop:no-subsets}
For all sufficiently large $n$, effectively and uniformly in $s\ge1$
with $H_{n,s}$ irreducible, its root action is not a subgroup of the
action of $S_m$ on $\ell$-element subsets with $2\le\ell<m/2$.
\end{proposition}

\begin{proof}
The small-parameter case follows from
Proposition~\ref{prop:cubic-range}, since a proper subset action has
order at most $m!<|A_n|$ for sufficiently large $n=\binom m\ell$.
Suppose $s>n^3/(10\log n)$ and put $k=\lfloor n/64\rfloor$.
At a width $N=4,8,\ldots,4k$, an odd valuation $e$ at a prime
$p>n$ gives inertia with
\[
 g=\gcd(N,e)\text{ odd},\qquad q=N/g\equiv0\pmod4.
\]
The inertia is nonidentity, homogeneous, and has fixed-point ratio
at least $15/16$. Lemma~\ref{lem:subset-cycles} excludes it.
Therefore every valuation at a prime above $n$ is even in every term of
\[
 s+n-4k,\ s+n-4k+4,\ldots,s+n-4.
\]
Divide by $h=\gcd(s+n,4)$. The resulting progression has step
$d=4/h\in\{1,2,4\}$, starting point
$A=(s+n-4k)/h$ coprime to $d$, and all terms squares times
$n$-smooth integers. For large $n$, we have $n\le65k$ and
$A\ge s/4>k^{5/2}$. Apply
Theorem~\ref{thm:dense-power-general}\textup{(i)} with $C=65$,
$r=2$, $\varepsilon=1/2$, and $\delta=1/102$.
\end{proof}

We can now apply the primitive-group theorem. The inequality in that
theorem is strict, so the numerical margin below also handles elements
of order two.


\begin{theorem}\label{thm:primitive}
For all effectively sufficiently large $n$ and every positive integer
$s$, if $G_{n,s}$ is primitive, then $A_n\le G_{n,s}$.
\end{theorem}
\begin{proof}

Primitivity includes transitivity, so $H_{n,s}$ is irreducible.
By Proposition~\ref{prop:small-support}, a sufficiently large
$G_{n,s}$ has a nonidentity element of support at most $n/3$.
A suitable nonidentity power has prime order $r$ and no larger support.
Its fixed-point ratio is at least $2/3>1/\sqrt{r+1}$.
Burness and Guralnick \cite[Corollary~2]{BG} imply that
the primitive action has the form
\[
 (A_m)^a\le G_{n,s}\le S_m\wr S_a,\qquad
 n=\left(\binom m\ell\right)^a,\qquad 1\le\ell<m/2,
\]
where each coordinate is the action on $\ell$-element subsets and
the wreath product acts on the Cartesian product of the coordinate
sets. The middle-subset action $\ell=m/2$ is imprimitive,
since complementary subsets form blocks, so it does not occur here.
Proposition~\ref{prop:no-cartesian} excludes $a\ge2$.
Proposition~\ref{prop:no-subsets} excludes $\ell\ge2$.
The remaining action has $a=\ell=1$, so $G_{n,s}\supseteq A_n$.
\end{proof}


\subsection{Invariant partitions}\label{sec:blocks}

We exclude invariant partitions in three steps.  Progressions of cubes
rule out blocks of size two.  Linear forms in logarithms bound the
parameter when both the block size and the number of blocks are at least
three.  The partition energy then excludes all partitions in the
complementary range. Only partitions into two large blocks remain
for the next section.

\subsubsection{Blocks of size two}
For blocks of size two, an odd packet cannot support inertia whose
nontrivial cycles have length divisible by three.

\begin{lemma}\label{prop:pair-blocks}
There is an effective absolute integer $n_0$ such that, if $n\ge n_0$
and $s>n^3/(10\log n)$, then $G_{n,s}$ preserves no partition of the
roots into blocks of size two.
\end{lemma}
\begin{proof}
Suppose such a partition exists. Put $k=\lfloor n/12\rfloor$ and use
the odd widths $N_t=3(1+2t)$, $0\le t<k$, all smaller than $n/2$.
If $p>n$ divides $s+n-N_t$ to an exponent $e$ not divisible by $3$,
Proposition~\ref{prop:exact-inertia} gives an inertia permutation
whose nontrivial cycles all have length
$q=N_t/\gcd(N_t,e)$, with $3\mid q$.
A block fixed by this permutation cannot contain a moved root, since
$q>2$. Every moved block consists entirely of moved roots. Its support
is therefore a union of pairs, contradicting its odd size $N_t$.
Thus every valuation above $n$ at these nodes is divisible by $3$.

In reverse order the nodes form $A,A+6,\ldots,A+6(k-1)$, with $A>s$.
Put $g=\gcd(A,6)$, $A'=A/g$, and $d=6/g$. Then $\gcd(A',d)=1$,
and every term $A'+jd$ is a cube times an $n$-smooth positive integer.
For all sufficiently large $n$, we have $n\le13k$ and
\[
 A'>s/6>\frac{n^3}{60\log n}>k^{5/2}.
\]
Theorem~\ref{thm:dense-power-general}\textup{(i)}, with $C=13$,
$r=3$, $\varepsilon=1/2$, $\delta=1/102$, and
$d\in\{1,2,3,6\}$, contradicts the fact that all $k$ terms qualify.
The four possible steps give a single effective threshold.
\end{proof}
\subsubsection{A parameter bound for invariant partitions}\label{sec:block-cap}

It is not necessary to discard a whole node for each small prime.
Instead, assign each small prime to one node where its valuation is
maximal. A pair of nodes with a common large exponent then has only
a bounded number of exceptional prime factors.

\begin{proposition}\label{prop:block-parameter-cap}
There are effective absolute constants $C_1,n_3>0$ such that, whenever
$n\ge n_3$ and $G_{n,s}$ preserves a partition with block size
\[
 3\le b\le n/3,
\]
one has
\[
 \log s\le C_1(\log n)^9.
\]
Irreducibility is not required for this assertion.
\end{proposition}

\begin{proof}
Put $a=n/b$, $u=\log n$, and $L=\log s$. We may assume $L>2u$, since otherwise
the conclusion holds after increasing $C_1$. For every prime
\[
 n/3<P\le n/2
\]
use the two widths $P$ and $2P$. They are distinct across all prime
groups. Since $a,b\ge3$ and $ab=n$, we have $\max\{a,b\}\le n/3<P$.
Thus $P$ does not divide $(b!)^a a!$, the order of the partition
stabilizer.
If a prime $p>n$ occurs in either corresponding node to an exponent
not divisible by $P$, Proposition~\ref{prop:exact-inertia} supplies
an inertia element whose order is divisible by $P$, a contradiction.
Therefore
\[
 z_{P,\ell}:=s+n-\ell P=a_{P,\ell}x_{P,\ell}^{P},
 \qquad P(a_{P,\ell})\le n,\qquad \ell\in\{1,2\},
\]
where each $a_{P,\ell}$ is $P$th-power-free.

For every prime $p\le n$ dividing at least one selected node, choose
one node $z_{P,\ell}$ where $v_p(z_{P,\ell})$ is maximal, and assign
$p$ to that node. Let $\mathcal E_{P,\ell}$ be its assigned primes.
Define
\[
 r_{P,\ell,p}=v_p(a_{P,\ell})\quad(p\in\mathcal E_{P,\ell}),
 \qquad
 d_{P,\ell}
 =a_{P,\ell}\prod_{p\in\mathcal E_{P,\ell}}p^{-r_{P,\ell,p}}.
\]
These are positive integers, and $0\le r_{P,\ell,p}<P$. The
maximal-valuation argument gives
\begin{equation}\label{eq:assigned-cofactor-product}
 \prod_{P,\ell}d_{P,\ell}\mid(n-1)!.
\end{equation}
This is the allocation assertion of Lemma~\ref{ene:deletion}:
the remaining cofactor at each node divides its unassigned
small-prime part.

Let $R$ be the number of prime groups. The effective prime number
theorem gives
\[
 R=\left(\frac16+o(1)\right)\frac n u,\qquad
 \pi(n)=(1+o(1))\frac n u.
\]
At most $\pi(n)/7$ groups have seven or more assigned primes on their
two nodes. By \eqref{eq:assigned-cofactor-product}, at most
$n/(1000u)$ groups satisfy
\[
 \log d_{P,1}+\log d_{P,2}>1000u^2.
\]
Since
\[
 \frac16-\frac17-\frac1{1000}>0,
\]
there is, for all sufficiently large $n$, a prime group with at most
six assigned primes and
\[
 \log d_{P,1}+\log d_{P,2}\le1000u^2.
\]
Write its assigned primes as $p_1,\ldots,p_h$, where $h\le6$.
They are distinct, because every small prime was assigned to only one
node. For suitable integers $c_j$ with $|c_j|<P$ we have
\[
 \frac{z_{P,1}}{z_{P,2}}
 =\frac{d_{P,1}}{d_{P,2}}
  \left(\frac{x_{P,1}}{x_{P,2}}\right)^P
  \prod_{j=1}^h p_j^{c_j}.
\]
Taking real logarithms gives a nonzero linear form
$\Lambda=\log(z_{P,1}/z_{P,2})$ satisfying
\[
 0<|\Lambda|\le n/s.
\]
The rational bases have logarithmic heights at most
\[
 1000u^2,\qquad
 \frac{\log(s+n)}P\le\frac{2L}P,\qquad
 u,\ldots,u,
\]
respectively. Omit any base equal to one and any term with zero
coefficient. The number of logarithms is at most eight and all
integer coefficients have modulus at most $P$.

If only one logarithm remains, its elementary rational-height bound
gives the inequality below directly. Otherwise Matveev's theorem
\cite[Corollary~2.3]{Matveev}, in degree one with coefficient bound
$B^*\le P$, gives
\[
 L-u\le C_2 u^9\max\left\{1,\frac{L}{P}\right\}
 \le C_3u^9\max\left\{1,\frac{L}{n}\right\}.
\]
Here the factors $u^2$, at most six factors $u$, and the coefficient
factor $\log(eP)=O(u)$ account for the exponent nine. The maximum
also covers the case $x_{P,1}=x_{P,2}$, when that logarithm is omitted.
Thus no assumption that a large-power part is nontrivial is needed.

Since $L>2u$, after changing the absolute constant we obtain
\[
 L\le C_4u^9\max\left\{1,\frac{L}{n}\right\}.
\]
For $n>2C_4u^9$, the alternative $L\ge n$ is impossible. Hence
$L<C_4u^9$, proving the assertion. The prime number theorem and the
logarithmic-form estimate both have effective constants.
\end{proof}

In particular, for partitions whose block size and number are both at least three, it now suffices to treat a parameter range of fixed polynomial
growth in $\log n$. This reduction is uniform in the number and size
of the blocks. It does not by itself exclude such partitions in that
remaining parameter range.


\subsubsection{The energy of a partition}\label{be:section}

The matrix energy identity applies even when individual blocks are
not defined over $\mathbb Q$. This is the reason to use the entire
invariant partition in place of a single block.

Suppose now that the Galois group preserves a partition into $a$ blocks
of size $b$, where $n=ab$ and $a,b\ge2$. Index the roots by their
positions in this partition and define
\[
 Q_{ij}=\frac{\boldsymbol1_{\{i,j\text{ in the same block}\}}}{b}
         -\frac1n.
\]
This is the orthogonal projection onto the block-constant vectors
orthogonal to the constant vector. It has rank $a-1$, constant diagonal
$(a-1)/n$, and is invariant under the Galois group.

\begin{lemma}\label{be:energy}
Put $u=\log n$, $T=\log s$, and
\[
 E_p=\sum_{i\ne j}Q_{ij}v_p(\alpha_i-\alpha_j)
       +\sum_iQ_{ii}\bigl(v_p(\alpha_i)+\epsilon_p\bigr).
\]
Then $E_p\ge0$. For $s\ge n$ and $\log n\ge10000$,
\begin{equation}\label{be:upper}
 \mathcal E:=\sum_pE_p\log p
 <(a-1)\left(\frac T2+\frac{3u}{2}+\log48\right).
\end{equation}
If $p>n$ divides $s+j$, set $N=n-j$ and $e=v_p(s+j)$, and let
$k_i$ count the roots of positive valuation in the $i$th block. Then
\begin{equation}\label{be:packet}
 E_p=\frac eN\left(\frac1b\sum_{i=1}^ak_i^2-\frac{N^2}{n}\right).
\end{equation}
\end{lemma}

\begin{proof}
The displayed matrix $Q$ is rational, positive semidefinite, and
Galois invariant, and it annihilates the constant vector.
Apply Lemma~\ref{lem:matrix-energy}, with $\operatorname{tr}Q=a-1$.
It gives nonnegativity and the upper bound \eqref{be:upper}.
For a packet $C$ of size $N$,
$\mathbf1_C^{\mathsf T}Q\mathbf1_C
=b^{-1}\sum_i k_i^2-N^2/n$.
The exact packet identity \eqref{eq:matrix-packet} proves
\eqref{be:packet}.
\end{proof}


An order-$P$ power of a homogeneous inertia permutation has the same
support. When $P$ exceeds the number of blocks, it fixes every block
and hence forces packet occupations to be multiples of $P$.

\begin{lemma}\label{be:prime-occupation}
Suppose $p>n$ divides the node corresponding to a width $N=\ell P$,
where $P$ is prime and $P\nmid e=v_p(s+n-N)$.
If $P>a$ and $a\nmid\ell$, then
\begin{equation}\label{be:occupation}
 E_p\ge\frac{eP^2(a-1)}{nN}.
\end{equation}
If $P>b$ and $b\nmid N$, such a prime contribution is impossible.
\end{lemma}
\begin{proof}
The homogeneous inertia cycle length $q=N/\gcd(N,e)$ is divisible by
$P$. Raising its generator to the power $q/P$ gives an order-$P$
permutation with the same support. If $P>a$, it fixes every block.
Write $k_i=Pt_i$, where $\sum_i t_i=\ell$. If $\ell=ah+r$ with
$0<r<a$, minimization over integral occupations gives
\[
 \frac1b\sum_i k_i^2-\frac{N^2}{n}
 \ge\frac{P^2r(a-r)}n\ge\frac{P^2(a-1)}n.
\]
Equation~\eqref{be:packet} proves \eqref{be:occupation}.
For the last assertion, $q\ge P>b$, so a fixed block contains no moved
point. The support is therefore a union of entire blocks, forcing
$b\mid N$, a contradiction.
\end{proof}

We next choose many widths to which this occupation bound applies.
Fix $\theta=101/200$ and put $\delta=\log(1/\theta)$; thus
$\delta/2>1/3$. For any integer $d\ge2$, the number of widths
\[
 N=\ell P\le n,\qquad P>n^\theta\text{ prime},\qquad d\nmid\ell
\]
is at least $(\delta(1-1/d)-o(1))n$, uniformly in $d$. Indeed, the prime
$P>\sqrt n$ is unique, and this number equals
\[
 \sum_{n^\theta<P\le n}
 \left(\left\lfloor\frac nP\right\rfloor-
       \left\lfloor\frac n{dP}\right\rfloor\right)
 \ge n(1-1/d)\sum_{n^\theta<P\le n}\frac1P-\pi(n).
\]
The prime reciprocal sum tends effectively to $\delta$, and
$\pi(n)=o(n)$. These are the usual effective consequences of the prime
number theorem; all errors here are independent of $d$.

\begin{proposition}\label{be:subpower-range}
For all effectively sufficiently large $n$, if
\[
 s>\frac{n^3}{10\log n},\qquad
 \frac{\log(s+n)}{\log n}<n^{101/200},
\]
then $G_{n,s}$ preserves no partition into equal blocks with at least
two blocks and at least two roots per block.
\end{proposition}
\begin{proof}
Suppose there are $a$ blocks of size $b$. First let $a\le n^\theta$.
Choose the widths just counted with $d=a$. Every prime $p>n$ dividing
a selected node has exponent $e< n^\theta<P$. Lemma~\ref{be:prime-occupation}
therefore gives $E_p>e(a-1)n^{2\theta-2}$.

Delete one selected node with maximal valuation for every prime at
most $n$. At most $\pi(n)$ nodes are lost, and the product of the
small-prime parts of the remaining nodes divides $(n-1)!$. This is Lemma~\ref{ene:deletion}.
Consequently the logarithmic mass $W$ of the retained prime powers
with primes exceeding $n$ satisfies
\[
 W\ge(\delta/2-o(1))nT-nu\ge c nT
\]
for an absolute $c>0$. Here $T>3u-\log(10u)$ and $\delta/2>1/3$.
Nonnegativity of the other local energies gives
$\mathcal E>c(a-1)n^{2\theta-1}T$, contradicting
\eqref{be:upper}, which is less than $2(a-1)T$ in this range.

