\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,microtype}
\usepackage[a4paper,margin=27mm]{geometry}
\usepackage{enumitem}
\usepackage[hidelinks]{hyperref}
\hypersetup{bookmarksdepth=1,pdfauthor={Bogdan Grechuk},pdftitle={Polynomial parametrizations for two classes of Diophantine equations},pdfsubject={Two monomial constructions and five complete worked examples},pdfkeywords={Diophantine equations, polynomial parametrization, monomials, coprimality}}
\setlength{\emergencystretch}{3em}
\numberwithin{equation}{section}
\newtheorem{lemma}[equation]{Lemma}
\newtheorem{proposition}[equation]{Proposition}
\theoremstyle{definition}\newtheorem{definition}[equation]{Definition}
\theoremstyle{remark}\newtheorem{remark}[equation]{Remark}
\theoremstyle{plain}\newtheorem{mastertheorem}{Theorem}
\renewcommand{\themastertheorem}{\Alph{mastertheorem}}
\newcommand{\Z}{\mathbb Z}\newcommand{\Q}{\mathbb Q}
\newcommand{\N}{\mathbb Z_{\ge0}}
\newcommand{\R}{\mathcal R}\newcommand{\OO}{\mathcal O}
\newcommand{\SL}{\operatorname{SL}}\newcommand{\Nm}{\operatorname{N}}
\newcommand{\diag}{\operatorname{diag}}\newcommand{\Res}{\operatorname{Res}}\newcommand{\vp}{v_p}
\newcommand{\bvec}[1]{\boldsymbol{#1}}

\title{Polynomial parametrizations for two classes of Diophantine equations}
\author{Bogdan Grechuk}
\date{26 October 2026}
\begin{document}
\maketitle

\begin{abstract}
We prove that the integer solution sets of two classes of Diophantine
equations are finite unions of polynomial families with integer
coefficients and unrestricted integer parameters. In the first class,
each monomial, with at most one exception, contains a variable that
occurs to the first power and in no other monomial. The second class
consists of equations $M=f(t)$, where $M$ is a monomial independent of
$t$ with a variable occurring to the first power, and $f\in\Z[t]$
splits completely over $\Q$. If $M$ involves at most two variables,
$f$ may instead be an arbitrary quadratic polynomial. We also give
complete formulas for five equations listed as open in the monograph
\emph{Polynomial Diophantine Equations: A Systematic Approach} (2024).
\end{abstract}

\section{Introduction}
For a polynomial Diophantine equation
\begin{equation}\label{eq:gen}
 P(x_1,\ldots,x_n)=0,
\end{equation}
a natural goal is to describe all integer solutions by finitely many
explicit formulas. Polynomial parametrizations provide a particularly
useful form of such a description: every parameter choice produces a
solution, and every solution is attained.

We use the terminology of \cite{Vas} and
\cite[Definition~1.4]{Grechuk}.
\begin{definition}\label{def:families}
A \emph{polynomial family} in $\Z^n$ is the image of a map
$\Phi\in\Z[t_1,\ldots,t_\nu]^n$ evaluated at all of $\Z^\nu$, where $\nu\geq0$.
The map $\Phi$ is a \emph{polynomial parametrization} of its image.
A set $S\subseteq\Z^n$ is a \emph{finite union of polynomial families} if
$S=\bigcup_{j=1}^h\Phi_j(\Z^{\nu_j})$ for finitely many such maps.
We allow $h=0$, representing the empty set, and $\nu_j=0$, representing
a single point. The families need not be disjoint.
\end{definition}

We ask when the integer solution set of \eqref{eq:gen} is a finite union
of polynomial families. The coefficients in Definition~\ref{def:families}
are integers, and the parameters are independent and unrestricted.
Rational functions with variable denominators, or formulas with
unresolved coprimality conditions, do not by themselves exhibit the
solution set as a finite union of polynomial families.

This question can be difficult even for simple equations. The existence
of a polynomial parametrization of
\[
 x_1x_2-x_3x_4=1
\]
was a longstanding open problem, resolved by Vaserstein
\cite[Theorem~1]{Vas}. Chapter~2 of the monograph~\cite{Grechuk}
uses this theorem to represent the solution sets of many other equations
as finite unions of polynomial families. Several small examples were left
open there and appear in \cite[Table~1]{GrechukOriginal}, including
\begin{gather}
 x^2y+zt+1=0,\qquad x_1x_2x_3+x_4x_5+1=0,\label{eq:2open}\\
 x^2y+z^2\pm1=0,\qquad xyz+t^2-1=0.\label{eq:3open}
\end{gather}
The present paper develops two general constructions that apply to these equations and supplies
complete formulas and proofs in Section~\ref{sec:examples}.


Our main results are the following two theorems. A monic monomial is
$x^e=\prod_jx_j^{e_j}$ with nonnegative integer exponents;
the constant monomial $1$ is allowed.

\begin{mastertheorem}\label{main:shared}
Let $M_0,\ldots,M_s$ be monic monomials in
$x=(x_1,\ldots,x_n)$, with $s\ge1$, and fix integers
$\alpha_1,\ldots,\alpha_s\ne0$ and $c$. Let $y_1,\ldots,y_s$ be
distinct variables absent from every $M_j$; the monomial supports may
overlap arbitrarily. Then the integer solutions of
\begin{equation}\label{eq:monomial-rhs}
\sum_{i=1}^s\alpha_iM_i(x)y_i=cM_0(x)
\end{equation}
form a finite union of polynomial families.
\end{mastertheorem}

\begin{mastertheorem}\label{main:values}
Fix $\kappa\in\Z\setminus\{0\}$, $f\in\Z[t]$, a nonnegative integer
$r$, and positive integers $e_1,\ldots,e_r$. Let
$z,x_1,\ldots,x_r,t$ be distinct variables. Then all integer solutions of
\[
\kappa z\prod_{i=1}^r x_i^{e_i}=f(t)
\]
form a finite union of polynomial families if $f$ splits completely over $\Q$, or
if $r\le1$ and $\deg f\le2$. Constant and zero polynomials are included.
\end{mastertheorem}

The first class in the abstract follows from Theorem~\ref{main:shared}:
in each non-exceptional monomial choose a variable that occurs to the
first power and in no other monomial, call it $y_i$, and collect all
remaining variables in $x$. Move the exceptional monomial to the
right-hand side, or take $c=0$ if there is none. Equations with at most
one nonzero monomial are immediate: their solution sets are unions of
coordinate hyperplanes, the empty set, or the whole space.

The equations in \eqref{eq:2open} are covered by
Theorem~\ref{main:shared}, and those in \eqref{eq:3open} by
Theorem~\ref{main:values}. In the latter theorem, the arbitrary quadratic
alternative allows only one powered variable in addition to the
linear factor.

The proofs use Vaserstein's parametrization of $\SL_2(\Z)$
\cite[Theorem~1]{Vas} and his factorization of congruence subgroups
\cite[Proposition~5.9 and Corollary~5.11]{Vas}. Section~\ref{sec:matrix}
establishes the required polynomial dependence on the congruence level.
We also use the representation of rational
polynomial families as finite unions of polynomial families
\cite[Lemmas~2--3]{Frisch} (see also \cite[Proposition~1.28]{Grechuk}),
and Gordan's lemma
\cite[Proposition~2.1.1(a)]{BGT}.
Related work of Frisch and Lettl \cite[Theorem~1]{FrischLettl}
gives integer-valued parametrizations for the affine cones over rational
projective plane curves, apart from specified singular-point exceptions.
Our constructions use coprime block products and valuation semigroups;
for quadratic polynomials irreducible over $\Q$, we also use standard ideal arithmetic.

The paper is organized as follows. Section~\ref{sec:basic} collects
polynomial operations and algebraic inputs. Section~\ref{sec:matrix}
adapts Vaserstein's published congruence parametrization to a variable
level, including levels $0$ and $1$. Section~\ref{sec:multipartite}
constructs polynomial parametrizations of pairwise coprime block
products, together with polynomial B\'ezout identities.
Section~\ref{sec:shared} proves Theorem~\ref{main:shared}.
Sections~\ref{sec:splitfull} and~\ref{sec:nonsplit} prove the split and
quadratic alternatives of Theorem~\ref{main:values}, respectively.
Section~\ref{sec:examples} gives complete solution formulas for the
five equations above.

\section{Polynomial operations and algebraic inputs}\label{sec:basic}

We take empty products to be $1$.
Gcds are nonnegative, with $\gcd(0,n)=|n|$; a tuple is primitive (or unimodular) if
its coordinates have gcd $1$. For $a\ne0$, $v_p(a)$ is the exponent of $p$
in $|a|$. For a quadratic number field $K$, its ring of integers is
$\OO_K$, conjugation is $\alpha\mapsto\bar\alpha$, and
$\Nm(\alpha)=\alpha\bar\alpha$. Ideal norms extend multiplicatively from
$\Nm(\mathfrak a)=[\OO_K:\mathfrak a]$, so
$\Nm((\alpha))=|\Nm(\alpha)|$.

