\documentclass[11pt,reqno]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern,mathtools,amssymb,microtype} \usepackage[margin=1in]{geometry} \usepackage{xurl,xcolor} \usepackage[colorlinks=true,linkcolor=blue!40!black,citecolor=blue!40!black,urlcolor=blue!40!black]{hyperref} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{question}[theorem]{Question} \newcommand{\Q}{\mathbb Q} \newcommand{\Z}{\mathbb Z} \newcommand{\F}{\mathbb F} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Li}{Li} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\lc}{lc} \DeclareMathOperator{\disc}{disc} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\GSp}{GSp} \DeclareMathOperator{\Hdg}{Hdg} \DeclareMathOperator{\MT}{MT} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\CH}{CH} \urlstyle{same} \newcommand{\doi}[1]{\href{https://doi.org/#1}{\nolinkurl{doi:#1}}} \numberwithin{equation}{section} \emergencystretch=2em \raggedbottom \title[Arithmetic of Jacobi polynomials]{Arithmetic of Jacobi 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, 11R32; Secondary 11N13, 14K15, 20B15, 31B05, 33C45.} \keywords{Jacobi polynomial, Legendre polynomial, Kaneko--Zagier polynomial, Galois group, resultant, Newton polygon} \hypersetup{pdftitle={Arithmetic of Jacobi polynomials},pdfauthor={Khai-Hoan Nguyen-Dang}} \begin{document} \begin{abstract} We study the arithmetic of Jacobi polynomials. As applications, we prove the Mahlburg--Ono conjecture on the irreducibility and Galois groups of the four Kaneko--Zagier families, and the Cullinan--Hajir conjecture on the Galois groups of Legendre polynomials. Other applications concern common zeros of Jacobi polynomials, harmonic functions on cones, and hyperelliptic Jacobians. \end{abstract} \maketitle \section{Introduction}\label{sec:introduction} The zeros of classical orthogonal polynomials are among the most familiar algebraic numbers in study. Orthogonality places them on a real interval and makes them simple. However, their arithmetic is much less known. An important class of orthogonal polynomials are Jacobi polynomials. More precisely, we use the normalization \begin{equation}\label{eq:jacobi-definition} P_n^{(\alpha,\beta)}(x)=2^{-n}\sum_{j=0}^n \binom{n+\alpha}{n-j}\binom{n+\beta}{j} (x-1)^j(x+1)^{n-j}, \qquad n\ge0, \end{equation} where $\binom zj=z(z-1)\cdots(z-j+1)/j!$ and $\binom z0=1$. For $\alpha,\beta>-1$, this is the degree-$n$ polynomial with $P_n^{(\alpha,\beta)}(1)=\binom{n+\alpha}{n}$ that is orthogonal to every polynomial of degree less than $n$ for the weight $(1-x)^\alpha(1+x)^\beta$ on $(-1,1)$. The Legendre polynomials are $P_n=P_n^{(0,0)}$, so $P_n(1)=1$. We also use Jacobi polynomials outside this orthogonal range in the Galois arguments. Standard identities are collected in \cite[Chapter~IV]{Szego} and \cite[Chapter~18]{DLMF}. For a nonconstant separable polynomial $f\in K[x]$, we write $\Gal(f/K)$ for the Galois group of its splitting field, acting on its roots. The polynomial is irreducible precisely when this action is transitive. The irreducibility problem of Legendre polynomials goes back to Stieltjes in 19th century. In his letter to Hermite of 2 October 1890, he asked whether $P_{2m}$ and $P_{2m+1}/x$ are irreducible over $\Q$ \cite[Letter~275]{Stieltjes}; see also \cite[\S1]{CHL}. The division by $x$ removes the zero forced by odd parity. Schur's work on Hermite and Laguerre polynomials showed how coefficient valuations can determine factorization and Galois groups \cite{Schur}. For Legendre polynomials, Wahab obtained substantial infinite families of irreducible examples by Newton polygons \cite{Wahab}. Related questions for Bessel polynomials were settled through the work of Filaseta and Filaseta--Trifonov \cite{Filaseta,FT}. Deng has recently proved Stieltjes' conjecture in every degree \cite[Theorem~1.1]{Deng}. His argument compares the real size of a factor resultant with the size permitted by its p-adic valuations. This comparison is our starting point. We first give it a common Jacobi form, then develop the arithmetic estimates needed for the different parameter families. The Galois arguments use another local input: ramification at suitable primes produces long cycles on the roots. The interaction of these two methods is also what gives the results on higher derivatives and number fields. \subsection{Galois groups} The first family in our consideration comes from modular forms. Kaneko and Zagier constructed polynomials over $\Q$ whose reductions at suitable primes describe the supersingular $j$-invariants of elliptic curves \cite{KZ}. After an affine change of variable and normalization, their Jacobi factors give the four families \[ P_n^{(\lambda/3,\mu/2)},\qquad \lambda,\mu\in\{-1,1\} \] \cite[Eq.~(3)]{CHK}. Thus their rational factorization describes the algebraic relations among the characteristic-zero lifts of these modular zeros. \begin{theorem}[The Mahlburg--Ono conjecture]\label{thm:kz-all} For every $n\ge1$ and $\lambda,\mu\in\{-1,1\}$, $P_n^{(\lambda/3,\mu/2)}$ is irreducible over $\Q$ and \[ \Gal(P_n^{(\lambda/3,\mu/2)}/\Q)=S_n. \] \end{theorem} Mahlburg and Ono conjectured both assertions in 2004 and proved irreducibility in infinitely many degrees \cite[p.~40 and Theorem~1.1]{MO}. Cullinan and Hajir proved further infinite families and established the implication from irreducibility to $S_n$ through modular index $\ell\le10^9$ \cite[Theorems~1.8--1.9]{CHK}. Cullinan and Gajek-Leonard studied the associated Newton polygons \cite{CGL}. Theorem~\ref{thm:kz-all} treats all four families in every positive degree. In Section~\ref{sec:modular-interpretation} we will give the precise normalization and explain the relation with modular differential equations. Appendix~\ref{sec:local-irred} gives stronger local irreducibility statements in prime-power degrees and in degrees $5\cdot3^a$. For Legendre polynomials, the symmetry of the zeros determines the largest possible group. Write \[ L_{2m+\delta}(x)=P_{2m+\delta}(x)/x^\delta, \qquad m\ge1,\quad\delta\in\{0,1\}. \] The $2m$ roots occur in opposite pairs. Every Galois automorphism preserves these pairs, so the group is contained in \[ W_m=C_2\wr S_m=\F_2^m\rtimes S_m. \] Here $S_m$ permutes the pairs and $\F_2^m$ independently exchanges the two members of each pair. Deng's theorem gives transitivity of the group action. Our next result determines all these permutations. \begin{theorem}[The Legendre Galois groups]\label{thm:legendre-all} For every $m\ge1$ and $\delta\in\{0,1\}$, \[ \Gal(P_{2m+\delta}(x)/x^\delta\,/\Q)=W_m. \] \end{theorem} This proves both parts of \cite[Conjecture~1.5]{CHL}: the induced group on the pairs is $S_m$, and the kernel of this action is all of $\F_2^m$. Cullinan and Hajir proved the first part, conditional on irreducibility, through reduced degree $10^9$ \cite[Theorem~1.9(a)]{CHL}. Our argument removes that bound and determines the kernel in every degree. \subsection{Integral parameters} Now we consider parameters $(\alpha,\beta)$ of Jacobi polynomials to be integers. \begin{theorem}\label{thm:main} Fix distinct nonnegative integers $a,b$. There is an effectively computable integer $N(a,b)$ such that, for every $n\ge N(a,b)$, $P_n^{(a,b)}$ is irreducible over $\Q$ and has Galois group $S_n$. \end{theorem} Moreover, for every fixed number field $E$, splitting field of $P_n^{(a,b)}$ is linearly disjoint from $E$ over $\Q$ for all sufficiently large $n$, effectively in $a,b,E$. Consequently the polynomial remains irreducible over $E$ with group $S_n$. The restriction to sufficiently large degree is necessary, for example, take $P_2^{(4,1)}(x)=3x(3x+2)$. Also, one can note that negative integral parameters give endpoint factors. After removing them, the same conclusion holds for each fixed integral pair with unequal absolute values (Corollary~\ref{cor:integral-parameters}). In our statements the degree varies and the parameters are fixed, while the Jacobi families studied in \cite{CHS} have a different variation: the degree is fixed and a rational parameter varies. Equal parameters connect the problem to derivatives. Define the Gegenbauer polynomials by \[ (1-2xt+t^2)^{-\lambda} =\sum_{j\ge0}C_j^\lambda(x)t^j. \] For an integer $r\ge0$, the polynomial $C_n^{r+1/2}$ is a nonzero rational multiple of $P_n^{(r,r)}$ \cite[(18.7.1)]{DLMF}, and \begin{equation}\label{eq:derivative-gegenbauer} \frac{d^r}{dx^r}P_{n+r}(x) =(2r-1)!!\,C_n^{r+1/2}(x), \qquad (2r-1)!!=\prod_{j=1}^r(2j-1). \end{equation} The empty product is $1$. A single superscript $P_\ell^{(r)}$ will denote a derivative; two superscripts denote Jacobi parameters. After removing the zero forced by parity, put \begin{equation}\label{eq:reduced-gegenbauer} F_{m,r,\delta}(x)=\frac{C_{2m+\delta}^{r+1/2}(x)}{x^\delta}, \qquad m\ge1,\quad\delta\in\{0,1\}. \end{equation} This polynomial has degree $2m$ in $x$. All irreducibility assertions for $F_{m,r,\delta}$ refer to that full degree. \begin{theorem}\label{thm:symmetric} For integers $r\ge0$, $m\ge1$, and $\delta\in\{0,1\}$, the polynomial $F_{m,r,\delta}$ is irreducible over $\Q$ whenever \[ m\ge\max\{100r,1000\}. \] There is also an effective absolute integer $M_0$ such that the same conclusion holds whenever $m\ge\max\{8r,M_0\}$. \end{theorem} The conclusion is uniform in the parameter: the derivative order may grow with $m$. In particular, the second assertion covers every $0\le r\le m/8$ once $m$ exceeds one absolute threshold. For derivative orders $r=0$ and $r=1$, the conclusion holds without any restriction on $m$. \begin{theorem}\label{thm:all-degrees} For every $m\ge1$, the polynomials \[ P_{2m}(x),\qquad P_{2m+1}(x)/x,\qquad P'_{2m+1}(x),\qquad P'_{2m+2}(x)/x \] are irreducible over $\Q$. \end{theorem} The first two assertions are Deng's theorem~\cite{Deng}. The last two say that the nonzero critical points of each Legendre polynomial $P_\ell$, $\ell\ge3$, form a single rational Galois orbit. Rolle's theorem places these critical points between consecutive zeros; the arithmetic assertion makes them conjugate over $\Q$. It also gives a common-zero theorem: distinct members of $\{P_\ell,P_\ell':\ell\ge1\}$ have no common nonzero root (Corollary~\ref{cor:common-roots}). This extends Deng's common-zero result for the Legendre polynomials themselves \cite{DengRoots}. \subsection{Rational parameters} We consider parameters $(\alpha,\beta)$ to be rational with certain conditions. One can observe that local ramification determines the Galois group on almost all roots even when irreducibility of the whole polynomial is not known. The condition below concerns the reduced denominators of the parameters and their sum. It ensures that the relevant Jacobi coefficients have one long Newton edge at suitably chosen primes. \begin{theorem}\label{thm:jacobi} Fix $\alpha=a/u$ and $\beta=b/v$ in lowest terms, with $u,v>0$, and let $D$ be the reduced denominator of $\alpha+\beta$. Assume \[ D\text{ is even},\qquad D\ge\max\{u,v\},\qquad \alpha,\beta\notin\{-1,-2,\ldots\}. \] For all sufficiently large $n$, $f_n=P_n^{(\alpha,\beta)}$ has a unique monic irreducible factor $h_n$ of degree $M_n>n/2$, with \[ \Gal(h_n/\Q)=S_{M_n},\qquad n-M_n=O_{\alpha,\beta,A}\!\left(\frac{n}{(\log n)^A}\right) \quad\text{for every fixed }A>0. \] Write $f_n=\kappa_nh_ng_n$, with $g_n$ monic and $\kappa_n\in\Q^\times$. The splitting fields of $h_n$ and $g_n$ are linearly disjoint over $\Q$, and restriction gives \begin{equation}\label{eq:general-product} \Gal(f_n/\Q)=S_{M_n}\times\Gal(g_n/\Q). \end{equation} For every fixed $c\in\Q^\times$ and all sufficiently large $n$, $h_n(1+cx^2)$ is irreducible with group $W_{M_n}$. Its splitting field is linearly disjoint from that of $g_n(1+cx^2)$, and \begin{equation}\label{eq:general-quadratic-product} \Gal(f_n(1+cx^2)/\Q)=W_{M_n}\times\Gal(g_n(1+cx^2)/\Q). \end{equation} The splitting fields of $h_n$ and $h_n(1+cx^2)$ are also each linearly disjoint from every fixed number field once $n$ is sufficiently large. \end{theorem} The threshold may depend on the fixed parameters, on $c$, and on the number field; in the degree estimate it may also depend on $A$. A constant polynomial has splitting field $\Q$ and trivial Galois group. The theorem separates a large symmetric part from a small complementary part and proves that their splitting fields have no common extension of $\Q$. No orthogonality hypothesis is needed here. Some restriction on the parameters is necessary: the specialization $\alpha=\beta=-1/2$ gives Chebyshev polynomials \cite[(18.7.3)]{DLMF}, whose zeros lie in cyclotomic fields and whose Galois groups are abelian. The small complementary factor $g_n$ in the theorem is not known to be constant so eventual irreducibility under the stated denominator condition remains a natural question. The integral irreducibility theorem and the rational-parameter Galois theorem meet through a classical quadratic transformation. For $\delta=0,1$, the polynomial $F_{m,r,\delta}(x)$ is a nonzero rational multiple of \[ P_m^{(\delta-1/2,r)}(1-2x^2). \] The parameters $\delta-1/2$ and $r$ satisfy the denominator condition with $D=2$. Write $F_{m,r,\delta}(x)=Q_{m,r,\delta}(x^2)$, where $Q_{m,r,\delta}\in\Q[t]$ has degree $m$. \begin{corollary} \label{cor:derivative-galois} Fix an integer $r\ge0$ and a number field $E$. For all sufficiently large $m$, and for both $\delta\in\{0,1\}$, the polynomials $F_{m,r,\delta}$ and $Q_{m,r,\delta}$ are irreducible over $E$, with \[ \Gal(F_{m,r,\delta}/E)=W_m,\qquad \Gal(Q_{m,r,\delta}/E)=S_m. \] Their splitting fields over $\Q$ are each linearly disjoint from $E$ over $\Q$. \end{corollary} The two groups describe the two natural sets of roots: $S_m$ acts on their squares, while $W_m$ also records the independent changes of sign. Here $r$ and $E$ are fixed. For $r=0$ and $E=\Q$, Theorem~\ref{thm:legendre-all} gives every degree. For $r=1$, irreducibility holds in every degree, whereas the full wreath-product conclusion is proved for sufficiently large degree. \subsection{The resultant method} The idea is to estimate the resultant in two ways: the real geometry of the roots gives a lower bound, while their local arithmetic gives an upper bound. Section~\ref{sec:comparison} develops the common Jacobi form of this argument. It extends Deng's comparison of real and $p$-adic factor resultants for Legendre polynomials \cite[\S1.2]{Deng}. Fix rational parameters $\alpha,\beta>-1$ and put $y_n(t)=P_n^{(\alpha,\beta)}(1-2t)$. This change of variable preserves irreducibility over $\Q$ and places the roots in $(0,1)$. Suppose that a primitive integral multiple $H=\kappa y_n$ admits a factorization \[ H=AB,\qquad A,B\in\Z[t]\ \text{primitive},\qquad 1\le d=\deg A\le e=\deg B. \] The roots are simple by orthogonality \cite[\S18.2(vi)]{DLMF}, so the resultant $R=\Res(A,B)$ is a nonzero integer. Consequently, \begin{equation}\label{eq:intro-product-formula} \log|R|=\sum_p v_p(R)\log p, \end{equation} where $v_p(p)=1$. We seek incompatible bounds for the two sides, uniformly in the possible factor degree $d$. The common input is an auxiliary polynomial constructed from orthogonality. Let $I$ denote integration against the Jacobi weight $t^\alpha(1-t)^\beta\,dt$ on $(0,1)$, normalized so that $I(1)=1$. Its moments are \[ \mu_j=I(t^j) =\frac{(\alpha+1)_j}{(\alpha+\beta+2)_j}\in\Q, \] where $(u)_0=1$ and $(u)_j=u(u+1)\cdots(u+j-1)$ for $j\ge1$; see \eqref{eq:jacobi-moments}. Define \[ Q_H(t)=I_s\!