If $a>n^\theta$, then $b<n^{1-\theta}<n^\theta$. Choose the widths
with $d=b$. Since $P>b$, $b\nmid\ell$ implies $b\nmid\ell P$.
The last assertion of Lemma~\ref{be:prime-occupation} says that every
selected node has no prime divisor exceeding $n$. This contradicts
the same lower bound $W\ge cnT$.
\end{proof}


\subsection{Laurent interpolation and quadratic block fields}\label{sec:quadratic}

Let $E$ be the splitting field of an irreducible polynomial, and
suppose that its Galois group preserves a partition into two blocks.
The setwise stabilizer $G_B$ of either block has index two; its
fixed field $K=E^{G_B}$ is the \emph{quadratic block field}.
Every root stabilizer is contained in $G_B$, so $K$ lies in each
root field. A bounded trace treats the real quadratic case.
For the imaginary quadratic case, we construct integral
polynomials whose block traces become small at infinity.

\subsubsection{The real quadratic case}
The centered block trace is bounded at both real embeddings.  Its norm
allows us to combine this bound with the fibers at $0$ and $-1$.

\begin{lemma}\label{be:real-quadratic-bound}
Let $n=2b\ge4$ and suppose that $H_{n,s}$ is irreducible. If its
Galois group preserves two blocks of size $b$ and the corresponding
quadratic field is real, then
\[
 \log s<2n\log n.
\]
\end{lemma}
\begin{proof}
Let $K$ be the quadratic block field. For one block $B$, put
$t=\sum_{\alpha\in B}(\alpha+s)$. This is an algebraic integer
in $K$, whose conjugate is $-t$, since all centered roots sum to zero.
Both traces are real, and Proposition~\ref{geo:trace} gives
$|t|<b^2$.

If $t\ne0$, then $A=N_{K/\mathbb Q}(t)=-t^2$ is a nonzero integer
with $|A|<b^4$. For $p^e\Vert s$, the congruence
$H_{n,s}(Y-s)\equiv Y^n\pmod s$ implies that every centered root has
valuation at least $e/n$, at every extension of $v_p$. The two
conjugate traces therefore give $v_p(A)\ge2e/n=e/b$. Thus
$s\mid A^b$ and
\[
 \log s\le b\log|A|<4b\log b<2n\log n.
\]
If $t=0$, the uncentered block trace is the integer $-bs$.
The congruence
$H_{n,s}(X)\equiv X^{n-1}(X-n)\pmod{s+1}$ and
Lemma~\ref{geo:two-cluster}, applied with $A=b$ and $c=n=2b$,
give $s+1\mid b^n$. That lemma requires only an integral selected
trace, not a rational factor. Hence
$\log s<n\log b<2n\log n$ in this case as well.
\end{proof}
\subsubsection{Laurent interpolation and cyclic intermediate fields}
\label{sec:general-odd-jets}

The imaginary quadratic case requires integral polynomials whose
negative odd Laurent coefficients vanish and whose first positive
odd coefficient is prescribed. We first prove the required
surjectivity over the ground field; we then bound denominators and
coefficient sizes. The next theorem gives a general criterion for
the first step.
At a fiber with a common ramification index $q$, one may prescribe
a finite string of Laurent coefficients in a chosen nonzero
congruence class modulo $q$.
The obstruction is a cyclic extension between intermediate fields
of the cover.

Let $k$ be an algebraically closed field of characteristic zero,
let $C$ be a smooth projective connected curve of genus $g$, and
let $T:C\to\mathbb P^1_k$ have degree $n=qb$, where $q,b\ge2$.
Suppose
\[
 (T)_\infty=qE,\qquad E=Q_1+\cdots+Q_b,
\]
with distinct points $Q_i$.  Choose formal local parameters $v_i$ with
$T=v_i^{-q}$.  Write $K=k(C)$ and $F=k(T)$, and put
\[
 L(DE)=\{h\in K:(h)+DE\ge0\}\cup\{0\}.
\]
For a finite branch value $t$, let $e_P$ be the ramification index
at $P$ above $t$.  For $m\ge1$ define
\begin{equation}\label{eq:laurent-pole-degree}
 d_m=\sum_{t\text{ finite branch value}}
       \max_{P\mid t}\left\lfloor\frac{m(e_P-1)}{e_P}\right\rfloor.
\end{equation}

\begin{theorem}
\label{thm:abstract-odd-jets}\label{thm:laurent-cyclic-obstruction}
Fix $1\le a<q$, and put $e=q/\gcd(a,q)$.
Let $D\ge0$ be divisible by $q$, with $Db>2g-2$, and suppose
\begin{equation}\label{eq:laurent-cutoff}
 \left\lceil\frac{D+q+m}{q}\right\rceil>d_m
 \qquad(1\le m\le n-1,\ e\nmid m).
\end{equation}
If the coefficient map
\begin{equation}\label{eq:laurent-coefficient-map}
 \begin{split}
 \mathcal J_{D,a}:L(DE)&\longrightarrow k^{b(D/q+1)},\\
 h&\longmapsto
 \bigl([v_i^{-D+a+qj}]h\bigr)_{\substack{1\le i\le b\\0\le j\le D/q}}
 \end{split}
\end{equation}
is not surjective, then there are intermediate fields
\[
 F\subseteq M\subsetneq L\subseteq K
 \quad\text{with}\quad L/M\text{ cyclic of degree }e.
\]
In particular, the map is surjective if the monodromy action of
$T$ is primitive.

If $B$ is the number of finite branch values, the choice
$D=qB(n-2)$ satisfies both numerical hypotheses, for every
$1\le a<q$.
\end{theorem}

\begin{proof}
We first describe the obstruction to prescribing Laurent coefficients.
Let $V$ be the space of all coefficients of
$v_i^{-D},\ldots,v_i^{q-1}$, for $1\le i\le b$, and let
$\mathcal J:L(DE)\to V$ be the full coefficient map.
Its kernel is $L(-qE)=0$. Since $Db>2g-2$, Riemann--Roch gives
\[
 \dim\operatorname{coker}\mathcal J
 =b(D+q)-(Db+1-g)=g+qb-1.
\]
The same theorem gives
$\dim H^0(C,\Omega_C(qE))=g+qb-1$.
Each differential in this space defines a functional on $V$ by
\[
 (h_i)_i\longmapsto
 \sum_{i=1}^b\operatorname{Res}_{Q_i}(h_i\omega).
\]
This functional vanishes on global functions by the residue theorem.
It determines $\omega$: if it vanishes on every coefficient in $V$,
then the Laurent coefficients of $\omega$ from degree $-q$ through
degree $D-1$ vanish at every $Q_i$. Thus
$\omega\in H^0(C,\Omega_C(-DE))=0$.
The dimension count therefore identifies
$H^0(C,\Omega_C(qE))$ with the dual of
$\operatorname{coker}\mathcal J$.

Suppose now that $\mathcal J_{D,a}$ is not surjective. Extend a
nonzero functional annihilating its image to $V$ by making it zero
on the other congruence classes of coefficients. The preceding
identification supplies a nonzero differential
$\omega\in H^0(C,\Omega_C(qE))$. Set $f=\omega/dT\in K$.
At $Q_i$ we have $dT=-qv_i^{-q-1}\,dv_i$, so $f=O(v_i)$.
The coefficient restrictions on $\omega$ say that, through degree
$D+q$, the expansion of $f$ uses only exponents congruent to $-a$
modulo $q$. Hence, for a primitive $q$th root of unity $\zeta$,
\begin{equation}\label{eq:laurent-character}
 f(\zeta v_i)=\zeta^{-a}f(v_i)+O(v_i^{D+q+1}).
\end{equation}
This is a relation between the local branches over $T$; it does
not assert that $v_i\mapsto\zeta v_i$ extends to an automorphism of $C$.

Let $1\le m\le n-1$ with $e\nmid m$. Since $f=O(v_i)$,
raising \eqref{eq:laurent-character} to the $m$th power and summing
over the $q$ local branches gives
\[
 \sum_{j=0}^{q-1}f(\zeta^jv_i)^m=O(v_i^{D+q+m}).
\]
The sum is a series in $v_i^q=T^{-1}$. Consequently
$\operatorname{Tr}_{K/F}(f^m)$ vanishes at infinity to order at least
$\lceil(D+q+m)/q\rceil$.

At a finite point $P$ of ramification index $e_P$, the differential
$\omega$ is regular, whereas $dT$ has order $e_P-1$.
In a local parameter with $T-T(P)=u^{e_P}$, the trace retains only
powers of $u$ divisible by $e_P$. Its pole order at $T(P)$ is
therefore at most $\lfloor m(e_P-1)/e_P\rfloor$.
Thus the total finite pole degree of
$\operatorname{Tr}_{K/F}(f^m)$ is at most $d_m$.
The cutoff hypothesis forces
\[
 \operatorname{Tr}_{K/F}(f^m)=0
 \qquad(1\le m\le n-1,\ e\nmid m).
\]

Since $e\mid n$, Newton's identities now show that
\[
 \operatorname{Norm}_{K/F}(W-f)\in F[W^e].
\]
Let $\mu_f$ be the monic minimal polynomial of $f$ over $F$.
The norm is a power of $\mu_f$, so both polynomials have the same
set of roots. This set is stable under multiplication by the
$e$th roots of unity. Moreover, none of its roots is zero, since
$f\ne0$. Every such orbit therefore has $e$ elements, and taking
the product over these orbits gives $\mu_f\in F[W^e]$.

Put $L=F(f)$ and $M=F(f^e)$. For each $e$th root of unity $\xi$,
the map $f\mapsto\xi f$ defines an $F$-automorphism of $L$ fixing
$M$: the image satisfies the same minimal polynomial and already
belongs to $L$. These are $e$ distinct automorphisms, while
$[L:M]\le e$ because $f$ satisfies $W^e-f^e$ over $M$.
Thus $L/M$ is cyclic of degree $e$.

If the monodromy is primitive, $K/F$ has no proper nontrivial
intermediate field. The identity $\operatorname{Tr}_{K/F}(f)=0$
excludes $f\in F$, so $L=K$. But $2\le e\le q<n$ then gives
$F\subsetneq M\subsetneq K$, a contradiction.

For the final assertion, note that
$d_m\le B(m-1)\le B(n-2)$.
Taking $D=qB(n-2)$ makes every left side in
\eqref{eq:laurent-cutoff} strictly larger than $B(n-2)$.
The finite ramification contribution is at most $B(n-1)$,
and the contribution at infinity is $n-b$.
Riemann--Hurwitz therefore gives
\[
 2g-2\le B(n-1)-n-b<nB(n-2)=Db,
\]
as required.
\end{proof}

Failure does occur for covers with a cyclic intermediate extension.
On $C=\mathbb P^1_z$, put $U=z+z^{-1}$ and $T=U^q$.
Then $n=2q$, $E=(0)+(\infty)$, and $v=U^{-1}$ is a local parameter
at both points of $E$. For $1\le a<q$, the differential
\[
 \omega=d(U^a)=-a v^{-a-1}\,dv
 \in H^0(C,\Omega_C(qE))
\]
and the residue theorem give
\[
 [v_0^a]h+[v_\infty^a]h=0\qquad(h\in L(DE)).
\]
Thus $\mathcal J_{D,a}$ is not surjective for any nonnegative
multiple $D$ of $q$. If $g_0=\gcd(a,q)$, the intermediate extension
$k(U^{g_0})/k(U^q)$ is cyclic of degree $q/g_0$, as predicted.

For $q=2$ and $a=1$, the prescribed powers are
$-D+1,-D+3,\ldots,-1,1$.
The condition \eqref{eq:laurent-cutoff} becomes
\begin{equation}\label{eq:odd-jet-budget}
 Db>2g-2,\qquad (D+m+3)/2>d_m
 \quad(m=1,3,\ldots,n-1).
\end{equation}
In this case the function $f$ constructed in the proof satisfies
\begin{equation}\label{eq:odd-jet-obstruction}
 f(v_i)=O(v_i),\qquad f(v_i)+f(-v_i)=O(v_i^{D+4}).
\end{equation}
We use this case below.  The theorem asserts a necessary field
obstruction to failure of interpolation; the presence of a cyclic
intermediate extension alone does not imply failure.


\subsubsection{Separating branches at infinity}\label{sec:jet-separation}

Write $n=2b\ge4$ and $C_n(Z,S)=C_{n,S}(Z)=H_{n,S}(Z-S)$,
and use $h_n$ from \eqref{geo:hermite-normalization}.
At infinity set $S=v^{-2}$. The roots have convergent expansions
$Z(v,\zeta)=\zeta/v+O(1)$, indexed by the simple roots $\zeta$ of
$h_n$, with $Z(-v,\zeta)=Z(v,-\zeta)$. Let
$\mathcal A^-\subset\mathbb Q[T]/(h_n)$ denote the odd subspace,
of dimension $b$.

We seek polynomials whose values on opposite branches have small
differences at infinity. The following theorem supplies enough such
polynomials to distinguish every selection of one branch from each
pair. Its proof uses traces of odd powers and the generic symmetric
group.

\begin{theorem}\label{thm:quadratic-cutoff-jet-rank}
Let $D=2n^2$. Consider the rational linear map from polynomials
$P(Z,S)$ of $Z$-degree less than $n$ and weighted degree at most $D$,
where $\operatorname{wt}(Z)=1$ and $\operatorname{wt}(S)=2$, to the
coefficients of the odd Laurent powers
\[
 v^{-D+1},v^{-D+3},\ldots,v^{-1},v^1
\]
in $P(Z(v,\zeta),v^{-2})$. Its target is
$(\mathcal A^-)^{D/2+1}$, and the map is surjective.

In particular there exist $b$ rational polynomials $P_0,\ldots,P_{b-1}$
of the stated degrees for which all negative odd coefficients vanish
and whose respective coefficients of $v^1$ are
$\zeta,\zeta^3,\ldots,\zeta^{2b-1}$. Terms independent of $Z$ may be
removed without changing these assertions.
\end{theorem}

\begin{proof}
Let $C$ be the smooth completion of $C_n(Z,S)=0$ and let $E$ be the
sum of its $b=n/2$ points at infinity.  The simple Hermite roots show
that $(S)_\infty=2E$.  The affine curve is smooth: its possible
multiple-root fibers occur at $S=0,-1,\ldots,-(n-2)$, the repeated
root is $X=Z-S=0$, and
$\partial H_{n,S}(0)/\partial S=((S)_n)'\neq0$ there.
Its geometric group is $S_n$. Eisenstein at $S=0$ gives transitivity.
The $(n-1)$-cycle at $S=-1$ lies in the stabilizer of its fixed
root and acts transitively on the other roots. Transitivity of
the full group therefore gives double transitivity. Conjugating
the transposition at $S=-(n-2)$ then gives every transposition,
and hence the group is $S_n$.

There is one finite branch point of each index $N=2,\ldots,n$.
Riemann--Hurwitz gives $g(C)=(b-1)^2$, while the pole budget of
Theorem~\ref{thm:abstract-odd-jets} is
\[
 d_m=\sum_{N=2}^n\left\lfloor\frac{m(N-1)}N\right\rfloor
     \leq(n-1)(m-1).
\]
The choice $D=2n^2$ satisfies both hypotheses
\eqref{eq:odd-jet-budget}.  Apply
Theorem~\ref{thm:abstract-odd-jets} with $q=2$ and $a=1$ to obtain
surjectivity on $L(DE)$.

The affine curve is smooth, so every function regular there belongs
to its coordinate ring. Because $C_n$ is monic in $Z$, such a
function has a unique polynomial representative of $Z$-degree less
than $n$. We claim that this representative lies in $L(DE)$ exactly
when its weighted degree is at most $D$.
One implication follows from the pole orders of $Z$ and $S$.
For the other, suppose its largest weight is $w>D$ and let $p_w(T)$
be the corresponding leading polynomial after substituting
$Z=T/v$, $S=v^{-2}$. Then $\deg p_w<n$, and every exponent of
$p_w$ has the same parity as $w$.
The absence of a pole of order $w$ at each point of $E$ forces
$p_w$ to vanish at one root from each opposite pair of Hermite
roots. Its parity then makes it vanish at all $n$ roots of $h_n$.
This is impossible for a nonzero polynomial of degree less than $n$.
Thus the intrinsic coefficient map is exactly the stated polynomial
map over $\mathbb C$. Its matrix over the monomial basis of
$\mathbb Q[T]/(h_n)$ has rational entries; its full rank over
$\mathbb C$ therefore gives full rank over $\mathbb Q$.
Prescribing the coefficient at $v$ yields the polynomials $P_j$.
Terms independent of $Z$ contribute only even Laurent powers and
may be removed.
\end{proof}


\subsubsection{Bounding the imaginary quadratic obstruction}\label{sec:imaginary-height}

We need quantitative bounds for this construction. Bounds for its denominators
and coefficients gives a uniform upper bound for a parameter admitting
an imaginary quadratic block field.