For $P\in\Q[t_1,\ldots,t_\nu]^n$, the set
$P(\Z^\nu)\cap\Z^n$ is a \emph{rational polynomial family}
\cite[Definition~1.27]{Grechuk}.
\begin{lemma}[Fixed denominators and congruences]\label{lem:filter}
Let $\mathbf t=(t_1,\ldots,t_\nu)$ and
$P\in D^{-1}\Z[\mathbf t]^n$, with $D\geq1$ fixed. The integral
values of $P$ on $\Z^\nu$ satisfying finitely many fixed coordinate
congruences form a finite union of polynomial families. The same holds
for fixed congruences on integer-polynomial expressions in the retained
coordinates and parameters. If $\Psi(\mathbf t,P(\mathbf t))=0$ is a
polynomial identity, it remains an identity after each residue-class
substitution used to construct these families.
\end{lemma}
\begin{proof}
This is the residue-class argument of \cite[Lemmas~2--3]{Frisch};
the case without congruences is also \cite[Proposition~1.28]{Grechuk}.
Let $m_0>0$ be a common multiple of the congruence moduli. For every
$a\in\Z^\nu$, expansion gives
\[
 P(a+Dm_0\mathbf u)-P(a)\in m_0\Z[\mathbf u]^n.
\]
Thus integrality and the required residues depend only on
$a\bmod Dm_0$. On each admissible class, $P(a+Dm_0\mathbf u)$ has
integer coefficients. The finitely many admissible classes give exactly
the desired values. For auxiliary expressions, append them as extra
coordinates, enlarge $D$ if necessary, apply the same argument, and
project to the desired coordinates.
\end{proof}


\begin{remark}[Polynomial quotients]
An expression such as $(A-a)/P$ below denotes a quotient proved to lie in
the integer polynomial ring, \emph{before} parameter specialization. For
example, if a construction gives $A-a=P^2H$, we define $(A-a)/P=PH$.
It then has value zero when $P=0$. All parameter substitutions are
allowed, including those at which a displayed polynomial denominator
vanishes. Congruences between polynomials mean ideal divisibility in
their polynomial ring.
\end{remark}

\subsection{B\'ezout identities}
For $A,B\in\R=\Z[\bvec u]$, we use B\'ezout coefficients
$R,S\in\R$ satisfying $AR+BS=1$ as a polynomial identity.
For a tuple $(A_1,\ldots,A_s)$, the corresponding identity is
$\sum_i A_iR_i=1$ with $R_i\in\R$.
These identities mean that the polynomials generate the unit ideal;
having no common nonunit factor alone would not suffice.
The following elementary identities will be used repeatedly. From
$AR+BS=1$ and $A'=A+BH$ one obtains
\begin{equation}\label{eq:witness-shift}
A'R+B(S-RH)=1.
\end{equation}
From $AR+BS=1$ and $AU+CV=1$ one obtains
\begin{equation}\label{eq:witness-product}
A(R+U-ARU)+BCSV=1.
\end{equation}
The second identity follows by expanding
$(1-AR)(1-AU)=BCSV$. It combines the identities for $(A,B)$ and $(A,C)$ into one for
$(A,BC)$; swapping the two roles gives the analogous operation for a
product in the first slot.
If $A=A_1A_2$, a B\'ezout identity for $(A,B)$ gives one for $(A_1,B)$ by replacing
$R$ with $A_2R$.

The same B\'ezout identity gives an inverse modulo a square:
\begin{equation}\label{eq:squareinverse}
A\bigl(R(2-AR)\bigr)=1-B^2S^2.
\end{equation}
Thus all inverses modulo variable squares used later are polynomials, not
chosen numerical inverse functions.


\begin{lemma}[Raising prescribed powers]\label{lem:powers}
If $X=\prod_i x_i$, $Y=\prod_j y_j$ and $XR+YS=1$ polynomially, then
$A=\prod_i x_i^{e_i}$ and $B=\prod_j y_j^{f_j}$ have polynomial
B\'ezout identities for any fixed positive exponents.
\end{lemma}
\begin{proof}
For nonempty products put $E=\max_i e_i$, $F=\max_j f_j$.
Expand $1=(XR+YS)^{E+F-1}$: every term is divisible by $A$ or by $B$,
so grouping terms gives the B\'ezout identities. An empty product equals $1$.
\end{proof}
\subsection{A congruence-compatible matrix completion}
\begin{lemma}[Lifting a primitive column]\label{lem:column}
Let $m_0>0$ and $H_0\in\SL_2(\Z)$. If the primitive column $(a,c)^T$ is
congruent modulo $m_0$ to the first column of $H_0$, it has a completion
$H_1\in\SL_2(\Z)$ satisfying $H_1\equiv H_0\pmod{m_0}$.
\end{lemma}

\begin{proof}
Choose any determinant-one completion $H$ of $(a,c)^T$. Modulo $m_0$ the
matrix $H^{-1}H_0$ has first column $(1,0)^T$ and determinant 1, so it has
form $\left(\begin{smallmatrix}1&h\\0&1\end{smallmatrix}\right)$.
Choose an integer representative of $h$ and set
$H_1=H\left(\begin{smallmatrix}1&h\\0&1\end{smallmatrix}\right)$.
This preserves the first column and gives the required congruence.
For $m_0=1$ the condition is vacuous and the proof still applies.
\end{proof}


\begin{lemma}[Gordan's lemma {\cite[Proposition~2.1.1(a)]{BGT}}]\label{lem:cone}
For a rational polyhedral cone $C\subseteq\mathbb R_{\geq0}^n$, the
additive monoid $C\cap\Z^n$ has finitely many generators.
\end{lemma}

\begin{lemma}[Solutions of a linear equation]\label{lem:rowcompletion}
Let $s\geq1$, let $\mathbf t=(t_1,\ldots,t_\nu)$, and suppose
$A_i,R_i,H\in\Z[\mathbf t]$ satisfy $\sum_{i=1}^s A_iR_i=\Lambda$,
where $\Lambda>0$ is fixed. The set
\[
 \left\{(\mathbf t,\mathbf y)\in\Z^\nu\times\Z^s:
       \sum_{i=1}^s A_i(\mathbf t)y_i=H(\mathbf t)\right\}
\]
is a finite union of polynomial families. More explicitly, it consists
of the integral values of the map
$(\mathbf t,\mathbf v)\mapsto(\mathbf t,\mathbf y)$ defined by
\begin{equation}\label{eq:rowcompletion}
 y_i=\frac{HR_i}{\Lambda}+v_i
       -\frac{R_i}{\Lambda}\sum_{j=1}^s A_jv_j,
 \qquad \mathbf v=(v_1,\ldots,v_s)\in\Z^s,
\end{equation}
with $\mathbf t\in\Z^\nu$.
\end{lemma}
\begin{proof}
Summing verifies the equation. For any prescribed solution take $v_i=y_i$;
the formula returns the target. Lemma~\ref{lem:filter} retains integral
outputs and removes the fixed denominator $\Lambda$.
\end{proof}


\section{A polynomial congruence map with variable level}\label{sec:matrix}
Vaserstein's Theorem~13 parametrizes each fixed principal congruence
subgroup. We need polynomial dependence on the level as well. We prove
this from his fixed factorization in \cite[Proposition~5.9]{Vas}.

\subsection{Polynomial matrices}
Put
\[
 U(v)=\begin{pmatrix}1&v\\0&1\end{pmatrix},\qquad
 L(v)=\begin{pmatrix}1&0\\v&1\end{pmatrix},\qquad
 J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.
\]
Let $D_j(v_1,\ldots,v_j)$ be the alternating product of $j$ elementary
matrices beginning with $U(v_1)$, and let $C_j$ begin with $L(v_1)$.
These are Vaserstein's polynomial matrices $\Delta_j,\Gamma_j$,
respectively.
We use the following explicit versions of his polynomial matrices from
\cite[Section~1, definitions preceding Lemma~1.1]{Vas}:
\begin{equation}\label{eq:phi-four}
 F(a,c)=\begin{pmatrix}1-ac&a^2\\-c^2&1+ac\end{pmatrix},\qquad
 \Phi_4(a,b,c,d)=F(b,d)F(a,c)F(b,d)J^{-1}.
\end{equation}
Every factor has determinant $1$, and $\Phi_4(1,0,0,1)=I_2$.
If $\alpha=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$
has determinant $1$, conjugating $L(-1)U(1)L(-1)=J$ by $\alpha$ gives
\begin{equation}\label{eq:phi-four-identity}
 \Phi_4(a,b,c,d)=\alpha\alpha^T.
\end{equation}
For five free arguments define
\begin{equation}\label{eq:phi-five}
 W=\Phi_4(1+ae,be,ce,1+de),\qquad
 \Phi_5(a,b,c,d,e)=
 \begin{pmatrix}W_{11}&eW_{12}\\W_{21}/e&W_{22}\end{pmatrix}.
\end{equation}
Here $W|_{e=0}=I_2$, so $W_{21}/e$ is an integer polynomial,
defined also at $e=0$. The determinant identity for $W$ gives
$\det\Phi_5=1$ as a polynomial identity.

With $q$ an indeterminate and all $v_i$ free, set
\begin{align}
 D_j(q;\mathbf v)&=D_j(qv_1,\ldots,qv_j),&
 C_j(q;\mathbf v)&=C_j(qv_1,\ldots,qv_j),\notag\\
 \widehat X_4(q;\mathbf v)&=
 \Phi_4(1+q^2v_1,qv_2,qv_3,1+q^2v_4),\label{eq:level-blocks}\\
 \widehat X_5(q;\mathbf v)&=
 \Phi_5(q^2v_1,qv_2,qv_3,q^2v_4,v_5),&
 \widehat Y_5(q;\mathbf v)&=\widehat X_5(q;\mathbf v)^T.\notag
\end{align}
All entries belong to $\Z[q,\mathbf v]$. To check their congruences,
expand \eqref{eq:phi-four} modulo $q^2$:
\[
 \widehat X_4\equiv
 \begin{pmatrix}1&q(v_2+v_3)\\q(v_2+v_3)&1\end{pmatrix},\qquad
 \widehat X_5\equiv
 \begin{pmatrix}1&qv_5^2(v_2+v_3)\\q(v_2+v_3)&1\end{pmatrix}
 \pmod{q^2}.
\]
For the lower-left entry of the second matrix, the remainder before
division is divisible by both $q^2$ and $v_5$, hence by $q^2v_5$.
The elementary products have the same diagonal and off-diagonal
divisibilities, since these are preserved under multiplication.
Consequently every matrix takes values in
\[
 G(q)=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\SL_2(\Z):
       q\mid b,c,\quad q^2\mid a-1,d-1\right\}
 \quad(q\geq2),
\]
and every matrix is $I_2$ at $q=0$.

\subsection{Coverage of the fixed factorization}
The sets $X_4(q),X_5(q),Y_5(q)$ in Vaserstein's
Proposition~5.9 are contained in the respective images in
\eqref{eq:level-blocks}.
Indeed, $X_4(q)$ consists of $\alpha\alpha^T$ with $\alpha\in G(q)$.
Equation~\eqref{eq:phi-four-identity} gives the inclusion by taking
the four free arguments to be the normalized entries of $\alpha$.

