[proofplan] The factorization hypothesis turns every integer solution into a factor pair of the fixed element $C$ in the unique factorization domain $R$. Unique factorization implies that the divisors of $C$ have only finitely many associate classes, so the possible values of $A(x)$ and $B(x)$ are finite up to multiplication by units. The assumed gcd restriction discards all factor pairs whose common divisor type is not allowed. What remains is exactly the coordinate compatibility problem: determine which of the admissible factor pairs actually lies in the image of the map $x\mapsto (A(x),B(x))$. [/proofplan]

[step:Translate each integer solution into a factor pair of $C$] Let $x=(x_1,\dots,x_n)\in S$. By the assumed equivalence defining $S$, \begin{align*} A(x)B(x)=C. \end{align*} Define \begin{align*} a&:=A(x),\\ b&:=B(x). \end{align*} Then $(a,b)\in R^2$ satisfies $ab=C$. Since $C\ne 0$ and $R$ is an integral domain, neither $a$ nor $b$ is zero. Thus $a$ and $b$ are divisors of $C$ in $R$. By the gcd hypothesis, every greatest common divisor of $A(x)$ and $B(x)$ is associate to an element of $\Gamma$. Since $a=A(x)$ and $b=B(x)$, every greatest common divisor of $a$ and $b$ is associate to an element of $\Gamma$. Therefore every solution $x\in S$ belongs to one of the sets \begin{align*} S_{a,b}=\{y\in \mathbb{Z}^n:A(y)=a,\ B(y)=b\} \end{align*} with $ab=C$ and with gcd type allowed by $\Gamma$. [/step]

[step:Show that only finitely many divisor pairs occur up to units] Because $R$ is a unique factorization domain and $C\ne 0$, choose a factorization \begin{align*} C=u p_1^{\alpha_1}\cdots p_m^{\alpha_m}, \end{align*} where $m$ is a nonnegative integer, $u\in R^\times$, the elements $p_1,\dots,p_m\in R$ are irreducible and pairwise non-associate, and each exponent $\alpha_i$ is a positive integer. Let $d\in R$ be a divisor of $C$. Then there exists $e\in R$ such that $de=C$. Choose factorizations of $d$ and $e$ into irreducibles. By uniqueness of factorization up to units and order, every irreducible factor occurring in either factorization must be associate to one of $p_1,\dots,p_m$, since their product is associate to $p_1^{\alpha_1}\cdots p_m^{\alpha_m}$. For each $i\in\{1,\dots,m\}$, let $\beta_i$ denote the exponent of the associate class of $p_i$ in $d$. Comparing exponents in the equality $de=C$ gives $0\le \beta_i\le \alpha_i$. Hence $d$ is associate to an element of the form \begin{align*} p_1^{\beta_1}\cdots p_m^{\beta_m}, \end{align*} where $0\le \beta_i\le \alpha_i$ for each $i\in\{1,\dots,m\}$. There are exactly \begin{align*} \prod_{i=1}^m(\alpha_i+1) \end{align*} possible exponent vectors $(\beta_1,\dots,\beta_m)$, so the divisors of $C$ have finitely many associate classes. Therefore the factor pairs $(a,b)$ with $ab=C$ have finitely many orbits under the unit action \begin{align*} u\cdot(a,b)=(ua,u^{-1}b),\qquad u\in R^\times. \end{align*} This action preserves the product because \begin{align*} (ua)(u^{-1}b)=ab=C. \end{align*} [/step]

[step:Filter the factor pairs by the prescribed gcd condition] Let $(a,b)\in R^2$ satisfy $ab=C$. Since $R$ is a unique factorization domain, a greatest common divisor of $a$ and $b$ exists and is unique up to multiplication by a unit. Thus the condition “a greatest common divisor of $a$ and $b$ is associate to an element of $\Gamma$” is independent of the chosen gcd. The theorem only retains those pairs $(a,b)$ whose gcd associate class is represented in $\Gamma$. By hypothesis, every solution-derived pair $(A(x),B(x))$ passes this filter. Conversely, if a pair fails this filter, then no $x\in S$ can satisfy $A(x)=a$ and $B(x)=b$, because such an $x$ would contradict the assumed gcd restriction on solutions. [/step]

[step:Recover precisely the integer solutions from the compatible factor pairs] Define $\mathcal{P}$ to be the set of all pairs $(a,b)\in R^2$ such that \begin{align*} ab=C \end{align*} and a greatest common divisor of $a$ and $b$ is associate to an element of $\Gamma$. We prove \begin{align*} S=\bigcup_{(a,b)\in\mathcal{P}} S_{a,b}. \end{align*} First let $x\in S$. The first step shows that $(A(x),B(x))\in\mathcal{P}$, and by definition $x\in S_{A(x),B(x)}$. Hence \begin{align*} S\subset \bigcup_{(a,b)\in\mathcal{P}} S_{a,b}. \end{align*} Conversely, let $x\in S_{a,b}$ for some $(a,b)\in\mathcal{P}$. Then $A(x)=a$, $B(x)=b$, and $ab=C$. Therefore \begin{align*} A(x)B(x)=ab=C. \end{align*} By the assumed equivalence between the original integer equation and the factorization equation, this implies $x\in S$. Hence \begin{align*} \bigcup_{(a,b)\in\mathcal{P}} S_{a,b}\subset S. \end{align*} The two inclusions give the desired equality. Thus every solution is obtained by choosing an admissible divisor pair of $C$, modulo the unit ambiguity, and then imposing the coordinate compatibility conditions $A(x)=a$ and $B(x)=b$. [/step]