\left(\frac{H(t)-H(s)}{t-s}\right), \qquad \sigma(t)=t(1-t). \] The divided difference is a polynomial, so $Q_H\in\Q[t]$. Orthogonality and the Jacobi differential equation give the identity of Lemma~\ref{lem:jacobi-wronskian}, which at a root $z$ of $H$ becomes \[ \sigma(z)H'(z)Q_H(z)=C_H, \qquad C_H=\kappa^2C_n\in\Q_{>0}. \] Here $C_n$ is the explicit constant in \eqref{eq:jacobi-wronskian}. Because the moments and the constant are rational, this identity can be used both over $\mathbb R$ and over $\Q_p$. At the real place, the Gaussian quadrature weight $\omega_z$ attached to a root $z$ satisfies $Q_H(z)=H'(z)\omega_z$. These weights are positive and sum to one \cite[\S3.5(v), equations (3.5.18) and (3.5.20.1)]{DLMF}. Since $04(\ell-1) \] (Corollary~\ref{cor:angular}). This gives a simultaneous restriction on degree and order in Agranovsky's question concerning homogeneous harmonic polynomials divisible by this cone \cite[\S6, Question~4]{Agranovsky}. It complements the theorem of Mangoubi and Weller Weiser~\cite[Theorem~2]{MWW}: the order must be even, and each fixed order admits at most one degree. The new estimate restricts the possible modes without giving a complete classification. The third application concerns hyperelliptic Jacobians. The roots of a separable polynomial $f$ are the finite branch points of the double cover with affine equation $y^2=f(x)$. Their permutation action determines the Galois action on the $2$-torsion of the Jacobian, connecting the arithmetic of the polynomial with that of an abelian variety. Let $J_f$ be the Jacobian of the smooth projective model. If $f\in\Q[x]$ has degree $d\ge9$ and $\Gal(f/\Q)=S_d$, Zarhin's theorem \cite[Theorem~2.4]{Zarhin02} gives \[ \End_{\overline{\Q}}(J_f)=\Z. \] His results also establish the Hodge and Tate conjectures on every power of $J_f$, with the corresponding classes generated by divisor classes, and the Mumford--Tate conjecture for $J_f$ and its powers \cite[Theorems~8.1 and~9.1, and Remark~9.2]{Zarhin02}. Corollary~\ref{cor:jacobians} records these consequences for our families. In particular, Theorem~\ref{thm:kz-all} makes them available for the Jacobians associated with all four Kaneko--Zagier families in every degree at least $9$. \subsection{Organization} Section~\ref{sec:comparison} develops the common resultant comparison. Section~\ref{sec:integral} treats integral parameters and derivatives, and Section~\ref{sec:kz-irreducibility} treats Kaneko--Zagier irreducibility. Section~\ref{sec:galois-groups} proves the Galois and splitting-field results. Section~\ref{sec:applications} gives the applications. The appendices contain local irreducibility theorems, the explicit cutoff, and the finite calculations needed for the all-degree statements. The finite Galois inputs are the published conditional results in \cite{CHL,CHK}, used with their stated degree and modular-index ranges. \subsection*{Disclosure of AI assistance} This work was developed by the author with assistance from ChatGPT (OpenAI). The author developed the overall approach and methods, building on the prior work cited in the manuscript. During the research and preparation of the manuscript, ChatGPT was used to discuss and revise mathematical arguments, improve the exposition, and search for relevant literature. It also assisted with all computations presented in the appendices. \section{A common resultant comparison}\label{sec:comparison} We now establish the estimates used in the irreducibility proofs. The aim is to bound a factor resultant from below at the real place and from above through its prime divisors. Both estimates must be proportional to the smaller factor degree, so that the comparison excludes every proper factor at once. The common input is a Jacobi identity obtained from orthogonality. We first use it to bound derivatives at the real roots and distances between roots over a $p$-adic field. We then specialize to integral weights, where the moment denominators admit a particularly simple bound. The final two lemmas turn these estimates into a form ready for the arithmetic comparison. For a prime $p$, we normalize the valuation by $v_p(p)=1$, extend it to $\overline{\Q}_p$, and set $v_p(0)=+\infty$. The Gauss norm is the maximum of the absolute values of the coefficients. A polynomial in $\Z_p[t]$ is primitive when its Gauss norm is $1$. Globally, a primitive integral polynomial has coefficient gcd $1$. All logarithms are natural, and empty products equal $1$. Write $(z)_j=z(z+1)\cdots(z+j-1)$, with $(z)_0=1$. \subsection{The Jacobi identity} Fix real $\alpha,\beta>-1$, put $c=\alpha+\beta$, and let $y_n(t)=P_n^{(\alpha,\beta)}(1-2t)$, where $n\ge1$. The endpoint expansion is \begin{equation}\label{eq:irred-coefficients} y_n(t)=\sum_{j=0}^n(-1)^j \binom{n+\alpha}{n-j}\binom{n+c+j}{j}t^j. \end{equation} Let $I$ denote integration against the probability measure \[ d\omega(t)=\frac{t^\alpha(1-t)^\beta}{B(\alpha+1,\beta+1)}\,dt, \qquad 00. \end{equation} At every root $t_i$ of $y_n$, \begin{equation}\label{eq:root-derivative-lower} |y_n'(t_i)|\ge2\sqrt{C_n}, \qquad C_n\longrightarrow \frac{\Gamma(c+2)}{\Gamma(\alpha+1)\Gamma(\beta+1)}>0. \end{equation} \end{lemma} \begin{proof} Write $\tau=\sigma'+\rho=\alpha+1-(c+2)t$ and $\Lambda=n(n+c+1)$. The Jacobi equation is \begin{equation}\label{eq:unit-interval-jacobi-ode} \sigma y_n''+\tau y_n'+\Lambda y_n=0, \end{equation} Writing $d\omega=w(t)\,dt$, the density satisfies $(\sigma w)'=\tau w$. For real $t>1$, put \[ m(t)=I_s((t-s)^{-1}),\qquad q(t)=I_s(y_n(s)/(t-s)). \] Integration by parts gives \[ \sigma m'-\rho m=c+1, \qquad \sigma q''+(\sigma'-\rho)q'+(\Lambda+c)q=0. \] The endpoint terms vanish because $\alpha,\beta>-1$ and $\sigma w$ vanishes at both endpoints. Substituting $Q_n=y_nm-q$ gives \[ \sigma Q_n''+(\sigma'-\rho)Q_n'+(\Lambda+c)Q_n =2(c+1)y_n'. \] Differentiating the left side of \eqref{eq:jacobi-wronskian}, and using these two differential equations, shows that it is constant. Orthogonality gives \[ q(t)=\frac{I(y_n^2)}{\operatorname{lc}(y_n)}t^{-n-1} +O(t^{-n-2}). \] Consequently the constant is $(2n+c+1)I(y_n^2)$, which equals $C_n$ by the Jacobi norm formula. The identity is polynomial, so its validity for $t>1$ proves it everywhere. For a root $t_i$, let \[ \ell_i(t)=\frac{y_n(t)}{(t-t_i)y_n'(t_i)}. \] Since $\ell_i^2-\ell_i$ is divisible by $y_n$, with quotient of degree at most $n-2$, orthogonality yields \[ \omega_i:=I(\ell_i)=I(\ell_i^2)>0. \] Also $\sum_i\ell_i=1$, so $\omega_i\le1$. The definition of $Q_n$ gives $Q_n(t_i)=y_n'(t_i)\omega_i$. Thus \[ C_n=t_i(1-t_i)y_n'(t_i)^2\omega_i \le\tfrac14y_n'(t_i)^2. \] Finally, the stated limit follows by expressing the rising factorials in $C_n$ as gamma quotients. \end{proof} The numbers $\omega_i$ in the proof are the Gaussian quadrature weights. Indeed, if $\deg q\le2n-1$, division by $y_n$ gives $q=y_ns+r$ with $\deg s,\deg r0$. Extend $I$ linearly to the local field containing $b$. The polynomial \[ q(t)=\frac{H(t)}{1-bt} \] has integral coefficients in that field: if $H_j$ and $q_j$ denote the coefficients of $H$ and $q$, respectively, coefficient comparison gives $q_j=H_j+bq_{j-1}$ for $0\le j\gamma>0$ at every root. Assume that the projective row sums for $h(2X+1)$ at $2$ are at most $D_2$. Choose $\kappa>0$ so that $H=\kappa h$ is primitive integral, and suppose \[ H=FG,\qquad 1\le d=\deg F\le e=\deg G,\qquad d+e=N, \] with $F,G\in\Z[x]$ primitive and with positive leading coefficients. Put \[ R=\Res(F,G),\quad \mathcal R=|R|/2^{v_2(R)},\quad u=\lc F,\quad c_F=\min_jv_2([X^j]F(2X+1)), \] and $V_F=\prod_{i2^{c_F}$ and \begin{equation}\label{eq:master-resultant} \frac{\log\mathcal R}{d}> e\log2-\frac{\log V_F}{d}+\log\frac{\gamma}{b_h} -D_2\log2+\frac{e-d}{d}\log\frac{u}{2^{c_F}}. \end{equation} In particular, \begin{equation}\label{eq:master-coarse} \frac{\log\mathcal R}{d}> (N-3)\log2-\log d+\log\frac{\gamma}{b_h}-D_2\log2. \end{equation} \end{lemma} \begin{proof} The derivative and resultant formulas give \[ |R|=\frac{u^{e-d}}{V_F}\prod_{F(\alpha)=0}|H'(\alpha)| >\frac{u^{e-d}}{V_F}(\kappa\gamma)^d. \] Define $c_G$ as for $c_F$. The affine resultant formula and the normalized row bound give \[ v_2(R)\le ec_F+dc_G-de+dD_2. \] Both $F(1)$ and $F(-1)$ are nonzero integers divisible by $2^{c_F}$. Therefore \[ u^2=\frac{|F(1)F(-1)|}{\prod_i(1-\alpha_i^2)}>2^{2c_F}. \] After subtracting the binary estimate, the first two displays yield the right side of \eqref{eq:master-resultant}, with $-\log b_h$ replaced by $\log\kappa-(c_F+c_G)\log2$. Write $\kappa=2^k/c$, with $c$ positive and odd. Since $h\in\Z[1/2][x]$ and $H$ is primitive, $c$ divides the odd part of $b_h$. If $t_2$ is the minimum binary coefficient valuation of $h(2X+1)$, Gauss multiplicativity gives $c_F+c_G=k+t_2$, whereas $t_2\le v_2(b_h)$. Thus \[ \log\kappa-(c_F+c_G)\log2 =-\log c-t_2\log2\ge-\log b_h. \] This proves \eqref{eq:master-resultant}. Finally, rescaling Lemma~\ref{lem:interval-vandermonde} from $[0,1]$ to $[-1,1]$ gives \[ V_F\le d^d2^{-(d-1)(d-2)},\qquad -\frac{\log V_F}{d}\ge(d-3)\log2-\log d. \] The same inequality holds for $d=1$. Dropping the nonnegative last term in \eqref{eq:master-resultant} proves \eqref{eq:master-coarse}. \end{proof} \subsection{A bound for the prime powers} The local moment bound contains $\lfloor\log_p S\rfloor$. After summing over primes, these terms are naturally expressed through least common multiples. The following observation groups the higher prime powers into a single smaller interval. For an integer $B\ge1$ and a real number $S\ge1$, write \[ \Psi_B(S)=\sum_{p\le B}\lfloor\log_pS\rfloor\log p. \] \begin{lemma}\label{lem:lcm} If $B\ge9$ and $S\le5B$, then \begin{equation}\label{eq:lcm-bound} \Psi_B(S)\le \log\lcm(1,\ldots,B)+\log\lcm(1,\ldots,\lfloor\sqrt S\rfloor) <(B+\sqrt S)\log3. \end{equation} \end{lemma} \begin{proof} For $p\le B$, let $l=\lfloor\log_pB\rfloor$ and $k=\lfloor\log_pS\rfloor$. Since $B\ge9$, one has $p^l\ge5$, including $p=2,3$. Hence \[ S\le5B<5p^{l+1}\le p^{2l+1}, \] so $k\le2l$ and $k\le l+\lfloor k/2\rfloor=l+\lfloor\log_p\sqrt S\rfloor$. Multiplying these primewise inequalities proves the first bound. The second follows from Hanson's estimate $\lcm(1,\ldots,t)<3^t$ for integers $t\ge1$ \cite{Hanson}. \end{proof} \section{Integral parameters and derivatives}\label{sec:integral} The local bounds of Section~\ref{sec:comparison} involve primes up to $2n+O(1)$ when the parameters are fixed. Summing over all these primes gives an upper bound too large for the real estimate to contradict. We must therefore show that most primes above the half-degree $m=\lfloor n/2\rfloor$ do not divide a factor resultant. For unequal integral parameters, the exceptional primes above $m+O_{a,b}(1)$ divide values of fixed polynomials. Their total logarithmic contribution is only $O_{a,b}(\log n)$. For equal parameters, in the range of the criterion below, parity excludes every odd prime above $m$. In that case, retaining the exact valuations of a factorial ratio also allows the parameter to grow with the degree. These two arguments prove the irreducibility assertions of Theorems~\ref{thm:main} and~\ref{thm:symmetric}, respectively. \subsection{Unequal parameters}\label{sec:asymmetric} By reflection, $P_n^{(b,a)}(-x)=(-1)^nP_n^{(a,b)}(x)$, we may suppose $a>b\ge0$. Write $q=a-b$, $s=a+b$ and $n=2m+\delta$, where $\delta\in\{0,1\}$, and put $b_m=4^{-m}\binom{2m}{m}$. The value at zero supplies the main arithmetic input. For each parity, Lemma~\ref{lem:center} will express it as $(-1)^mb_m$ times a rational function $A_\delta(m)/D_\delta(m)$ whose numerator and denominator have degrees depending only on $a,b$. The numerator identifies the additional large primes that must be retained in the resultant estimate, while the denominator enters the real lower bound. Both therefore cost only $O_{a,b}(\log n)$ in the final comparison, away from the finitely many zeros of the numerator. The following formulas make this dependence explicit: \begin{align} H_\delta&=\left\lfloor\frac{q-\delta}{2}\right\rfloor, &K_\delta&=b+\delta+H_\delta,\label{eq:HK}\\ D_\delta(z)&=2^\delta\prod_{j=1}^{K_\delta}(z+j),\label{eq:D}\\ A_\delta(z)&=\left(\prod_{j=1}^{b+\delta}(2z+j)\right) \sum_{h=0}^{H_\delta}(-1)^h\binom q{\delta+2h} z^{\underline h}\prod_{j=b+\delta+h+1}^{K_\delta}(z+j),\label{eq:A} \end{align} where $z^{\underline h}=z(z-1)\cdots(z-h+1)$ and $z^{\underline0}=1$. The polynomials $A_\delta,D_\delta$ have integer coefficients and degree at most $K_\delta$, and $A_\delta(0)>0$. Recall from Section~\ref{sec:comparison} that \[ B=\binom{n+a}{a},\qquad S=2n+s+1,\qquad T=\prod_{j=1}^{s}(n+j). \] Put \begin{equation}\label{eq:asymmetric-constants} M=m+s+2,\qquad c_{a,b}=\frac{(a+b+1)^{a+b+1}} {(2a+1)^{a+1/2}(2b+1)^{b+1/2}}. \end{equation} \begin{theorem}\label{thm:criterion} Suppose $m\ge2s+10$ and $A_\delta(m)\ne0$. If \begin{equation}\label{eq:criterion} \begin{split} n\log2>{}&(M+\sqrt S)\log3+3\log T+\log|A_\delta(m)|\\ &+\log\left(\frac{8(m+1)SBD_\delta(m)}{c_{a,b}\sqrt m}\right), \end{split} \end{equation} then $P_n^{(a,b)}$ is irreducible over $\Q$. \end{theorem} For fixed $a,b$, the right side is $(n/2)\log3+O_{a,b}(\sqrt n)$, while the left side is $n\log2$. The positive difference $\log2-\tfrac12\log3$ proves eventual irreducibility once the exceptional primes have been controlled. \subsubsection{Central values and reduction} We first prove the central-value formula. Its exact numerator is needed to determine the exceptional primes; a lower bound for its absolute value will also give a lower bound for derivatives at roots. \begin{lemma}\label{lem:center} For $m\ge0$, \begin{equation}\label{eq:center} P_{2m+\delta}^{(a,b)}(0) =(-1)^mb_m\frac{A_\delta(m)}{D_\delta(m)}. \end{equation} If $m\ge1$ and $A_\delta(m)\ne0$, then \[ |P_{2m+\delta}^{(a,b)}(0)| \ge\frac1{2\sqrt m\,D_\delta(m)}. \] In particular, each parity has only finitely many zero central values. \end{lemma} \begin{proof} Rodrigues' formula \cite[(4.3.1)]{Szego} gives \[ P_n^{(a,b)}(0)=(-1)^n2^{-n}[t^n](1-t)^{n+a}(1+t)^{n+b}. \] Expanding $(1-t)^q(1-t^2)^{n+b}$ yields \[ P_n^{(a,b)}(0)=(-1)^m2^{-2m-\delta} \sum_{h=0}^{H_\delta}(-1)^h\binom q{\delta+2h} \binom{2m+\delta+b}{m-h}. \] The identity \[ \frac{\binom{2m+\delta+b}{m-h}}{\binom{2m}{m}} =\frac{\prod_{j=1}^{b+\delta}(2m+j)\,m^{\underline h}} {\prod_{j=1}^{b+\delta+h}(m+j)} \] proves \eqref{eq:center}. It also holds for $h>m$, with both sides zero. At $m=0$, the summands with $h>0$ vanish and the remaining summand is positive, so $A_\delta$ is nonzero. Finally, $|A_\delta(m)|\ge1$ when nonzero, and $b_m\ge1/(2\sqrt m)$ for $m\ge1$. The latter inequality follows by induction from $b_1=1/2$ and $b_{m+1}/b_m=(2m+1)/(2m+2)$. \end{proof} For example, when $(a,b)=(1,0)$ the formulas reduce to \[ A_0(z)=D_0(z)=1,\qquad A_1(z)=2z+1,\qquad D_1(z)=2(z+1). \] Thus the central value never vanishes in either parity in this case. The next lemma controls roots whose reduction is nonzero. For an odd prime $p$, a bar denotes coefficientwise reduction modulo $p$, and roots of reduced polynomials are taken in an algebraic closure of $\F_p$. \begin{lemma}\label{lem:reduction} Suppose $m\ge2s+10$, and let $p>M$ be an odd prime with $p\nmid T$. Write $n=\epsilon p+j$, with $\epsilon\in\{0,1\}$ and $0\le jM,\quad p\nmid TA_\delta(m)\quad\Longrightarrow\quad v_p(R)=0. \end{equation} To see this, $f(1)=B$ is a unit and $f$ is integral at $p$, so $\kappa$ is a unit. Lemma~\ref{lem:reduction} makes every unit root simple modulo $p$. Since $p^2>2n+s$, the leading coefficient $2^{-n}\binom{2n+s}{n}$ has valuation $0$ or $1$. Formula \eqref{eq:center} gives the same alternatives for $f(0)$, because $p>m+K_\delta$ and $p\nmid A_\delta(m)$. For a polynomial $U$ of Gauss norm $1$ with nonzero constant term, Gauss multiplicativity gives \begin{equation}\label{eq:root-masses} v_p(U(0))=\sum_\alpha\max\{v_p(\alpha),0\},\qquad v_p(\lc U)=-\sum_\alpha\min\{v_p(\alpha),0\}. \end{equation} For each primitive integral factor, both sums are nonnegative integers. A total of at most $1$ therefore cannot be positive in both $F$ and $G$. Thus the two factors cannot both have a root reducing to zero, and they cannot both have a root reducing to infinity. Every other residue class contains at most one root of $f$. Every cross projective distance vanishes, proving \eqref{eq:asymmetric-prime-support}. The terms $v_p(T)$ contribute at most $\log T$ after division by $d$. For $p\le M$, Lemma~\ref{lem:lcm} bounds the remaining contribution by $(M+\sqrt S)\log3$. For $p>M$ we have $p^2>S$, and a surviving prime divides $TA_\delta(m)$. Hence \begin{equation}\label{eq:asymmetric-upper} \frac{\log\mathcal R}{d} <(M+\sqrt S)\log3+2\log T+\log|A_\delta(m)|. \end{equation} We now need a lower bound for $|f'(\alpha)|$ at every root $\alpha$ of $f$. The central-value formula gives information at one point; the differential equation extends it to the whole interval. Set $\lambda=n(n+s+1)$ and define the energy \[ E(x)=(1-x^2)f'(x)^2+\lambda f(x)^2. \] At a root $\alpha$, this is $(1-\alpha^2)f'(\alpha)^2$, so a positive lower bound for $E$ gives the required derivative bound. The differential equation gives \[ E'(x)=2\{q+(s+1)x\}f'(x)^2. \] The minimum on $[-1,1]$ occurs at $x_0=-q/(s+1)$. On $[x_0,0]$, \[ 0\le E'(x)/E(x)\le\frac{2\{q+(s+1)x\}}{1-x^2}. \] The integral of the last expression shows that $\sqrt{E(x_0)}\ge c_{a,b}\sqrt{E(0)}$, with the explicit constant in \eqref{eq:asymmetric-constants}. Lemma~\ref{lem:center} yields \[ \sqrt{E(x_0)}\ge c_{a,b}\sqrt\lambda\,|f(0)| \ge\frac{c_{a,b}\sqrt m}{D_\delta(m)}=:\gamma. \] At every root $\alpha$, $\alpha\ne0$ and $|f'(\alpha)|>\gamma$. Apply Lemma~\ref{lem:real-resultant} with $b_h=B$ and $D_2=\mathcal D_2$. Since $d\le m+1$ and $\mathcal D_2\log2\le\log S+\log T$, it gives \begin{equation}\label{eq:asymmetric-lower} \frac{\log\mathcal R}{d}> n\log2-\log T- \log\left(\frac{8(m+1)SBD_\delta(m)}{c_{a,b}\sqrt m}\right). \end{equation} The two bounds contradict \eqref{eq:criterion}. \end{proof} \begin{proof}[Irreducibility in Theorem~\ref{thm:main}] Fix $a>b$. A polynomial root bound gives an effective point beyond which neither $A_0(m)$ nor $A_1(m)$ vanishes. Their coefficients give explicit constants such that \[ |A_\delta(m)|\le C(m+1)^{K_\delta},\quad D_\delta(m)\le2(m+K_\delta)^{K_\delta},\quad T\le(n+s)^s,\quad B\le(n+a)^a. \] For sufficiently large $m$, the difference between the left and right sides of \eqref{eq:criterion} is at least \[ m\log(4/3)-C_1\sqrt m-C_2\log(m+1)-C_3, \] with effective nonnegative constants depending on $a,b$. After division by $m$, every error term decreases to zero. Successive integer doubling therefore gives an effective bound valid for all subsequent $m$. The constants and logarithms can be bounded by rational interval estimates; take the larger bound for the two parities. Reflection treats $b>a$. \end{proof} \subsection{Symmetric parameters}\label{sec:symmetric} Equal parameters give an additional restriction on the reductions. At an odd prime, parity pairs every nonzero root with its distinct negative. A multiple root away from the central point and the endpoints must therefore occur in an opposite pair. The differential equation forces these multiplicities to be large, and the degree then excludes them at the primes under consideration. This is what removes the exceptional large primes present in the unequal-parameter argument. Let $r,m$ be integers with $r\ge0$, $m\ge1$, and put $n=2m+\delta$ for $\delta\in\{0,1\}$. We use the derivative normalization from \eqref{eq:derivative-gegenbauer}: \[ f(x)=P_{n+r}^{(r)}(x)=(2r-1)!!\,C_n^{r+1/2}(x). \] Set \begin{equation}\label{eq:symmetric-constants} S=2n+2r+1,\quad M=m+r,\quad A=\frac{(n+r)!^2}{n!(n+2r)!},\quad K=\binom{m+r}{r},\quad B=\binom{n+r}{r}, \end{equation} and put $A_{\le m}=\prod_{p\le m}p^{v_p(A)}$. We retain the sign of every exponent $v_p(A)$: cancellation between the numerator and denominator will determine the uniform range in $r$. Although $A\le1$, the truncated product $A_{\le m}$ need not be at most $1$. For example, $r=1$, $m=4$, and $\delta=0$ give $A=9/10$ but $A_{\le m}=9/2$. \subsubsection{The exact criterion} \begin{theorem}\label{thm:symmetric-exact} Suppose $m\ge\max\{2r,4\}$. If \begin{equation}\label{eq:symmetric-exact} m\log4>\Psi_m(S)+\log(8\sqrt m)+\log K+\log A_{\le m}, \end{equation} then $C_{2m+\delta}^{r+1/2}(x)/x^\delta$ is irreducible over $\Q$. When also $M\ge9$, the following elementary condition is sufficient: \begin{equation}\label{eq:symmetric-elementary} m\log4>(M+\sqrt S)\log3+\log(8\sqrt m) +\log K+2\log B-\log\binom{2r}{r}. \end{equation} \end{theorem} \begin{proof} Apply Lemma~\ref{lem:local} to $Q(X)=B^{-1}P_n^{(r,r)}(2X+1)$. It gives the row bound \begin{equation}\label{eq:symmetric-row} D_p=L_p+v_p(A),\qquad L_p=\lfloor\log_pS\rfloor. \end{equation} This number is nonnegative, since it bounds a nonnegative row sum. Removing the zero forced by odd parity removes only nonnegative distances, so the same bound holds for $h=f/x^\delta$ after the shift. Equivalently, the removed factor $2X+1$ has Gauss norm $1$ at every prime, including $2$. Differentiation of the Legendre coefficient formula gives \begin{equation}\label{eq:gegenbauer-coefficients} f(x)=2^{-n-r}\sum_{k=0}^m \frac{(-1)^k(2n+2r-2k)!}{k!(n+r-k)!(n-2k)!}\,x^{n-2k}. \end{equation} Its nonzero central coefficient is \begin{equation}\label{eq:gegenbauer-center} [x^\delta]f(x)=\frac{(-1)^m}{2^mm!} (2m+2r+2\delta-1)!!. \end{equation} The other roots are simple and lie in $(-1,1)$. Choose $\kappa>0$ so that $H=\kappa h$ is primitive integral, and suppose $H=FG$ with $1\le d=\deg F\le e=\deg G$, $d+e=2m$. As before, put $R=\Res(F,G)$ and $\mathcal R=|R|/2^{v_2(R)}$. We claim that \begin{equation}\label{eq:symmetric-prime-support} p>m\quad\Longrightarrow\quad v_p(R)=0. \end{equation} Such a prime is odd, exceeds $2r$, and satisfies $p^2>S$. The primitive polynomial $x^\delta H=\kappa f$ has parity $n$ and satisfies the Gegenbauer equation. Its nonzero reduction has degree at most $n<2p$. A multiple root away from $0,1,-1$ would have multiplicity at least $p$, by the lowest-order term of the differential equation. Parity gives the same multiplicity at its distinct negative, forcing degree at least $2p$. Thus no such multiple root exists. For the four remaining residue classes $0,1,-1,\infty$, each total root-valuation sum is at most $1$. Indeed, before primitive normalization, the central coefficient in \eqref{eq:gegenbauer-center} has valuation at most $1$: its odd numerator factors are smaller than $3p$, and $p>m$. The leading coefficient is \[ \lc(f)=\frac{(2n+2r-1)!!}{n!}. \] Its odd numerator factors are smaller than $5p$. If $p\le n$, at most $p$ and $3p$ occur and $n!$ contributes one $p$; if $p>n$, the numerator factors are smaller than $3p$ and $n!$ is a unit. The valuation is at most $1$ in either case. Finally, \begin{equation}\label{eq:symmetric-endpoint} f(1)=\frac{(n+2r)!}{2^rr!n!} =\binom{n+2r}{2r}(2r-1)!!, \qquad |f(-1)|=f(1). \end{equation} The numerator product contains $2r0$. At each nonzero root $\alpha$, $|h'(\alpha)|=|f'(\alpha)|/|\alpha|^\delta>\gamma$. The exact central and endpoint formulas imply \begin{equation}\label{eq:symmetric-energy} \frac{\gamma}{f(1)}= \begin{cases} \sqrt{2m(2m+2r+1)}\,b_m/K,&\delta=0,\\ (2m+1)b_m/K,&\delta=1, \end{cases} \qquad \frac{\gamma}{f(1)}>\frac{\sqrt m}{K}. \end{equation} Lemma~\ref{lem:real-resultant}, with $N=2m$ and $d\le m$, therefore gives \begin{equation}\label{eq:symmetric-lower} \frac{\log\mathcal R}{d}> m\log4-\log(8\sqrt m)-\log K-D_2\log2. \end{equation} Comparing \eqref{eq:symmetric-upper} and \eqref{eq:symmetric-lower} proves \eqref{eq:symmetric-exact}. In particular the binary contributions cancel exactly. For the elementary criterion, retain the nonnegativity of $D_p$. Every prime occurring in $A$ divides $\prod_{j=1}^{2r}(n+j)$. A prime $p>M$ in this product satisfies $2p>n+2r$ and therefore belongs to $(n,n+2r]$. Thus \begin{equation}\label{eq:symmetric-enlarge} \Psi_m(S)+\log A_{\le m} \le\Psi_M(S)+\sum_{nm$, one has $p^2>n+2r$ for large $m$, and $3p>n+2r$. The only possible multiples in \[ A=\frac{\prod_{j=1}^r(n+j)}{\prod_{j=r+1}^{2r}(n+j)} \] are $p$ and $2p$. Consequently \begin{align} \log A-\log A_{\le m} ={}&2\vartheta(n+r)-\vartheta(n)-\vartheta(n+2r)\notag\\ &+2\vartheta((n+r)/2)-\vartheta(n/2) -\vartheta((n+2r)/2).\label{eq:large-A} \end{align} Every argument is between fixed positive multiples of $m$. The prime number theorem gives a uniform $o(m)$ error, and the linear terms cancel. An explicit prime number theorem, for example \cite[Theorem~5.2]{Dusart10}, makes this error effective. The elementary factorial estimate $\log(k!)=k\log k-k+O(\log(k+1))$, obtained by integrating $\log x$, gives uniformly \[ \log K+\log A=r\,g(m/r)+O(\log m). \] Indeed, for $t=m/r$, the two main terms divided by $r$ are \[ (t+1)\log(t+1)-t\log t \] and \[ 2(2t+1)\log(2t+1)-2t\log(2t)-(2t+2)\log(2t+2). \] Their sum is $g(t)$; changing $\delta$ changes the error by $O(\log m)$. Combining the estimates proves \eqref{eq:uniform-asymptotics}. Finally, \[ tg'(t)-g(t)=\log\frac{t+1}{(t+1/2)^2}<0\qquad(t\ge2). \] \end{proof} \begin{proposition}\label{prop:asymptotic-range} Let $c\ge2$ satisfy \begin{equation}\label{eq:asymptotic-condition} c(\log4-1)>g(c). \end{equation} For all sufficiently large $m$, uniformly for $0\le r\le m/c$, both polynomials $C_{2m+\delta}^{r+1/2}(x)/x^\delta$ are irreducible over $\Q$. For rational $c$, the threshold is effectively computable. In particular, $c=8$ is admissible. \end{proposition} \begin{proof} For $r\ge1$, put $t=m/r\ge c$. Lemma~\ref{lem:uniform-asymptotics} and monotonicity give \[ \Psi_m(S)+\log K+\log A_{\le m} \le m\left(1+\frac{g(c)}c\right)+o(m). \] The strict margin in \eqref{eq:asymptotic-condition} absorbs this error and $\log(8\sqrt m)$ in Theorem~\ref{thm:symmetric-exact}. For $r=0$, $K=A=1$ and the conclusion follows directly from that criterion. All errors above are effective; for rational $c$ the strict margin can be bounded below by rational approximation. Finally, \[ 8(\log4-1)-g(8)>0.0097. \] An exact rational verification is recorded in Appendix~\ref{app:finite-derivatives}. This also proves the remaining assertion of Theorem~\ref{thm:symmetric}. \end{proof} \subsubsection{Legendre polynomials and their first derivatives} \begin{proof}[Proof of Theorem~\ref{thm:all-degrees}] The assertions for $P_{2m}$ and $P_{2m+1}/x$ are \cite[Theorem~1.1]{Deng}. To see their relation to the present argument, specialize Theorem~\ref{thm:symmetric-exact} to $r=0$. Then $K=A=1$, and Lemma~\ref{lem:lcm} gives the sufficient inequality \[ m\log(4/3)>\sqrt{4m+3}\log3+\log(8\sqrt m). \] This holds for $m\ge100$: after division by $m$, the right side decreases, and at $m=100$ it is less than $0.275<2/7<\log(4/3)$. Thus the general comparison also recovers these two families in sufficiently large degree. For the first derivatives, take $r=1$ and write $n=2m+\delta$. The parameters in \eqref{eq:symmetric-constants} become \[ K=m+1,\qquad A=\frac{n+1}{n+2},\qquad S=2n+3. \] Since $A_{\le m}\le n+1\le2m+2$, it suffices that \[ m\log(4/3)> \sqrt{4m+5}\log3+\log(8\sqrt m)+\log\bigl(2(m+1)^2\bigr). \] After division by $m$, the right side decreases. At $m=200$ it is less than $(29\cdot1.1+5+12)/200=0.2445<2/7$, so the inequality holds for every $m\ge200$. Lemma~\ref{lem:first-derivative-finite} treats $1\le m\le199$ by exact integer comparisons and reduction modulo primes. This proves both first-derivative assertions in every degree. \end{proof} \section{The Kaneko--Zagier families}\label{sec:kz-irreducibility} \label{sec:resultant-estimates} We turn to irreducibility of the four Kaneko--Zagier families. The comparison of Section~\ref{sec:comparison} requires two further estimates here: one for the common divisor introduced by clearing denominators, and one for the clustering of roots at $2$ and $3$. Both contribute to the leading constants. We calculate them below, then combine the bounds to obtain a positive gap uniform in the possible factor degree. Appendix~\ref{sec:explicit-cutoff} makes the comparison explicit and completes the finite range. \subsection{Primitive normalization} \label{sec:irred-prelim} In this section and Appendix~\ref{sec:explicit-cutoff}, unless stated otherwise, $n\ge1$ and we fix \[ \alpha=\lambda/3,\quad \beta=\mu/2,\quad \lambda,\mu\in\{-1,1\},\quad c=\alpha+\beta, \qquad y_n(t)=P_n^{(\alpha,\beta)}(1-2t). \] All constants implicit in asymptotic notation are uniform in these four choices. The functional $I$, moments $\mu_j$, and constant $C_n$ are those of \eqref{eq:jacobi-moments} and \eqref{eq:jacobi-wronskian}, evaluated at these parameters. Put \begin{equation}\label{eq:primitive-normalization} \kappa_n=2^{n+v_2(n!)}3^{n+v_3(n!)},\qquad g_n=\operatorname{cont}(\kappa_n y_n),\qquad H_n=\frac{\kappa_n}{g_n}y_n. \end{equation} Here $\operatorname{cont}$ is the positive gcd of the integral coefficients. The factor $\kappa_n$ clears the denominators at $2$ and $3$. Removing the content $g_n$ is necessary before applying Gauss's lemma and the local resultant bounds. It also changes the real derivative estimate by the factor $g_n^{-1}$. We shall therefore estimate $g_n$ as carefully as the local contributions to the resultant. \begin{lemma}\label{lem:primitive-kz} The polynomial $H_n$ is primitive and integral. Its leading coefficient is prime to $6$, and $(g_n,6)=1$. Moreover, \[ \log\kappa_n= \left(2\log2+\frac32\log3\right)n+O(\log n). \] \end{lemma} \begin{proof} For $p\ge5$, both generalized binomial coefficients in \eqref{eq:irred-coefficients} are $p$-integral. At $2$, the first is integral and the second has valuation $-j-v_2(j!)