An element of $X_5(q)$ has the form
\begin{equation}\label{eq:vas-X5-target}
 \begin{pmatrix}1+aq^2e&bqe^2\\cq&1+dq^2e\end{pmatrix}
 \begin{pmatrix}1+aq^2e&cqe^2\\bq&1+dq^2e\end{pmatrix},
\end{equation}
where the first matrix has determinant $1$. The entry $cqe^2$ in the
second factor follows from \cite[Section~1]{Vas}; the extra factor $q$
printed on p.~1000 appears to be a misprint.
For $e\ne0$, put $D_e=\diag(e,1)$ and
$\alpha=\left(\begin{smallmatrix}1+aq^2e&bqe\\cqe&1+dq^2e\end{smallmatrix}\right)$.
The two factors in \eqref{eq:vas-X5-target} are
$D_e\alpha D_e^{-1}$ and $D_e\alpha^TD_e^{-1}$. Since $\det\alpha=1$,
\eqref{eq:phi-four-identity}--\eqref{eq:phi-five} give
\[
 D_e\alpha\alpha^TD_e^{-1}
   =\Phi_5(aq^2,bq,cq,dq^2,e)=\widehat X_5(q;a,b,c,d,e).
\]
At $e=0$ the product in \eqref{eq:vas-X5-target} is $L(q(b+c))$.
Here the same arguments cannot in general be used, since
\[
 \widehat X_5(q;a,b,c,d,0)=L\bigl(q(b+c)-2q^2(a+d)\bigr).
\]
Instead, the preimage $(0,b,c,0,0)$ gives $L(q(b+c))$.
Replacing $(a,b,c,d)$ by $(d,-c,-b,a)$ reverses and inverts the two
factors, so $X_5(q)$ is closed under inversion. Hence Vaserstein's
$Y_5(q)=(X_5(q)^{-1})^T$ is covered by $\widehat Y_5$.

Suppressing the level and the free arguments, form the following
polynomial matrix, with independent arguments at each occurrence:
\begin{align}\label{eq:uniform-G-word}
 \mathcal P(q;\theta)={}&
 C_6\widehat X_5D_4\widehat Y_5C_6\widehat X_5C_6\widehat X_4\notag\\
 &\cdot C_5\widehat Y_5C_4\widehat X_5D_6\widehat Y_5D_6\widehat X_4\notag\\
 &\cdot C_3\widehat X_5D_2\widehat X_5U(q)\widehat Y_5C_2.
\end{align}
The elementary products use $50$ parameters, the nine five-parameter
matrices use $45$, and the two four-parameter matrices use $8$.
Thus $\theta\in\Z^{103}$. For each integer $q\geq2$,
\cite[Proposition~5.9]{Vas} expresses every element of $G(q)$ as this
word in the original sets. The inclusions just proved supply preimages
in our free polynomial matrices. Conversely, every factor in our word
belongs to $G(q)$. Hence
\begin{equation}\label{eq:uniform-G-image}
 \mathcal P(q;\Z^{103})=G(q)\qquad(q\geq2).
\end{equation}
The primes and exponents chosen in the proof of that proposition
determine preimages in its factor sets. The fixed maps
\eqref{eq:level-blocks} and word \eqref{eq:uniform-G-word} give joint
polynomial dependence on the level and the parameters. We use the
resulting $103$-parameter map; Vaserstein obtains smaller counts for
fixed levels in \cite[Corollaries~5.10--5.11]{Vas}.

\subsection{The full congruence subgroup and levels zero and one}
Set $\Gamma(0)=\{I_2\}$ and
$\Gamma(q)=\{g\in\SL_2(\Z):g\equiv I_2\pmod q\}$ for $q>0$.
For a free integer $h$ put
\[
 T(q,h)=L(1)U(qh)L(-1)=
 \begin{pmatrix}1-qh&qh\\-qh&1+qh\end{pmatrix},\qquad
 Q(q;\theta,h)=\mathcal P(q;\theta)T(q,h).
\]
Then $Q\in\SL_2(\Z[q,\theta,h])$, $Q-I_2$ is divisible by $q$,
and $Q(q;\Z^{104})=\Gamma(q)$ for $q\geq2$. To see surjectivity,
write the diagonal of $g\in\Gamma(q)$ as
$(1+q\lambda,1+q\mu)$. Its determinant gives
$\lambda+\mu\equiv0\pmod q$. Choose $h\equiv-\lambda\pmod q$.
The diagonal of $gT(q,h)^{-1}$ is then $(1,1)$ modulo $q^2$, so
this matrix belongs to $G(q)$ and is attained in
\eqref{eq:uniform-G-image}. This verifies directly the extra-parameter
step underlying \cite[Corollary~5.11]{Vas}.

Fix a matrix $V(\xi)\in\SL_2(\Z[\xi_1,\ldots,\xi_{46}])$
parametrizing $\SL_2(\Z)$ by \cite[Theorem~1]{Vas}.
Replacing it by $V(\xi)V(0)^{-1}$ ensures $V(0)=I_2$.
\begin{lemma}[Uniform congruence parametrization]\label{lem:uniform}
There is $K\in\SL_2(\Z[q,\vartheta_1,\ldots,\vartheta_{150}])$
such that $K-I_2$ is polynomially divisible by $q$ and
$K(q;\Z^{150})=\Gamma(q)$ for every integer $q\geq0$.
\end{lemma}
\begin{proof}
For $\vartheta=(\theta,h,\xi)$ define
\begin{equation}\label{eq:uniform-K}
 K(q;\vartheta)=Q(q;\theta,h)V(q\xi).
\end{equation}
The determinant is $1$ and $K-I_2$ is divisible by $q$ in the
integer polynomial ring. For $q\geq2$, setting $\xi=0$ gives the
whole of $\Gamma(q)$ by the preceding argument. There is a fixed
$\theta_*$ with $\mathcal P(q;\theta_*)=I_2$ identically:
set all arguments to zero except the second argument of the final
$C_2$, which is $-1$. The remaining factors are $U(q)U(-q)$.
At $q=1$, take $\theta=\theta_*$ and $h=0$; then $K=V(\xi)$.
At $q=0$, every factor is $I_2$.
\end{proof}

\section{Pairwise coprime block products}\label{sec:multipartite}
The next construction replaces coprimality conditions by unrestricted
polynomial parameters. All identities in this section take place in an
integer polynomial ring. In particular, inverses modulo a variable
polynomial are supplied by B\'ezout identities.

\begin{lemma}[Insertion with preserved congruences]\label{lem:protected}
Let $\R=\Z[s_1,\ldots,s_N]$. Suppose
$a,B,F,P,Q,w,\rho,\sigma,\iota_P,\iota_w,v,h_0\in\R$, with $P=BF$,
satisfy
\begin{equation}\label{eq:insertion-assumptions}
a\rho+P\sigma=1,\quad P\iota_P\equiv w\iota_w\equiv1\pmod{Q^2},
\quad Qv+Bh_0=1.
\end{equation}
Suppose also that $B$ and $w$ have a polynomial B\'ezout identity.
With additional unrestricted parameters $\theta\in\Z^{150}$ and
$\tau\in\Z$, one can construct $A,u,\rho',\sigma'\in\R[\theta,\tau]$ such that
\begin{equation}\label{eq:insertion-conclusions}
\begin{gathered}
A\rho'+Pu\sigma'=1,\\
u\equiv w\pmod{B^2},\qquad u\equiv w\pmod{Q^2}.
\end{gathered}
\end{equation}
All congruences here are ideal congruences in the integer polynomial
ring. The output pairs $(u,B)$ and $(u,Q)$ have polynomial B\'ezout identities.
For every integer specialization of $s_1,\ldots,s_N$ at which
$\gcd(a,Pw)=1$, there is $(\theta,\tau)\in\Z^{151}$ giving $A=a$
and $u=w$, including when $P=0$ or $Q=0$.
\end{lemma}

\begin{proof}
The elementary product
$\mathcal D(s,t)=U(s-1)L(1)U(t-1)L(-s)$ equals
$\diag(s,t)$ modulo $st-1$. Set $\lambda=P^2\iota_P^2$ and
\begin{align}\label{eq:correctionE}
E={}&U(\lambda a\iota_w)U\bigl(\lambda(P\iota_w-1)\bigr)L(\lambda)
 U\bigl(\lambda(w\iota_P-1)\bigr)L(-\lambda P\iota_w)U(-\lambda a\iota_P).
\end{align}
Then $E\equiv I_2\pmod{P^2}$. Modulo $Q^2$ it equals
$U(a\iota_w)\mathcal D(P\iota_w,w\iota_P)U(-a\iota_P)$, which sends $(a,P)^T$ to
$(a,w)^T$. Define
\begin{equation}\label{eq:insertionM}
H=\begin{pmatrix}a&-\sigma\\P&\rho\end{pmatrix},\qquad
EH K((PQ)^2;\theta)=\begin{pmatrix}A&B_0\\C&D\end{pmatrix},
\end{equation}
using the polynomial matrix of Lemma~\ref{lem:uniform}. Consequently
$(A,C)\equiv(a,P)\pmod{P^2}$ and
$(A,C)\equiv(a,w)\pmod{Q^2}$.

The construction makes $A-a$ polynomially divisible by $P^2$.
Thus $h=(A-a)/P$ is an integer polynomial even at $P=0$, and
$\sigma_0=\sigma-\rho h$ satisfies $A\rho+P\sigma_0=1$.
Put
\begin{align}\label{eq:protected-shear}
\zeta&=\bigl(v(2-Qv)\bigr)^2,& \iota_A&=\rho(2-A\rho),\notag\\
\chi&=Q^2\bigl((w-P)\iota_A\zeta+B^2\tau\bigr),&
u&=C+\chi A,\qquad D_0=D+\chi B_0,
\end{align}
where $\tau$ is free. Formula~\eqref{eq:squareinverse} gives
$A\iota_A\equiv1\pmod{B^2}$ and $Q^2\zeta\equiv1\pmod{B^2}$.
It follows that $u\equiv w\pmod{B^2}$; the congruence modulo
$Q^2$ is immediate. The determinant identity is
$AD_0-uB_0=1$. Therefore
\[
\rho'=\rho+P\sigma_0D_0,\qquad \sigma'=-\sigma_0B_0
\]
satisfy $A\rho'+Pu\sigma'=1$.
Writing $w\iota_w-1=Q^2s_w$ gives
$w\iota_w-Q(Qs_w)=1$. Thus $(w,Q)$ has a B\'ezout identity, as does
$(w,B)$ by hypothesis. Formula~\eqref{eq:witness-shift} transfers
these identities to $(u,Q)$ and $(u,B)$.