We begin with a denominator estimate for the root expansions. The
Hermite differential equation makes the estimate independent of a
discriminant denominator.

\begin{lemma}\label{lem:factorial-root-denominators}
Let $F_n$ and $h_n$ be as in \eqref{geo:hermite-scaling}, and put
$\mathcal A=\mathbb Z[T]/(h_n)$.
The formal branch
\[
 T(v)=T+\sum_{k\ge1}a_kv^k\in
                   (\mathcal A\otimes\mathbb Q)[[v]],
 \qquad F_n(T(v),v)=0,
\]
satisfies $(3k)!a_k\in\mathcal A$ for every $k\ge1$.
Moreover, $(3k)![v^k]T(v)^j\in\mathcal A$ for every $j\ge0$.
\end{lemma}

\begin{proof}
All quotients in this proof are first interpreted in
$\mathbb Q[T]/(h_n)$; the simplicity of the Hermite roots makes
$h_n'$ invertible there. The specific quotients $h_n^{(r)}/h_n'$
have integral representatives. Indeed, their representatives $A_r$
can be defined by
\[
 A_0=0,\qquad A_1=1,\qquad
 A_{r+2}=-T A_{r+1}+(n-r)A_r,
\]
using the Hermite differential equation. Thus these quotients do not
introduce the denominators of a general inverse of $h_n'$. Write
$F_n(T,v)=\sum_{l\ge0}F_l(T)v^l$. Expansion of the exponential gives
\[
 F_l^{(j)}(T)=
 \sum_{\substack{r_3,r_4,\ldots\ge0\\
                  \sum_{m\ge3}(m-2)r_m=l}}
 \frac{h_n^{(M+j)}(T)}{\prod_{m\ge3}m^{r_m}r_m!},
 \qquad M=\sum_{m\ge3}m r_m\le3l.
\]
The denominator in each summand divides $M!$, by the formula counting
permutations of a specified cycle type, and hence divides $(3l)!$.
Consequently
\[
 (3l)!\,F_l^{(j)}/h_n'\in\mathcal A.
\]
This includes $l=0$, using the derivative quotients above.

We prove the denominator bound by induction on $k$. Write
$A(v)=\sum_{i\ge1}a_i v^i$ and expand
$F_n(T+A(v),v)=0$ in the ring
$\mathbb Q[T]/(h_n)[[v]]$.
After division by $h_n'$, the only term involving $a_k$ is $a_k$
itself. Every other contribution to the coefficient of $v^k$ has
the form
\[
 \frac{F_l^{(j)}(T)}{h_n'(T)}
 \prod_{i\ge1}\frac{a_i^{r_i}}{r_i!},
 \qquad j=\sum_i r_i,\qquad
 l+\sum_i i r_i=k,
\]
and uses only coefficients already covered by the induction.
Its denominator divides
\[
 (3l)!\prod_{i\ge1}((3i)!)^{r_i}\prod_{i\ge1}r_i!.
\]
This integer divides $(3k)!$: the quotient counts partitions of a
labeled set of size $3k$ into one distinguished block of size $3l$
and $r_i$ indistinguishable blocks of size $3i$ for each $i$.
Hence $(3k)!a_k\in\mathcal A$.
The same multinomial argument applied to $(T+A(v))^j$ proves
$(3k)![v^k]T(v)^j\in\mathcal A$ for every $j\ge0$.
\end{proof}

The following analytic estimates give a common disk for all root branches.

\begin{lemma}\label{lem:analytic-hermite-branches}
Let $n\ge4$, $F(t,v)=F_n(t,v)$, and $h=h_n$.
The roots $\zeta_i$ of $h$ are simple and purely imaginary, satisfy
$|\zeta_i|\le2\sqrt n$, and are separated by more than
$1/(6\sqrt n)$. On a neighborhood of $|v|\le r=(1000n^2)^{-1}$,
they extend to holomorphic roots $t_i(v)$ of $F$ with
$|t_i(v)-\zeta_i|\le2n|v|$ and $|t_i(v)|<3\sqrt n$.
\end{lemma}
\begin{proof}
The roots
$\zeta_1,\ldots,\zeta_n$ of $h$ are simple and purely imaginary,
with $|\zeta_i|\le2\sqrt n$.  One way to see these facts is to use
the skew-symmetric tridiagonal matrix whose adjacent entries are
$\sqrt j$ and $-\sqrt j$, for $1\le j<n$; its characteristic
polynomial is $h$.  Multiplication by $i$, followed by a diagonal
unitary similarity, gives an irreducible real symmetric tridiagonal
matrix.  Its eigenvalues are simple, and its norm is at most
$2\sqrt n$.

The equation $h''+t h'-nh=0$ also gives
\begin{equation}\label{eqn:hermite-spacing}
 |\zeta_i-\zeta_j|>\frac1{6\sqrt n}\quad(i\ne j).
\end{equation}
Indeed, at a root put $A_k=h^{(k)}(\zeta)/h'(\zeta)$.  The recurrence
$A_{k+2}=-\zeta A_{k+1}+(n-k)A_k$, with $A_0=0,A_1=1$, gives
$|A_k|\le(3\sqrt n)^{k-1}$. For $0<|z|\le1/(6\sqrt n)$,
\[
 \left|\frac{h(\zeta+z)}{h'(\zeta)z}-1\right|
 \le\sum_{k=2}^n\frac{2^{1-k}}{k!}
 <2(e^{1/2}-3/2)<1,
\]
so the Taylor quotient cannot vanish.

Set $r=(1000n^2)^{-1}$.  The polynomial $F(t,v)$ is the
characteristic polynomial of the analytic tridiagonal matrix with
diagonal $v(2j-n-1)$ and adjacent entries
\[
 \sqrt{j}\sqrt{1+(n-j)v^2},\qquad
 -\sqrt{j}\sqrt{1+(n-j)v^2}.
\]
Here the square roots are the branches equal to $1$ at $v=0$.
For $|v|\le r$, its distance in operator norm from the matrix at
$v=0$ is at most $2n|v|$.  Applying the resolvent identity on the
circles of radius $1/(18\sqrt n)$ about the $\zeta_i$ shows that
each circle contains exactly one eigenvalue.  Consequently there
are holomorphic branches $t_i(v)$ on $|v|\le r$ satisfying
\[
 t_i(0)=\zeta_i,\qquad |t_i(v)|<3\sqrt n.
\]
The assertions remain valid on an open neighborhood of this closed
disk, since all the preceding norm inequalities are strict.

The bound $|t_i(v)-\zeta_i|\le2n|v|$ also follows from the
resolvent bound for the normal matrix at $v=0$: every perturbed
eigenvalue lies within the perturbation norm of its spectrum, and
the disjoint disks identify the corresponding root.
\end{proof}

\paragraph{Eliminating the odd monomials}

The large jet system has a triangular part which can be removed
before taking determinants. This reduces the number of rows from
order $n^3$ to order $n^2$ without changing the proved cutoff.

Put $n=2b$, let $D=2m\ge n$, and write $u=v^2$. Let
$\mathcal A=\mathbb Q[T]/(h_n)$. Define the $\mathbb Q$-linear map
$\mathcal T(v):\mathcal A\to\mathcal A[[v]]$ on the monomial
basis by
\[
 \mathcal T(v)(T^j)=Y(v)^j\pmod{h_n},\qquad 0\le j<n,
\]
where $Y(v)=vZ(v,T)$ is the Hermite-limit root series. Split
$\mathcal A=\mathcal A^+\oplus\mathcal A^-$ into even and odd
polynomials, with bases $1,T^2,\ldots,T^{2b-2}$ and
$T,T^3,\ldots,T^{2b-1}$. Parity gives block series
\[
 \mathcal T_{--}(v)=\mathcal O(u),\qquad
 \mathcal T_{-+}(v)=v\mathcal B(u),\qquad \mathcal O(0)=I.
\]
Define
\begin{equation}\label{eq:triangular-R}
 R(u)=u\mathcal O(u)^{-1}\mathcal B(u)
     =v\mathcal O(v^2)^{-1}\mathcal T_{-+}(v).
\end{equation}
Its column corresponding to the constant polynomial is zero.

\begin{lemma}\label{lem:triangular-odd-jets}
Consider a polynomial $P$ with $\deg_ZP<n$ and weighted degree at
most $2m$, with no term independent of $Z$. Write its even and odd
monomials as Laurent-polynomial vectors
\[
 A_i(u)=p_{2i}(u^{-1})u^{-i}\quad(1\le i<b),\qquad A_0=0,
\]
and
\[
 B_i^{\rm in}(u)=p_{2i+1}(u^{-1})u^{-i}\quad(0\le i<b).
\]
Thus $A_i$ has powers $u^{-m},\ldots,u^{-i}$, whereas
$B_i^{\rm in}$ has powers $u^{-(m-1)},\ldots,u^{-i}$.
Fix a vector $y\in\mathbb Q^b$ in the odd monomial basis.
There exists an odd part $B^{\rm in}$ for which the negative odd
Laurent coefficients of $P(Z(v,T),v^{-2})$ vanish and its coefficient
at $v^1$ is $y$ if and only if
\begin{equation}\label{eq:triangular-residual}
 [u^k](R(u)A(u))_i=0\quad(1-i\le k\le0),
 \qquad [u^1](R(u)A(u))_i=y_i,
 \quad 0\le i<b.
\end{equation}
When it exists, the odd part is unique and is obtained by taking the
negative of the terms of $(RA)_i$ with exponents
$-(m-1),\ldots,-i$. The reduced system \eqref{eq:triangular-residual} has exactly
$b(b+1)/2$ rows.

For $D=2n^2$, this reduced map is surjective.
\end{lemma}

\begin{proof}
The odd part, in both the Laurent parameter and the Hermite variable,
of the evaluated polynomial is
\[
 v\bigl(\mathcal B(u)A(u)+u^{-1}\mathcal O(u)B^{\rm in}(u)\bigr)
 =v^{-1}\mathcal O(u)\bigl(B^{\rm in}(u)+R(u)A(u)\bigr).
\]
It equals $vy+O(v^3)$ exactly when
\[
 B^{\rm in}(u)+R(u)A(u)=uy+O(u^2),
\]
since $\mathcal O(0)=I$. The series $R$ starts at order $u$, so $RA$ has
no power below $u^{-(m-1)}$. The allowed coefficients of
$B_i^{\rm in}$ cancel precisely the powers through $u^{-i}$;
the remaining conditions are \eqref{eq:triangular-residual}. There are $i+1$ conditions in
component $i$, giving $b(b+1)/2$ in all.

For $D=2n^2$, the full odd-jet map of
Theorem~\ref{thm:quadratic-cutoff-jet-rank} is surjective.
Multiplication by $\mathcal O(u)^{-1}$ is an invertible triangular change
of jet coordinates. The odd-monomial columns occupy exactly the
coefficients canceled above, with unit pivots. Passing to the
quotient by these columns gives the reduced map
\eqref{eq:triangular-residual}, which is therefore surjective.
\end{proof}

The inverse matrix in \eqref{eq:triangular-R} has a uniform analytic disk. Working in a
Hermite quadrature norm avoids a loss caused by the condition number
of a monomial Vandermonde matrix.

\begin{lemma}\label{lem:quadrature-composition-bound}
Put $r=(1000n^2)^{-1}$. The matrix $R(u)$ is holomorphic on a
neighborhood of $|u|\le r^2$. In the norm induced by Gaussian
Hermite quadrature on the even and odd polynomial spaces,
\[
 \|\mathcal O(v^2)^{-1}\|\le2\qquad(|v|\le r).
\]
In the monomial bases, its entries also satisfy
\[
 |R_{ij}(u)|\le\exp(6n\log n)\qquad(|u|\le r^2).
\]
Moreover every coefficient of $u^k$ in $R(u)$, in the monomial
bases above, has denominator dividing $(6k-3)!$ for $k\ge1$.
\end{lemma}

\begin{proof}
Write the Hermite roots as $\zeta_i=i\eta_i$ with $\eta_i\in\mathbb R$.
On polynomials of degree below $n$, use the norm
\[
 \|q\|^2=\frac1{\sqrt{2\pi}}
  \int_{\mathbb R}|q(ix)|^2e^{-x^2/2}\,dx.
\]
Gaussian quadrature at the $n$ Hermite nodes identifies this norm
with a weighted Euclidean norm on the values $q(\zeta_i)$, with
positive, symmetric weights. This elementary quadrature identity
follows by division by the monic Hermite polynomial and its
orthogonality; the weights are positive because they are the
integrals of the squares of the cardinal polynomials.

The polynomials $h_k(T)/\sqrt{k!}$ are orthonormal in this norm.
Their derivative relation shows that differentiation on
degree-below-$n$ polynomials has norm at most $\sqrt n$, and its
$k$th power has norm at most $n^{k/2}$.

Lemma~\ref{lem:analytic-hermite-branches} gives root branches
$t_i(v)$ on $|v|\le r$ with
\[
 \delta_i(v)=t_i(v)-\zeta_i,
 \qquad \max_i|\delta_i(v)|\le2n|v|.
\]
Let $\mathcal D$ denote differentiation in node coordinates.
Taylor expansion of each polynomial, which is finite here,
therefore gives the operator identity on its node values
\[
 \mathcal T(v)=I+
 \sum_{k=1}^{n-1}\frac{\operatorname{diag}(\delta_i(v)^k)\mathcal D^k}{k!}.
\]
It follows that
\[
 \|\mathcal T(v)-I\|
 \le\exp(2n^{3/2}|v|)-1<\tfrac12.
\]
The projection onto odd polynomials is orthogonal because the
quadrature weights are symmetric. Hence the same bound holds for
$\mathcal O(v^2)-I$, and the geometric-series inverse has norm at most two.
Equation~\eqref{eq:triangular-R} proves the stated holomorphy of $R$.
The same estimates give
\[
 \|R(u)\|\le6n^{3/2}|u|\qquad(|u|\le r^2).
\]
To pass to monomial coordinates, each input monomial has norm at
most $(2n)^{n/2}$.  Each coefficient of a normalized Hermite
polynomial $h_k/\sqrt{k!}$ has absolute value at most $\sqrt{n!}$;
therefore each monomial coefficient of an output polynomial is at
most $n\sqrt{n!}$ times its norm.  These estimates imply
the displayed bound $\exp(6n\log n)$ for all $n\ge4$.

By Lemma~\ref{lem:factorial-root-denominators}, every coefficient
of $v^k$ in $\mathcal T(v)$ becomes integral after multiplication
by $(3k)!$. The substitution $v=w^3$ turns this into integrality
of the exponential-generating coefficients. If
$U(w)=\sum_{j\ge0}U_jw^j/j!$ has integral matrix coefficients
and $U_0=I$, the coefficients of its inverse satisfy
\[
 V_0=I,\qquad
 V_j=-\sum_{i=1}^j\binom ji U_iV_{j-i}.
\]
They are therefore integral. Products have the same property by
the binomial convolution formula. Applying these observations to
the odd diagonal block gives the asserted integrality for its
inverse and for its product with $\mathcal T_{-+}$.
The coefficient of $u^k=v^{2k}$ in \eqref{eq:triangular-R} is the coefficient of
$v^{2k-1}$ in $\mathcal O(v^2)^{-1}\mathcal T_{-+}(v)$, proving the
denominator bound $(6k-3)!$.
\end{proof}

\paragraph{The parameter bound}

We construct integral polynomials with small differences on paired
branches and apply them to the two blocks.

\begin{proposition}\label{eqn:imaginary-upper-bound}
Let $n=2b\ge4$ and $s\in\mathbb Z_{>0}$.  If the Galois group of
$H_{n,s}$ preserves two blocks of size $b$, and its action on these
blocks defines an imaginary quadratic field, then
\[
 \log s<1000n^4\log n.
\]
\end{proposition}

\begin{proof}
Write $C(Z,S)=C_{n,S}(Z)$, put $D=2n^2$, and use
$F=F_n$, $h=h_n$ from \eqref{geo:hermite-scaling}.
The generating function
\[
 \sum_{m\ge0}C_m(Z,S)\frac{w^m}{m!}
 =\exp\left(Zw+S\sum_{j\ge2}\frac{w^j}{j}\right)
\]
shows that $C$ has nonnegative integer coefficients, weighted degree
$n$ for weights $1,2$, and coefficient sum $C(1,1)=n!$.
In particular $F\in\mathbb Z[t,v]$.

We use the branches and radius $r$ from
Lemma~\ref{lem:analytic-hermite-branches}.