$. At $3$, the coefficient has valuation \[ -n-v_3(j!)-v_3((n-j)!) =-n-v_3(n!)+v_3\binom nj. \] The minimum at either prime is attained by the leading coefficient. This proves the integral and unit assertions. The last formula follows from $v_p(n!)=(n-s_p(n))/(p-1)$, where $s_p(n)$ is the sum of the base-$p$ digits of $n$. \end{proof} \subsection{The resultant away from \texorpdfstring{$2$ and $3$}{2 and 3}} \label{sec:good-resultant} Suppose that \[ H_n=A B,\qquad A,B\in\Z[t]\text{ primitive},\qquad k=\deg A\le e=\deg B, \] and put \begin{equation}\label{eq:prime-to-six-resultant} R=|\operatorname{Res}(A,B)|>0, \qquad R^{(6)}=\frac{R}{2^{v_2(R)}3^{v_3(R)}}. \end{equation} The main estimates in this section are \begin{equation}\label{eq:two-arithmetic-asymptotics} \log g_n=(\mathfrak c+o(1))n, \qquad \log R^{(6)}\le(\mathfrak u+o(1))kn, \end{equation} where \begin{equation}\label{eq:arithmetic-constants} \mathfrak c=\frac{\pi}{2\sqrt3}, \qquad \mathfrak u=\frac{43}{20}-\frac{\pi}{\sqrt3}. \end{equation} \subsubsection{A valuation calculation} The content and the moment denominators come from closely related products. Estimating them separately loses the cancellation needed in the resultant comparison. We instead count divisibility by each prime power $q$ at the same time. The indicator $m_q$ records the common divisibility forced on all coefficients at that level; $e_q$ records the contribution that remains in the local bound after the primitive normalization. The crucial fact will be $m_q+e_q\le1$. Fix $p\ge5$ and $q=p^h$. Let $a_q,b_q$ be the least residues of $\alpha,\beta$ modulo $q$, and put \[ c_q=(a_q+b_q)\bmod q,\qquad d_q=\lfloor(a_q+b_q)/q\rfloor, \qquad s=n\bmod q. \] Thus $1\le a_q,b_q\le q-1$ and $d_q\in\{0,1\}$. Define \[ E_a=\mathbf1_{s+a_q\ge q},\quad E_b=\mathbf1_{s+b_q\ge q},\quad E_c=\mathbf1_{s+c_q\ge q}, \] \[ m_q=\mathbf1_{E_a=E_b=1,\ d_q+E_c=1},\quad \eta_q=\mathbf1_{c_q=q-1},\quad b_q^*=\begin{cases}0,&c_q=q-1,\\1-d_q,&c_q\ne q-1,\end{cases} \] and \begin{equation}\label{eq:local-budget-definition} e_q=\eta_q+E_a+E_b-E_c+b_q^*-2m_q. \end{equation} The notation $b_q^*$ is a numerical bound for moments, distinct from the residue $b_q$. \begin{lemma}\label{lem:carry-budget} For every $q=p^h$, $p\ge5$, one has $e_q\in\{0,1\}$ and $m_q+e_q\le1$. Writing $M=12n+5$ and \[ E_p(n)=\sum_{\substack{h\ge1\\p^h\le M}}e_{p^h}, \qquad m_p=v_p(g_n), \] one has \begin{equation}\label{eq:moment-content-budget} v_p(C_n)-2m_p-\min_{0\le j\le2n-1}v_p(\mu_j) \le E_p(n)\le\lfloor\log_p M\rfloor. \end{equation} If $p>\sqrt M$, then $m_p=m_p^{\rm lev}$, where $m_p^{\rm lev}$ means $m_q$ at the single level $q=p$. \end{lemma} \begin{proof} For a rational $p$-adic integer $z$, the contribution at level $q=p^h$ to $v_p\binom zj$ is \[ \mathbf1_{j\bmod q>z\bmod q}. \] This follows by counting multiples of $q$ in the numerator and denominator. Hence the contribution to the coefficient of $t^j$ in \eqref{eq:irred-coefficients}, with $u=j\bmod q$, is \begin{equation}\label{eq:coefficient-level-indicators} \mathbf1_{(s-u\bmod q)>(s+a_q\bmod q)} +\mathbf1_{u+(s+c_q\bmod q)\ge q}. \end{equation} Its minimum over residues $u$ is $m_q$. To see this, if $E_a=0$ the value at $u=0$ is zero. If $E_a=1$, the first indicator is zero on $[q-a_q,s]$, and the second is zero on $[0,q-(s+c_q\bmod q)-1]$. These intervals are disjoint exactly when $E_b=1$ and $d_q+E_c=1$. The same minimum is attained for $0\le j\le n$: if $n\sqrt M$ there is only one level, so the minimum in \eqref{eq:coefficient-level-indicators} is the actual content valuation. \end{proof} The moment bound alone allows primes as large as $12n+5$. For primes exceeding the degree, the differential equation gives more information. A repeated reduced root can occur only at an endpoint or at infinity, and the endpoint and leading-coefficient valuations are too small for two primitive factors to meet there. For $n\ge11$, this removes every prime $p>n$ with $p\ge5$ from the resultant. \begin{lemma} \label{lem:projective-cluster-exclusion} Let $H\in\mathbb Z_p[t]$ be primitive of degree $nn$, then $v_p(R)=0$. \end{lemma} \begin{proof} Fix $p\ge5$. Orthogonality and \eqref{eq:jacobi-wronskian} allow Lemma~\ref{lem:projective-moment-budget} to be applied to $H_n$ with moments \eqref{eq:jacobi-moments} and $C=(\kappa_n/g_n)^2C_n$. Since $\kappa_n$ is a $p$-adic unit, Lemma~\ref{lem:carry-budget} gives $\mathcal E_p\le E_p(n)$, proving \eqref{eq:good-prime-bound}. Now suppose $n\ge11$ and $p>n$. Each of \[ y_n(0)=\frac{(\alpha+1)_n}{n!},\qquad y_n(1)=(-1)^n\frac{(\beta+1)_n}{n!},\qquad \operatorname{lc}(y_n)=(-1)^n\frac{(n+c+1)_n}{n!} \] has valuation at most one: $n!$ is a unit, and each cleared numerator product has length $n\sqrt{12n+5}$ there is only one valuation level. In the class $p\equiv r\pmod6$, where $r=2\lambda+3\mu$, one has $a_p/p=1/3+O(1/p)$; in the other reduced class, $a_p/p=2/3+O(1/p)$. In both classes $b_p/p=1/2+O(1/p)$. The indicator calculation in Lemma~\ref{lem:carry-budget} gives the following intervals, up to endpoint changes of size $O(1/p)$: \[ \begin{array}{c|c|c} a_p/p&\{n/p\}\text{ for }e_p=1&\{n/p\}\text{ for }m_p=1\\ \hline 1/3&[0,1/6)\cup[1/2,2/3)&[2/3,1)\\ 2/3&[1/3,1/2)\cup[5/6,1)&[1/2,5/6). \end{array} \] Small primes and higher prime powers contribute $O(\sqrt n\log n)$. For the content this follows also from $v_p(g_n)\le v_p(\operatorname{lc}(y_n)) \le\lfloor\log_p(12n+5)\rfloor$. Apply the prime number theorem in the two classes modulo $6$ to the intervals in the table; the fixed-modulus consequence of \cite[Corollary~1.7]{BMOR} suffices. For each fixed bound on $\ell=\lfloor n/p\rfloor$ only finitely many intervals occur. The remaining primes contribute $O(n/\ell)$, so the resulting convergent sums may be obtained by first fixing this bound and then letting it tend to infinity. Endpoint perturbations contribute $o(n)$; after removing $p\le\sqrt{12n+5}$ there are only $O(\sqrt n)$ endpoints, each shifted by $O(1)$. Lemma~\ref{lem:good-prime-resultant} permits restriction to $p\le n$. The upper bound for the resultant is therefore \begin{align*} \frac1n\sum_{5\le p\le n}E_p(n)\log p\longrightarrow \frac12\sum_{\ell\ge1}\bigg(&\frac1\ell-\frac1{\ell+1/6} +\frac1{\ell+1/3}-\frac1{\ell+2/3}\\ &+\frac1{\ell+5/6}-\frac1{\ell+1}\bigg) =\frac{43}{20}-\frac{\pi}{\sqrt3}. \end{align*} For the content all primes are included, so the corresponding sum starts at zero: \[ \frac{\log g_n}{n}\longrightarrow \frac12\sum_{\ell\ge0}\left( \frac1{\ell+2/3}-\frac1{\ell+1} +\frac1{\ell+1/2}-\frac1{\ell+5/6}\right) =\frac{\pi}{2\sqrt3}. \] These evaluations follow from the partial-fraction expansion of $\pi\cot(\pi z)$, or from the reflection and duplication formulas for the logarithmic derivative of $\Gamma$ \cite[Eqs.~5.5.4 and 5.5.8]{DLMF}. \end{proof} \subsection{The primes \texorpdfstring{$2$ and $3$}{2 and 3}}\label{sec:bad-resultant} The monic endpoint expansions place every root in the ball $v_2(t)\ge1$ at $2$, and in the ball $v_3(1-t)\ge1$ at $3$. The endpoint values of a factor measure the total valuations of its roots in these balls. If two roots have different endpoint valuations, their difference has the smaller valuation. The only additional contribution comes from roots with equal valuations. We bound it by expanding the differential equation at one root. To sum the resulting errors, we use the Newton polygon. For a polynomial $f(X)=\sum_j c_jX^j$ over $\Q_p$ with nonzero constant term, this is the lower convex hull of the points $(j,v_p(c_j))$, omitting zero coefficients. A side of slope $-r$ and horizontal length $d$ corresponds to exactly $d$ roots in $\overline{\Q}_p$ of valuation $r$, counted with multiplicity. Thus inequalities for coefficient valuations can bound the total number and size of exceptional root valuations. For the factorization $H_n=AB$, put \[ D_{2,A}=v_2(A(0)),\quad D_{2,B}=v_2(B(0)),\qquad D_{3,A}=v_3(A(1)),\quad D_{3,B}=v_3(B(1)). \] The leading coefficients of both factors are units at $2$ and $3$. \begin{proposition}\label{prop:bad-prime-resultant} Let $s_p(m)$ denote the sum of the base-$p$ digits of $m$, and put $\delta=(\mu+1)/2$. Then \begin{align*} v_2(R)&\le\min(eD_{2,A},kD_{2,B}) +\frac{ke}{2}+\frac{k}{2}s_2(n),\\ v_3(R)&\le\min(eD_{3,A},kD_{3,B}) +\frac{ke}{3}+\frac{k}{6}s_3(2n+\delta). \end{align*} These bounds hold for every factorization; no restriction on $k$ is needed. In particular, the error beyond the displayed bilinear terms is $O(k\log n)$. \end{proposition} \begin{proof} The proof has two parts. First, the Taylor coefficients at a root give a uniform upper bound for the extra closeness of another root with the same endpoint valuation. The error in that bound measures how far the root valuation lies below $2$ at the prime $2$, or below $3/2$ at the prime $3$. Second, the coefficient valuations and the Newton polygon show that the total of these deficits is controlled by a digit sum, hence is $O(\log n)$. At $p=2$ use the endpoint coordinate $z=t$ and $b=\alpha$. At $p=3$ use $z=1-t$ and $b=\beta$. In either case the corresponding Jacobi polynomial $P(z)$ satisfies \[ \sigma P''+\tau P'+\Lambda P=0, \qquad \sigma=z(1-z),\quad \tau=b+1-(c+2)z, \quad \Lambda=n(n+c+1). \] Every root $z$ has $v_p(z)\ge1$. Indeed, the monic endpoint coefficients are given by \begin{equation}\label{eq:monic-endpoint} \frac{n!P_n^{(a,b)}(2X-1)}{(n+a+b+1)_n} =\sum_{j=0}^n(-1)^{n-j}\binom nj \frac{(b+j+1)_{n-j}}{(n+a+b+j+1)_{n-j}}X^j. \end{equation} Here $b\in\mathbb Z_p$, while every factor in the denominator has valuation $-1$. Thus their valuations are at least $n-j$. Fix a root $z$, put $r=v_p(z)\ge1$, and let \[ h=\min(1,r-1),\qquad u_j=z^{j-1}\frac{P^{(j)}(z)}{P'(z)}. \] Thus $u_0=0$ and $u_1=1$. Differentiating the differential equation $m$ times gives \begin{equation}\label{eq:taylor-valuation-recurrence} \begin{gathered} u_{m+2}=-a_mu_{m+1}-b_mu_m,\\ a_m=\frac{b+1+m-(c+2+2m)z}{1-z},\qquad b_m=\frac{z(n-m)(n+m+c+1)}{1-z}. \end{gathered} \end{equation} Since $v_p(c+2+2m)=v_p(n+m+c+1)=-1$, we have $v_p(b_m)\ge r-1\ge h$ and $v_p(a_m)\ge0$. At $2$, $\alpha+1$ is even in $\mathbb Z_2$, so $v_2(a_m)\ge h$ whenever $m$ is even. At $3$, $\beta+1$ is congruent to $2$ if $\delta=0$ and to $0$ if $\delta=1$, so $v_3(a_m)\ge h$ whenever $m\equiv1-\delta\pmod3$. Induction in \eqref{eq:taylor-valuation-recurrence} therefore gives \begin{equation}\label{eq:taylor-coefficient-gain} v_2(u_j)\ge\left\lfloor\frac j2\right\rfloor h, \qquad v_3(u_j)\ge\left\lfloor\frac{j+\delta}{3}\right\rfloor h. \end{equation} For example, in the $3$-adic induction the first term gains an extra $h$ precisely when the target floor increases, and the second term gains at least $h$ at every step. The initial values satisfy the stated bounds. For $p=2$ or $3$, $j\ge2$, and $0\le h\le1$, Legendre's formula implies \begin{equation}\label{eq:factorial-radius-bound} \frac{v_p(j!)-\lfloor j/p\rfloor h}{j-1} \le\frac1{p-1}-\frac hp. \end{equation} To verify it, write $j=pq+s$, $0\le s1/(p-1)-h/p$. Thus, writing $\rho_p=p/(p-1)$, we obtain the convenient weaker bound \begin{equation}\label{eq:paired-prime-separation} v_p(w-z)-\min(v_p(w),v_p(z)) \le\frac1p+\frac{(\rho_p-v_p(z))_+}{p} \quad\text{if }v_p(w)=v_p(z). \end{equation} The excess is zero when the root valuations differ. Here $(x)_+=\max(x,0)$; the bound follows from \[ \frac1{p-1}-\frac{\min(1,r-1)}p \le\frac1p+\frac{(\rho_p-r)_+}{p}. \] It remains to control the total endpoint deficits. Let $c_j$ be the coefficients of the monic normalization of $y_n(t)$. At $2$ their exact valuations are \[ v_2(c_j)=2(n-j)+\Psi(j),\qquad \Psi(j)=s_2(j)-s_2(n)+v_2\binom{n+\alpha}{n-j}. \] The generalized binomial coefficient is $2$-integral, so $\Psi(j)\ge-s_2(n)$, and $\Psi(n)=0$. If $W$ is the lower convex hull of the points $(j,\Psi(j))$, then the Newton-polygon theorem gives \[ \sum_i(2-v_2(t_i))_+ =\int_0^n(W'(x))_+\,dx =W(n)-\min W\le s_2(n). \] The middle equality follows from convexity: a convex function first decreases and then increases. At $3$, the monic polynomial in $z=1-t$ has coefficient valuations \[ \frac32(n-j)+\frac12\Phi(j),\qquad \Phi(j)=s_3(2j+\delta)+s_3(n-j)-s_3(2n+\delta). \] Here $\Phi(j)\ge-s_3(2n+\delta)$ and $\Phi(n)=0$. Applying the same positive-variation argument gives \[ \sum_i\left(\frac32-v_3(1-t_i)\right)_+ \le\frac12s_3(2n+\delta). \] The leading coefficients of $A,B$ are units at $2,3$, so the valuation of their resultant is the sum of the cross-factor root difference valuations. The sum of the minimum valuations is at most both $eD_{p,A}$ and $kD_{p,B}$. Sum \eqref{eq:paired-prime-separation}, using the root of $B$ for the deficit term. The sum is at most $ke/p$ plus $k/p$ times the total endpoint deficit among the roots of $B$, and hence among all roots of $H_n$. The two asserted inequalities follow. \end{proof} \subsection{The archimedean comparison} \label{sec:irred-proof} The real roots lie in $(0,1)$. Their small Vandermonde product therefore makes the resultant large once the derivatives are bounded below. The leading coefficient of a factor contributes another positive term, because its values at $0$ and $1$ are nonzero integers. These are the same endpoint values that occur in the local bounds at $2$ and $3$. Combining the bounds before optimizing eliminates the unknown way in which the two factors share the endpoint valuations. Retaining the Vandermonde term makes the final estimate uniform in the factor degree. Write $\gamma_n=\kappa_n/g_n$, and set \begin{align} D_{2,n}&=n+\sum_{j=1}^n v_2(3j+\lambda),& T_{2,n}&=D_{2,n}\log2/n,\label{eq:exact-two-total}\\ D_{3,n}&=n+\sum_{j=1}^n v_3(2j+\mu),& T_{3,n}&=D_{3,n}\log3/n.