For coverage, first assume $PQ\ne0$. At the target value $A=a$, let
$\chi_0=Q^2(w-P)\rho(2-a\rho)\zeta$. The inverse identities give
$w-a\chi_0\equiv P\pmod{B^2}$. Choose $\tau$ such that
\[
c=w-a\chi_0-aQ^2B^2\tau\equiv P\pmod{P^2},\qquad P^2=B^2F^2.
\]
After division by $B^2$ this is soluble modulo $F^2$, because
$\gcd(aQ,P)=1$. We also have $c\equiv w\pmod{Q^2}$ and
$\gcd(a,c)=\gcd(a,w)=1$. The first column $(a,c)^T$ is therefore
primitive and congruent to that of $EH$ modulo $(PQ)^2$.
Lemma~\ref{lem:column} and the surjectivity of $K((PQ)^2)$ supply
a matrix in~\eqref{eq:insertionM} with that column. The final shear
then gives $u=w$.

If $P=0$, the input identities force $a,Q\in\{1,-1\}$ and
$a\rho=1$. Also $E=I_2$ and $K(0)=I_2$. Hence $A=a$, $C=0$, and
$u=w\zeta+aB^2\tau$. If $B=0$, then $Qv=1$ and $\zeta=1$.
If $B\ne0$, then $\zeta\equiv1\pmod{B^2}$, so a choice of $\tau$
again gives $u=w$. Finally, if $Q=0$, then $P\iota_P=w\iota_w=1$;
the matrix $E$ sends $(a,P)^T$ exactly to $(a,w)^T$, while
$K(0)=I_2$ and $\chi=0$. The two zero cases cannot occur together.
\end{proof}

\begin{proposition}[Pairwise coprime products]\label{thm:multipartite}
Partition $N$ coordinates into blocks $I_1,\ldots,I_h$, and put
$X_j(x)=\prod_{\nu\in I_j}x_\nu$. There is an integer-polynomial map
$\Phi$ on unrestricted integer parameters whose image is exactly
\[
 \{x\in\Z^N:\gcd(X_i(x),X_j(x))=1\text{ for }i<j\}.
\]
There are also polynomial outputs $R_{ij},S_{ij}$ such that
$X_i(\Phi)R_{ij}+X_j(\Phi)S_{ij}=1$ identically for every $i<j$.
\end{proposition}

\begin{proof}
For two singleton blocks, the assertion follows from
\cite[Theorems~1 and~2.2]{Vas}. We construct the map for arbitrary block
sizes, including the condition $\gcd(x_1x_2,x_3)=1$.

Induct on the total number of coordinates, filling the blocks in order.
The first block has unrestricted coordinates; empty blocks have product
$1$ and trivial B\'ezout coefficients. The induction hypothesis supplies
a polynomial map onto all tuples in the old blocks, together with
polynomial B\'ezout coefficients for each pair of block products.
Write these products as $X_1,\ldots,X_g,Z$, where $Z$ belongs to the
block currently being filled. Adjoining a new coordinate keeps the
coordinates of that block fixed and replaces those in blocks
$1,\ldots,g$ by new polynomials. Start the new coordinate as a free
parameter $a$ and initialize
$P=1$, $\rho=0$, $\sigma=1$, so $a\rho+P\sigma=1$.

Process the coordinates of blocks $1,\ldots,g$ in order. At an old
coordinate $w$ of block $j$, let $B$ be the product of the completed
new blocks and let $F$ be the product of the processed new coordinates
in the current block. Put
\[
 P=BF,\qquad Q=Z\prod_{\ell>j}X_\ell,
\]
where future blocks retain their old coordinates. The invariants at
this step are:
\begin{enumerate}[label=(\roman*),leftmargin=2.3em]
\item every processed coordinate is congruent to its old value modulo
      the current $Q^2$;
\item $B$ is congruent to its old product modulo $X_j^2$, and hence
      modulo $w^2$;
\item $a\rho+P\sigma=1$ is a polynomial identity.
\end{enumerate}
At an earlier step in block $j'\le j$, the modulus was
$Q_{j'}=Z\prod_{\ell>j'}X_\ell$, with $Z$ and the future $X_\ell$
still the old polynomials. Since $Q_j\mid Q_{j'}$, the congruence
modulo $Q_{j'}^2$ gives~(i). For $j'<j$, also $X_j\mid Q_{j'}$;
taking products over completed blocks gives~(ii). Taking products of
the congruences in~(i) also gives $P\equiv P_{\mathrm{old}}\pmod{Q^2}$,
where $P_{\mathrm{old}}$ is the product of the corresponding old coordinates.
These are ideal divisibilities and remain valid when a factor vanishes.

The original pairwise B\'ezout identities, factored and combined using
\eqref{eq:witness-product}, give B\'ezout identities between each old processed
product and $Q$. Transfer them by \eqref{eq:witness-shift} and
invariant~(i). This explicitly supplies polynomial identities
\[
 P r_P+Q s_P=1,\qquad w r_w+Q s_w=1,
 \qquad Qv+Bh_0=1.
\]
Set
\[
 \iota_P=r_P(2-Pr_P),\qquad
 \iota_w=r_w(2-wr_w).
\]
Equation~\eqref{eq:squareinverse} gives the required inverses modulo
$Q^2$. Invariant~(ii), the old pairwise B\'ezout identities and
\eqref{eq:witness-shift} also give a B\'ezout identity for $(B,w)$.
Together with invariant~(iii), these are all the inputs to
Lemma~\ref{lem:protected}. Apply it, replace $a$ by $A$, and replace
$P$ by $Pu$ after recording the new coordinate $u$.
The conclusion maintains~(iii). Its congruences modulo $Q^2$ and $B^2$
maintain~(i)--(ii) as we move through the current block and then pass
to the next block.

Write $X_i'$ for the new completed block products. For $i<j$,
the protected congruences give
\[
 X_i'\equiv X_i\pmod{X_j^2},\qquad
 X_j'\equiv X_j\pmod{(X_i')^2}.
\]
Applying \eqref{eq:witness-shift} twice gives a B\'ezout identity for
$(X_i',X_j')$. Also $X_i'\equiv X_i\pmod{Z^2}$, so $(X_i',Z)$
has a B\'ezout identity. The final identity
$a\rho+\bigl(\prod_iX_i'\bigr)\sigma=1$ gives a B\'ezout identity for
$(X_i',a)$; combining these gives one for $(X_i',Za)$.
Thus all the new block products are pairwise coprime.

For a prescribed target, the old coordinates satisfy the induction
hypothesis, so choose old parameters attaining them. Set $a$ equal to
the desired new coordinate. At every insertion the target satisfies
$\gcd(a,Pw)=1$. The coverage part of Lemma~\ref{lem:protected} retains
$a$ and the old value $w$. Hence the final output is the entire target.
This proves surjectivity at the induction step. Every operation is
polynomial, so zero coordinates are included throughout.
\end{proof}

\section{Proof of Theorem~\ref{main:shared}: monomial coefficients}\label{sec:shared}
\begin{lemma}[Monomials omitting a block]\label{lem:omitblock}
Suppose the products $P_1,\ldots,P_h$ of disjoint blocks of polynomial
coordinates have polynomial pairwise B\'ezout identities. Let
$B_1,\ldots,B_s$, with $s\geq1$, be monomials with coefficients $1$ or $-1$ in these coordinates. Suppose that for
each block $j$ at least one $B_{i(j)}$ contains no coordinate of block $j$.
Then $B_1,\ldots,B_s$ have polynomial B\'ezout identities.
\end{lemma}

\begin{proof}
If there are no blocks, every $B_i$ is $1$ or $-1$, so the result is
immediate. Otherwise choose a fixed $E\geq1$ at least as large as
every exponent in every $B_i$. Set
$Q_j=P_j^E$ and $C_j=\prod_{\ell\ne j}Q_\ell$. Raising powers and combining
pairwise B\'ezout identities gives
$1=U\prod_{\ell<h}Q_\ell+VQ_h$.
The complements $C_j$ generate the unit ideal by induction, starting
with the single-block complement $1$. For the induction step, if
$1=\sum_{j<h}W_j\prod_{\ell<h,\ell\ne j}Q_\ell$, substitute this
identity in the term $VQ_h$ of the preceding identity. The result is
a polynomial combination of all $C_j$.
Each $C_j$ is a monomial multiple
of $B_{i(j)}$; substituting those formal monomial quotients yields the desired
B\'ezout identities. The argument is an identity in a polynomial ring, so it does not
exclude zeros.
\end{proof}


\begin{proof}[Proof of Theorem~\ref{main:shared}]
Absorb the fixed $\alpha_i$ into the variables $y_i$ and impose their
divisibilities at the end. First take a target with every $x_j\ne0$,
and let $e_0,e_1,\ldots,e_s$ be the monomial exponent vectors.
For each coefficient index $1\leq i\leq s$, define
\[
 E_i=\{v\geq0:(e_j-e_i)\cdot v\geq0\ (1\leq j\leq s)\}.
\]
If $c=0$, put $C_i=E_i$; these cones cover the nonnegative orthant.
If $c\ne0$, put
$C_i=E_i\cap\{v:(e_0-e_i)\cdot v\geq0\}$.
For $c\ne0$, solvability after absorbing the $\alpha_i$ requires
$\gcd_i M_i(x)\mid cM_0(x)$. Thus, for the target valuation vector
$v=(v_p(x_1),\ldots,v_p(x_n))$,
\[
 \min_{1\leq i\leq s} e_i\cdot v
      \leq e_0\cdot v+v_p(c).
\]
At every prime $p\nmid c$, this places $v$ in some $C_i$.
At $p\mid c$, with $\mu=v_p(c)$, it places $v$ in some
\[
 \mathcal A_i(\mu)=E_i\cap\{v:(e_0-e_i)\cdot v\geq-\mu\}.
\]

There is a finite set $F_i(\mu)\subseteq\N^n$ with
\begin{equation}\label{eq:affine-cone-module}
 \mathcal A_i(\mu)\cap\Z^n
   =\bigcup_{b\in F_i(\mu)}\bigl(b+(C_i\cap\Z^n)\bigr).
\end{equation}
Indeed, homogenize the last inequality to
$(e_0-e_i)\cdot v+\mu t\geq0$, $t\geq0$, and apply Gordan's lemma.
The slice $t=0$ is exactly $C_i$, so the generators with last
coordinate zero generate $C_i\cap\Z^n$. Take $F_i(\mu)$ to be the
projections of the generators with last coordinate one. In a
decomposition with $t=1$, exactly one generator has last coordinate
one and all others have last coordinate zero. Conversely,
$b+C_i\subseteq\mathcal A_i(\mu)$ for every $b\in F_i(\mu)$.