We now apply the triangular elimination in
Lemma~\ref{lem:triangular-odd-jets}, with $m=n^2$, $D=2m$,
and $u=v^2$.  Write $\mathcal T(v)$ for the substitution matrix
$t^j\mapsto t(v)^j$ in $\mathbb Q[t]/(h)$, split into its even
and odd parts.  In the notation of that lemma,
\[
 \mathcal O(u)=\mathcal T_{--}(v),\qquad
 \mathcal R(u)=v\mathcal O(u)^{-1}\mathcal T_{-+}(v).
\]
The latter is a power series in $u$ with zero constant term.
Lemma~\ref{lem:quadrature-composition-bound} shows that its
entries, in monomial coordinates, satisfy
\[
 |\mathcal R_{ij}(u)|\le\exp(6n\log n)
 \qquad(|u|=r^2).
\]
Consequently, all coefficients through degree $m+1$ have absolute
value at most
\begin{equation}\label{eqn:reduced-raw-height}
 \exp(20n^2\log n).
\end{equation}
Indeed $\log r^{-1}\le7\log n$, so Cauchy's inequality bounds
their logarithms by
$6n\log n+14(n^2+1)\log n\le20n^2\log n$.

Lemma~\ref{lem:quadrature-composition-bound} gives the
denominator bound $(6k-3)!$ for the coefficient of $u^k$ in
$\mathcal R$. Hence the integer
\[
 q=(6m+3)!
\]
clears every coefficient used below and satisfies
\begin{equation}\label{eqn:factorial-denominator-bound}
 \log q\le36n^2\log n.
\end{equation}

The reduced map in Lemma~\ref{lem:triangular-odd-jets} has
\[
 R=b(b+1)/2\le n^2
\]
rows.  It sends the coefficients of the even monomials $Z^{2i}S^k$
of weighted degree at most $D$ to the residual coefficients
$[u^l](\mathcal R E)_i$ with $1-i\le l\le1$. Here
$E_i(u)=u^{-i}p_{2i}(u^{-1})$ and $E_0=0$; this is the vector
denoted by $A$ in that lemma.
It has full row rank, by Theorem~\ref{thm:quadratic-cutoff-jet-rank}
and the triangular elimination.  Let $J$ be its rational matrix.
Equations \eqref{eqn:reduced-raw-height} and
\eqref{eqn:factorial-denominator-bound} give
\begin{equation}\label{eqn:jet-height}
 \max|(qJ)_{ij}|\le\exp(60n^2\log n).
\end{equation}
Choose $R$ columns of $qJ$ forming an invertible matrix $B$.
For $1\le k\le b$, let $r_k$ be the row corresponding to the
coefficient at $u^1$ in the $k$th odd component.
Let $E^{(k)}$ be the even coefficient vector with entries
\[
 q\operatorname{adj}(B)e_{r_k}
\]
on these selected columns and zero on all other columns.
Its entries are integers divisible by $q$, and multiplication by
$J$ gives
\[
 JE^{(k)}=q^{-1}B\bigl(q\operatorname{adj}(B)e_{r_k}\bigr)
 =\det(B)e_{r_k}.
\]
Thus every residual coefficient vanishes except the prescribed one,
whose value is the same nonzero integer $c=\det(B)$ for every $k$.
Hadamard's inequality and \eqref{eqn:jet-height} bound the even
coefficients by $\exp(70n^4\log n)$.

The odd coefficients are obtained by canceling the allowed Laurent
polynomial part of $\mathcal R E$, as in
Lemma~\ref{lem:triangular-odd-jets}.
This step introduces no denominators: each even coefficient is
divisible by $q$, and $q$ clears every coefficient of $\mathcal R$
used in the reconstruction. Indeed, the largest possible index is
$m+1$, already covered by the definition $q=(6m+3)!$.
The resulting odd terms also have the required degrees, since a
power $u^{-j}$ in the $i$th odd component corresponds to
$Z^{2i+1}S^{j-i}$, with $i\le j\le m-1$.

Each reconstructed coefficient is a sum of at most $2n^3$ products
of an even coefficient with an entry bounded by
\eqref{eqn:reduced-raw-height}.
We have therefore constructed integer polynomials
$P_1,\ldots,P_b$, of $Z$-degree less than $n$ and weighted degree
at most $D$, such that every negative odd Laurent coefficient
vanishes and
\begin{equation}\label{eqn:prescribed-jets}
 [v^1]P_k(t(v)/v,v^{-2})=c\,t^{2k-1},
 \qquad c=\det B\in\mathbb Z\setminus\{0\}.
\end{equation}
Their coefficients satisfy
\begin{equation}\label{eqn:polynomial-height}
 H:=\max_{k,j,l}|[Z^jS^l]P_k|
 \le\exp(80n^4\log n).
\end{equation}

Let $\rho=r/2$ and put
\[
 M=2n^3H(3\sqrt n)^n\rho^{-D}.
\]
Then
\begin{equation}\label{eqn:M-bound}
 M\le\exp(90n^4\log n).
\end{equation}
For a branch with initial root $\zeta$, write
\[
 G_{k,\zeta}(v)=\frac12\left(
 P_k(t_\zeta(v)/v,v^{-2})-
 P_k(t_\zeta(-v)/(-v),v^{-2})\right).
\]
The parity of the equation shows that this is the odd Laurent part.
Its negative powers have been canceled, so it is holomorphic at
$0$.  On $|v|=\rho$ its absolute value is at most $M$.
Cauchy's inequality and \eqref{eqn:prescribed-jets} therefore give,
for $0<v\le\rho/2$,
\begin{align}
 |G_{k,\zeta}(v)|&\le2Mv/\rho,\label{eqn:odd-value}\\
 |G_{k,\zeta}(v)-c\zeta^{2k-1}v|
 &\le2Mv^3/\rho^3.\label{eqn:odd-error}
\end{align}

Suppose now that the two-block field is imaginary quadratic, and
assume, for a contradiction, that
$T=\log s\ge1000n^4\log n$. Then $v=s^{-1/2}<\rho/2$.
Write the positive imaginary Hermite roots as
$i\lambda_1,\ldots,i\lambda_b$, where
$0<\lambda_1<\cdots<\lambda_b$.
At this specialization, complex conjugation pairs the branches
starting at $i\lambda_j$ and $-i\lambda_j$ and exchanges the two blocks.
Each block therefore contains exactly one root from each pair.
There are signs $\epsilon_j\in\{1,-1\}$ such that the
difference between the two block traces of $P_k$ is
\[
 \Delta_k=2\sum_{j=1}^b\epsilon_jG_{k,i\lambda_j}(v).
\]
The sum of $P_k$ over either block is fixed by the kernel of the
action on the two blocks, so it belongs to the quadratic block
field. It is an algebraic integer because $P_k$ has integer
coefficients and the centered roots are algebraic integers.
Complex conjugation exchanges the blocks, hence
$\overline{\Delta_k}=-\Delta_k$.
If $\Delta_k\ne0$, its norm is
\[
 N_{K/\mathbb Q}(\Delta_k)
 =-\Delta_k^2=|\Delta_k|^2\in\mathbb Z_{>0},
\]
and therefore $|\Delta_k|\ge1$.
This is the arithmetic reason for constructing the polynomials
with integer, rather than merely rational, coefficients.

Equations
\eqref{eqn:M-bound} and \eqref{eqn:odd-value} imply
$|\Delta_k|\le4bMv/\rho<1$.  Thus $\Delta_k=0$ for every $k$.
Using \eqref{eqn:odd-error} and $|c|\ge1$ gives
\begin{equation}\label{eqn:signed-moments-small}
 \left|\sum_{j=1}^b\epsilon_j(i\lambda_j)^{2k-1}\right|
 \le2bMv^2/\rho^3\qquad(1\le k\le b).
\end{equation}

These inequalities contradict an elementary Vandermonde bound.
Indeed, \eqref{eqn:hermite-spacing} implies
\[
 \lambda_j>\frac1{12\sqrt n},\qquad
 |\lambda_i^2-\lambda_j^2|>\frac1{36n}\quad(i\ne j),
 \qquad \lambda_j^2\le4n.
\]
For the matrix $V_{kj}=\lambda_j^{2k-1}$, Lagrange interpolation
at the points $\lambda_j^2$ therefore gives
\[
 \|V^{-1}\|_\infty
 \le12b\sqrt n(180n^2)^{b-1}.
\]
Since $\|\epsilon\|_\infty=1$, at least one signed moment on the
left of \eqref{eqn:signed-moments-small} has absolute value at least
\[
 \eta=\frac1{12b\sqrt n(180n^2)^{b-1}}.
\]
But $\log\eta^{-1}\le5n\log n$, while
\eqref{eqn:M-bound}, $\rho^{-1}=2000n^2$, and the assumed lower
bound on $T$ give $2bMv^2/\rho^3<\eta$.  This contradiction proves
the proposition.
\end{proof}


\subsection{Completion of the uniform Galois theorem}\label{sec:final-galois}

The preceding bounds reduce a possible failure of alternating
containment to two blocks with $\sqrt n<\log s<n^5$.
The nearby-node criterion excludes this last range. We then use
the square-discriminant classification to determine the group.

\begin{theorem}\label{thm:conditional-alternating}
There is an effectively computable absolute integer $N_{\rm Gal}$
such that, for every $n\ge N_{\rm Gal}$ and every positive integer
$s$, irreducibility of $H_{n,s}$ implies $A_n\le G_{n,s}$.
\end{theorem}
\begin{proof}
Choose an effective absolute threshold for the results used below,
including Corollary~\ref{fea:polynomial-height-exclusion} with
$\kappa=6$ and $D=5$. Enlarge it so that
\[
 \max\{C_1(\log n)^9,(\log n)^6\}<\sqrt n,\qquad
 2n\log n<1000n^4\log n<n^5,\qquad
 n^{101/200}\log n-\log2>\sqrt n,
\]
where $C_1$ is the constant in
Proposition~\ref{prop:block-parameter-cap}.

Suppose $H_{n,s}$ is irreducible and $G_{n,s}$ omits $A_n$.
Proposition~\ref{prop:cubic-range} gives
$s>n^3/(10\log n)>n$, and Theorem~\ref{thm:primitive} shows
that the root action is imprimitive. Choose a partition into
$a$ blocks of size $b$, with $ab=n$ and $a,b\ge2$.
Proposition~\ref{be:subpower-range} gives
\[
 \log(s+n)\ge n^{101/200}\log n,
\]
so $L=\log s>\sqrt n$.
Lemma~\ref{prop:pair-blocks} excludes $b=2$.
If $a\ge3$ as well, Proposition~\ref{prop:block-parameter-cap}
gives $L\le C_1(\log n)^9<\sqrt n$, a contradiction.
Hence $a=2$.

The corresponding block field is quadratic. If it is real,
Lemma~\ref{be:real-quadratic-bound} gives $L<2n\log n$;
if it is imaginary, Proposition~\ref{eqn:imaginary-upper-bound}
gives $L<1000n^4\log n$. Thus in either case
\[
 n^{1/2}<L<n^5,
\]
contradicting Corollary~\ref{fea:polynomial-height-exclusion}.
Every threshold used here is effective and independent of $s$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:uniform-combined}]
Take $N=\max\{N_0,N_{\rm Gal}\}$, with $N_0$ from
Proposition~\ref{prop:qualitative-uniform-irreducibility} and
$N_{\rm Gal}$ from Theorem~\ref{thm:conditional-alternating}.
For $n\ge N$, the first result gives irreducibility for every
positive integer $s$, and the second gives alternating containment.
Both constants are effective and independent of $s$.
\end{proof}

\subsubsection{An explicit description of the symmetric specializations}
\label{sec:explicit-symmetric-specializations}

The square-discriminant classification of
Banerjee--Filaseta--Finch--Leidy \cite[Theorem~2.1]{BFFL2013}
turns uniform alternating containment into a short description of every
Galois group in sufficiently large degree.  The classification of
alternating Laguerre specializations was further developed by
Banerjee--Bera \cite{BanerjeeBera2022}.  The square-discriminant families
used below are those of \cite{BFFL2013}; the uniform theorem determines
the symmetric group throughout their complement in the region
$\alpha=-n-s$, $s>0$.

For $k\geq0$, define the integers
\[
 a_k=\frac{(2+\sqrt3)^k+(2-\sqrt3)^k}{2},\quad
 b_k=\frac{(4+\sqrt{15})^k+(4-\sqrt{15})^k}{2},\quad
 c_k=\frac{(1+\sqrt2)^k+(1-\sqrt2)^k}{2}.
\]

\begin{corollary}
\label{cor:ten-parameter-lines}
There is an effectively computable absolute integer $N_1$ such that,
for every $n\geq N_1$ and every positive integer $s$, one has
$G_{n,s}=A_n$ exactly in the cases in
Table~\ref{tab:negative-alternating-cases}.  In all other cases
$G_{n,s}=S_n$.  In particular,
\[
 s\notin\{1,2,3,4,5,6,n-4,n,n+2,n+4\}
 \quad\Longrightarrow\quad G_{n,s}=S_n.
\]
\end{corollary}

\begin{table}[htbp]
\centering
\renewcommand{\arraystretch}{1.16}
\begin{tabular}{c|l}
$s$ & condition for $G_{n,s}=A_n$ \\
\hline
$1$ & $n\equiv0\pmod4$ \\
$2$ & $n\equiv1\pmod4$ \\
$3$ & $n\equiv0\pmod4$ and $n+1$ is a square \\
$4$ & $n\equiv1\pmod{24}$ and $(n+2)/3$ is a square \\
$5$ & $n+2=a_{2k+1}$ for some $k\geq1$ \\
$6$ & $n+3=b_{2k+1}$ for some $k\geq1$ \\
$n-4$ & $(2n-5)/3=c_{4k}$ for some $k\geq1$ \\
$n$ & $n$ is an odd square \\
$n+2$ & $n\equiv0\pmod4$, or $(n+1)/2$ is an odd square \\
$n+4$ & $n+1=c_{2k+1}^{\,2}$ for some $k\geq1$, or $n+3$ is an even square
\end{tabular}
\par\smallskip
\caption{The negative-parameter part of the square-discriminant
classification in \cite[Section~2, items (v)--(vii) and
(xx)--(xxx)]{BFFL2013}, after imposing $\alpha=-n-s$ and taking
$n\geq N_1$.}
\label{tab:negative-alternating-cases}
\end{table}

\begin{proof}
The cited classification gives thirty explicit families and an
effectively computable finite exceptional set $\mathcal A_0$.
Take $N_1$ larger than $10$, the constant in
Theorem~\ref{thm:uniform-combined}, and every degree occurring in
$\mathcal A_0$.  Under $\alpha=-n-s<-n$, the families of unbounded
degree that remain are precisely items (v)--(vii), (xx)--(xxv), and
(xxviii)--(xxx) of \cite[Section~2]{BFFL2013}.  Substitution gives
the ten rows of the table.  The discriminant is nonzero for $s>0$,
and Theorem~\ref{thm:uniform-combined} gives $A_n\subseteq G_{n,s}$.
Hence a square discriminant gives $A_n$, and every nonsquare
discriminant gives $S_n$.
\end{proof}


\section{Effective exceptional specializations}
\label{sec:general-infinity}

Runge's theorem gives effective specialization results when an
intermediate cover has several closed points above infinity.  Their
number is the number of orbits of the arithmetic decomposition group
on the corresponding coset space.  We use this observation to turn the
cycle type of geometric inertia into an effective criterion.

\begin{proposition}
\label{prop:decomposition-runge}
Let $K/\mathbb Q(T)$ be a regular finite Galois extension with group $G$,
and let $D_\infty\le G$ be the stabilizer of a place above
$T=\infty$ in the arithmetic Galois group. Let $S_f$ be a finite
set of rational primes, and let
$\mathcal H$ be a finite collection of proper subgroups of $G$.
Suppose that
\[
 \#(D_\infty\backslash G/H)>|S_f|+1
 \qquad(H\in\mathcal H).
\]
Then the parameters $t\in\mathbb Z[S_f^{-1}]$ outside the branch locus
whose specialization group is contained in a conjugate of a member of
$\mathcal H$ form an effectively computable finite set.

A sufficient condition is
$[G:H]>|D_\infty|(|S_f|+1)$ for every $H\in\mathcal H$.
For integral parameters, corresponding to $S_f=\varnothing$, it also
suffices that for each $H\in\mathcal H$ some prime divides $[G:H]$
without dividing $|D_\infty|$.
\end{proposition}

\begin{proof}
Let $X_H$ be the smooth projective curve with function field $K^H$.
Its closed points above infinity correspond to the double cosets
$D_\infty\backslash G/H$.  Thus the support of the pole divisor of
$T$ has more than $|S_f|+1$ irreducible components over $\mathbb Q$.
Apply Runge's theorem in the form of \cite[Theorem~1.2]{Levin}, with
$S=S_f\cup\{\infty\}$, to obtain an effective finite set
\[
 \{P\in X_H(\mathbb Q):T(P)\in\mathbb Z[S_f^{-1}]\}.
\]
At an unramified rational parameter, the specialization group is
contained in a conjugate of $H$ precisely when it fixes a point of
$G/H$, or equivalently when $X_H$ has a rational point above that
parameter.  Taking the union over $\mathcal H$ proves the assertion.

