In this article in AoPS the author gives a parametric solution to
$$ \begin{equation} aX^2+bY^2=cZ^2+q \tag 1 \end{equation} $$
in terms of the solutions to the equation
$$ax^2+by^2=cz^2 \tag 2$$ as
$$ \begin{cases} \begin{align} X=\frac{x}{2}(ck^2-as^2-bp^2+q)+s \\ Y=\frac{y}{2}(ck^2-as^2-bp^2+q)+p \\ Z=\frac{z}{2}(ck^2-as^2-bp^2+q)+k \end{align} \end{cases} \tag 3 $$
where $s,p,k$ are solutions of the linear diophantine equation
$$axs+byp-czk=1. \tag 4$$
Question: Can we generate all solutions of $(1)$ from all solutions of Eqn. $(2)$ or is just a single particular solution of $(2)$ sufficient?
As an example, I was trying to solve $X^2 + 3Y^2 = Z^2 + 144832$. The equation $x^2 + 3y^2 = z^2$ has particular solutions $(1,0,1), (1,1,2), \cdots$.
Are the solutions $(X,Y,Z)$ generated from $(1,0,1)$ equivalent to the solutions generated from $(1,1,2)$?
Here is output as I indicAted, using $u,v$ positive only. One may then negate $x$ or $z$ or both...
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
middle of the program
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$