For each prime divisor of nonzero $c$, choose an index and a base
vector from the finite sets above and put $h_j=\prod_{p\mid c}p^{b_{p,j}}$.
For $c=0$, take every $h_j=1$. Set $x_j=h_jX_j$.
Choose generators $g_{i,\ell}$ of the additive monoids $C_i\cap\Z^n$.
Parametrize their variable blocks $(u_{i,\ell})_\ell$ with pairwise coprime
block products using Proposition~\ref{thm:multipartite}. For each of
the finitely many signs $\varepsilon_j\in\{-1,1\}$, write
$\varepsilon^e=\prod_j\varepsilon_j^{e_j}$ and set
\begin{align*}
 X_j&=\varepsilon_j\prod_{i,\ell}u_{i,\ell}^{(g_{i,\ell})_j},&
 G&=\prod_{i,\ell}u_{i,\ell}^{e_i\cdot g_{i,\ell}},\\
 B_j&=\varepsilon^{e_j}
       \prod_{i,\ell}u_{i,\ell}^{(e_j-e_i)\cdot g_{i,\ell}}
       \quad(1\leq j\leq s).
\end{align*}
If $c\ne0$, also set
$D=\varepsilon^{e_0}\prod_{i,\ell}u_{i,\ell}^{(e_0-e_i)\cdot g_{i,\ell}}$;
if $c=0$, set $D=0$. All exponents are nonnegative, and
$M_j(X)=GB_j$, with $M_0(X)=GD$ when $c\ne0$.
For block $i$, $B_i$ omits that block, so Lemma~\ref{lem:omitblock}
supplies $\sum_jB_jW_j=1$. With $H=cM_0(h)D$, apply
Lemma~\ref{lem:rowcompletion} with $\Lambda=1$ to obtain
\[
 Y_j=HW_j+v_j-W_j\sum_k B_kv_k,\qquad
 y_j=Y_j/(\alpha_jM_j(h)),
\]
retaining the fixed divisibilities $\alpha_jM_j(h)\mid Y_j$ by
Lemma~\ref{lem:filter}. These give polynomial solutions even when
some parameter values make $G=0$.

For coverage, assign each prime of a target to a cone containing its
residual valuation vector and decompose in its generators. Distinct
cone blocks have disjoint prime supports. Choose all $u_{i,\ell}>0$
for this preimage and put the signs in $\varepsilon$.
Proposition~\ref{thm:multipartite} attains these parameters, recovering
the target $X$ with $G\ne0$.
Taking $v_j=\alpha_jM_j(h)y_j$ then recovers every $y_j$
by formula~\eqref{eq:rowcompletion}.

Finally fix a zero pattern among the $x_j$ and delete the vanishing
coefficient monomials. If $M_0$ vanishes, use the $c=0$ construction
just given. If $M_0$ survives, use the same proof on the remaining
nonzero coordinates. When no coefficient survives, either the
right-hand side also vanishes and all $y_i$ are free,
or there are no targets in that pattern. The variables $y_i$ belonging
to deleted summands are free.
Every deleted monomial vanishes identically once the specified zero
coordinates are fixed. Consequently, outputs of a reduced family still
solve the original equation if further coordinates vanish. The families
are indexed by finitely many zero patterns, sign vectors, choices
$(i_p,b_p)_{p\mid c}$ when $c\ne0$, and residue classes from
Lemma~\ref{lem:filter}.
\end{proof}

\section{Proof of Theorem~\ref{main:values}: split polynomials}\label{sec:splitfull}
We first prove the completely split alternative, namely
\begin{equation}\label{eq:split}
 \kappa z\prod_{i=1}^r x_i^{e_i}=f(t),
\end{equation}
for fixed $\kappa\ne0$ and $f\in\Z[t]$ split over $\Q$.

\subsection{Monomial columns from Gordan's lemma}
Fix $m\geq1$ and positive $e_i$. The valuation vectors for
\begin{equation}\label{eq:homogeneous-column}
 Z\prod_i P_i^{e_i}=F^m
\end{equation}
form the monoid
$\{(z,p_1,\ldots,p_r,f)\in\N^{r+2}:z+\sum_i e_ip_i=mf\}$.
By Gordan's lemma it has a finite Hilbert basis, consisting of its
indecomposable nonzero elements. Its only basis element
with every $p_i=0$ is $g_0=(m,0,\ldots,0,1)$; write $\mathcal G$ for
the other elements. With free $u,v_g$ and signs $\varepsilon_i=\pm1$, set
\begin{align}
 P_i&=\varepsilon_i\prod_{g\in\mathcal G}v_g^{g_{p_i}},&
 A&=\prod_{g\in\mathcal G}v_g^{g_f},\notag\\
 F&=uA,&
 Z&=\Bigl(\prod_i\varepsilon_i^{e_i}\Bigr)u^m
                  \prod_{g\in\mathcal G}v_g^{g_z}.
 \label{eq:homogeneous-parametrization}
\end{align}
For a nonzero target, decompose each prime valuation in the Hilbert
basis. Choose $v_g>0$, let $\varepsilon_i$ carry the sign of $P_i$,
and let $u$ carry the sign of $F$. The equation then forces the sign
of $Z$ to be $\bigl(\prod_i\varepsilon_i^{e_i}\bigr)\operatorname{sign}(u)^m$,
as in \eqref{eq:homogeneous-parametrization}. Thus every nonzero target is attained.
Every $g\in\mathcal G$ has a positive $p_i$ and positive $f$ coordinate.
Consequently a prime absent from all target $P_i$ can be assigned entirely
to $u$, leaving every $v_g$ prime to it. For $r=0$ the formulas are
$F=u,Z=u^m$.

\subsection{Normalizing distinct integral roots}
First take
\begin{equation}\label{eq:monic-split}
f(t)=\prod_{j=1}^{s_f}(t-r_j)^{m_j},
\end{equation}
where $s_f\geq1$, the $r_j$ are distinct integers, and $m_j>0$.
Write $d_f=\sum_j m_j$. If $s_f\geq2$ set
\begin{equation}\label{eq:root-normalization}
\Delta_{\mathrm{rt}}=\prod_{j<\ell}|r_j-r_\ell|,\qquad B_{\mathrm{rt}}=\Delta_{\mathrm{rt}}^{d_f};
\end{equation}
if $s_f=1$, set $\Delta_{\mathrm{rt}}=B_{\mathrm{rt}}=1$.
For the moment take a target of \eqref{eq:split} with $f(t)\ne0$.
All $z,x_i$, and $F_j=t-r_j$ are then nonzero. Put
\[
h_i=\gcd(|x_i|,B_{\mathrm{rt}}),\qquad x_i=h_iX_i,\qquad H=\prod_i h_i^{e_i}.
\]
The tuples $(h_1,\ldots,h_r)$ belong to a finite set.

\begin{lemma}[Allocation to separate root factors]\label{lem:rootallocation}
For the normalized target there are integers $P_{ij}$ such that
\begin{equation}\label{eq:rootallocation}
X_i=\prod_j P_{ij},\qquad
\prod_i P_{ij}^{e_i}\mid F_j^{m_j},
\end{equation}
and each prime divides entries $P_{ij}$ in at most one column $j$.
\end{lemma}

\begin{proof}
If a prime $p$ does not divide $\Delta_{\mathrm{rt}}$, at most one $F_j$ is divisible by
$p$; allocate all residual $p$-powers to that column. If $p\mid\Delta_{\mathrm{rt}}$,
put $\ell=\vp(\Delta_{\mathrm{rt}})$. At most one of the $\vp(F_j)$ can exceed $\ell$,
since $F_j-F_{j'}=r_{j'}-r_j$ and each difference has valuation at most
$\ell$. Choose a column $j$ of maximal valuation.
If a residual $X_i$ contains $p$, some original $x_i$ has valuation greater
than $d_f\ell$, and hence the total removed weighted valuation is at least
$d_f\ell$. Therefore
\begin{align*}
\sum_i e_i\vp(X_i)
&=\sum_h m_h\vp(F_h)-\vp(\kappa)-\vp(z)
  -\sum_i e_i\min(\vp(x_i),d_f\ell)\\
&\leq m_j\vp(F_j)+\sum_{h\ne j}m_h\ell-d_f\ell
\leq m_j\vp(F_j).
\end{align*}
Allocate all residual powers at this prime to the single column $j$.
If no residual $X_i$ contains $p$, nothing is allocated. The allocations
produce \eqref{eq:rootallocation}; signs can be distributed arbitrarily
among the columns to give the prescribed signs of the $X_i$.
\end{proof}

Put $Z_j=F_j^{m_j}/\prod_iP_{ij}^{e_i}$. Then
\begin{equation}\label{eq:columns}
Z_j\prod_iP_{ij}^{e_i}=F_j^{m_j},\qquad
\prod_j Z_j=\kappa Hz.
\end{equation}
Represent each column by \eqref{eq:homogeneous-parametrization}. By its support
property and Lemma~\ref{lem:rootallocation}, the blocks of non-$g_0$ parameters
$v_{jg}$ can be chosen with pairwise coprime block products.
Their coefficient monomials
\begin{equation}\label{eq:Aj}
A_j=\prod_g v_{jg}^{g_f}
\end{equation}
are therefore pairwise coprime, and $F_j=u_jA_j$.