Every $D_\infty$-orbit has at most $|D_\infty|$ elements, proving
the numerical sufficient condition.  When $S_f=\varnothing$, the
hypothesis is intransitivity.  A transitive action would imply
$[G:H]\mid |D_\infty|$, which proves the final test.
\end{proof}

\begin{lemma}
\label{lem:cyclic-normalizer}
Let $\sigma\in S_n$ have $a_d$ cycles of length $d$, including
$a_1$ fixed points, and put $m=\operatorname{ord}(\sigma)$.  Then
\[
 |N_{S_n}(\langle\sigma\rangle)|
 =\varphi(m)\prod_{d\geq1} d^{a_d}a_d!.
\]
If $p$ is an odd prime with $n/2<p<n$, then
\[
 p\nmid |N_{S_n}(\langle\sigma\rangle)|
 \quad\Longleftrightarrow\quad a_p=0\ \text{and}\ a_1<p.
\]
\end{lemma}

\begin{proof}
The centralizer has order $\prod_d d^{a_d}a_d!$.  Every automorphism
$\sigma\mapsto\sigma^u$ of $\langle\sigma\rangle$ is realized by
simultaneously multiplying the cyclic coordinates in each $d$-cycle by
$u\bmod d$.  The quotient of the normalizer by the centralizer therefore
has order $\varphi(m)$.

We claim that $p\nmid\varphi(m)$.  Otherwise either $p^2\mid m$, or
some prime $q\mid m$ satisfies $p\mid q-1$.  The first possibility
requires a cycle of length at least $p^2>n$.  In the second, $p$ is odd,
so $q\geq2p+1>n$, whereas every prime divisor of $m$ divides a cycle
length at most $n$.  Both are impossible.  Also $p\mid d$ for a cycle
length $d\leq n$ precisely when $d=p$.  For $d\geq2$ one has
$a_d\leq n/2<p$, so the only factorial that can contribute $p$ is
$a_1!$.  This proves the equivalence.
\end{proof}

\begin{lemma}
\label{lem:jordan-index}
Let $n/2<p\leq n-3$ be prime.  If $H\leq S_n$ is transitive and
$A_n\nleq H$, then $p\nmid |H|$ and $p\mid[S_n:H]$.
\end{lemma}

\begin{proof}
If $p\mid |H|$, Cauchy's theorem supplies an element of order $p$.
Since $2p>n$, this element is a single $p$-cycle.  Any nontrivial block
system has $a$ blocks of size $b$, with $a,b\leq n/2<p$.
An element of order $p$ must fix every block and then every point in
each block, a contradiction.  Thus $H$ is primitive.  Jordan's
prime-cycle theorem \cite[Theorem~3.3E]{DM}, with $p\leq n-3$, now
gives $A_n\leq H$.
\end{proof}

\begin{theorem}
\label{thm:infinity-cycle}
Let $F(X,T)\in\mathbb Q(T)[X]$ have degree $n\geq8$ in $X$ and
geometric Galois group $S_n$.  Let $\sigma\in S_n$ generate inertia
at infinity in its splitting field.  Suppose that there is a prime
\[
 n/2<p\leq n-3
\]
such that $\sigma$ has no $p$-cycle and has fewer than $p$ fixed points.
Then the integers $t$ for which $F(X,t)$ is defined, has degree $n$,
is irreducible over $\mathbb Q$, and has Galois group not containing
$A_n$ form an effectively computable finite set.
\end{theorem}

\begin{proof}
The arithmetic group is also $S_n$ and the splitting field is regular.
In characteristic zero the inertia group at infinity is cyclic and is
normal in the arithmetic decomposition group.  Consequently
\[
 D_\infty\leq N_{S_n}(\langle\sigma\rangle).
\]
Lemma~\ref{lem:cyclic-normalizer} shows that $p\nmid |D_\infty|$.
By Lemma~\ref{lem:jordan-index}, $p$ divides the index of every
transitive subgroup omitting $A_n$.  Apply
Proposition~\ref{prop:decomposition-runge} to representatives of their
finitely many conjugacy classes.  The finitely many bad parameters
coming from coefficient poles, degree drops, and branch points can be
handled directly.
\end{proof}

\subsection{Effective exceptional sets for the Laguerre family}
\label{sec:runge-imprimitive}

\begin{corollary}\label{thm:runge-alternating}
For every fixed $n\ge5$, the positive integers $s$ for which
$H_{n,s}$ is irreducible and $A_n\not\le G_{n,s}$ form a finite,
effectively computable set.
\end{corollary}
\begin{proof}
The generic splitting field is a regular $S_n$-extension of
$\mathbb Q(s)$.  Indeed, the inertia permutations at
$s=0,-1,-(n-2)$ have an $n$-cycle, an $(n-1)$-cycle, and a
transposition, respectively \cite[Lemma~15]{HW}.  The first two
give double transitivity, and the transposition then gives $S_n$.
At infinity, the Hermite expansions from
Section~\ref{geo:section} have inertia cycle type
$2^{\lfloor n/2\rfloor}1^{n\bmod2}$.  For $n\ge8$ and a prime
$n/2<p\le n-3$, this permutation has no $p$-cycle and has fewer
than $p$ fixed points. Such a prime exists for every $n\ge8$:
for $n\ge51$ use Nagura's theorem \cite{Nagura}, and for
$8\le n\le50$ the primes $5,7,11,13,17,23,29$ suffice.
Apply Theorem~\ref{thm:infinity-cycle}.

It remains to treat $n=5,6,7$.  Write $\sigma$ for the inertia
involution at infinity and $C_n=C_{S_n}(\sigma)$.  Since
$\sigma$ has order two, $D_\infty\le C_n$, and
\[
 |C_5|=8,\qquad |C_6|=|C_7|=48.
\]
The classification of transitive groups of these degrees
\cite[pp.~869--871]{ButlerMcKay1983} shows that a transitive
subgroup omitting $A_5$ is contained in a conjugate of
$\operatorname{AGL}_1(\mathbb F_5)$, of index $6$ in $S_5$.
In degree seven, a transitive subgroup omitting $A_7$ is
contained in a conjugate of
$\operatorname{AGL}_1(\mathbb F_7)$ or
$\operatorname{GL}_3(\mathbb F_2)$, of indices $120$ and $30$
in $S_7$.  None of these indices divides the corresponding
$|C_n|$.  Thus $C_n$ is intransitive on each of these coset
spaces, and also on the coset space of every subgroup of the
listed groups.

In degree six, an imprimitive subgroup is contained in a conjugate
of $S_3\wr S_2$ or $S_2\wr S_3$, of indices $10$ and $15$.
These indices do not divide $48$, so the same argument applies.
Every primitive subgroup omitting $A_6$ is contained in a
conjugate of $M=\operatorname{PGL}_2(\mathbb F_5)$, acting on
$\mathbb P^1(\mathbb F_5)$.  The map $z\mapsto2/z$ in $M$ has
cycle type $2^3$: it acts as
$(0\ \infty)(1\ 2)(3\ 4)$.  Consequently $\sigma$ fixes a
point of $S_6/M$.  The action of $S_6$ on $S_6/M$ is faithful,
because $M$ contains no nontrivial normal subgroup of $S_6$.
Hence $\sigma$ does not fix every point.  Its fixed-point set
is therefore nonempty and proper; it is $C_6$-invariant because
$C_6$ centralizes $\sigma$.  This proves that $C_6$ is
intransitive on $S_6/M$, and therefore on $S_6/H$ whenever $H$
is contained in a conjugate of $M$.

In all three degrees, $D_\infty$ is thus intransitive on
$S_n/H$ for every transitive subgroup $H$ omitting $A_n$.
Proposition~\ref{prop:decomposition-runge} completes the proof.
All positive parameters are outside the branch locus.
\end{proof}

The conclusion is effective in each fixed degree.  The uniform
Galois theorem supplies the separate bound on the degrees that
can remain exceptional.

\begin{corollary}\label{cor:finite-exceptions}
The positive integral pairs $(n,s)$ for which $H_{n,s}$ is
reducible or $A_n\not\le G_{n,s}$ form a finite, effectively
computable set.
\end{corollary}
\begin{proof}
There are no exceptions in degrees $n\le4$, by
\cite[Theorem~6.3(i)]{Hajir}.
Theorem~\ref{thm:uniform-combined} bounds every remaining
exceptional degree by an effective absolute constant.
In each such degree, Proposition~\ref{geo:parameter-cap}
bounds every reducible specialization by
\[
 s<\left\lfloor n^2/4\right\rfloor^n.
\]
Exact factorization determines those exceptions.
For the irreducible specializations, Corollary~\ref{thm:runge-alternating}
gives a finite effective list containing every exception to
alternating containment.  Exact Galois-group computation
determines which entries are exceptions.
\end{proof}

This reduces Hajir's two conjectures to an effective finite
verification.  It does not give a practical enumeration: the
estimates obtained in the uniform proof are very large.

\section{Applications and further questions}\label{sec:applications}

The preceding arguments have two kinds of applications. The
finite-fiber and prime-valuation criteria apply to other polynomial
families. The uniform Galois theorem also gives consequences for
splitting fields, algebraic relations among roots, and products
of hyperelliptic Jacobians.

\subsection{An Appell family with two scales}\label{sec:appell}

We apply the finite-fiber criterion of Section~\ref{sec:finite-fibers}
to a convolution of two Laguerre families. This application uses the
centered root strip, but does not require the uniform Galois theorem.
For an integer $c\ge2$, define
\begin{equation}\label{new:two-scale-definition}
 P^{(c)}_{n,s}(X)
 =n![t^n]e^{Xt}(1-t)^{-s}(1-ct)^{-s},
 \qquad P_{n,s}=P^{(2)}_{n,s},\qquad \kappa_c=1+c.
\end{equation}
For real $s>0$, this is the moment polynomial
$\mathbb E(X+Y_1+cY_2)^n$ of two independent gamma variables
of shape $s$ and scale $1$. Its coefficient of $X^{n-k}$ is
\[
 \sum_{j=0}^{k}
 \frac{n!}{(n-k)!j!(k-j)!}(s)_j(s)_{k-j}c^{k-j}.
\]
Thus $P^{(c)}_{n,s}\in\mathbb Z[s,X]$, and this coefficient has
$s$-degree at most $k$.

\begin{theorem}\label{new:two-scale-irreducibility}
For every $\varepsilon>0$, there is an effective integer
$N(\varepsilon)$ such that $P_{n,s}$ is irreducible over $\mathbb Q$
for all integers
\[
 n\ge N(\varepsilon),\qquad s\ge1,
 \qquad\log s\ge n^\varepsilon.
\]
\end{theorem}

The same argument permits rational parameters and a scale that grows
with the degree.

\begin{corollary}\label{new:two-scale-variable-scale}
There is a finite effectively computable set
$\mathcal E\subset\mathbb Z_{\ge3}$ with the following property.
For every $0<\eta<1/3472$, $B\ge0$, and $\varepsilon>0$, there
is an effective integer $N(B,\varepsilon,\eta)$, independent of $c$,
such that $P^{(c)}_{n,a/b}$ is irreducible over $\mathbb Q$ whenever
\[
 n\ge N(B,\varepsilon,\eta),\qquad
 2\le c\le n^\eta,\qquad c\in\mathbb Z\setminus\mathcal E,
\]
\[
 a,b\in\mathbb Z_{>0},\qquad \gcd(a,b)=1,
 \qquad b\le n^B,\qquad \log a\ge n^\varepsilon.
\]
\end{corollary}

The set $\mathcal E$ will be defined by an explicit polynomial;
we do not assert that it is empty. Since $2\notin\mathcal E$, the
corollary implies the theorem.

\begin{proof}[Proof of the theorem and corollary]
We check the two inputs to the finite-fiber criterion: a strip
bound from polynomial convolution, and trace separation at four
fibers. We then make both estimates uniform in $c$.

\smallskip\noindent\emph{Convolution and the root strip.}

For monic polynomials $p,q$ of degree $n$, write
$q(X)=\sum_{k=0}^{n}q_kX^{n-k}$ and define
\[
 (p\boxplus_nq)(X)=
 \sum_{k=0}^{n}\frac{(n-k)!}{n!}q_kp^{(k)}(X).
\]
This is Walsh's additive convolution, also called symmetric additive
finite free convolution; see \cite[Section~1]{MarcusSpielmanSrivastava2022}.
We need its preservation of half-planes, including for polynomials
whose coefficients are not real.

Suppose the roots of $p$ and $q$ lie in the closed left half-plane.
The polynomials
\[
 \widetilde p(Z)=i^np(Z/i),\qquad
 \widetilde q(Z)=i^nq(Z/i)
\]
have their roots in the closed lower half-plane. Consider
\[
 T_{\widetilde q}=\sum_{k=0}^{n}
 \frac{(n-k)!}{n!}\widetilde q_k\frac{d^k}{dX^k}.
\]
Its algebraic symbol is
$T_{\widetilde q}[(X+Y)^n]=\widetilde q(X+Y)$, which is nonzero
when both variables lie in the open upper half-plane.
By \cite[Theorem~4, p.~471]{BorceaBranden2009},
$T_{\widetilde q}\widetilde p$ has no root there. The output is
monic, so the zero-polynomial alternative is excluded. Since
\[
 (\widetilde p\boxplus_n\widetilde q)(Z)
       =i^n(p\boxplus_nq)(Z/i),
\]
$p\boxplus_nq$ has its roots in the closed left half-plane.
Translation covariance adds upper bounds on real parts; reflection
adds lower bounds. Both covariance identities follow directly from
the definition.

For formal power series $f,g$ with constant term $1$, coefficient
comparison gives
\[
 \bigl(n![t^n]e^{Xt}f(t)\bigr)\boxplus_n
 \bigl(n![t^n]e^{Xt}g(t)\bigr)
 =n![t^n]e^{Xt}f(t)g(t).
\]
Consequently
\[
 P^{(c)}_{n,s}
 =H_{n,s}\boxplus_n\bigl(c^nH_{n,s}(X/c)\bigr).
\]
Proposition~\ref{geo:trace} and the half-plane bounds just proved
show that, for every real $s>0$ and every root $\alpha$,
\begin{equation}\label{new:two-scale-strip}
 |\operatorname{Re}(\alpha+\kappa_c s)|\le\kappa_c(n-1).
\end{equation}

\smallskip\noindent\emph{Four fibers and the trace condition.}

Introduce indeterminates $N,C$, and put
\[
 f_k(N)=N(N-1)\cdots(N-k+1),\qquad f_0(N)=1,
 \qquad A_C(t)=(1-t)(1-Ct).
\]
For $m=1,2,3$, define
\[
 R_{m,C}(N,X)=\sum_{k=0}^{2m}f_k(N)[t^k]A_C(t)^mX^{2m-k}.
\]
These belong to $\mathbb Z[N,C,X]$, and, for $n\ge6$,
\[
 P^{(c)}_{n,0}=X^n,\qquad
 P^{(c)}_{n,-m}=X^{n-2m}R_{m,c}(n,X).
\]
Let $U_{m,C}(N,Z)$ be the subset-trace polynomial of $R_{m,C}$,
and put $U_{0,C}(N,Z)=Z$. The fundamental theorem of symmetric
polynomials gives
\[
 U_{m,C}\in\mathbb Z[N,C,Z],\qquad \deg_ZU_{m,C}=2^{2m}.
\]
The full subset has sum $m(1+C)N$. For $m=2,3$, remove the
empty and full subsets by putting
\[
 V_{m,C}(N,Z)=
 \frac{U_{m,C}(N,Z)}{Z\bigl(Z-m(1+C)N\bigr)}.
\]
These are monic polynomials in $\mathbb Z[N,C,Z]$ of degrees
$14$ and $62$. When $C=2$, omit the subscript $C$ from
$R_{m,C}$, $U_{m,C}$, and $V_{m,C}$.
For an integral factor at a positive integral parameter $s$, with
degree $d$ and centered trace $A=\tau+\kappa_c ds$, trace
specialization gives
\begin{equation}\label{new:two-scale-fiber-trace}
 p\mid s+m\quad\Longrightarrow\quad
 p\mid U_{m,c}(n,A+m\kappa_c d),\qquad 0\le m\le3.
\end{equation}

The first residual polynomial is
\[
 R_{1,c}(n,X)=X^2-\kappa_c nX+cn(n-1).
\]
Its discriminant $(c-1)^2n^2+4cn$ is not a square for $n\ge5$.
Indeed, if it equals $q^2$, the integers
\[
 U=(c-1)^2n+2c,\qquad V=(c-1)q
\]
satisfy $U>V\ge0$ and $(U-V)(U+V)=4c^2$.
For positive integers with fixed product $4c^2$, their sum is at
most $4c^2+1$. Hence $2U\le4c^2+1$, and integrality gives
$U\le2c^2$. This would imply $n\le2c/(c-1)\le4$.
It follows that the only rational roots of $U_{1,c}(n,Z)$ are
$0$ and $\kappa_c n$.