\label{eq:exact-three-total} \end{align} These are the endpoint valuations of $H_n$, and \begin{equation}\label{eq:endpoint-asymptotics} D_{2,n}=2n+O(\log n),\qquad D_{3,n}=\tfrac32n+O(\log n). \end{equation} Define \begin{equation}\label{eq:arch-constants-finite} \begin{split} L_n&=\log(e^{T_{2,n}}+e^{T_{3,n}}),\qquad E=\tfrac12\log2+\tfrac13\log3,\\ J_n&=\tfrac12s_2(n)\log2+\tfrac16s_3(2n+\delta)\log3, \qquad \delta=(\mu+1)/2. \end{split} \end{equation} \begin{proposition}\label{prop:arch-resultant} For every factorization $H_n=AB$ with $1\le k=\deg A\le n/2$, \begin{equation}\label{eq:exact-arch-resultant} \begin{split} \frac{\log R^{(6)}}k\ge{}& \log(2\gamma_n\sqrt{C_n})-J_n +(n-2k)L_n\\ &-(n-k)(T_{2,n}+T_{3,n}+E) +\frac{2(k-1)^2}{k}\log2-\log k. \end{split} \end{equation} Consequently, uniformly in all such factorizations, \begin{equation}\label{eq:arch-resultant-lower} \log R^{(6)}\ge(\mathfrak l-o(1))kn, \quad \mathfrak l=\log(4+3\sqrt3)-\tfrac12\log2-\tfrac13\log3 -\frac{\pi}{2\sqrt3}. \end{equation} \end{proposition} \begin{proof} Let $a=|\operatorname{lc}(A)|$, $e=n-k$, and let $t_1,\ldots,t_k$ be the roots of $A$. Differentiating $H_n=AB$ at these roots gives the exact identity \[ R=a^{e-k}\frac{\prod_{i=1}^k|H_n'(t_i)|} {\prod_{i0. \] We may discard this term. Now use Lemma~\ref{lem:primitive-kz}, Proposition~\ref{prop:arithmetic-asymptotics}, \eqref{eq:root-derivative-lower}, and \eqref{eq:endpoint-asymptotics}. The remaining expression tends to $\mathfrak l$, proving \eqref{eq:arch-resultant-lower}. \end{proof} The two uniform bounds already imply eventual irreducibility, since \begin{equation}\label{eq:positive-gap} \mathfrak l-\mathfrak u =\log(4+3\sqrt3)-\tfrac12\log2-\tfrac13\log3 -\frac{43}{20}+\frac{\pi}{2\sqrt3} >\frac14. \end{equation} For reference, $\mathfrak l=0.5991078130\ldots$, $\mathfrak u=0.3362006358\ldots$, and their difference is $0.2629071772\ldots$. For an elementary verification of the rational bound, use \[ \frac{1732}{1000}<\sqrt3<\frac{1733}{1000},\qquad \pi>\frac{157}{50},\qquad \log3>\frac{549}{500},\qquad \log2<\frac7{10}. \] Since $4+3\sqrt3>9\cdot2299/2250$ and $\log(1+t)>t/(1+t)$ for $t>0$, these bounds give \[ \mathfrak l-\mathfrak u> \frac53\frac{549}{500}-\frac7{20}-\frac{43}{20} +\frac{49}{2299}+\frac{1570}{1733}>\frac14. \] The logarithmic bounds follow, for example, from the positive-term series used in Appendix~\ref{sec:finite-verification}. Thus a proper factorization is impossible once $n$ is sufficiently large, with no restriction on the degree of a possible factor. To obtain an all-degree theorem, we retain the finite terms in the same comparison. \subsection{An exact finite criterion}\label{sec:finite-criterion} The resultant estimates give a finite criterion that does not require a factorization of the Jacobi polynomial. Use the quantities in \eqref{eq:exact-two-total}--\eqref{eq:arch-constants-finite} and the residue bounds of Lemma~\ref{lem:carry-budget}, and put \[ S_n=\sum_{\substack{p\text{ prime}\\5\le p\le n}}E_p(n)\log p. \] For $z\ge1$, define \begin{equation}\label{eq:finite-criterion-function} \begin{split} \mathcal H_n(z)={}&\log(2\gamma_n\sqrt{C_n})-J_n +(n-2z)L_n-(n-z)(T_{2,n}+T_{3,n}+E)\\ &+\frac{2(z-1)^2}{z}\log2-\log z. \end{split} \end{equation} \begin{proposition}\label{prop:finite-criterion} Let $n\ge11$. If \[ \min_{1\le k\le\lfloor n/2\rfloor}\mathcal H_n(k)>S_n, \] then $y_n$ is irreducible over $\Q$. More generally, the minimum may be restricted to any set known to contain all possible degrees of a nonconstant factor of degree at most $n/2$. \end{proposition} \begin{proof} A nontrivial factorization would give $S_n\ge\log R^{(6)}/k\ge\mathcal H_n(k)$ by Lemma~\ref{lem:good-prime-resultant} and \eqref{eq:exact-arch-resultant}, a contradiction. \end{proof} The minimization is elementary. Set \[ a_n=T_{2,n}+T_{3,n}+E-2L_n+2\log2. \] The part of $\mathcal H_n(z)$ that depends on $z$ is $a_nz+2\log2/z-\log z$, which is strictly convex for $z>0$. If $a_n\le0$ it is decreasing. If $a_n>0$, its minimum is at \[ q_n=\frac{1+\sqrt{1+8a_n\log2}}{2a_n}. \] Thus on an interval of integer degrees it suffices to check the integers adjacent to the clipped value of $q_n$. The criterion depends only on endpoint valuations, content, and finite sums over primes. Appendix~\ref{sec:finite-verification} uses a slightly weaker form in which the possible factor degree has been eliminated and every comparison reduces to an inequality between linear forms in logarithms. \subsection{Completion in every degree} Appendix~\ref{sec:explicit-cutoff} proves irreducibility for $n\ge250\,000$. The degrees $1\le n\le8332$ are covered by the published computation of Cullinan and Hajir \cite[Remark~4.6]{CHK}; the conversion from their index to the degree is given in Appendix~\ref{sec:finite-verification}. For $8333\le n\le250\,000$, two explicit inequalities from the resultant comparison are verified by integer arithmetic. Together these ranges give irreducibility in every degree. \begin{proof}[Irreducibility in Theorem~\ref{thm:kz-all}] Proposition~\ref{prop:explicit-kz-cutoff} treats $n\ge250\,000$, and Proposition~\ref{prop:finite-kz} treats $1\le n\le250\,000$. The Galois assertion will follow from Theorem~\ref{thm:kz}. \end{proof} \section{Galois groups and splitting fields} \label{sec:galois-groups} Irreducibility gives transitivity on the roots. The remaining task is to find enough permutations in the Galois group to determine it. We obtain these permutations from ramification at suitable primes. The first argument is common to the Legendre and Kaneko--Zagier families and extends to fixed rational parameters. For unequal integral parameters, a different local factorization gives a prime-length cycle. We also keep track of which roots each inertia element fixes. This support information will prove the splitting-field disjointness statements. \subsection{Inertia cycles and permutation groups}\label{sec:cycles} We call a permutation a \emph{single $d$-cycle} if it cyclically permutes $d$ points and fixes every other point; for $d=1$ this means the identity. All valuations $v_p$ are normalized by $v_p(p)=1$. \subsubsection{A local criterion} Under the hypotheses of the next lemma, the Newton polygon has a final edge of slope $1/d$: exactly $d$ roots have valuation $-1/d$, and all other roots have valuation zero. Passing to the reciprocal polynomial isolates the first set in an Eisenstein factor. In a finite Galois extension of $\Q_p$, the inertia subgroup consists of the automorphisms acting trivially on the residue field. We want its action to be a single cycle on the isolated roots. The degree bound $dn,\qquad p\nmid uvD,\qquad 0<|uk+a|N/2$, then $b\mid d$. \end{lemma} \begin{proof} We have $b\le N/2b$. Thus the support of the cycle is a union of whole blocks, and $b\mid d$. \end{proof} Once primitivity is known, a single cycle with at least three fixed points forces the alternating group. We use Jones's theorem in this form; the cycle length need not be prime. \begin{theorem}[Jones {\cite[Corollary~1.3]{Jones}}]\label{thm:jones} A primitive permutation group of degree $N$ containing a nontrivial single cycle with at least three fixed points contains $A_N$. \end{theorem} This form of Jones's theorem uses the classification of finite simple groups. The prime-length case used later is Jordan's classical theorem. For roots occurring in opposite pairs, the largest possible group has two parts: permutations of the pairs and interchanges within each pair. Even after the group on pairs is known to be $S_N$, the interchanges may satisfy a relation. A single $2d$-cycle with odd $d\frac{Dn}{2\varphi(D)(L+\log(3D/2))} -\frac{0.0675Dn}{L^2}-10D\notag\\ &\ge c_D\frac nL \left(\frac{L}{L+\log9}-\frac{0.27}{L} -\frac{40L}{n}\right) >0.84c_D\frac nL.\label{eq:prime-total} \end{align} The expression in parentheses increases with $n$. At $n=10^6$, use $69/5<\log(10^6)<691/50$ and $\log9<11/5$ to bound it below by \[ \frac{69}{80}-\frac9{460}-\frac{691}{1250000}>\frac{21}{25}. \] For $b=2,3$ and any reduced class $a$ modulo $Db$, the same estimate gives \begin{align} &\#\{p:Dn+10D \frac{n}{2(L+\log9)}-\frac{0.405n}{L^2}-60 >0.4\frac nL.\label{eq:small-lift} \end{align} Indeed, $D/(2\varphi(Db))\ge1/2$ and $D\le6$. After multiplication by $L/n$, the lower bound increases with $n$. At $10^6$ it is greater than \[ \frac{69}{160}-\frac{27}{920}-\frac{2073}{2500000}>\frac25. \] In particular, both reduced classes $r$ and $r+D$ modulo $2D$ contain primes in the interval. Suppose that $G$ preserves blocks of size $b$, with $2\le b\le n/2$ and $b\mid n$. Since $|r|Dn$. If $b=2$ or $3$, reduction modulo $D$ has at least two reduced classes modulo $Db$ above $r$. In fact the fiber size $\varphi(Db)/\varphi(D)$ is $2$ or $3$. Choose a class different from \eqref{eq:forced-class}; the positive lower bound \eqref{eq:small-lift} is a contradiction. If $4\le b\le n^{1/4}$, then \[ \varphi(Db)\ge4\varphi(D). \] For $D=2$ this follows from $\varphi(2b)\ge4$; for $D=6$, use also $\varphi(6b)/2\ge\varphi(2b)$. The short-interval Brun--Titchmarsh inequality states that \begin{equation}\label{eq:BT} \pi(x+y;q,a)-\pi(x;q,a) <\frac{2y}{\varphi(q)\log(y/q)} \qquad(y>q,\ (a,q)=1); \end{equation} see \cite[Theorem~2, Eq.~(1.10)]{MV}. With $y=Dn/2$ and $q=Db$, it yields \begin{align*} T&<\frac{Dn}{\varphi(Db)\log(n/(2b))}\\ &\le\frac{Dn}{4\varphi(D)(3L/4-\log2)} =c_D\frac nL\frac{L}{2(3L/4-\log2)} <0.75c_D\frac nL, \end{align*} contradicting \eqref{eq:prime-total}. The last normalized factor decreases with $n$. At $10^6$, the bounds $L>69/5$ and $\log2<7/10$ make it less than $138/193<3/4$. Finally, if $b>n^{1/4}$, counting integers in the class \eqref{eq:forced-class} gives \[ T\le\frac n{2b}+1 <\frac12n^{3/4}+1 <0.25\frac nL \le0.25c_D\frac nL, \] again a contradiction. Here $L/(2n^{1/4})+L/n$ decreases for $n\ge10^6$. At the left endpoint, $n^{1/4}>158/5$ and $L<691/50$ give the upper bound \[ \frac{691}{3160}+\frac{691}{50000000}<\frac14. \] Thus $G$ is primitive. Every cycle under consideration is nontrivial and fixes at least nine points. Theorem~\ref{thm:jones} gives $A_n\le G$. Choose a prime in the class $r+D$ modulo $2D$, using \eqref{eq:small-lift}. Then $(p-r)/D$ is odd and $d_p$ is even. Its cycle is an odd permutation, so $G=S_n$. \end{proof} \begin{remark} The strict comparisons needed in this proof follow from elementary monotonicity and the bounds \[ 13.8<\log(10^6)<13.82,\quad \log9<2.2,\quad \log2<0.7,\quad (10^6)^{1/4}>31.6. \] Thus the numerical threshold requires no additional finite computation. \end{remark} \subsection{The Legendre Galois group}\label{sec:legendre} We first replace each opposite pair of nonzero Legendre roots by its common square. The resulting degree-$n$ polynomial determines the group on pairs. The local quadratic cycles then recover the independent interchanges within the pairs. We use the variable $T=-X^2$ for the quotient polynomial, which gives the following convenient integral normalization. For $\epsilon=2\delta-1$, put \begin{equation}\label{eq:legendre-J} J_n^\epsilon(T)=\sum_{j=0}^{n}\binom nj \prod_{k=1}^{n}(2j+2k+\epsilon)T^j. \end{equation} Coefficient comparison gives \begin{equation}\label{eq:legendre-transform} J_n^\epsilon(-X^2)=(-2)^nn!L_{2n+\delta}(X). \end{equation} \begin{theorem}\label{thm:legendre} Let $n\ge1$, $\delta\in\{0,1\}$, and $\epsilon=2\delta-1$. If $J_n^\epsilon$ is irreducible over $\Q$, then \[ \Gal(J_n^\epsilon/\Q)=S_n, \qquad \Gal(L_{2n+\delta}/\Q)=W_n. \] In particular, $J_n^\epsilon$ and $L_{2n+\delta}$ are irreducible over $\Q$ simultaneously. \end{theorem} These are the two conditional Galois assertions in \cite[Conjecture~1.5]{CHL}. The irreducibility equivalence was proved in \cite[Lemma~1.3]{CHL}; the proof below also establishes it directly. The integral polynomial in \eqref{eq:legendre-J} is the specialization of \eqref{eq:integral-Jacobi} with $D=2$: \begin{equation}\label{eq:legendre-jacobi} J_n^\epsilon(T)=2^n n!P_n^{(\epsilon/2,0)}(2T+1). \end{equation} Indeed, the two products in \eqref{eq:integral-Jacobi} concatenate to the product in \eqref{eq:legendre-J}. Together with the classical quadratic transformation, this proves \eqref{eq:legendre-transform}; see \cite[Lemma~1.2]{CHL}. \begin{lemma}\label{lem:legendre-cycle} Let $N=2n+\delta$, $\epsilon=2\delta-1$, and let $p$ be a prime with $Nx$. If $\epsilon=-1$, the upper endpoint is $4n-1$. If $\epsilon=1$, it is $4n+3$, and the congruence forces $p\le4n+1$. Thus \[ 2n+2+\epsilon0$. For every reduced class $a$ modulo $q$ and all sufficiently large $x$, the interval \[ \left(x,x+\frac{x}{(\log x)^A}\right] \] contains a prime congruent to $a$ modulo $q$. The threshold can be chosen uniformly in $a$. \end{lemma} \begin{proof} Choose a fixed power $Q=q^m\ge10^5$. Every representative $a$ prime to $q$ is also prime to $Q$. Proposition~6.19 of \cite{BMOR}, with $Z=A+1$, gives \[ \pi(t;Q,a)=\frac{\Li(t)}{\varphi(Q)} +O_{q,A}\!\left(\frac{t}{(\log t)^{A+2}}\right) \qquad(t\longrightarrow\infty). \] Indeed, for each fixed $Z$, the three auxiliary parameters in \cite[Eq.~(6.23)]{BMOR} can be chosen sufficiently large; the resulting threshold is fixed and uniform in the reduced classes. Set $H=x/(\log x)^A$. Then \[ \pi(x+H;Q,a)-\pi(x;Q,a) =\left(\frac1{\varphi(Q)}+o_{q,A}(1)\right) \frac{x}{(\log x)^{A+1}}>0 \] for all sufficiently large $x$. Such a prime is also congruent to $a$ modulo $q$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:jacobi}] Write $\alpha=a/u$ and $\beta=b/v$ as in the statement and put $\alpha+\beta=r/D$ in lowest terms. By hypothesis, \[ (r,D)=1,\qquad D\ge\max\{u,v\},\qquad D\ge2\text{ is even}. \] In particular, $\alpha+\beta=r/D$ is not an integer, so the degree of $P_n^{(\alpha,\beta)}$ is $n$. Its endpoint values are nonzero because neither parameter is a negative integer. Equation~\eqref{eq:jacobi-ode}, over a field of characteristic zero, then also shows that all its roots are simple. Choose a fixed constant \[ B>|r|+|a|+|b|+4D+10, \] and consider primes in \begin{equation}\label{eq:general-primes} p\equiv r\pmod D,\qquad Dn+B0$ and apply Lemma~\ref{lem:fixed-class-gap} with $q=D$ and $x=Dn+B$. It gives a prime satisfying \eqref{eq:general-primes} with \[ p=Dn+O_{\alpha,\beta,A}\!\left(\frac{n}{(\log n)^A}\right). \] The associated cycle has length $n-O_{\alpha,\beta,A}(n/(\log n)^A)$. Hence \[ 0\le n-M =O_{\alpha,\beta,A}\!\left(\frac{n}{(\log n)^A}\right), \qquad M:=|O|, \] for every fixed $A>0$. The orbit $O$ does not depend on $A$; the implied constant and threshold in this estimate may depend on $A$. It gives the unique monic irreducible factor $h_n$ of degree greater than $n/2$. \smallskip\noindent\emph{Primitivity on the large orbit.} Let $G_O$ be the induced transitive group on $O$. Suppose it preserves blocks of size $s$, where $2\le s\le M/2$. Lemma~\ref{lem:blocks} applies because $d_p>n/2\ge M/2$. It gives $s\mid d_p$ for every prime in \eqref{eq:general-primes}, and hence \begin{equation}\label{eq:general-forced} p\equiv D(2n+1)+r\pmod{Ds}. \end{equation} Here $s$ divides $M$, but it need not divide $n$; we therefore retain the full residue in \eqref{eq:general-forced}. Let $T_n$ be the number of primes in \eqref{eq:general-primes}. The prime number theorem in arithmetic progressions, with fixed modulus $D$, gives \begin{equation}\label{eq:general-count} T_n\sim\frac{D}{2\varphi(D)}\frac n{\log n}. \end{equation} Choose a fixed integer $s_0$ such that $\varphi(Ds)\ge4\varphi(D)$ for every $s\ge s_0$. Such a choice exists, for instance from $\varphi(q)\ge\sqrt{q/2}$. If $s_0\le s\le n^{1/4}$, the uniform bound \eqref{eq:BT} applied to \eqref{eq:general-forced} gives \[ T_n<\frac{Dn}{\varphi(Ds)\log(n/(2s))} \le\frac{Dn}{4\varphi(D)(\tfrac34\log n-\log2)} \sim\frac{D}{3\varphi(D)}\frac n{\log n}, \] contradicting \eqref{eq:general-count}. If $s>n^{1/4}$, counting integers gives \[ T_n\le\frac n{2s}+1\le\frac12n^{3/4}+1 =o(n/\log n), \] another contradiction. In both cases a nonreduced forced class is already impossible, since its primes would divide $Ds\le Dn/2$. It remains to exclude the finitely many values $2\le s-1$ and reduced denominator $v$ that is odd or divisible by $4$. For $\delta\in\{0,1\}$, define $Q_m^{a,\delta}$ by \[ P_{2m+\delta}^{(a,a)}(X)/X^\delta=Q_m^{a,\delta}(X^2). \] For all sufficiently large $m$, irreducibility of $Q_m^{a,\delta}$ implies \[ \Gal(Q_m^{a,\delta}/\Q)=S_m,\qquad \Gal(P_{2m+\delta}^{(a,a)}(X)/X^\delta\,/\Q) =C_2\wr S_m. \] \end{corollary} \begin{proof} The quadratic transformations \cite[Eqs.