\subsection{The Chinese remainder theorem}
Use Proposition~\ref{thm:multipartite} to parametrize the blocks $(v_{jg})_g$ so
that their block products are pairwise coprime, with polynomial B\'ezout identities.
By Lemma~\ref{lem:powers} and \eqref{eq:witness-product}, for each $j$ we can
construct polynomial B\'ezout identities
\begin{equation}\label{eq:CRT-bezout}
A_jR_j+B_jS_j=1,\qquad B_j=\prod_{\ell\ne j}A_\ell.
\end{equation}
For $s_f=1$, use $B_1=1,R_1=0,S_1=1$.
Let $\Pi=\prod_jA_j$ and introduce one free parameter $w$.
Define
\begin{align}
T_0&=\sum_j r_jB_jS_j,& T&=T_0+\Pi w,\label{eq:CRT-T}\\
u_j&=-r_jR_j+
\sum_{\ell\ne j}r_\ell\left(\prod_{h\ne\ell,j}A_h\right)S_\ell
+B_jw.\label{eq:CRT-u}
\end{align}
Using \eqref{eq:CRT-bezout}, expansion gives the polynomial identities
\begin{equation}\label{eq:CRT-identity}
T-r_j=A_ju_j\qquad(1\leq j\leq s_f).
\end{equation}
The formulas contain no division by $A_j$ and are valid when some $A_j=0$.
For a nonzero target representation all $A_j$ are nonzero and pairwise
coprime. The integer $t-T_0$ is divisible by each $A_j$, hence by their
product, so some integer $w$ makes $T=t$. The formulas then give exactly
the target $u_j=(t-r_j)/A_j$.

Substitute these $u_j$ in the parametrization~\eqref{eq:homogeneous-parametrization} of each column. Retain
the one fixed congruence
\begin{equation}\label{eq:split-fixed-filter}
\prod_jZ_j\equiv0\pmod{\kappa H}.
\end{equation}
After the finite parameter subdivision of Lemma~\ref{lem:filter}, define
\begin{equation}\label{eq:split-output}
t=T,\qquad x_i=h_i\prod_jP_{ij},\qquad
z=\frac{\prod_jZ_j}{\kappa H}.
\end{equation}
All these coordinates are integer polynomials. Multiplying
\eqref{eq:columns} and using \eqref{eq:CRT-identity} proves the original
equation for every parameter value. Conversely, the allocation and column
representation above put every nonzero target in one of these families;
its parameters satisfy \eqref{eq:split-fixed-filter} because
$\prod_j Z_j=\kappa Hz$. The choices of $h_i$ and the column sign vectors
are finite. There are no remaining variable coprimality or divisibility
conditions on the parameters.

For a target with $f(t)=0$, fix $t$ to one of its finitely many integral
roots and fix one of $z,x_i$ to zero. Leave all other variables free.
These finitely many linear families cover the remaining targets.

\subsection{Multipliers and rational roots}
We first allow a fixed nonzero integer multiplier $\lambda$ on the right:
\begin{equation}\label{eq:lambda-split}
\kappa z\prod_i x_i^{e_i}=\lambda p_0(t),
\end{equation}
where $p_0$ is monic with integral roots as in \eqref{eq:monic-split}.
Targets with $p_0(t)=0$ are covered by the linear root families above.
For a target with $p_0(t)\ne0$, set
\[
h_0=\gcd(|z|,|\lambda|),\qquad h_i=\gcd(|x_i|,|\lambda|).
\]
Among the finitely many divisor choices, retain those for which
\begin{equation}\label{eq:kappaprime}
\kappa'=\frac{\kappa h_0\prod_i h_i^{e_i}}{\lambda}
\end{equation}
is an integer. Every target supplies such a choice. Indeed, for
$\mu=\vp(\lambda)$ the original equation gives
\[
 \vp(\kappa)+\vp(z)+\sum_i e_i\vp(x_i)\geq\mu.
\]
Keep $\vp(\kappa)$ unchanged and truncate each coordinate valuation
at $\mu$. If no coordinate is truncated, the sum is unchanged;
otherwise a truncated coordinate alone contributes at least $\mu$.
Thus $\vp(\kappa)+\vp(h_0)+\sum_i e_i\vp(h_i)\geq\mu$,
proving that \eqref{eq:kappaprime} is integral.
Apply the result already proved to
$\kappa' Z\prod_i X_i^{e_i}=p_0(t)$ and rescale
$z=h_0Z,x_i=h_iX_i$. Every output solves \eqref{eq:lambda-split}, and
every target is covered. Coefficient signs cause no change to the argument.

Finally let $f\in\Z[t]$ split over $\Q$ and have degree $d_f>0$ and
leading coefficient $\lambda\ne0$. Choose a positive common denominator
$d_{\mathrm{rt}}$ for its rational roots, and put $v=d_{\mathrm{rt}}t$. Then, for distinct integral
$r_j$ and positive multiplicities $m_j$,
\[
d_{\mathrm{rt}}^{d_f} f(t)=\lambda\prod_j(v-r_j)^{m_j}.
\]
Apply the preceding result with coefficient $d_{\mathrm{rt}}^{d_f}\kappa$ on the left,
retain the fixed congruence $v\equiv0\pmod{d_{\mathrm{rt}}}$, and output $t=v/d_{\mathrm{rt}}$ using
Lemma~\ref{lem:filter}. Every target and every output correspond under this
substitution.

If $f=0$, the solution set is the union of the coordinate hyperplanes
$z=0$ or $x_i=0$, with $t$ free. If $f$ is a nonzero constant, there are
finitely many choices for the present monomial variables (or none), and
$t$ is free. This proves the split alternative of Theorem~\ref{main:values}.
\section{Proof of Theorem~\ref{main:values}: irreducible quadratics}\label{sec:nonsplit}
It remains to treat quadratic polynomials irreducible over $\Q$. We use unique
factorization of nonzero ideals \cite[Corollary~3.20]{Dedekind} and
finiteness of the ideal class group
\cite[Theorem~14.16 and Corollary~14.18]{Sutherland}.


\begin{lemma}[A divisor ideal with prescribed norm]\label{lem:idealdivisor}
Let $k\geq1$ and let $\Delta$ be a nonsquare integer. Set $K=\Q(\delta)$ with
$\delta^2=\Delta$, and let $\OO$ be its full ring of integers.
Suppose $x\ne0$ and $x^ky=T^2-\Delta$ with integers $x,y,T$.
Set $B_\Delta=4|\Delta|$, $d=\gcd(|x|,B_\Delta)$ and $X=|x|/d$.
There is an integral ideal $\mathfrak a$ with
\begin{equation}\label{eq:idealdivisor}
\Nm\mathfrak a=X,\qquad
\mathfrak a^k\mid (T+\delta).
\end{equation}
Only the finitely many divisors $d$ of the fixed $B_\Delta$ occur.
\end{lemma}

\begin{proof}
Put $\alpha=T+\delta$, so $\overline\alpha=T-\delta$ and
$\alpha-\overline\alpha=2\delta$. Neither $\alpha$ nor its norm is zero.
Use ideal valuations normalized so a prime ideal has valuation 1 in itself.

At an inert or ramified rational prime $p$ there is a single prime ideal
$\mathfrak p$ above $p$, fixed by conjugation. Therefore
$v_{\mathfrak p}(\alpha)=v_{\mathfrak p}(\overline\alpha)
 \leq v_{\mathfrak p}(2\delta)$.
Multiplying by its residue degree gives
$v_p(|\Nm\alpha|)\leq v_p(|\Nm(2\delta)|)=v_p(B_\Delta)$.
Since $kv_p(x)\leq v_p(|\Nm\alpha|)$, no such prime divides $X$.

At a split prime, write $(p)=\mathfrak p\overline{\mathfrak p}$ and let
$R_p\geq S_p$ be the two valuations of $\alpha$, relabeling the ideals if
necessary. The difference identity implies
$S_p\leq v_p(B_\Delta)$: at the chosen prime the minimum of the valuations of
$\alpha,\overline\alpha$ is at most that of $2\delta$, which is at most
$v_p(B_\Delta)$. Write $v=v_p(x)$ and $m=v_p(B_\Delta)$.
If $v>m$, then
\[
R_p\geq kv-S_p\geq kv-m\geq k(v-m)=kv_p(X).
\]
If $v\leq m$, there is nothing to allocate. For every prime of $X$ put
$\mathfrak p^{v_p(X)}$ at the side with the larger valuation into
$\mathfrak a$. Split prime ideals have norm $p$, so this gives exactly
\eqref{eq:idealdivisor}. For $p\nmid B_\Delta$ the argument simply says that one
side has valuation zero and all the required valuation lies on the other.
\end{proof}

\subsection{Finitely many ideal bases and two homogeneous forms}

\begin{proposition}[The equation $x^kY=T^2-\Delta$]\label{prop:nonsplit}
For every nonsquare integer $\Delta$ and every integer $k\geq1$,
the integer solutions of
\[
 x^kY=T^2-\Delta
\]
form a finite union of polynomial families.
\end{proposition}
\begin{proof}
Put $K=\Q(\delta)$, $\delta^2=\Delta$, let $\OO$ be its ring
of integers, and put $B_\Delta=4|\Delta|$.
Choose a finite list of nonzero integral ideals $\mathfrak c$ representing
all ideal classes of $\OO$. For each choose a $\Z$-basis
$\omega_1,\omega_2$ of $\mathfrak c$ and a $\Z$-basis
$\eta_1,\eta_2$ of the fractional ideal $\mathfrak c^{-k}$. These are fixed
rank-two lattices. Write $n=\Nm\mathfrak c>0$.
For unrestricted integers $a,b,u,v$ put
\[
\gamma=a\omega_1+b\omega_2,\qquad \beta=u\eta_1+v\eta_2.
\]
Let $\ell:K\longrightarrow\Q$ extract the coefficient of $\delta$ in the
basis $(1,\delta)$. With a fixed common denominator $D>0$, define integer
homogeneous forms of degree $k$ by
\[
F(a,b)=D\ell(\gamma^k\eta_1),\qquad
G(a,b)=D\ell(\gamma^k\eta_2).
\]
These forms have no common nonzero algebraic zero. Indeed,
$\ell(z)=(\iota_1(z)-\iota_2(z))/(2\delta)$ for the two embeddings.
The embedding matrix of the basis $\eta_1,\eta_2$ is invertible, so
$F=G=0$ forces $\iota_1(\gamma)^k=\iota_2(\gamma)^k=0$.
The embedding matrix of $\omega_1,\omega_2$ then gives $a=b=0$.