Thus fibers $0,1$ already give two nonzero evaluations unless the
centered trace is $0$, $-\kappa_c d$, or $\kappa_c(n-d)$.
For these three possibilities, it is enough to eliminate the
single variable $d$. Put $\kappa=1+C$ and define
\begin{align*}
 D_0(N,C)&=\operatorname{Res}_D\bigl(
 V_{2,C}(N,2\kappa D),V_{3,C}(N,3\kappa D)\bigr),\\
 D_1(N,C)&=\operatorname{Res}_D\bigl(
 V_{2,C}(N,\kappa D),V_{3,C}(N,2\kappa D)\bigr).
\end{align*}
Both pairs have degrees $14,62$ in $D$. Their leading coefficients
are powers of $\kappa$ times positive integers, independent of $N$;
specializing $N$ or a positive integer $C$ therefore preserves
these degrees. Write $\mathcal D_i(N)=D_i(N,2)$.
The exact calculation in Appendix~\ref{app:appell-certificate} gives
\begin{equation}\label{new:two-scale-resultants-nonzero}
 \mathcal D_0(11)\mathcal D_1(11)\ne0.
\end{equation}
It therefore proves nonvanishing of the two resultants as
polynomials, rather than merely a trace calculation in one degree.

Suppose $n\ge6$, $c\ge2$, and $D_0(n,c)D_1(n,c)\ne0$.
For every $z\in\mathbb Q$ and integer $0<d<n$, we claim that
at least two of
\begin{equation}\label{new:scale-rational-evaluations}
 U_{m,c}(n,z+m\kappa_c d),\qquad m=0,1,2,3,
\end{equation}
are nonzero. Outside the three exceptional traces above, this
follows from fibers $0,1$. If $z=0$, fiber $1$ is nonzero.
The empty and full subset factors in fibers $2,3$ are nonzero
because $0<d<n$, so $D_0(n,c)\ne0$ prevents both evaluations
from vanishing. If $z=-\kappa_c d$, fiber $0$ is nonzero.
The factors removed from fibers $2,3$ can vanish only at
$d=0$, $d=2n$, or $d=3n/2$, all outside $0<d<n$.
Thus $D_1(n,c)\ne0$ again gives a second nonzero evaluation. For
$z=\kappa_c(n-d)$, replace $(z,d)$ by $(-z,n-d)$.
Complementation of subsets gives
\[
 U_{m,c}(n,m\kappa_c n-Z)=U_{m,c}(n,Z)\quad(m\ge1),
 \qquad U_{0,c}(n,-Z)=-U_{0,c}(n,Z).
\]
This replacement therefore preserves which evaluations vanish
and reduces the last case to the preceding one.

\smallskip\noindent\emph{Uniformity in the scale.}

Set
\[
 \mathcal D(N,C)=D_0(N,C)D_1(N,C),\qquad
 E(C)=\mathcal D(11,C),\qquad
 \mathcal E=\{c\in\mathbb Z_{\ge2}:E(c)=0\}.
\]
By \eqref{new:two-scale-resultants-nonzero}, $E(2)\ne0$.
Thus $\mathcal E$ is finite and effectively computable, and
$\mathcal E\subset\mathbb Z_{\ge3}$.

Put $K=\max\{1,\deg_C\mathcal D\}$. We have $K\le3472$.
Indeed, put $d_m=2^{2m}-2$ for $m=2,3$. The coefficient of
$X^{2m-k}$ in $R_{m,C}$ has $C$-degree at most $k$.
The coefficient of $Z^{d_m-j}$ in $V_{m,C}$ is symmetric and
homogeneous of degree $j$ in the residual roots. Giving the
$k$th elementary symmetric function weight $k$ therefore bounds
its $C$-degree by $j$. After substituting $Z=\ell(1+C)D$,
every coefficient in $D$ has $C$-degree at most $d_m$.
The bidegrees of the Sylvester resultant in its two coefficient
lists now give
\[
 \deg_C D_i\le62\cdot14+14\cdot62=1736
 \qquad(i=0,1),
\]
and hence $K\le3472$.

There is an effective absolute $C_0\ge1$ such that every coefficient
of $\mathcal D(N,c)$, as a polynomial in $N$, is at most
$C_0(1+c)^K$ in absolute value. For $c\notin\mathcal E$ this
polynomial is nonzero; its highest nonzero coefficient is an integer
of absolute value at least $1$, even if its degree drops with $c$.
Cauchy's bound puts every complex root in the disk of radius
$1+C_0(1+c)^K$. If $c\le n^\eta$ and $0<\eta<1/3472$, then
\[
 1+C_0(1+c)^K\le1+2^KC_0n^{K\eta}<n
\]
for all sufficiently large $n$, effectively in terms of $\eta$.
Consequently $D_0(n,c)D_1(n,c)\ne0$ uniformly over all permitted
scales, and the trace condition holds.

To finish, apply Corollary~\ref{new:rational-finite-fiber-criterion}.
For each sufficiently large $n$, select any permitted scale $c=c(n)$
and use the family $P_n(T,X)=P^{(c(n))}_{n,T}(X)$ with center
$c_n(T)=-\kappa_{c(n)}T$.
The coefficient-degree condition follows from the displayed coefficient
formula. Equation~\eqref{new:two-scale-strip} gives the required
real-parameter bound, since
$\kappa_c(n-1)\le2n^{1+\eta}$.
The four residual degrees are at most $6$, and their coefficients
are fixed polynomials in $n,c$. Their heights and the values
$|c_n(-m)|$ are therefore bounded by a fixed power of $n$.
Equation~\eqref{new:scale-rational-evaluations} supplies the trace
condition. All constants are independent of the choice of $c(n)$,
so the resulting threshold $N(B,\varepsilon,\eta)$ works
simultaneously for every permitted scale. This proves the corollary
and, with $c=2$ and $b=1$, the theorem.
\end{proof}

\paragraph{Further Appell families.}
Let $F(t)\in\mathbb Z[t]$ be nonconstant with $F(0)=1$ and all
zeros positive. Write $F(t)=\prod_{j=1}^{r}(1-c_jt)$ with $c_j>0$,
and put $\mu=\sum_jc_j=-F'(0)\in\mathbb Z_{>0}$.
For $Q_{n,s}(X)=n![t^n]e^{Xt}F(t)^{-s}$, repeated convolution gives
\[
 |\operatorname{Re}(\alpha+\mu s)|\le\mu(n-1)
\]
for every root and every real $s>0$.
Expanding $F(t)^{-s}=\sum_{j\ge0}(s)_j(1-F(t))^j/j!$ shows that
$Q_{n,s}\in\mathbb Z[s,X]$, with the coefficient of $X^{n-k}$
of $s$-degree at most $k$: for $j\le k\le n$, the multiplier
$n!/((n-k)!j!)$ is integral.
At $s=-m$, the residual degree is $rm$ for $n\ge rm$, and its
coefficients are polynomial in $n$.
Thus the strip and height hypotheses hold for any fixed list of
negative-integer fibers. The trace condition remains to be proved
for each new $F$.

Generic irreducibility does not settle this verification. Indeed,
Eisenstein at $s=0$ already gives absolute irreducibility of
$Q_{n,s}(X)$ as a two-variable polynomial: every nonleading
coefficient is divisible by $s$, while
$[s]Q_{n,s}(0)=(n-1)!\sum_jc_j^n>0$.
The two-scale calculation above supplies the additional trace
separation needed for arithmetic specializations.
An all-parameter result would also require an argument in the
complementary range, as in Corollary~\ref{new:finite-fiber-bootstrap}.

\subsection{A deterministic application to trimmed exponentials}
\label{sec:trimmed-exponentials}

For positive integers $n,u$, define the monic integral polynomial
\begin{equation}\label{eq:trimmed-definition}
 E_{n,u}(X)=(u+n)!\sum_{j=0}^n\frac{X^j}{(u+j)!}
 =\sum_{j=0}^n\left(\prod_{r=j+1}^n(u+r)\right)X^j.
\end{equation}
Bary-Soroker and Ben-Porath \cite[Conjecture~1]{BarySorokerBenPorath2021}
conjectured that every such trimmed exponential is irreducible
with Galois group $S_n$, and suggested just before that conjecture
that its discriminant is nonsquare when $u>1$ and $n\ge2$.
They proved \cite[Theorem~1]{BarySorokerBenPorath2021}
that, if $k_{\max}(x)=x^{o(1)}$ and the endpoint $m$ is chosen
uniformly from $\{1,\ldots,x\}$, then with probability tending to
one, $m>k_{\max}(x)$ and all $E_{n,m-n}$ with
$8\le n\le k_{\max}(x)$ have Galois group $S_n$.
The preceding arithmetic criteria give a deterministic result
for the discriminant, a conditional $S_n$ theorem throughout a
parameter window, and an unconditional $S_n$ theorem under an
explicit endpoint condition.

\subsubsection{Local packets and the discriminant}

\begin{lemma}
\label{lem:trimmed-local}
Let $1\le N\le n$, let $p>n$ be prime, and suppose that
$e=v_p(u+N)>0$. Then $E_{n,u}$ is separable, its Newton polygon
has an edge from $(0,e)$ to $(N,0)$, and
\begin{equation}\label{eq:trimmed-local-discriminant}
 v_p\!\left(\Disc E_{n,u}\right)=e(N-1).
\end{equation}
Its local inertia group is cyclic, and a generator has cycle type
$(N/g)^g1^{n-N}$, where $g=\gcd(N,e)$.
\end{lemma}
\begin{proof}
Only $u+N$ among $u+1,\ldots,u+n$ is divisible by $p$.
The coefficient formula \eqref{eq:trimmed-definition} therefore gives
\[
 v_p([X^j]E_{n,u})=
 \begin{cases}
 e,&0\le j<N,\\
 0,&N\le j\le n.
 \end{cases}
\]
This proves the Newton-polygon assertion. Put $m=n-N$ and
$e_m(X)=\sum_{j=0}^mX^j/j!$. Since $u\equiv-N\pmod p$,
\[
 \overline{E}_{n,u}(X)=X^N R_m(X),\qquad
 R_m(X)=m!e_m(X)\in\mathbb F_p[X].
\]
Here $R_m(0)=m!\ne0$ and $R_m-R_m'=X^m$, so $R_m$ is
squarefree; the case $m=0$ means $R_m=1$.
Let $A\in\mathbb Z_p[X]$ be the monic degree-$N$ Hensel factor
lifting $X^N$. The complementary factor has unit constant term.
Comparison of coefficients in increasing degree shows that every
nonleading coefficient of $A$ is divisible by $p^e$, and that its
constant coefficient has valuation exactly $e$.
Lemma~\ref{lem:general-tame-packet} now gives separability and the
stated inertia action. It also gives valuation $e/N$ for every
difference between two roots in the packet and valuation zero
for every other root difference. Taking the product of the
squares of these differences gives
$2\binom N2e/N=e(N-1)$ in
\eqref{eq:trimmed-local-discriminant}.
\end{proof}

\begin{proposition}
\label{prop:trimmed-discriminant}
For each fixed $\eta>0$, there is an effectively computable
$n_0(\eta)$ such that $E_{n,u}$ is separable and has nonsquare
discriminant whenever
\[
 n\ge n_0(\eta),\qquad
 u\ge\bigl(\gcd(u,2)-1+\eta\bigr)n.
\]
In particular, $u\ge(1+\eta)n$ suffices, and for odd $u$ the
weaker condition $u\ge\eta n$ suffices.
In addition, for all effectively sufficiently large $n$ and
$u\ge n^{21/10}$, more than $\lfloor n/2\rfloor/102$ distinct
primes $p>3\lfloor n/2\rfloor$ have odd valuation in
$\Disc E_{n,u}$.
\end{proposition}
\begin{proof}
Put $k=\lfloor n/2\rfloor$, $h=\gcd(u,2)$,
$A=(u+2)/h$, and $d=2/h$. Then $d\in\{1,2\}$,
$\gcd(A,d)=1$, and
\begin{equation}\label{eq:trimmed-even-product}
 \prod_{j=0}^{k-1}(A+jd)
 =h^{-k}\prod_{j=1}^k(u+2j).
\end{equation}
If a prime $p>n$ has odd valuation in this product, it divides
exactly one node $u+N$ with $N$ even, and
$e=v_p(u+N)$ is odd. Lemma~\ref{lem:trimmed-local} then proves
separability and gives the odd valuation
$v_p(\Disc E_{n,u})=e(N-1)$.

For the first assertion, set
$\theta=\min(\eta,1)/4$ and $C=2+\theta$.
Filaseta--Laishram--Saradha \cite[Theorem~1.1]{FLS} imply that,
for fixed $d$, all sufficiently large $k$, and every $A$ with
\[
 \gcd(A,d)=1,\qquad A\ge(C-d+\theta d)k,
\]
the product in \eqref{eq:trimmed-even-product} has a prime
$p>Ck$ of odd valuation. Indeed, otherwise that product is
$by^2$, with every prime divisor of $b$ at most $Ck$, contrary
to their effective finiteness theorem. Here $C\ge d$ and
$0<\theta<1$, as required. If $h=2$, our hypothesis gives
\[
 A\ge(1+\eta)k\ge(1+2\theta)k
   =(C-d+\theta d)k.
\]
If $h=1$, it gives
$A\ge2\eta k\ge3\theta k=(C-d+\theta d)k$.
Take $k$ beyond the effective exceptional bounds for both
$d=1$ and $d=2$, and also require $k>1/\theta$.
Then $p>Ck>n$, so the preceding local argument applies.

For the quantitative assertion, $u\ge n^{21/10}$ implies
$A\ge k^{21/10}$. Apply
Theorem~\ref{thm:dense-power-general}\textup{(ii)} with
$C=3$, $r=2$, $\varepsilon=1/10$, and $\delta=1/102$.
It gives more than $k/102$ primes $p>3k$, each with odd
valuation in \eqref{eq:trimmed-even-product}.
For $n\ge5$ these primes exceed $n$, so each has odd
discriminant valuation.
\end{proof}

\subsubsection{Symmetric groups in a parameter window}

\begin{theorem}
\label{thm:trimmed-deterministic}
Fix $D>0$ and $\kappa>5$. There is an effectively computable
$N=N(D,\kappa)$ such that, for all integers
\[
 n\ge N,\qquad u\ge1,\qquad
 (\log n)^\kappa\le\log u\le n^D,
\]
irreducibility of $E_{n,u}$ over $\mathbb Q$ implies
\[
 \operatorname{Gal}(E_{n,u}/\mathbb Q)=S_n.
\]
In particular, for fixed $0<\varepsilon\le D$ the same conclusion
holds whenever $n^\varepsilon\le\log u\le n^D$ and $n$ is
sufficiently large in terms of $\varepsilon,D$.
\end{theorem}
\begin{proof}
For each prime $n/2<P\le n-3$, take the node $u+P$.
There are at least $2n/(5\log n)$ such primes for all effectively
sufficiently large $n$. Lemma~\ref{lem:trimmed-local} supplies
the required Newton edge at each prime $p>n$ dividing $u+P$.
Theorem~\ref{thm:newton-node-criterion}, with
$s=u$, $C=1$, $c=2/5$, and $j_P=P$, proves alternating
containment whenever $E_{n,u}$ is irreducible, since
$\lfloor C/c\rfloor+3=5$.
For sufficiently large $n$, the lower bound
$\log u\ge(\log n)^\kappa$ implies $u\ge2n$.
Proposition~\ref{prop:trimmed-discriminant}, with $\eta=1$,
excludes containment in $A_n$, and hence the group is $S_n$.
For the final assertion take $\kappa=6$ and use
$(\log n)^6\le n^\varepsilon$ for sufficiently large $n$.
\end{proof}

\begin{corollary}
\label{cor:trimmed-endpoint}
Let $n,u$ be positive integers. Suppose that primes
$q_1,\ldots,q_t>n$, with $t\ge1$, satisfy
\[
 e_i=v_{q_i}(u+n)>0,\qquad \gcd(n,e_1,\ldots,e_t)=1.
\]
Then $E_{n,u}$ is irreducible over $\mathbb Q$.
If, in addition, $n,u$ satisfy the hypotheses of
Theorem~\ref{thm:trimmed-deterministic}, its Galois group is $S_n$.
In particular, the endpoint condition holds if some prime $q>n$
divides $u+n$ with $\gcd(n,v_q(u+n))=1$, and hence whenever
$u+n$ is prime.
\end{corollary}
\begin{proof}
Every nonleading coefficient in \eqref{eq:trimmed-definition}
has $q_i$-adic valuation $e_i$, while the leading coefficient is $1$.
Thus the Newton polygon is the segment joining $(0,e_i)$ to $(n,0)$,
and every root has valuation $e_i/n$.
If a monic rational factor has degree $d$, its nonzero constant
coefficient has integral $q_i$-adic valuation $de_i/n$.
Consequently $n/\gcd(n,e_i)$ divides $d$ for every $i$.
The least common multiple of these integers is
\[
 \operatorname{lcm}_{1\le i\le t}\frac{n}{\gcd(n,e_i)}
 =\frac{n}{\gcd(n,e_1,\ldots,e_t)}=n.
\]
There is therefore no proper rational factor. This argument works
for all positive $n,u$. In the stated parameter range,
Theorem~\ref{thm:trimmed-deterministic} gives $S_n$.
If $u+n$ is prime, the irreducibility assertion is
Eisenstein's criterion.
\end{proof}

For example, if $6\mid n$ and $u+n=q_1^2q_2^3$ for distinct
primes $q_1,q_2>n$, the two endpoint valuations together prove
irreducibility, although neither is individually coprime to $n$.
In the parameter range of the theorem the group is $S_n$.