~18.7.13--18.7.14]{DLMF}, followed by Jacobi symmetry, identify $P_{2m+\delta}^{(a,a)}(X)/X^\delta$ with a nonzero rational multiple of \[ P_m^{(\delta-1/2,a)}(1-2X^2). \] The reduced denominator of $\delta-1/2+a$ is $2v$ if $v$ is odd and $v$ if $4\mid v$. In both cases it is even and at least $\max\{2,v\}$. Apply Theorem~\ref{thm:jacobi}, with $c=-2$, and use the affine change $T\mapsto1-2T$ for the quotient polynomial. \end{proof} These are symmetric Jacobi families, related to Gegenbauer polynomials with parameter $a+1/2$ by the usual scalar normalization. The case $a=0$ is the Legendre family, for which Theorem~\ref{thm:legendre} supplies every degree. For nonnegative integral parameters, Theorem~\ref{thm:symmetric} supplies the required eventual irreducibility, as used below. Unconditionally, $Q_m^{a,\delta}$ has a unique monic irreducible factor of degree greater than $m/2$. Its degree is $m-O_{a,A}(m/(\log m)^A)$ for every fixed $A>0$, and its Galois group is symmetric. Substituting $T=X^2$ in that factor gives an irreducible factor of twice the degree with the full quadratic wreath-product group. \subsection{Unequal integral parameters}\label{sec:galois} Return to the notation of Section~\ref{sec:asymmetric}, with $a>b\ge0$, $s=a+b$, and $n=2m+\delta$. Here we choose $p\in(3n/4,7n/8]$, so the desired cycle has the prime length $p$. Reduction separates a factor $x^p$ from a squarefree factor that is nonzero at the origin. If the original constant coefficient has valuation one, the degree-$p$ lift is Eisenstein. Its inertia group has order divisible by $p$, and $n<2p$ forces an element of order $p$ to be a single $p$-cycle. This argument permits wild ramification. The central-value numerators identify the primes for which the construction works, and the same prime gives an odd discriminant valuation. The residual factor has degree congruent to $1-\delta$ modulo $2$, so its value at zero involves the central numerator for the opposite parity. Accordingly, define \begin{equation}\label{eq:opposite-numerator} C_\delta(z)=2^{K_{1-\delta}} A_{1-\delta}\left(z+\delta-\frac12\right), \qquad \mathcal W_\delta(z)=A_\delta(z)C_\delta(z). \end{equation} Since $\deg A_\varepsilon\le K_\varepsilon$ and $K_0+K_1=s$, these are nonzero integral polynomials and $\deg \mathcal W_\delta\le s$. \begin{lemma}\label{lem:galois-prime} Suppose $m\ge2s+10$ and $P_n^{(a,b)}$ is irreducible over $\Q$. If a prime $p$ satisfies \begin{equation}\label{eq:galois-prime} \frac{3n}{4}M,\qquad p3\,594\,641). \] In particular, $0.99x<\vartheta(x)<1.01x$ in this range. Thus \[ \vartheta(7n/8)-\vartheta(3n/4) >\left(0.99\frac78-1.01\frac34\right)n =0.10875n>n/10. \] If every prime in this interval divided $\mathcal W_\delta(m)$, the difference would be at most $\log|\mathcal W_\delta(m)|$. A prime satisfying \eqref{eq:galois-prime} therefore exists, and Lemma~\ref{lem:galois-prime} applies. \end{proof} \begin{proof}[The rational Galois assertion of Theorem~\ref{thm:main}] For each parity, a root bound excludes the zeros of the fixed polynomial $\mathcal W_\delta$ beyond an effective point, and its coefficients give $|\mathcal W_\delta(m)|\le C(m+1)^s$ with effective $C$. Hence \eqref{eq:explicit-galois} holds for all sufficiently large $m$, effectively. Combine this with Theorem~\ref{thm:criterion}, and use reflection when $b>a$. \end{proof} \begin{proof}[The fixed-field assertion of Theorem~\ref{thm:main}] Let $E'$ be the Galois closure of $E$ over $\mathbb Q$. By the rational Galois assertion just proved and the construction in Lemma~\ref{lem:galois-prime}, for all sufficiently large $n$ the polynomial $f=P_n^{(a,b)}$ has group $S_n$ and there is an odd prime $p$ such that \[ \frac{3n}{4}R$. One may take the maximum of $1$ and the prime divisors of $|\operatorname{Disc}(E)|$, because a Galois closure introduces no new ramified rational primes. This also covers $E=\Q$. \end{proof} \subsection{Fixed derivatives} The quadratic transformation now brings the symmetric integral families within the denominator hypotheses of the rational-parameter theorem, and their established irreducibility makes its large factor the whole quotient polynomial. \begin{proof}[Proof of Corollary~\ref{cor:derivative-galois}] Put $J_m=P_m^{(\delta-1/2,r)}$. The quadratic transformations \cite[Eqs.~18.7.15--18.7.16]{DLMF}, followed by Jacobi reflection, give \begin{equation}\label{eq:gegenbauer-jacobi-bridge} F_{m,r,\delta}(X)= (-1)^m2^\delta \frac{(r+1/2)_{m+\delta}}{(\delta+1/2)_m} J_m(1-2X^2). \end{equation} The scalar is a nonzero rational number. By Theorem~\ref{thm:symmetric}, $F_{m,r,\delta}$ is irreducible over $\mathbb Q$ for all sufficiently large $m$. Consequently $J_m$ is irreducible, since any factorization of $J_m$ would give one of $J_m(1-2X^2)$. The parameters $\delta-1/2$ and $r$ have reduced denominators $2$ and $1$, and their sum has reduced denominator $D=2$. Neither parameter is a negative integer. Thus Theorem~\ref{thm:jacobi} and Corollary~\ref{cor:fixed-field} apply, with $c=-2$. Since $J_m$ is irreducible, the large factor in that theorem is its monic normalization. Its splitting field and that of $J_m(1-2X^2)$ are therefore linearly disjoint from $E$ for all sufficiently large $m$, and their groups over $E$ are $S_m$ and $C_2\wr S_m$. Equation~\eqref{eq:gegenbauer-jacobi-bridge} identifies the second polynomial with $F_{m,r,\delta}$ up to a scalar and identifies $Q_{m,r,\delta}$ with $J_m(1-2T)$ up to the same scalar. Multiplication by a nonzero rational scalar and the affine change $T\mapsto1-2T$ preserve the respective splitting fields. The two group actions are transitive, proving irreducibility over $E$. Taking the larger threshold for $\delta=0,1$ makes it valid for both parities. Finally, $P_{2m+\delta+r}^{(r)}=(2r-1)!!\,C_{2m+\delta}^{r+1/2}$, with $(-1)!!=1$. After division by $X^\delta$, the derivative has the same roots as $F_{m,r,\delta}$. \end{proof} \section{Applications and further questions}\label{sec:applications} The results have consequences for both individual algebraic zeros and geometric objects defined by the polynomials. Irreducibility controls common and prescribed zeros; the full symmetric group provides the additional information needed for the Jacobians below. \subsection{Integral parameters and common roots} Negative integral parameters contribute endpoint factors. Removing them extends Theorem~\ref{thm:main} to all fixed integral parameters whose absolute values differ. \begin{corollary}\label{cor:integral-parameters} Fix integers $a,b$ with $|a|\ne|b|$, and put $a_- =\max\{-a,0\}$ and $b_- =\max\{-b,0\}$. For all sufficiently large $n$, effectively in $a,b$, the polynomial \[ \frac{P_n^{(a,b)}(x)}{(x-1)^{a_-}(x+1)^{b_-}} \] is irreducible over $\Q$ and has Galois group $S_{n-a_--b_-}$. \end{corollary} \begin{proof} For $A,b\ge0$ and $n\ge A$, the binomial formula gives \[ P_n^{(-A,b)}(x) =\frac{(n-A)!(n+b)!}{n!(n-A+b)!} \left(\frac{x-1}{2}\right)^A P_{n-A}^{(A,b)}(x). \] Reflection treats a negative second parameter. For $A,B\ge0$ and $n\ge A+B$, the same formula gives \[ P_n^{(-A,-B)}(x) =\left(\frac{x-1}{2}\right)^A\left(\frac{x+1}{2}\right)^B P_{n-A-B}^{(A,B)}(x). \] The remaining factor is a nonzero rational multiple of the Jacobi polynomial with parameters $|a|,|b|$, so Theorem~\ref{thm:main} applies. \end{proof} Irreducibility also controls how the root sets vary with the degree. If two rational polynomials share an algebraic root, its minimal polynomial divides both. Thus an irreducible member cannot meet the root set of a lower-degree member. For the Legendre polynomials and their derivatives, the only additional comparison is between the four reduced polynomials of the same degree. \begin{corollary}\label{cor:common-roots} For fixed distinct nonnegative integers $a,b$, only finitely many pairs of different degrees give Jacobi polynomials $P_n^{(a,b)}$ with a common root, and all such pairs lie in an effective range. Moreover, no two distinct members of $\{P_\ell,P_\ell':\ell\ge1\}$ have a common nonzero root. \end{corollary} \begin{proof} In the asymmetric case, once the larger degree is at least $N(a,b)$, its irreducible polynomial cannot share a root with a polynomial of smaller degree. For the second assertion, remove any forced zero. Theorem~\ref{thm:all-degrees} settles different positive reduced degrees. At reduced degree $2m$, the four possibilities are \[ P_{2m},\quad P_{2m+1}/x,\quad P'_{2m+1},\quad P'_{2m+2}/x. \] Their absolute ratios of the coefficient of $x^{2m-2}$ to the leading coefficient are, in increasing order, \[ \frac{m(2m-1)}{4m+1} <\frac{m(2m-1)}{4m-1} <\frac{m(2m+1)}{4m+3} <\frac{m(2m+1)}{4m+1}. \] They are therefore pairwise nonproportional. Irreducibility again excludes a common root. Reduced constants have no roots. \end{proof} The same argument gives a degree bound for any prescribed algebraic number $\xi$: \[ P_n^{(a,b)}(\xi)\ne0 \quad\text{for }n>\max\{N(a,b),[\Q(\xi):\Q]\}. \] \subsection{Excluding a prescribed factor} The proof of irreducibility must allow every possible factor. When a candidate factor $F$ is specified, its leading coefficient and discriminant provide additional information about the real resultant. Keeping these quantities in the comparison gives a stronger exclusion criterion. This is useful when a proposed zero is a fixed algebraic number: its minimal polynomial supplies $F$. The resulting estimates apply in parameter ranges larger than the range in which we prove complete irreducibility. \begin{proposition}\label{prop:fixed-factor} Let $F\in\Z[x]$ be primitive, of degree $d\ge1$, with positive leading coefficient $u$, distinct roots in $(-1,1)$, and $F(0)\ne0$. Put \[ c_F=\min_jv_2([X^j]F(2X+1)),\qquad \beta(F)=\log4+\frac2d\log\frac{u}{2^{c_F}}. \] Let $c\ge2$ satisfy \begin{equation}\label{eq:fixed-factor-condition} c\bigl(\beta(F)-1\bigr)>g(c). \end{equation} For all sufficiently large $m$, uniformly for $0\le r\le m/c$ and $\delta\in\{0,1\}$, \[ F\nmid C_{2m+\delta}^{r+1/2}(x)/x^\delta. \] For rational $c$, the threshold is effective in $F,c$. \end{proposition} \begin{proof} Suppose $F\mid H$, with $H$ the primitive polynomial in the proof of Theorem~\ref{thm:symmetric-exact}. For large $m$, write $H=FG$ with $e=2m-d\ge d$. Set $z=u/2^{c_F}>1$ and \[ V_F=\prod_{i m\beta(F)+\tfrac12\log m-\log K-C_F-D_2\log2. \] Comparison with \eqref{eq:symmetric-upper} forces \begin{equation}\label{eq:fixed-factor-necessary} m\beta(F)<\Psi_m(S)+\log K+\log A_{\le m} +C_F-\tfrac12\log m. \end{equation} For $r\ge1$, Lemma~\ref{lem:uniform-asymptotics} bounds the right side by $m(1+g(c)/c)+o_F(m)$, uniformly in the stated range. This contradicts \eqref{eq:fixed-factor-condition}. The case $r=0$ follows with $K=A=1$. The effective bounds in that lemma also prove effectivity here. \end{proof} Render \cite[Theorem~6]{Render} proved that $P_n^{(r,r)}(\sqrt{b/c})\ne0$ for coprime positive integers $b,c$ with $b\notin\{1,3\}$ and integers $n,r\ge0$. In particular, any nonzero rational root has numerator $\pm1$ in lowest terms. The next corollary excludes these remaining possibilities uniformly when $m\ge2r$ and $m$ is sufficiently large. Its proof uses the uniform lower bound for $u/2^{c_F}$ among primitive linear polynomials. \begin{corollary}\label{cor:rational-roots} There is an effective absolute integer $M_1$ such that, for $m\ge M_1$ and $0\le r\le m/2$, neither $C_{2m}^{r+1/2}$ nor $C_{2m+1}^{r+1/2}/x$ has a rational root. \end{corollary} \begin{proof} A nonzero rational root in $(-1,1)$ gives a primitive factor $F=ux+v$, with $0<|v|0. \] The last inequality follows from $e<3$ and $3^5 2^{16}>5^{10}$. \end{proof} \subsection{Vanishing modes on a circular cone} In spherical coordinates $(\rho,\theta,\varphi)$, a homogeneous harmonic mode of degree $\ell$ and order $0\le r\le\ell$ is, up to a nonzero constant, \[ \rho^\ell e^{ir\varphi}(\sin\theta)^r P_\ell^{(r)}(\cos\theta); \] see \cite[Eqs.~14.7.8 and 14.30.1, and \S14.30(iv)]{DLMF}. For $\alpha=\cos\theta\in(-1,1)$, vanishing on that latitude is therefore equivalent to $P_\ell^{(r)}(\alpha)=0$. The cone $x^2+y^2-2z^2=0$ meets the unit sphere at $\alpha=\pm1/\sqrt3$; parity makes vanishing at these two latitudes equivalent. The arithmetic question is thus whether the fixed polynomial $3x^2-1$ can divide a Legendre derivative. The preceding estimates restrict the degree and order for which this can happen. \begin{corollary}\label{cor:angular} For every integer $D\ge1$, there is an effective integer $L_D$ such that, for every nonzero algebraic $\alpha\in(-1,1)$ of degree at most $D$, \[ \ell\ge L_D,\quad 0\le r\le\ell,\quad P_\ell^{(r)}(\alpha)=0 \quad\Longrightarrow\quad 17r>\ell-1. \] For $\alpha=1/\sqrt3$, the stronger conclusion \begin{equation}\label{eq:cone-bound} 23r>4(\ell-1) \end{equation} holds for all sufficiently large $\ell$, with an effective threshold. For nonzero rational $\alpha\in(-1,1)$, one may replace the conclusion by $5r>\ell-1$, with an absolute effective threshold in $\ell$. \end{corollary} \begin{proof} Write $\ell-r=2m+\delta$. If $17r\le\ell-1$, then $m\ge8r$ and $m\ge8(\ell-1)/17$. Choose $L_D$ so that $\ell\ge L_D$ forces $m\ge M_0$ and $2m>D$ under this inequality. Theorem~\ref{thm:symmetric} makes $P_\ell^{(r)}/x^\delta$ irreducible of degree $2m$. If it vanishes at $\alpha$, this degree must equal $[\Q(\alpha):\Q]\le D$, a contradiction. For $\alpha=1/\sqrt3$, its primitive integral minimal polynomial is $F=3x^2-1$. Here $d=2$, $u=3$, and $c_F=1$, because $F(2X+1)=12X^2+12X+2$. Thus $\beta(F)=\log6$. The choice $c=19/8$ is admissible in Proposition~\ref{prop:fixed-factor}, since \[ \frac{19}{8}(\log6-1)-g(19/8)>0.0042; \] an exact rational verification is given in Appendix~\ref{app:finite-derivatives}. If $23r\le4(\ell-1)$, then $m\ge19r/8$, so that proposition excludes the factor once $\ell$ is large. Finally, $5r\le\ell-1$ implies $m\ge2r$. Now Corollary~\ref{cor:rational-roots} applies, uniformly in the rational root. \end{proof} For this cone, Mangoubi and Weller Weiser show that $r$ is even and that each fixed order $r$ permits at most one degree $\ell$ \cite[Theorem~2]{MWW}. The bound \eqref{eq:cone-bound} restricts the two variables simultaneously: in large degree, a vanishing mode must have order greater than $4(\ell-1)/23$. These restrictions still leave infinitely many possible pairs, so the classification proposed by Agranovsky \cite[\S6, Question~4]{Agranovsky} remains