We need to express the solution set of the single coefficient equation
\begin{equation}\label{eq:quadraticlinear}
 F(a,b)u+G(a,b)v=D
\end{equation}
as a finite union of polynomial families. Homogeneity shows that $e^k\mid D$, where $e=\gcd(a,b)>0$.
There are finitely many such $e$. Write $a=eA,b=eB$ and parametrize
primitive $(A,B)$ with a polynomial B\'ezout identity $A\rho+B\sigma=1$ using
\cite[Theorem~1]{Vas}. The homogeneous resultant $\Res(F,G)$ is
nonzero, since $F,G$ have no common projective zero
\cite[Chapter~3]{CLO}. Put
$\Lambda=|\Res(F,G)|$ and $m=2k-1$. The adjugate of the Sylvester
matrix, applied to the coefficient vectors of $a^m$ and $b^m$, gives
\[
 \Lambda a^m,\ \Lambda b^m\in(F(a,b),G(a,b))\,\Z[a,b].
\]
Substitute the polynomial pair $(A,B)$ and expand
$1=(A\rho+B\sigma)^{2m-1}$. Each term is divisible by $A^m$ or $B^m$,
so $\Lambda$ belongs to the substituted ideal. Lemma~\ref{lem:rowcompletion}, with right-hand side
$D/e^k$, now gives every solution $(u,v)$. Scaling back yields the
required finite union of polynomial families for \eqref{eq:quadraticlinear}.
On every family write
$\gamma^k\beta=T+\delta$, where $T$ is a rational polynomial with a fixed
denominator. For each positive divisor $d\mid B_\Delta$ and each
$\varepsilon\in\{-1,1\}$ define
\begin{equation}\label{eq:idealoutput}
x=\varepsilon d\frac{\Nm\gamma}{n},\qquad
Y=\varepsilon^k\frac{n^k\Nm\beta}{d^k}.
\end{equation}
The forms $\Nm\gamma/n$ and $n^k\Nm\beta$ are integer
quadratic polynomials. For the first assertion, $(\gamma)\subseteq
\mathfrak c$ implies that the norm of each integer linear combination of
the basis elements is divisible by $n$. Applying this to $(a,b)=(1,0)$,
$(0,1)$ and $(1,1)$ verifies divisibility of its three coefficients.
The same argument with $(\beta)\mathfrak c^k\subseteq\OO$ proves
the second assertion, including the fractional ideal basis.
Retain the fixed integrality conditions on $T,Y$ and eliminate them by
Lemma~\ref{lem:filter}. Equations \eqref{eq:idealoutput} and
\eqref{eq:quadraticlinear} then give unrestricted integer polynomials with
\[
x^kY=(\Nm\gamma)^k\Nm\beta=T^2-\Delta.
\]

To prove coverage, take an arbitrary target $(x,Y,T)$, put
$\alpha=T+\delta$, and take the ideal $\mathfrak a$ from
Lemma~\ref{lem:idealdivisor}. Choose the representative $\mathfrak c$
whose class is inverse to that of $\mathfrak a$. Then
$\mathfrak a\mathfrak c=(\gamma)$ is principal. It is contained in
$\mathfrak c$, so its generator has the form $a\omega_1+b\omega_2$ with
integers $a,b$. Its norm satisfies $|\Nm\gamma|=nX$.
Since $\mathfrak a^k\mid(\alpha)$, the element
$\beta=\alpha/\gamma^k$ belongs to $\mathfrak c^{-k}$, and therefore has
integer coordinates $u,v$ in its fixed basis. Its coefficient equation
is exactly \eqref{eq:quadraticlinear}. Choose $\varepsilon$ so that the
first formula in \eqref{eq:idealoutput} equals the target signed $x$.
The norm identity then forces the output $Y$ to equal the target value,
because $x\ne0$.
All the fixed integrality conditions hold for these target data, so the
polynomial families attain every target.

There is no omitted branch $x=0$: a nonsquare $\Delta$ makes
$T^2-\Delta\ne0$. Conversely \eqref{eq:quadraticlinear} prevents
$\gamma=0$ at integer parameter values. Nonprincipal ideals are included through the representative list.
The families are indexed by the finite choices
$(\mathfrak c,e,d,\varepsilon)$, with $e^k\mid D(\mathfrak c)$ and
$d\mid B_\Delta$, and by finitely many residue classes from
Lemma~\ref{lem:filter}.
\end{proof}


\begin{proof}[Completion of Theorem~\ref{main:values}]
When $r=0$, the equation is $\kappa z=f(t)$. Retain
$f(t)\equiv0\pmod\kappa$ and set $z=f(t)/\kappa$ using
Lemma~\ref{lem:filter}; this works for every $f\in\Z[t]$.

Now let $r=1$ and write the equation as $\kappa x^ky=f(t)$.
The split case, including degree at most one, constants and the zero
polynomial, was proved in Section~\ref{sec:splitfull}.
For a quadratic $f(t)=At^2+Bt+C$ irreducible over $\Q$, put
\[
 \Delta=B^2-4AC,\qquad T=2At+B,\qquad Y=4A\kappa y.
\]
Then $\Delta$ is nonsquare and
$T^2-\Delta=4Af(t)=x^kY$.
Apply Proposition~\ref{prop:nonsplit}, retain the fixed congruences
\[
 T\equiv B\pmod{2A},\qquad Y\equiv0\pmod{4A\kappa},
\]
and output $t=(T-B)/(2A)$, $y=Y/(4A\kappa)$ by
Lemma~\ref{lem:filter}. These substitutions preserve all targets and
produce only solutions.

For $k=1$, \cite[Example~7]{Vas} already expresses the solution set
of $xY=T^2-\Delta$, with $\Delta\ne0$, as finitely many orbits,
each a polynomial family. The same substitutions and fixed congruences
give $\kappa xy=f(t)$ in this case.
\end{proof}

\section{Five equations: complete solution formulas}\label{sec:examples}
The following five equations appear among the open cases in
\cite[Table~2.3 and Open Question~2.29]{Grechuk} and
\cite[Table~1]{GrechukOriginal}. The companion list records that their
solution sets are finite unions of polynomial families
\cite[Sections~8.6 and~8.9]{GrechukUpdate}. We give compact formulas and
prove that they attain every solution. The first two examples use
Theorem~\ref{main:shared}, and the other three use Theorem~\ref{main:values}.
Throughout this section, $n$ is an unrestricted integer parameter.

\subsection{Two polynomial parametrizations}
The formulas use the following polynomial parametrizations.
\begin{enumerate}[label=(\alph*),leftmargin=2em]
\item A matrix $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$
with $ad-bc=1$ may be taken from Vaserstein's published parametrization
\cite[Theorem~1]{Vas}. Fixed congruences on its entries are handled by
Lemma~\ref{lem:filter}.
\item Write
\[
 (a_1,\ldots,a_p\,;\,b_1,\ldots,b_q\,;\,R,S)
       =\mathcal B_{p,q}(\theta),\qquad
 \Bigl(\prod_i a_i\Bigr)R+\Bigl(\prod_j b_j\Bigr)S=1,
\]
for the polynomial map, together with its polynomial B\'ezout coefficients, constructed
in Proposition~\ref{thm:multipartite}. Its coordinate tuples cover
exactly those with coprime block products. All entries, including $R,S$,
are polynomial outputs on unrestricted integer parameters $\theta$.
\end{enumerate}
Composing the formulas below with these parametrizations and applying
Lemma~\ref{lem:filter} gives integer polynomials on unrestricted
parameters. Any displayed denominator is the fixed integer $1$ or $2$,
and the displayed congruences are resolved by finite residue subdivision.
The proofs establish completeness, including zero coordinates.
Using these two parametrizations keeps the displayed formulas short.
The total parameter counts are larger: with the unoptimized
$150$-parameter map of Lemma~\ref{lem:uniform}, each insertion adds
$151$ parameters. Filling the first block freely and then the second
uses at most $p+q+151pq$ parameters for $\mathcal B_{p,q}$.
Thus \eqref{eq:example-five} and \eqref{eq:example-split} use at most
$306$ and $609$ parameters per residue branch, respectively; the
other three formulas use at most $47$. These upper bounds come directly
from the construction.

\subsection{The equation $x^2y+zt+1=0$}
For $ad-bc=1$, all solutions are
\begin{equation}\label{eq:example-linear}
 \boxed{(x,y,z,t)=
 \bigl(a,\,-d^2+bn,\,b,\,c(2+bc)-a^2n\bigr).}
\end{equation}
Indeed,
$a^2d^2-bc(2+bc)=(ad-bc)(ad+bc)-2bc=1$.
Conversely, every solution has $\gcd(x,z)=1$, so choose a
determinant-one completion with $(a,b)=(x,z)$. The displayed particular
solution and the kernel vector $(b,-a^2)$ give every pair $(y,t)$.
This remains valid when $a=0$ or $b=0$.