Theorem~\ref{thm:trimmed-deterministic} is uniform in every
parameter in its window, conditional on irreducibility.
Corollary~\ref{cor:trimmed-endpoint} removes that condition on an
explicit arithmetic subfamily. These statements do not settle
the full conjecture of \cite{BarySorokerBenPorath2021}:
irreducibility outside the endpoint condition remains unproved
here. The local Newton-polygon step is classical.
Bary-Soroker and Ben-Porath already use an endpoint
valuation-one prime for irreducibility and a full cycle, and an
even-length cycle to exclude alternating groups
\cite[Propositions~5.1 and~5.2]{BarySorokerBenPorath2021}.
Theorem~\ref{thm:dense-nearby-nodes} supplies the prime
valuations needed for alternating containment throughout the
stated parameter window. The linear discriminant range uses
\cite[Theorem~1.1]{FLS}, while
Theorem~\ref{thm:dense-power-general} gives a positive
proportion of odd discriminant valuations in the larger-parameter
range. All these conclusions are deterministic.

\subsection{Splitting fields and Jacobians}\label{sec:arithmetic-geometry}

The uniform Galois theorem has consequences for relations among selected
roots and for products of hyperelliptic Jacobians. The general criteria
are due to Bary-Soroker--Kadets, Zarhin, and Lombardo; here they apply
with arbitrary positive integral parameters.
Increase $N$ in Theorem~\ref{thm:uniform-combined} so that $N\ge5$, and
write $J_{n,s}$ for the Jacobian of the smooth projective model of $Y^2=H_{n,s}(X)$.

\subsubsection{Splitting fields of distinct degrees}

\begin{proposition}\label{prop:splitting-independence}
Let $f_1,\ldots,f_r\in\mathbb Q[X]$ be separable of distinct degrees $n_i\ge5$,
with Galois groups $G_i=A_{n_i}$ or $S_{n_i}$.
Let $L_i$ be their splitting fields, $\delta_i\in\mathbb Q^*/\mathbb Q^{*2}$
their discriminant classes, and
$\mathcal R=\{(e_i)\in\mathbb F_2^r:\prod_i\delta_i^{e_i}=1\}$.
Restriction identifies the group of the compositum with
\[
 \operatorname{Gal}(L_1\cdots L_r/\mathbb Q)
 =\left\{(g_i)\in\prod_iG_i:
   \prod_i\operatorname{sgn}(g_i)^{e_i}=1
                 \text{ for every }(e_i)\in\mathcal R\right\}.
\]
If the $\delta_i$ span a space of dimension $a$, then $[L_1\cdots L_r:\mathbb Q]=2^a\prod_i(n_i!/2)$.
\end{proposition}
\begin{proof}
The compositum group $G$ is a subdirect product of the $G_i$, so $G'$
is a subdirect product of the pairwise nonisomorphic simple groups
$A_{n_i}$. Goursat's lemma gives $G'=\prod_iA_{n_i}$.
The sign quotient is the Galois group of
$\mathbb Q(\sqrt{\delta_1},\ldots,\sqrt{\delta_r})$, with relation space
$\mathcal R$. This proves both formulas.
\end{proof}

Thus, for distinct $n_i\ge N$ and arbitrary $s_i\ge1$, the discriminant
classes in \eqref{eq:intro-discriminant} determine the Galois group of
the compositum of the splitting fields of $H_{n_i,s_i}$.

\subsubsection{Algebraic relations among selected roots}

The criterion of Bary-Soroker--Kadets \cite[Definition~1.6 and
Theorem~1.7]{BarySorokerKadets2026} extends their application with fixed
Laguerre parameters as follows.

\begin{corollary}\label{cor:varying-parameter-special-set}
For any sequence of positive integers $(s_n)_{n\ge1}$, let
$\mathcal S=\{z\in\overline{\mathbb Q}:H_{n,s_n}(z)=0\text{ for some }n\ge1\}$.
For every $m\ge1$ and $B\subseteq\mathcal S^m$, the Zariski closure
of $B$ in $\mathbb A^m_{\mathbb C}$ is a finite union of affine linear
subspaces defined by equations
$X_i=X_j$ $(i\ne j)$ and $X_i=c$ $(c\in\mathcal S)$.
In particular, an irreducible plane curve contains infinitely many
points of $\mathcal S^2$ if and only if it is the diagonal, or a
horizontal or vertical line through an element of $\mathcal S$.
\end{corollary}
\begin{proof}
For $n\ge N$, the selected roots have degree $n$ and Galois group $A_n$ or $S_n$.
There are finitely many of bounded degree; nonconjugate roots have splitting fields
whose intersection has degree at most two, by Proposition~\ref{prop:splitting-independence}.
These are the hypotheses of the cited criterion. The finitely many
remaining roots may be adjoined by partitioning tuples according to
the coordinates taking these values. Since $\mathcal S$ is infinite,
the plane-curve assertion follows.
\end{proof}

\begin{remark}\label{rem:all-parameters-not-special}
The choice of one parameter per degree matters. For any fixed $d\ge2$,
ordered root tuples for all $s\ge1$ satisfy
\[
 (\textstyle\sum_i x_i)^2-d\sum_i x_i^2
                    +d(d-1)\sum_i x_i=0.
\]
This proper hypersurface contains infinitely many tuples. None lies
on a coordinate diagonal, and each equation $x_i=c$ holds
for only finitely many parameters, since $H_{d,s}(c)$ is monic of
degree $d$ in $s$. Their closure therefore cannot be a finite union
of the permitted proper subspaces. The corollary fails for the union
over all parameters, even after any finite range of degrees is omitted.
\end{remark}

\subsubsection{Products of Jacobians}

\begin{theorem}
\label{thm:mixed-jacobian-products}
Let $K$ be a number field and let $f_1,\ldots,f_r\in K[X]$ be separable of pairwise
distinct degrees $n_i\ge5$, with $\operatorname{Gal}(f_i/K)\supseteq A_{n_i}$.
Let $A_i$ be the Jacobian of $y^2=f_i(x)$ and $B=\prod_i A_i^{e_i}$,
where $e_i\ge1$. Then
\[
 \operatorname{End}_{\overline K}(A_i)=\mathbb Z,\qquad
 \operatorname{Hom}_{\overline K}(A_i,A_j)=0\ (i\ne j),\qquad
 \operatorname{End}_{\overline K}(B)\simeq\prod_iM_{e_i}(\mathbb Z).
\]
Fix $K\hookrightarrow\mathbb C$ and polarization forms $\psi_i$ on
$V_i=H_1(A_i(\mathbb C),\mathbb Q)$. In the diagonal action on repeated copies,
\[
 \operatorname{Hg}(B)=\prod_i\operatorname{Sp}(V_i,\psi_i),
\]
\[
 \operatorname{MT}(B)=\mathcal G:=
 \{(u_i)\in\prod_i\operatorname{GSp}(V_i,\psi_i):
                         \nu(u_1)=\cdots=\nu(u_r)\},
\]
where $\nu$ is the similitude multiplier. For every prime $\ell$, the algebraic
monodromy group is $\mathcal G_{\mathbb Q_\ell}$ and the image has Lie algebra
\[
 \mathbb Q_\ell\operatorname{Id}\oplus
       \bigoplus_i\mathfrak{sp}(V_\ell(A_i),\psi_{i,\ell}),
 \qquad V_\ell(A_i)=T_\ell(A_i)\otimes\mathbb Q_\ell.
\]
The Mumford--Tate conjecture holds for $B$. On $B$ and every power of $B$,
all rational Hodge classes and all $\ell$-adic Tate classes
over finite extensions of $K$ are generated by divisor classes.
In particular, the Hodge and Tate conjectures hold for these varieties.
\end{theorem}
\begin{proof}
The endomorphism and homomorphism assertions follow from
\cite[Theorem~1.1 and Corollary~1.4]{Zarhin2003}, including adjacent degrees
$2g+1,2g+2$, which have the same genus.
Each individual $\ell$-adic Lie algebra is symplectic plus scalars by \cite[Theorem~2.6]{Zarhin2002}.
Lombardo's product criterion \cite[Theorem~4.1]{Lombardo2016} now
applies: the symplectic algebras are simple, their standard modules
are irreducible, and every automorphism preserves the standard module.
Indeed, a proper semisimple subdirect product would identify two
simple factors, yielding an intertwiner of their Tate modules after
a finite extension. Faltings's homomorphism theorem \cite{Faltings1983}
would contradict geometric Hom-vanishing. Thus the derived Lie
algebra is the full direct sum. Polarizations force a common scalar
line, and the cyclotomic character supplies that line.

The algebraic monodromy group lies in the connected group
$\mathcal G_{\mathbb Q_\ell}$ and has its full Lie algebra, hence
equals it. The usual inclusion of the $\ell$-adic Lie algebra in the
Mumford--Tate Lie algebra, together with the bound from polarizations,
gives $\operatorname{MT}(B)=\mathcal G$ and the stated Hodge group;
compare \cite[proof of Theorem~9.1]{Zarhin2002}. Repeated factors merely
repeat the representations.

Finally, invariants of each symplectic factor are generated by
pairwise contractions. In cohomology these are divisor classes,
obtained from polarizations by pullback and addition between repeated
copies; see \cite[Theorems~8.1 and~9.1]{Zarhin2002}.
The same argument applies to every power. Finite extensions of $K$ do not
change the Lie algebra, so it covers all the stated Tate classes.
The equality $\mathcal G=\mathbb G_m\cdot\prod_i\operatorname{Sp}(V_i)$
is algebraic and need not give the corresponding factorization of rational points.
\end{proof}

\begin{corollary}\label{cor:uniform-geometric-applications}
For pairwise distinct $n_i\ge N$ and arbitrary positive integers
$s_i,e_i$, the product $\prod_iJ_{n_i,s_i}^{e_i}$ satisfies every
conclusion of Theorem~\ref{thm:mixed-jacobian-products}.
In particular, $J_{2g+1,s}$ and $J_{2g+2,t}$ are nonisogenous
Jacobians of the same dimension for every $s,t\ge1$ if $2g+1\ge N$.
\end{corollary}
\begin{proof}Apply Theorem~\ref{thm:uniform-combined}.\end{proof}


\subsection{Further questions}\label{sec:consequences}

Theorem~\ref{thm:uniform-combined} gives a uniform eventual form of
Hajir's conjectures.  Corollary~\ref{cor:finite-exceptions} reduces
both all-degree conjectures to an effective finite verification.
The remaining question is whether that finite exceptional set is empty.
The current bounds do not make its enumeration practical.
An all-degree proof needs estimates that remain useful when the
degree is comparable to the constants in the local and logarithmic
arguments.  Equivalently, one may ask for irreducibility and
alternating containment for every nonconstant numerator and
denominator in the exponential Pad\'e table.

The finite-fiber theorem suggests a broader irreducibility problem.
For fixed positive integers $c_1,\ldots,c_r$, consider
\[
 P_{n,s}(X)=n![z^n]e^{Xz}\prod_{\nu=1}^r(1-c_\nu z)^{-s}.
\]
Does there exist an effective $N(c_1,\ldots,c_r)$ such that
$P_{n,s}$ is irreducible for every $n\ge N(c_1,\ldots,c_r)$ and
every positive integer $s$?
Generic irreducibility is immediate: Eisenstein at $s=0$ applies
because
\[
 [s]P_{n,s}(0)=(n-1)!\sum_{\nu=1}^r c_\nu^n\ne0.
\]
The specialization problem needs more information.
For each fixed $c\ge2$ outside the finite set $\mathcal E$,
Corollary~\ref{new:two-scale-variable-scale} treats $P^{(c)}_{n,s}$
when $\log s\ge n^\varepsilon$, for every fixed $\varepsilon>0$.
A proof in $1\le s\le\exp(n^\delta)$ for one fixed $\delta>0$
would therefore complete the uniform argument for that scale.
The Laguerre energy proof does not transfer directly:
the two-scale degenerate fibers have cluster sizes $n-2j$,
so in even degree a factor of half the degree can have the same
occupation ratio in every such cluster.  One must break this
possible balance or find a different source of arithmetic control.
For three or more scales, a structural trace-separation theorem
would also be preferable to a separate elimination certificate
for each choice of scales.

Theorem~\ref{thm:dense-nearby-nodes} and its Newton-polygon
consequence isolate a different question.
For trimmed exponentials, the remaining irreducibility hypothesis
is removed here by an endpoint valuation condition.
A uniform irreducibility theorem without that condition would
give the full symmetric group throughout the same logarithmic
parameter range. The nonsquare-discriminant theorem already
has no upper bound on $u$. Can its lower restriction
$u\ge(\gcd(u,2)-1+\eta)n$ be removed for all sufficiently
large degrees?
These are partial steps toward the full symmetric-group conjecture
of Bary-Soroker--Ben-Porath \cite{BarySorokerBenPorath2021};
the current argument neither proves irreducibility for every
endpoint nor covers every relation between $u$ and $n$.
The lower bound $(\log n)^\kappa$, $\kappa>5$, comes from the
small-prime allocation and the height product in the simultaneous
logarithmic estimate. Replacing it by a bound of order $\log n$
would extend this criterion to parameters of polynomial size;
the present estimate does not reach that range.

Finally, Theorem~\ref{thm:laurent-cyclic-obstruction} is geometric,
whereas the application to quadratic subfields also uses
arithmetic bounds for the interpolants.
It would be useful to obtain such bounds for other covers from
explicit ramification and height data.
The Hermite argument here uses integral derivative identities
and Gaussian quadrature, so those bounds are not supplied by
the geometric theorem alone.
For higher cyclic degree, one must also control all relevant
embeddings and the possible effect of units; a small value at
one embedding no longer gives the imaginary-quadratic
contradiction used in Section~\ref{sec:imaginary-height}.
A quantitative interpolation theorem with a second arithmetic
application would substantially extend this part of the method.


\appendix


\section{The root-modulus representation}\label{app:modulus}

We prove Theorem~\ref{geo:bernstein} by continuing the roots through
the upper half of the parameter plane. At each parameter
$0,-1,\ldots,1-n$, one more branch reaches the positive real axis.
Keeping track of this order determines the boundary jump of the
holomorphic continuation of the positive-axis modulus. Its
expansion at infinity then permits Cauchy's formula to recover
the representing measure.

\begin{lemma}\label{geo:ray}
Let $1-n<s<0$ be real. Every root of $H_{n,s}$ that is not positive
real has modulus less than $n+s-1$. Two distinct nonzero roots cannot
lie on the same ray except on the positive real axis.
\end{lemma}
\begin{proof}
The matrix argument in Lemma~\ref{geo:annulus} remains valid for the
upper bound even when $D_0$ has negative diagonal entries. For a
nonzero root $x$ not on the positive real axis, choose $z^2=x$ with
$\Im z\ne0$. It gives $|x|=v^*D_0v<n+s-1$; strictness follows from
the same endpoint coupling. A zero root satisfies the bound as well.

Put $A=n+s-1\in(0,n-1)$ and $\kappa=n-A/2>A/2$.
The ray substitution used in Proposition~\ref{geo:hermite-order}
gives \eqref{geo:ray-ode}, which is valid for every real parameter.
Suppose two roots lie on a nonreal ray, at radii $\rho_1<\rho_2$.
Both are less than $A$. Multiply this equation by $\bar u$, integrate
between the two radii, and take imaginary parts. The endpoint terms
vanish, giving
\[
 \sin\theta\int_{\rho_1}^{\rho_2}
       \left(\frac\kappa\rho-\frac{\cos\theta}2\right)|u|^2\,d\rho=0.
\]
The parenthesized function is strictly positive, since
$\rho<A<2\kappa$, and $u$ is not identically zero. This is impossible.
On the negative real ray the potential in the differential equation
is strictly negative; the real part of the same integrated identity
is then $-\int|u'|^2+\int V|u|^2=0$ with $V<0$, also impossible.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{geo:bernstein}]
\emph{Continuation and boundary labels.}
The roots are simple and nonzero when $\Im s>0$. Monicity bounds
them on compact parameter sets, so each root continues along every
path in that half-plane. Since the domain is simply connected,
we obtain $n$ single-valued, nonvanishing holomorphic root functions,
each with a holomorphic logarithm. They extend locally across every
real parameter outside $\{0,-1,\ldots,1-n\}$.