open. The distinct generic existence problem of Agranovsky--Krasnov \cite{AK} was solved by Greilhuber \cite{Greilhuber}. For comparison, the associated Legendre equation gives the opposite constraint \[ r^2<\tfrac23\ell(\ell+1)\qquad(r\ge1) \] when $P_\ell^{(r)}(1/\sqrt3)=0$. Indeed, set $y(x)=(1-x^2)^{r/2}P_\ell^{(r)}(x)$. Integration by parts on $[1/\sqrt3,1]$ gives \[ \ell(\ell+1)\int_{1/\sqrt3}^1y^2 =\int_{1/\sqrt3}^1\left((1-x^2)(y')^2+ \frac{r^2y^2}{1-x^2}\right) >\frac{3r^2}{2}\int_{1/\sqrt3}^1y^2. \] Both endpoint terms vanish. Thus the necessary conditions still leave a nonempty linear range of derivative orders. \subsection{Modular interpretations}\label{sec:modular-interpretation} The fractional parameters have a modular origin. Kaneko and Zagier's characteristic-zero polynomials occur in modular-form solutions. At prime indices, adjoining the prescribed factors at $0$ and $1728$ gives a polynomial whose reduction is the supersingular $j$-polynomial \cite[Theorem~1]{KZ}. Theorem~\ref{thm:kz-all} determines the arithmetic of each positive-degree Jacobi factor: its roots form one Galois orbit over $\Q$, with symmetric splitting group. We record the normalization explicitly to relate this conclusion to the modular forms and differential equations in which the same polynomials occur. For $n\ge0$ and $\lambda,\mu\in\{-1,1\}$, put \begin{equation}\label{eq:kz-index} r=2\lambda+3\mu,\qquad e=6+r,\qquad \ell=12n+e, \end{equation} and define \begin{equation}\label{eq:kz-definition} \widetilde F_{\ell-1}(Y) =1728^nP_n^{(\lambda/3,\mu/2)}(1-Y/864). \end{equation} Here $e\in\{1,5,7,11\}$ and the leading coefficient is \begin{equation}\label{eq:kz-leading} A_{n,r}=(-1)^n\binom{2n+r/6}{n}. \end{equation} The affine substitution and nonzero rational scalar preserve irreducibility and splitting fields. Put $a=(\lambda+1)/2$ and $b=(\mu+1)/2$, so $k=\ell-1=12n+4a+6b$. The associated modular solution, with this normalization, is \[ F_k(z)=\Delta(z)^nE_4(z)^aE_6(z)^b\widetilde F_k(j(z)), \] where $E_4,E_6$ are the normalized Eisenstein series, $\Delta$ is the modular discriminant, and $j=E_4^3/\Delta$. Its Fourier constant term is $A_{n,r}$; division by this constant gives the solution with constant term one. When $\ell\ge5$ is prime, \[ Y^a(Y-1728)^b\widetilde F_{\ell-1}(Y) \] has $\ell$-integral coefficients and reduces to the monic polynomial whose roots are the supersingular $j$-invariants in $\overline{\F}_\ell$ \cite[Theorem~1]{KZ}. Indeed, $2n+r/6=\ell/6-1$ also gives $A_{n,r}\equiv1\pmod\ell$. The same four Jacobi families occur as pole polynomials in modular differential equations \cite[Theorem~4.2]{BSS}. The Legendre family has a related classical supersingular interpretation. For an odd prime $p=2m+1$, \[ (1-z)^mP_m\!\left(\frac{1+z}{1-z}\right) =\sum_{j=0}^m\binom mj^2z^j. \] The reduction is a Hasse polynomial for $y^2=x(x-1)(x-z)$: for $z\ne0,1$ it vanishes exactly at the supersingular Legendre curves \cite[\S1]{BM}. \subsection{Hyperelliptic Jacobians} For the curve $y^2=f(x)$, the roots of $f$ are the finite branch points of the double cover of the projective line. Divisor classes supported on the branch points describe the $2$-torsion of its Jacobian, linking the permutation group of the roots to the arithmetic of the abelian variety. Zarhin's theorems show that, in the degree range below, the full symmetric group forces the geometric endomorphism ring to be $\Z$, gives the Hodge group $\Sp$, and gives the $\ell$-adic Lie algebra $\Q_\ell\operatorname{Id}\oplus\mathfrak{sp}$ \cite{Zarhin02}. They also determine the Hodge and Tate classes on all powers. The following corollary collects these consequences under one Galois-group hypothesis. \begin{corollary}\label{cor:jacobians} Let $f\in\Q[x]$ be separable of degree $d\ge9$ with $\Gal(f/\Q)=S_d$, and let $J_f$ be the Jacobian of the smooth projective curve with affine equation $y^2=f(x)$. Then \[ \dim J_f=\lfloor(d-1)/2\rfloor,\qquad \End_{\overline\Q}(J_f)=\Z. \] For $V=H_1(J_f(\mathbb C),\Q)$ and the alternating form $\psi$ induced by the principal polarization, \[ \Hdg(J_f)=\Sp(V,\psi),\qquad \MT(J_f)=\GSp(V,\psi). \] For every number field $K\subset\overline\Q$ and every prime $\ell$, put $V_\ell=T_\ell(J_f)\otimes_{\Z_\ell}\Q_\ell$, and let $\psi_\ell$ be the alternating Weil pairing with values in $\Q_\ell(1)$ induced by the principal polarization. The image of $\Gal(\overline\Q/K)$ on $V_\ell$ has Lie algebra \[ \Q_\ell\operatorname{Id}\oplus\mathfrak{sp}(V_\ell,\psi_\ell). \] Every rational Hodge class on every power of $J_f$ is a rational linear combination of products of divisor classes. For every $h\ge1$ and $c\ge0$, the cycle-class map \[ \CH^c((J_f^h)_K)\otimes\Q_\ell\longrightarrow H^{2c}_{\mathrm{\acute et}} ((J_f^h)_{\overline\Q},\Q_\ell(c))^{\Gal(\overline\Q/K)} \] is surjective. The Hodge, Tate and Mumford--Tate conjectures therefore hold for $J_f$ and all its powers. \end{corollary} \begin{proof} These are consequences of Zarhin's theorems \cite{Zarhin02}: Theorem~2.4 gives the geometric endomorphism ring, Theorem~2.6 gives the scalar plus symplectic $\ell$-adic Lie algebra, Theorem~8.1 gives the Tate statement over every finite extension and on every power, and Theorem~9.1 and Remark~9.2 give the Hodge and Mumford--Tate conclusions. Passing to a number field preserves the Lie algebra because the Galois subgroup has finite index. For powers, the groups act diagonally on direct sums. A Galois-invariant class represented by algebraic cycles over a finite extension descends with rational coefficients by averaging over a finite Galois extension. \end{proof} For the Kaneko--Zagier family, take $f=P_d^{(\lambda/3,\mu/2)}$ and apply Theorem~\ref{thm:kz-all}. For fixed unequal nonnegative integers $a,b$, take $f=P_d^{(a,b)}$ with $d\ge\max\{9,N(a,b)\}$ and apply Theorem~\ref{thm:main}. For a dominant factor $h_n$ in Theorem~\ref{thm:jacobi}, use its actual degree $d=M_n$. For the derivative families, fix $r$ and take $f=Q_{m,r,\delta}$ with $m\ge9$ sufficiently large in Corollary~\ref{cor:derivative-galois}. The substitution $t=x^2$ also gives the map \[ \{y^2=F_{m,r,\delta}(x)\}\longrightarrow \{y^2=Q_{m,r,\delta}(t)\},\qquad (x,y)\longmapsto(x^2,y). \] This map extends to a degree-two morphism of the smooth projective models. It is the curve on the right whose defining polynomial has group $S_m$ and hence satisfies the preceding corollary. The polynomial $F_{m,r,\delta}$ has group $W_m$; that group does not satisfy the symmetric-group hypothesis. \subsection{Further questions} The fixed rational-parameter theorem leaves a precise irreducibility problem. \begin{question} Suppose $\alpha,\beta>-1$ are rational and satisfy the denominator conditions of Theorem~\ref{thm:jacobi}. Is $P_n^{(\alpha,\beta)}$ irreducible over $\Q$ for all sufficiently large $n$? \end{question} The theorem bounds the complementary degree by $O_{\alpha,\beta,A}(n/(\log n)^A)$ for every fixed $A>0$. This permits a bounded nonconstant factor. To exclude it, the resultant comparison would need sufficient control of the primitive content, the moment valuations, and the root distances at the denominator primes. An explicit criterion in the parameters for a positive comparison gap would identify further families to which the method applies. For integral parameters, the exceptional degrees form an effective finite set. Classifying them calls for more information about the central numerators and small factors. In the symmetric case, all-degree irreducibility cannot hold for every fixed $r$: \[ C_2^{7/2}(x)=\tfrac72(3x-1)(3x+1). \] The cases $r=0,1$ have no exceptions. Which other values of $r$ have this property, and how far can the uniform range $m\ge8r$ be enlarged? The constant $8$ is a sufficient bound and is not claimed to be optimal. Corollary~\ref{cor:derivative-galois} settles eventual maximality for each fixed $r$; it leaves the all-degree Galois groups of the first-derivative families and uniform Galois estimates for growing $r$ as separate questions. The prescribed-factor estimate suggests a related question: how does the arithmetic of a minimal polynomial constrain the occurrence of its roots in varying Jacobi families? The cone $x^2+y^2-2z^2=0$ gives one concrete instance. The bound $23r>4(\ell-1)$ is uniform in degree and order, but it leaves infinitely many possible pairs and does not complete the classification of harmonic modes on that cone. Finally, the local results in Appendix~\ref{sec:local-irred} come from a dilation of digit-sum Newton polygons. A general description of when multiplication of the degree by a denominator prime dilates the lower hull could yield further local irreducibility sequences, including cases where the global resultant comparison gives no contradiction. \appendix \section{Local irreducibility} \label{sec:local-irred} Rational irreducibility does not in general imply irreducibility over a completion. The following result gives the latter conclusion in two infinite sequences of degrees. \begin{theorem}\label{thm:kz-towers} For every integer $a\ge0$ and every $\lambda,\mu\in\{-1,1\}$, $P_{3^a}^{(\lambda/3,\mu/2)}$ is irreducible over $\Q_3$. The polynomial $P_{5\cdot3^a}^{(\lambda/3,\mu/2)}$ is irreducible over $\Q_3$ precisely when $\mu=-1$ or $a\ge2$. For $\mu=1$, the local factor degrees are $1,1,3$ when $a=0$, and $2,13$ when $a=1$. \end{theorem} The first assertion follows from a general prime-power criterion. The second follows from a dilation identity for the lower convex hull of a digit-sum function. \subsection{Irreducibility in prime-power degrees} \label{sec:prime-powers} A denominator in one parameter, together with an exact valuation in the other, forces a single Newton edge. \begin{theorem}\label{thm:prime-power-criterion} Let $p$ be a prime, $a\ge1$, $n=p^a$, and let $\alpha,\beta\in\Q$. Suppose \[ v_p(\alpha)=-e<0,\qquad v_p(\beta)\ge0, \qquad v_p((\beta+1)_n)=v_p(n!)=\frac{n-1}{p-1}. \] Then $P_n^{(\alpha,\beta)}$ is irreducible over $\Q_p$. More precisely, the monic polynomial \[ \frac{n!P_n^{(\alpha,\beta)}(2X-1)} {(n+\alpha+\beta+1)_n} \] has a single Newton edge joining \[ \left(0,en+\frac{n-1}{p-1}\right)\quad\text{and}\quad(n,0), \] with every intermediate coefficient point strictly above the edge. \end{theorem} \begin{proof} Every denominator factor in \eqref{eq:monic-endpoint} has valuation $-e$. Put $h_j=v_p((\beta+1)_j)$ and $h_0=0$. For $1\le i\le a$, exactly $n/p^i$ of $\beta+1,\ldots,\beta+n$ belong to $p^i\Z_p$. The hypothesis on $h_n$ therefore shows that none belongs to $p^{a+1}\Z_p$. For $1\le j0. \] The endpoint valuations are $H$ and $0$. Since $H\equiv1\pmod p$, one has $(H,n)=1$. Every root has valuation $H/n$, so a monic factor over $\Q_p$ of degree $d$ would have constant-term valuation $dH/n\in\Z$. Hence $n\mid d$, proving irreducibility. \end{proof} \begin{corollary}\label{cor:odd-prime-half} Let $p$ be an odd prime, let $a\ge1$ be an integer, and let $\alpha\in\Q$ satisfy $v_p(\alpha)<0$. Both $P_{p^a}^{(\alpha,-1/2)}$ and $P_{p^a}^{(\alpha,1/2)}$ are irreducible over $\Q_p$. In particular, for every $a\ge1$ all four $P_{3^a}^{(\lambda/3,\mu/2)}$ are irreducible over $\Q_3$. \end{corollary} \begin{proof} For $n=p^a$ and odd $p$, \[ (1/2)_n=\frac{(2n)!}{4^n n!},\qquad (3/2)_n=\frac{(2n+1)!}{4^n n!}, \] and $v_p((2n)!)=v_p((2n+1)!)=2(n-1)/(p-1)$. Apply Theorem~\ref{thm:prime-power-criterion}. The degree-one case in Theorem~\ref{thm:kz-towers} is immediate. \end{proof} \begin{remark}\label{rem:cgl-eisenstein} In the notation of \cite{CGL}, the monic endpoint polynomial is $S_n^{(\lambda,\mu)}(X-1)$. Corollary~\ref{cor:odd-prime-half} proves the irreducibility prediction in \cite[Conjecture~1(4)]{CGL}. The precise Newton edge has height $(3n-1)/2$ for $n=3^a$. Thus its height need not be one, as in the usual Eisenstein criterion. For example, \[ S_3^{(1,1)}(X-1) =X^3-\frac{63}{41}X^2+\frac{27}{41}X-\frac{81}{1189}. \] The single-edge criterion with relatively prime height and length is the exact local assertion used here. \end{remark} \subsection{A dilation law and the degrees \texorpdfstring{$5\cdot3^a$}{5 times powers of 3}}\label{sec:five-tower} Fix $n\ge1$ and $\lambda,\mu\in\{-1,1\}$. A dilation identity gives another sequence of irreducible degrees. All valuations in this subsection are normalized by $v_3(3)=1$, and $s_3(r)$ denotes the sum of the base-$3$ digits of a nonnegative integer $r$. Legendre's formula $v_3(k!)=(k-s_3(k))/2$ separates each coefficient valuation into a linear term and a digit correction. Only the correction affects the vertices of the Newton polygon. Multiplying the degree by $3$ shifts the base-$3$ digits; the lemma below shows exactly how this stretches the lower convex hull. For $\delta\in\{0,1\}$ set \[ \Phi_{n,\delta}(j) =s_3(2j+\delta)+s_3(n-j)-s_3(2n+\delta) \qquad(0\le j\le n), \] and let $H_{n,\delta}$ be the lower convex hull of these points, viewed as a piecewise linear function on $[0,n]$. The monic polynomial \[ T_n^{\lambda,\mu}(X) =\frac{n!}{(n+\lambda/3+\mu/2+1)_n} P_n^{(\lambda/3,\mu/2)}(2X-1) \] has coefficients \[ b_j=(-1)^{n-j}\binom nj \prod_{r=j+1}^n\frac{3(2r+\mu)}{6(n+r)+2\lambda+3\mu}. \] Since each denominator in this product is a $3$-adic unit, \begin{equation}\label{eq:five-tower-valuations} v_3(b_j) =n-j+v_3((2n+\delta)!)-v_3((2j+\delta)!)