\subsection{The equation $x_1x_2x_3+x_4x_5+1=0$}
Take $(a,b\,;\,c\,;\,R,S)=\mathcal B_{2,1}(\theta)$. Then all solutions are
\begin{equation}\label{eq:example-five}
 \boxed{(x_1,x_2,x_3,x_4,x_5)
       =(a,b,-R+cn,c,-S-abn).}
\end{equation}
The identity follows from $abR+cS=1$. Conversely, the equation forces
$\gcd(x_1x_2,x_4)=1$, so the map $\mathcal B_{2,1}$ attains
$(a,b,c)=(x_1,x_2,x_4)$. For any supplied B\'ezout coefficients $R,S$, the kernel
vector $(c,-ab)$ gives every pair $(x_3,x_5)$, including all zero cases.

\subsection{The equation $x^2y+z^2-1=0$}
For $ad-bc=1$, put
\[
 R=d^2(1-2bc),\qquad S=c^2(1+2ad),\qquad a^2R+b^2S=1.
\]
For either $\eta=1$ or $\eta=2$, all the following triples are solutions,
and together the two formulas cover every solution:
\begin{equation}\label{eq:example-minus}
 \boxed{\begin{aligned}
 x&=\eta ab,\\
 y&=\bigl((2/\eta)R+b^2n\bigr)
       \bigl((2/\eta)S-a^2n\bigr),\\
 z&=a^2R-b^2S+\eta a^2b^2n.
 \end{aligned}}
\end{equation}
Here $2/\eta$ is the fixed integer $2$ or $1$. The factors of $1-z^2$ are
\[
 z+1=a^2(2R+\eta b^2n),\qquad
 1-z=b^2(2S-\eta a^2n),
\]
which prove the equation.

For completeness, if $x$ is odd, allocate its prime powers between
$z+1$ and $z-1$, whose gcd divides $2$. This gives $x=ab$,
$\gcd(a,b)=1$, $a^2\mid z+1$, and $b^2\mid z-1$.
If $z=\pm1$, allocate all prime powers to the vanishing factor.
Choose $c,d$ completing $(a,b)$; the Chinese remainder theorem gives
$z=a^2R-b^2S+a^2b^2n$, the $\eta=1$ formula.
If $x\ne0$ is even, then $z$ is odd and $(z+1)/2,(z-1)/2$ are
coprime. Allocating the prime powers of $x/2$ gives
$x=2ab$, $a^2\mid(z+1)/2$, and $b^2\mid(z-1)/2$.
Put $z_0=a^2R-b^2S$. Since $z_0+1=2a^2R$ and
$1-z_0=2b^2S$, both $2a^2$ and $2b^2$ divide $z-z_0$.
Their least common multiple is $2a^2b^2$, giving the $\eta=2$ formula.
In both cases the equation determines $y$ as displayed. Finally,
for $x=0$, the $\eta=2$ formula with $(a,b)=(0,1)$ or $(1,0)$
gives $z=-1$ or $1$, respectively, and lets $y$ range over all integers.

\subsection{The equation $x^2y+z^2+1=0$}
Take $ad-bc=1$ with the fixed parity conditions $c\equiv d\equiv1\pmod2$,
and set
\[
 A=a^2+b^2,\qquad U=ac+bd,\qquad V=c^2+d^2,\qquad
 H=UV/2+An.
\]
Together, the choices $\varepsilon=1$ and $\varepsilon=-1$ cover all solutions:
\begin{equation}\label{eq:example-plus}
 \boxed{(x,y,z)=\bigl(\varepsilon A,\,-V^2-2Un-H^2,\,U+AH\bigr).}
\end{equation}
The denominator $2$ is fixed and $V$ is even. Since
$AV-U^2=(ad-bc)^2=1$, direct expansion gives
\[
 (U+AH)^2+1=A^2(V^2+2Un+H^2).
\]

For completeness, every solution has $x\ne0$ and $A=|x|$ odd.
Unique factorization in $\Z[i]$ supplies a Gaussian divisor
$a+bi$ of $z-i$ with norm $A$ and $\gcd(a,b)=1$:
each odd prime $p\mid A$ splits, and its two conjugate Gaussian
prime factors cannot both divide $z-i$, whose imaginary part is $-1$.
For the unique factor $\pi$ dividing $z-i$,
$v_\pi(z-i)=v_p(z^2+1)\geq2v_p(A)$. Taking the factor
$\pi^{v_p(A)}$ for each $p\mid A$ gives the required divisor. Write
\[
 z-i=(a+bi)(c_0-d_0i).
\]
Then $ad_0-bc_0=1$ and $ac_0+bd_0=z$.
The integers $a,b$ have opposite parity. Replacing
$(c_0,d_0)$ by $(c_0+a\tau,d_0+b\tau)$ for a suitable integer $\tau$
makes both entries odd and preserves the determinant, while making
$U\equiv z\pmod A$. At $n=0$, the displayed formula gives a root
$Z_0$ of $Z_0^2+1\equiv0\pmod{A^2}$ with $Z_0\equiv z\pmod A$.
Since $\gcd(2z,A)=1$, this root lifts uniquely from modulo $A$ to
modulo $A^2$: from $A^2\mid(Z_0-z)(Z_0+z)$ one obtains
$A^2\mid Z_0-z$. Varying $n$ changes $z$ by $A^2n$, so every target
$z$ is attained. The sign $\varepsilon$ gives the prescribed $x$,
and the original equation then fixes $y$.

\subsection{The equation $xyz+t^2-1=0$}
Take $(a,c\,;\,b,d\,;\,R,S)=\mathcal B_{2,2}(\theta)$ and put
\[
 P=ac,\qquad Q=bd,\qquad U=2R+Qn,\qquad V=2S-Pn.
\]
For each $(h,k)\in\{(1,1),(2,1),(1,2)\}$, retain the fixed congruence
$UV\equiv0\pmod{hk}$ and set
\begin{equation}\label{eq:example-split}
 \boxed{(x,y,z,t)=
       \bigl(hab,\,kcd,\,UV/(hk),\,PR-QS+PQn\bigr).}
\end{equation}
Add the six linear families obtained by setting $t=\pm1$ and one of
$x,y,z$ equal to zero, leaving the other two free. Together these families cover all solutions.
Indeed, $PR+QS=1$ gives $t+1=PU$ and $1-t=QV$, so
$1-t^2=PQ\,UV=xyz$. The denominator $hk$ is the fixed integer $1$ or $2$.

For completeness when $t\ne\pm1$, the three coordinates $x,y,z$ are nonzero.
All odd prime powers of $xy$ can be allocated wholly to $t+1$ or $t-1$.
At $2$, at most one of $t+1,t-1$ has valuation greater than $1$.
If $xy$ is even, remove one factor $2$ from an even coordinate of
$x,y$, choosing $(h,k)=(2,1)$ or $(1,2)$; otherwise choose $(1,1)$.
The remaining $2$-valuation of $xy/(hk)$ fits in one root factor.
Thus one can factor $x/h=ab$, $y/k=cd$ with
$\gcd(ac,bd)=1$, $ac\mid t+1$, and $bd\mid t-1$.
The map $\mathcal B_{2,2}$ attains these coordinates. For its supplied B\'ezout coefficients $R,S$,
the Chinese remainder theorem gives $t=PR-QS+PQn$.
The target equation then gives $UV=hkz$, including the required
fixed congruence. The six linear families cover $t=\pm1$.


\paragraph{Use of AI.}
OpenAI's ChatGPT and Codex assisted in developing the arguments,
checking identities, and drafting and revising this manuscript.
A review generated by Anthropic's Claude Opus~5.5 informed the revision.

\clearpage
\begin{thebibliography}{10}
\small
\setlength{\itemsep}{3pt}
\bibitem{Vas}
L.~N.~Vaserstein, \emph{Polynomial parametrization for the solutions of
Diophantine equations and arithmetic groups}, Ann. of Math. (2)
\textbf{171} (2010), no.~2, 979--1009.
\url{https://doi.org/10.4007/annals.2010.171.979}.

\bibitem{Frisch}
S.~Frisch, \emph{Remarks on polynomial parametrization of sets of integer
points}, Comm. Algebra \textbf{36} (2008), no.~3, 1110--1114.
\url{https://doi.org/10.1080/00927870701776938}.

\bibitem{FrischLettl}
S.~Frisch and G.~Lettl, \emph{Polynomial parametrization of the solutions
of Diophantine equations of genus~0}, Funct. Approx. Comment. Math.
\textbf{39} (2008), no.~2, 205--209.
\url{https://doi.org/10.7169/facm/1229696571}.

\bibitem{BGT}
W.~Bruns, J.~Gubeladze and N.~V.~Trung,
\emph{Problems and algorithms for affine semigroups}, Semigroup Forum
\textbf{64} (2002), 180--212.
\url{https://doi.org/10.1007/s002330010099}.

\bibitem{CLO}
D.~Cox, J.~Little and D.~O'Shea, \emph{Using Algebraic Geometry},
Graduate Texts in Mathematics~185, Springer, 1998.
\url{https://doi.org/10.1007/978-1-4757-6911-1}.

\bibitem{Dedekind}
A.~V.~Sutherland, \emph{Properties of Dedekind domains},
MIT~18.785, Lecture~3, Fall~2017, Corollary~3.20.
\url{https://math.mit.edu/classes/18.785/2017fa/LectureNotes3.pdf}.

\bibitem{Sutherland}
A.~V.~Sutherland, \emph{The Minkowski bound and finiteness results},
MIT~18.785, Lecture~14, Fall~2017.
\url{https://math.mit.edu/classes/18.785/2017fa/LectureNotes14.pdf}.

\bibitem{Grechuk}
B.~Grechuk, \emph{Polynomial Diophantine Equations: A Systematic Approach},
Springer, 2024.
\url{https://doi.org/10.1007/978-3-031-62949-5}.

\bibitem{GrechukOriginal}
B.~Grechuk, \emph{A systematic approach to Diophantine equations:
open problems}, arXiv:2404.08518v1, 12~April~2024.
\url{https://arxiv.org/abs/2404.08518v1}.

\bibitem{GrechukUpdate}
B.~Grechuk, \emph{A systematic approach to Diophantine equations:
open problems}, arXiv:2404.08518v10, 1~October~2026.
\url{https://arxiv.org/abs/2404.08518v10}.
\end{thebibliography}
\end{document}