We describe their boundary labels explicitly. For $0\le m\le n-1$
put $N=n-m$. Directly from the coefficient formula,
\begin{equation}\label{geo:singular-fiber}
 H_{n,-m}(X)=X^N J_{m,n}(X),\qquad
 J_{m,n}(0)=(-1)^m\frac{n!}{N!}.
\end{equation}
The $m$ zeros of $J_{m,n}$ are simple and positive. Indeed,
\[
 J_{m,n}(X)=(-1)^m e^X X^{-N}
               \frac{d^m}{dX^m}(e^{-X}X^n).
\]
Integration by parts shows that this monic degree-$m$ polynomial is
orthogonal to every polynomial of smaller degree for the weight
$X^Ne^{-X}$ on $(0,\infty)$. All boundary terms vanish. If it had
fewer than $m$ sign changes on that interval, multiplication by the
product of its sign-changing linear factors would give a polynomial
of smaller degree whose weighted product with $J_{m,n}$ has constant
nonzero sign, contradicting orthogonality. It therefore has exactly
$m$ distinct positive roots. The assertion is vacuous when $m=0$.

Write $s=-m+\delta$. The $N$ roots tending to zero satisfy
\begin{equation}\label{geo:crossing}
 X^N=-c_m\delta(1+o(1)),\qquad
 c_m=\frac{m!(N-1)!N!}{n!}>0.
\end{equation}
To see this, balance the coefficient $J_{m,n}(0)$ of $X^N$ against
$H_{n,-m+\delta}(0)=(-1)^m m!(N-1)!\delta+O(\delta^2)$.
More explicitly, with $\rho=\delta^{1/N}$, the polynomial
$\rho^{-N}H_{n,-m+\rho^N}(\rho w)$ extends holomorphically to
$\rho=0$, with value $J_{m,n}(0)(w^N+c_m)$. Its roots are simple.
The holomorphic implicit function theorem gives the asserted asymptotics
uniformly for $0\le\arg\delta\le\pi$, with a continuous choice of
$\delta^{1/N}$. On the right of $-m$ the limiting arguments are
$(2j+1)\pi/N$, $0\le j<N$. Moving around a small upper semicircle
increases each limiting argument by $\pi/N$. The last branch thus
reaches the positive axis with continuous argument $2\pi$.
This branch is positive real just to the left of $-m$: writing
$\delta=-r^N$ and $X=rw$ with $r>0$, its limiting $w$-root is
positive and simple, so the real implicit function theorem gives
a positive real branch.

On $(-m,-m+1)$, $1\le m\le n-1$, the coefficient signs in descending
powers alternate through $(s)_m$ and are thereafter constant.
There are exactly $m$ variations. Descartes' rule allows at most $m$ positive roots;
near the left endpoint \eqref{geo:singular-fiber} supplies $m$ such
simple roots. They remain positive throughout the interval: leaving
the real axis or changing sign would require a collision or a zero
root. There are consequently exactly $m$ positive roots. These are
precisely the branches tending to the nonzero endpoint roots; all
other branches tend to zero at the left endpoint. Indeed, any branch
approaching a nonzero positive endpoint root is real and positive nearby
by the implicit function theorem. A branch off the positive real axis
cannot meet that axis in the interval, because a
simple real root has a unique real local continuation. By
Lemma~\ref{geo:ray}, the remaining branches cannot exchange angular
order.

Near $s=0+$ label the roots $x_0,\ldots,x_{n-1}$ in increasing
argument order in $(0,2\pi)$, and choose their holomorphic logarithms
accordingly. For $s>0$ these continuous arguments remain in that
interval. Conjugation pairs $j$ with $n-1-j$, first near zero and
then everywhere on the positive axis by uniqueness and absence of
collisions. Their arguments sum to $2\pi$.

These observations give an induction across $0,-1,\ldots,1-n$.
On $(-m,-m+1)$ put $N=n-m$ and write
$\theta_r=\Im\log x_r$ for the continued arguments. Then
\begin{equation}\label{geo:negative-labels}
 \begin{gathered}
  0<\theta_0<\cdots<\theta_{N-1}<2\pi,\qquad
  \theta_r+\theta_{N-1-r}=2\pi\quad(0\le r<N),\\
  \theta_N=\cdots=\theta_{n-1}=2\pi.
 \end{gathered}
\end{equation}
Indeed, the $m$ positive branches have argument $2\pi$ and are the
labels already removed at earlier crossings. The other branches
preserve their angular order by Lemma~\ref{geo:ray}. At $s=-m$,
\eqref{geo:crossing} rotates their limiting arguments from
$(2r+1)\pi/N$ to $2(r+1)\pi/N$, so label $N-1$ becomes the next
positive branch. After $s=1-n$, all roots are positive with continued
argument $2\pi$. They remain so for $s<1-n$, since no further
collision or zero root is possible.

\emph{Continuing the modulus.}
For an original conjugate pair $j,k=n-1-j$, $j<k$, define in the
upper half-plane
\begin{equation}\label{geo:q-branch}
 q(s)=\exp\left(\frac{\log x_j(s)+\log x_k(s)-2\pi i}{2}\right).
\end{equation}
On $s>0$ this is positive and equals $|x_j(s)|$. The chosen logarithms
specify the square root throughout the parameter domain.
On the negative axis the continuous argument sum of $x_j,x_k$ lies
in $[2\pi,4\pi]$. To verify the lower bound when both labels still lie
off the positive real axis, let $N$ be the number of remaining such labels.
Conjugation gives $\theta_j+\theta_{N-1-j}=2\pi$, while
$k=n-1-j\ge N-1-j$ and angular order give
$\theta_k\ge\theta_{N-1-j}$. If one or both labels have become
positive, their arguments are $2\pi$, and the bound is immediate.
The upper bound follows because no individual argument exceeds
$2\pi$. Hence the boundary argument of $q$ is in $[0,\pi]$.

For odd $n$, the real root has label $j=(n-1)/2$. While this label
remains off the positive real axis, it lies at or above the middle of the remaining
ordered labels, so its continuous argument belongs to $[\pi,2\pi)$.
Afterwards its argument is $2\pi$. The function $q=-x_j$ therefore
has the same boundary-argument property. In both cases $q$ is
holomorphic and nonvanishing in the upper half-plane, and its boundary
imaginary part is nonnegative.

\emph{The expansion at infinity.}
Put $z=s^{-1/2}$, with $-\pi/2\le\arg z\le0$ for $s$ in the closed
upper half-plane. The scaled polynomial $F_n(w,z)$ in \eqref{geo:hermite-scaling}
has limiting polynomial $h_n(w)$ with simple purely imaginary zeros,
as proved in Section~\ref{geo:section}. Each zero therefore extends
to a holomorphic root germ $w_j(z)$ in a full disk about zero.

For positive $z$, conjugate roots have opposite limiting $w$-roots.
The identity $F_n(-w,-z)=(-1)^nF_n(w,z)$ and uniqueness
of the root germs imply $w_k(z)=-w_j(-z)$ for such a pair. The
uncentered roots $x_j(z)=z^{-1}w_j(z)-z^{-2}$ consequently satisfy
$x_k(z)=x_j(-z)$. Their product is an even Laurent series beginning
with $z^{-4}=s^2$. Its square root that is positive for large positive
$s$ has an expansion $s+\theta+O(s^{-1})$, with $\theta\in\R$.
For the real root in odd degree, the germ with limiting $w$-root zero
is odd in $z$, and its uncentered root is itself an even Laurent
series, giving the same expansion for $q=-x_j$.

The germs and the already continued functions agree on the positive
axis and extend across it; the identity theorem identifies them in
the exterior upper half-plane. Convergence of the full germ expansions
gives, uniformly in that closed half-plane as $|s|\to\infty$,
\begin{equation}\label{geo:q-infinity}
 q(s)=s+\theta+O(s^{-1}),\qquad \theta\in\R.
\end{equation}

\emph{Cauchy's formula and the measure.}
The root functions have convergent expansions in $(s+m)^{1/N}$
at each singular parameter $-m$, by the rescaling in
\eqref{geo:crossing}; the nonzero root branches are analytic there.
The function $q$ therefore has a finite,
continuous boundary value at each such point. If either root used
in \eqref{geo:q-branch} vanishes, then $q$ tends to zero. The same
statement holds for the single real branch. In particular,
$q(s)=O(|s|^{1/n})$ near zero in the closed upper half-plane.

The boundary values of $q$ are real for $s>0$ and for $s<1-n$;
in the latter interval they are negative, by the continued arguments
above. Schwarz reflection therefore gives a holomorphic function
on $\mathbb C\setminus[1-n,0]$, whose two boundary values on the
cut are conjugate. Put $\psi(t)=\Im q(-t+i0)$. The argument bounds
already proved give $\psi\ge0$. Also $\psi$ is bounded on the cut,
and $\psi(t)=O(t^{1/n})$ as $t\downarrow0$.

Apply Cauchy's formula to $q(s)-s-\theta$ around the cut. The
expansion \eqref{geo:q-infinity} removes the large-circle integral;
boundedness removes the small arcs about the finitely many singular
points. The jump across the cut is $2i\psi(t)$, so
\begin{equation}\label{geo:cauchy}
 q(s)=s+\theta-\frac1\pi\int_0^{n-1}\frac{\psi(t)}{s+t}\,dt
 \qquad\bigl(s\in\mathbb C\setminus[1-n,0]\bigr).
\end{equation}
Because $\psi(t)/t=O(t^{1/n-1})$ near zero, this quotient is
integrable. Letting $s\downarrow0$ and using dominated convergence
gives
\[
 \theta=\frac1\pi\int_0^{n-1}\frac{\psi(t)}t\,dt.
\]
Hence \eqref{geo:eq-bernstein} holds with the finite nonnegative
measure
\begin{equation}\label{geo:measure-density}
 d\mu(t)=\frac{\psi(t)}{\pi t}\,dt.
\end{equation}
There are no endpoint atoms: the formula is obtained from the full
contour, and every indentation integral tends to zero.
If two finite measures on $[0,n-1]$ gave the same representation,
expansion for large $s$ would equate all their moments. Density of
polynomials in $C[0,n-1]$ then makes the measures equal.
Finally, the integrand $s/(s+t)$ is nondecreasing in $s>0$, proving
the stated monotonicity.
\end{proof}
\begin{proposition}
\label{geo:exact-support}
Fix $n\ge2$. For $0\le j\le\lfloor(n-1)/2\rfloor$, label a root branch by
\[
 \alpha_j(s)=((n-1)!s)^{1/n}
       \exp\!\left(\frac{(2j+1)\pi i}{n}\right)
       +o(s^{1/n})\qquad(s\downarrow0),
\]
and continue it along the positive real axis. Put
$q_j(s)=|\alpha_j(s)|$ and $L_j=n-1-j$.
The nonnegative measure in the representation
\[
 q_j(s)=s+\int\frac{s}{s+t}\,d\mu_j(t)
\]
has support exactly $[0,L_j]$. It has a density that is strictly
positive on $(m,m+1)$ for every integer $0\le m<L_j$, and zero
for $t>L_j$.
In particular, if
\[
 M_j=\frac{n-1}{3}+\frac{\xi_j^2}{6},
\qquad
 \alpha_j(s)=-s+i\xi_j\sqrt{s}+O(1)
 \quad(s\longrightarrow\infty),
\]
then
\begin{equation}\label{geo:mass-bounds}
 M_j\frac{s}{s+L_j}<q_j(s)-s<M_j\qquad(s>0).
\end{equation}
For odd $n$ and $j=(n-1)/2$, this is the real root branch,
$\xi_j=0$, and $L_j=(n-1)/2$.
\end{proposition}
\begin{proof}
On $(-m,-m+1)$, $1\le m\le n-1$, put $N=n-m$ and use the
continued arguments in \eqref{geo:negative-labels}.
First suppose $j<k=n-1-j$. If $j,k<N$, then
$k>N-1-j$, because $m>0$. Thus
\[
 2\pi<\theta_j+\theta_k<4\pi.
\]
If $j<N\le k$, the sum is $\theta_j+2\pi$ and satisfies the same
strict inequalities. If $N\le j$, both arguments are $2\pi$.
The holomorphic modulus branch has continuous argument
$(\theta_j+\theta_k-2\pi)/2$. Its boundary imaginary part is
therefore strictly positive while $m\le n-1-j$, and is zero
once $m\ge n-j$. These are precisely the regular negative
parameters $s=-t$ with $0<t<L_j$ and $t>L_j$, respectively.

If $n$ is odd and $j=(n-1)/2$, the same conclusion follows from
$q_j=-\alpha_j$: while $j<N$, one has
$j>(N-1)/2$ and hence $\pi<\theta_j<2\pi$; once $N\le j$,
the root has argument $2\pi$. The transition is at $t=j=L_j$.

In either case the representation obtained from Cauchy's formula is
\[
 d\mu_j(t)=\frac{\operatorname{Im}q_j(-t+i0)}{\pi t}\,dt.
\]
The density is positive on every indicated open interval. It vanishes
past $L_j$, and there are no atoms at the finitely many singular
parameters. Consequently the support is exactly $[0,L_j]$.
Integrating
\[
 \frac{s}{s+L_j}\le\frac{s}{s+t}\le1
\]
and using the already established mass $\mu_j([0,L_j])=M_j$
gives the last inequalities. They are strict because the measure
has positive density on $(0,L_j)$ away from finitely many points.
\end{proof}


\section{An exact certificate for the Appell family}\label{app:appell-certificate}

The two resultants $\mathcal D_i(N)=D_i(N,2)$ eliminate the factor
degree before the degree parameter $N$ is specialized. To prove
$\mathcal D_0(11)\mathcal D_1(11)\ne0$, we work over
$\mathbb F_{479}$. This single exact calculation also proves that
$\mathcal D_0,\mathcal D_1$ are nonzero polynomials in $N$.
The residual polynomials at $N=11$ are
\begin{align*}
 R_2(11,X)&=X^4-66X^3+1430X^2-11880X+31680,\\
 R_3(11,X)&=X^6-99X^5+3630X^4-62370X^3
              +522720X^2-1995840X+2661120.
\end{align*}
The first polynomial has four distinct roots
\[
 \{72,180,353,419\}\subset\mathbb F_{479}.
\]
Its nonempty proper subsets have fourteen distinct sums, namely
\[
 S=\{12,54,72,120,126,180,192,252,293,353,365,419,425,473\}.
\]
These are the roots of $V_2(11,Z)$. Thus the first polynomials
defining $\mathcal D_0(11)$ and $\mathcal D_1(11)$ have roots
$z/6$ and $z/3$, respectively, for $z\in S$.
Their companion factors take the following values; $V_3$
abbreviates $V_3(11,\cdot)$.
\[
\begin{array}{r|rr}
 z&V_3(3z/2)&V_3(2z)\\\hline
12&442&466\\
54&442&222\\
72&222&452\\
120&391&238\\
126&78&27\\
180&181&347\\
192&382&320\\
252&313&457\\
293&313&60\\
353&382&112\\
365&181&74\\
419&78&151\\
425&391&262\\
473&222&53
\end{array}
\]
Every entry is nonzero. Each specialized pair is therefore
relatively prime. The four leading coefficients are
$6^{14},9^{62},3^{14},6^{62}$, all nonzero modulo $479$;
specialization consequently preserves the Sylvester resultants.
This proves their required nonvanishing.

The table can be reproduced using matrices of order at most $20$.
Let $C$ be the companion matrix of
\[
 X^6+380X^5+277X^4+379X^3+131X^2+153X+275
       \quad\text{over }\mathbb F_{479}.
\]
For $1\le k\le5$, let $C^{[k]}$ act on $\bigwedge^k\mathbb F_{479}^6$
by
\[
 C^{[k]}(v_1\wedge\cdots\wedge v_k)
 =\sum_{j=1}^{k}v_1\wedge\cdots\wedge Cv_j\wedge\cdots\wedge v_k.
\]
The eigenvalues of this additive exterior-power operator are the
$k$-element subset sums of the eigenvalues of $C$. Consequently
\[
 V_3(11,Z)=\prod_{k=1}^{5}\det(ZI-C^{[k]}).
\]
Triangularization proves the identity without a separability assumption.


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\end{document}