-v_3((n-j)!) =\frac32(n-j)+\frac12\Phi_{n,\delta}(j), \qquad \delta=\frac{\mu+1}{2}. \end{equation} Thus its Newton polygon is obtained from $H_{n,\delta}$ by a vertical scaling and the addition of a linear function. In particular it does not depend on $\lambda$. \begin{lemma}\label{lem:digit-dilation} For every integer $m\ge1$, \[ H_{3m,0}(x)=H_{m,0}(x/3)\quad(0\le x\le3m), \qquad H_{9m,1}(x)=H_{m,0}(x/9)\quad(0\le x\le9m). \] All points whose abscissae are not divisible by $3$, respectively by $9$, lie strictly above the indicated hull. \end{lemma} \begin{proof} First, all slopes of $H_{m,0}$ belong to $[-1,1]$. Digit subadditivity gives \[ \Phi_{m,0}(j)-\Phi_{m,0}(0) =s_3(2j)+s_3(m-j)-s_3(m)\ge-s_3(j)\ge-j. \] Writing $t=m-j$, it also gives \[ \Phi_{m,0}(j) \ge s_3(t)-s_3(2t)\ge-s_3(t)\ge-t. \] These inequalities bound the first and last slopes of the convex hull, and hence every slope, as asserted. For $0\le k0$, this ratio exceeds one. Otherwise $\alpha\beta=-1/6$, $C_1\ge2/3$, and, for $m\ge1$, \[ \frac{C_{m+1}}{C_m}\ge1-\frac1{6m(m+1)}. \] The product is at least $1-\sum_{m\ge1}1/(6m(m+1))=5/6$, so in that case $C_n\ge5/9$. In particular $2\sqrt{C_n}\ge1$. Let $a_n=T_2+T_3+E-2L_n+2\log2$. The displayed endpoint bounds give, for $n\ge N_0$, \[ |T_2-T_3| \le\tfrac32\log3-2\log2+\frac{2\log(9n)}n <\frac{27}{100}+\frac{30}{250000}<\frac3{10}. \] Here we used $\log3<1.1$, $\log2>0.69$, and $\log(9N_0)<15$, together with the decrease of $\log(9n)/n$. Since $\log\cosh u\le u^2/2$, it follows that \[ a_n=E-2\log\cosh((T_2-T_3)/2)>\frac12. \] Consequently $a_nk+2\log2/k-\log k>0$ for $k>0$: indeed $k/2-\log k\ge1-\log2>0$. The exact Vandermonde criterion therefore implies \begin{align*} S_n+\log g_n &\ge \log\kappa_n-J_n+n(L_n-T_2-T_3-E)-4\log2\\ &\ge n(L_*-E)\\ &\quad-\left\{\frac43\log(6n+3)+\log(3n+1)+2\log(3n) +\frac12\log(2n)+\frac12\log3+4\log2\right\}. \end{align*} The expression in braces is at most $6\log(12n)$ for $n\ge2$. For example, the difference is positive at $n=2$ and its derivative is greater than $7/(6n)$. This proves \eqref{eq:cutoff-arch}. \subsubsection{Content and the good-prime sum} Put $M=12n+5$ and \[ H(M)=\sum_{\substack{p\text{ prime},\ h\ge2\\p^h\le M}}\log p. \] For each prime $p\ge5$, choose a coefficient attaining the minimum $m_p^{(1)}$ of its first-level valuation. Every higher-level coefficient contribution is at most two. Thus \[ v_p(g_n)\le m_p^{(1)} +2\#\{h\ge2:p^h\le M\}. \] Since each exponent $e_{p^h}$ is at most one, \begin{equation}\label{eq:cutoff-combined} S_n+\log g_n\le T_0(n)+3H(M),\qquad T_0(n)=\sum_{p\ge5} \bigl(m_p^{(1)}+\mathbf1_{p\le n}e_p\bigr)\log p. \end{equation} The two indicators $m_p^{(1)},e_p$ cannot both equal one, so each summand in $T_0$ has multiplicity at most one. For $p\ge13$, the first-level carry calculation gives the following exact intervals. The first class is $p\equiv3+2\lambda\pmod6$, and the second is $p\equiv3-2\lambda\pmod6$. For $p\le n$, write $\ell=\lfloor n/p\rfloor\ge1$. The supported intervals are \[ \begin{array}{c|c} \text{first class}& \displaystyle\left(\frac{6n+r}{6\ell+1},\frac n\ell\right] \quad\text{and}\quad \displaystyle\left(\frac n{\ell+1},\frac{2n+\mu}{2\ell+1}\right] \\[6pt] \text{second class}& \displaystyle\left(\frac n{\ell+1},\frac{3n+\lambda}{3\ell+1}\right]. \end{array} \] For $p>n$, only content contributes, in the intervals \[ \left(n,\frac{3n+\lambda}{2}\right] \quad\text{in the first class},\qquad \left(\frac{6n+r}{5},2n+\mu\right] \quad\text{in the second class}. \] These formulas follow by substituting $a_p=(p+\lambda)/3$ or $(2p+\lambda)/3$ and $b_p=(p+\mu)/2$ into the indicators of Lemma~\ref{lem:carry-budget}. In the first case the combined indicator for $p\le n$ is $1+E_b-E_c$; in the second it is $E_a$. The small primes $5,7,11$ contribute at most $\log385$. We retain the first six groups of intervals. They contain enough of the main term for the required contradiction, while all discarded primes are at most $n/7$ and contribute at most $\theta(n/7)$. For the retained intervals, the square-root error in the prime estimate is useful near the cutoff, and the relative error estimate treats the unbounded range. The coefficient of $n$ in one half the sum of all retained interval lengths is \begin{align*} Q_6={}&\frac{13}{20} +\frac12\sum_{\ell=1}^{6}\left\{ \frac1\ell-\frac6{6\ell+1} +\frac2{2\ell+1}-\frac1{\ell+1} +\frac3{3\ell+1}-\frac1{\ell+1}\right\}\\ ={}&\frac{180556377763}{157066509600}<\frac{23}{20}. \end{align*} The constant shifts in the interval lengths contribute at most \[ \frac54+\frac56\sum_{\ell=1}^{6}\frac1\ell<5. \] Together with the small primes, this is less than $12$. We use two published estimates of Bennett, Martin, O'Bryant, and Rechnitzer \cite[Corollary~1.7 and Theorem~1.9]{BMOR}: for $a\in\{1,5\}$, \begin{align*} |\theta(x;6,a)-x/2|&\le2.072\sqrt x &&(1\le x\le4\cdot10^{13}),\\ |\theta(x;6,a)-x/2|&<0.024\,x/\log x &&(x\ge10^6). \end{align*} The range $4\cdot10^{13}$ in the first estimate follows from $x_2(6)=x_2(3)$ in their Theorem~1.9. For $N_0\le n\le10^{13}$, all retained endpoints are at most $2n+1<4\cdot10^{13}$. For each $1\le\ell\le6$, the six endpoints are at most $(n+1)/\ell$; the four remaining endpoints are at most $2n+1$. Their total error is therefore at most \[ 2.072\left\{6\sqrt{n+1} \sum_{\ell=1}^{6}\ell^{-1/2}+4\sqrt{2n+1}\right\} <61\sqrt n. \] To obtain the last inequality, use $\sum_{\ell=1}^{6}\ell^{-1/2}\le2\sqrt6-1<3.9$, $\sqrt{1+1/N_0}<1.000002$, and $\sqrt{2+1/N_0}<1.415$. For $n\ge10^{13}$, every retained endpoint is at least $n/7$. Since $\sum_{\ell=1}^{6}\ell^{-1}=49/20$, their sum is less than \[ 6(n+1)\frac{49}{20}+4(2n+1)<23n. \] Using $\log(10^{13}/7)>27$, the second prime estimate bounds the total error by \[ 0.024\frac{23n}{27}<\frac{21}{1000}n. \] Hanson's elementary theorem \cite[Theorem~1]{Hanson} states $\operatorname{lcm}(1,\ldots,N)<3^N$. Consequently $\theta(x)\le\psi(x)10^{13}. \end{cases} \end{equation} Hanson's estimate also gives \[ H(M)\le\log3\left\{\sqrt M+(\log_2M)M^{1/3}\right\} <11\sqrt n\qquad(n\ge N_0). \] For an explicit verification, use $M\le12.1n$ and note that $n^{-1/6}\log(12.1n)$ decreases for $n\ge N_0$. At the left endpoint the rational bounds \[ \sqrt{12.1}<3.5,\quad 12.1^{1/3}<2.3,\quad N_0^{-1/6}<0.127,\quad \log(12.1N_0)<15, \quad \log2>0.69,\quad \log3<1.1 \] give \[ \frac{H(M)}{\sqrt n} <1.1\{3.5+2.3\cdot0.127\cdot15/0.69\} =10.835<11. \] Combining this with \eqref{eq:cutoff-combined} and \eqref{eq:cutoff-level-one}, and rounding the first square-root coefficient upward, gives \begin{equation}\label{eq:cutoff-upper} S_n+\log g_n<\left(\frac{23}{20}+\frac{\log3}{7}\right)n+12+ \begin{cases} 95\sqrt n,&N_0\le n\le10^{13},\\ \dfrac{21}{1000}n+33\sqrt n,&n>10^{13}. \end{cases} \end{equation} \subsubsection{The contradiction} We first record the lower bound $L_*-E>3/2$. Indeed, $\sqrt3>1.73$ gives $4+3\sqrt3>9\cdot919/900$. Since $\log3>1.098$, $\log2<0.7$, and $\log(1+t)>t/(1+t)$ for $t>0$, \[ L_*-E> \frac53\log3-\frac12\log2+\log\frac{919}{900} >\frac53\frac{549}{500}-\frac7{20}+\frac{19}{919} =\frac{34478}{22975}>\frac32. \] For $N_0\le n\le10^{13}$, the first line of \eqref{eq:cutoff-upper}, the decrease of $\log(12n)/n$, and $\log(12N_0)<15$ give \[ \frac{23}{20}+\frac{\log3}{7} +\frac{95}{\sqrt n}+\frac{12+6\log(12n)}n <\frac{23}{20}+\frac{11}{70}+\frac{19}{100} +\frac{102}{250000} =\frac{1310357}{875000}<\frac32. \] For $n>10^{13}$, the decreasing error satisfies \[ \frac{33}{\sqrt n}+\frac{12+6\log(12n)}n <\frac{33}{10^6}+\frac{210}{10^{13}}<\frac1{1000}, \] where $\log(12\cdot10^{13})<33$. Hence the second line gives \[ \frac{23}{20}+\frac{\log3}{7}+\frac{21}{1000} +\frac{33}{\sqrt n}+\frac{12+6\log(12n)}n <\frac{23}{20}+\frac{11}{70}+\frac{22}{1000} <\frac43<\frac32. \] Both ranges contradict \eqref{eq:cutoff-arch}. This proves irreducibility for every $n\ge250\,000$ in all four families. The numerical inequalities used here follow from the displayed rational bounds and the positive-term series for the logarithm. \subsection{The finite range}\label{sec:finite-verification} We use a published computation for the small degrees. Cullinan and Hajir \cite[Remark~4.6]{CHK} checked the irreducibility of $K_\ell$ for every $13\le\ell\le10^5$ relatively prime to $6$. Their Definition~2.3 gives \[ \ell=12n+6+2\lambda+3\mu, \qquad K_\ell(X)=3^nn!P_n^{(\lambda/3,\mu/2)}(4X+1). \] Consequently their result covers all four parameter choices for $1\le n\le8332$, since $12\cdot8332+11=99\,995$. We use their irreducibility computation here, independently of the conditional Galois-group statement in \cite[Theorem~1.8]{CHK}. \begin{proposition}\label{prop:finite-kz} For every $1\le n\le250\,000$ and every $\lambda,\mu\in\{-1,1\}$, the polynomial $P_n^{(\lambda/3,\mu/2)}$ is irreducible over $\Q$. \end{proposition} \begin{proof} It remains to treat $8333\le n\le250\,000$. Retain $T_0(n)$ from \eqref{eq:cutoff-combined} and put \[ Q(n)=\sum_{\substack{p\ge5\text{ prime},\ h\ge2\\p^h\le12n+5}} \log p. \] The same content and moment calculation gives \begin{equation}\label{eq:fast-content-budget} S_n+\log g_n\le T_0(n)+3Q(n). \end{equation} Write $T_2=T_{2,n}$, $T_3=T_{3,n}$, $L=L_n$, and $J=J_n$. Since $C_n\ge1/3$, the exact real comparison implies, for every possible factor degree $1\le k\le n/2$, \begin{align*} \frac{\log R^{(6)}}k\ge{}& \log\kappa_n-\log g_n-J+n(L-T_2-T_3-E)-4\log2\\ &+ak+\frac{2\log2}{k}-\log k, \qquad a=T_2+T_3+E-2L+2\log2. \end{align*} If $|T_2-T_3|\le9/10$, then \[ a=E-2\log\cosh((T_2-T_3)/2) \ge E-\frac{81}{400}>\frac12. \] Thus the last three terms in the second line have positive sum, because $k/2-\log k\ge1-\log2>0$. Weighted AM--GM with weights $7/16$ and $9/16$ gives \[ L\ge\frac7{16}T_2+\frac9{16}T_3 +4\log2-\frac7{16}\log7-\frac98\log3. \] Define the integers \begin{align*} c_2&=216n+48v_2(n!)-24s_2(n)-27D_{2,n}-192,\\ c_3&=48v_3(n!)-22n-21D_{3,n}-8s_3(2n+\delta),\\ c_7&=-21n, \qquad \delta=(\mu+1)/2, \end{align*} and the real number \[ \Delta_n= \frac{c_2\log2+c_3\log3+c_7\log7}{48}-T_0(n)-3Q(n). \] Combining the preceding bounds shows that the two conditions \begin{equation}\label{eq:fast-finite-criterion} |D_{2,n}\log2-D_{3,n}\log3|\le\frac9{10}n, \qquad \Delta_n>0 \end{equation} exclude every proper factorization. For each of the four parameter pairs and every integer $8333\le n\le250\,000$, exact integer arithmetic verifies the first inequality in \eqref{eq:fast-finite-criterion} and the stronger bound \[ \Delta_n>\frac{n}{1000}. \] The calculation uses only the residue indicators of Lemma~\ref{lem:carry-budget}, digit sums, and rational bounds for logarithms. For completeness, these bounds follow by truncating \[ \log\frac{1+t}{1-t} =2\sum_{j=0}^{r-1}\frac{t^{2j+1}}{2j+1}+\mathcal E_r(t), \qquad 0\le\mathcal E_r(t)\le \frac{2t^{2r+1}}{(2r+1)(1-t^2)} \quad(0\le t\le1/3). \] Rescaling an integer by a power of $2$ puts its logarithm in this form. The accompanying self-contained verification recomputes these comparisons from the formulas, with outward rounding and without stored numerical data. This completes the finite range and proves the proposition. \end{proof} \section{The finite first-derivative calculation} \label{app:finite-derivatives} We give the remaining finite verification for the first derivatives, and record elementary rational bounds for three logarithmic constants. The coefficient formulas and prime bounds below specify the calculation completely. \subsection{A factor-degree criterion} \begin{lemma}\label{lem:finite-factor-degrees} Let $H\in\mathbb Z[x]$ be primitive of degree $D$. For a prime $p$ at which its reduction has degree $D$ and is squarefree, let $\mathcal D_p$ be the multiset of irreducible factor degrees, and let $\Sigma_p$ be the set of its subset sums, with each entry of the multiset used at most once. If \[ \{1,\ldots,\lfloor D/2\rfloor\} \cap\bigcap_{p\in\mathcal P}\Sigma_p=\varnothing \] for a finite set $\mathcal P$ of such primes, then $H$ is irreducible over $\mathbb Q$. \end{lemma} \begin{proof} By Gauss's lemma, a nontrivial rational factorization gives an integral factor of degree $1\le k\le\lfloor D/2\rfloor$. Since the leading coefficient of $H$ is nonzero modulo every listed prime, the factor retains its degree. Its reduction is a product of distinct irreducible factors, so $k\in\Sigma_p$ for every $p\in\mathcal P$, a contradiction. \end{proof} \subsection{The finite ranges} For $n=2m+\delta$, where $\delta\in\{0,1\}$, let $H_n$ be the primitive integral polynomial with positive leading coefficient proportional to \begin{equation}\label{eq:finite-derivative-coefficients} \sum_{k=0}^{m} \frac{(-1)^k(2n+2-2k)!} {k!(n+1-k)!(n-2k)!}\,x^{n-2k-\delta}. \end{equation} Thus $H_n$ is a nonzero rational multiple of $P'_{n+1}(x)/x^\delta$ and has degree $2m$. Put \[ U=\prod_{p\le m}p^{v_p(n+1)},\qquad V=\prod_{p\le m}p^{v_p(n+2)},\qquad Z=\prod_{p\le m}p^{\lfloor\log_p(2n+3)\rfloor}. \] For $m\ge4$, the sufficient condition of Theorem~\ref{thm:symmetric-exact} is exactly \begin{equation}\label{eq:finite-derivative-integer} 16^mV^2>64m(m+1)^2Z^2U^2. \end{equation} Indeed, $K=m+1$ and $A_{\le m}=U/V$; exponentiating and squaring the criterion gives \eqref{eq:finite-derivative-integer}. \begin{lemma}\label{lem:first-derivative-finite} For $1\le m\le199$ and $\delta\in\{0,1\}$, $H_{2m+\delta}$ is irreducible over $\mathbb Q$. \end{lemma} \begin{proof} For each parity, the indices $m<4$ together with the failures of \eqref{eq:finite-derivative-integer} for $4\le m\le199$ are \[ \begin{aligned} E_0={}&\{1,\ldots,25\}\cup \{28,30,31,32,34,37,40,42,43,45,46,47\},\\ E_1={}&\{1,\ldots,26\}\cup \{29,31,32,33,34,35,37,38,41,43,44,45,47,49,50,53,73\}. \end{aligned} \] The other $318$ pairs satisfy the strict integer inequality. For the $37+43=80$ pairs with $m\in E_\delta$, apply Lemma~\ref{lem:finite-factor-degrees} to the full-degree, squarefree reductions at primes $p\le317$. In every case the intersection in that lemma is empty. Here is a complete procedure for this finite calculation. The coefficients are obtained from \eqref{eq:finite-derivative-coefficients} by dividing by their positive gcd. The integers $U,V,Z$ use trial division and successive multiplication by each prime. For each remaining $H_n$, start with the candidate set $\{1,\ldots,m\}$ and visit the primes up to $317$ in increasing order. Discard a prime if the degree drops or if $\gcd(\overline H_n,\overline H_n')\ne1$. At every other prime, compute $\mathcal D_p$ as follows. Make $f=\overline H_n/\operatorname{lc}(\overline H_n)$ monic, and initialize $R=f$ and $h=x$. For $j=1,2,\ldots$, as long as $2j\le\deg R$, replace $h$ by $h^p\bmod R$, put $G=\gcd(R,h-x)$, record $j$ exactly $\deg(G)/j$ times, and replace $R$ by $R/G$ and $h$ by its remainder modulo this new $R$. If a nonconstant $R$ remains at the end, record its degree once. This gives $\mathcal D_p$: the polynomial $x^{p^j}-x$ is the product of the monic irreducibles whose degrees divide $j$, and all smaller degrees have already been removed. The final polynomial $R$ has at most one irreducible factor: each remaining factor has degree at least $j$, while $\deg R<2j$. To compute $\Sigma_p$, start with $\{0\}$ and, for every entry $d$ of $\mathcal D_p$, adjoin its translate by $d$. Intersect the candidate set with $\Sigma_p$, stopping when it is empty. Retaining only reductions that strictly decrease that set gives $254$ reductions in total, with largest prime $317$. These operations use integer arithmetic and Euclidean division over finite fields. Lemma~\ref{lem:finite-factor-degrees} proves the remaining $80$ cases and completes the proof. \end{proof} \subsection{Rational logarithm bounds} For $0\frac{97}{10000} \] and \[ 3T(1/3)+8T(1/5)+57T(3/35)+92T(1/47)-19 >\frac{42}{1250}. \] Both are direct comparisons of rational numbers; the second right side is $8\cdot21/5000$. Finally, \[ 2(\log9-1)-g(2) =\log\left(\frac{3^7 2^{16}}{5^{10}}\right)-2>0, \] because $e<3$ and $3^5 2^{16}>5^{10}$. \begin{thebibliography}{99} \bibitem{Agranovsky} M.~L.~Agranovsky, \emph{On a problem of injectivity for the Radon transform on a paraboloid}, Contemp. Math. \textbf{251} (2000), 1--14. \href{https://doi.org/10.1090/conm/251/03856}{doi:10.1090/conm/251/03856}. \bibitem{AK} M.~L.~Agranovsky and Y.~Krasnov, \emph{Quadratic divisors of harmonic polynomials in $\mathbb R^n$}, J. Anal. Math. \textbf{82} (2000), 379--395. \href{https://doi.org/10.1007/BF02791236}{doi:10.1007/BF02791236}. \bibitem{BMOR} M.~A. Bennett, G. Martin, K. O'Bryant, and A. 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