<s>[INST] Solving $X^2-6Y^2=Z^3$ in positive integers I’m trying to solve the Diophantine equation
$$X^2-6Y^2=Z^3  \tag{$\star$}$$
in positive integers $x,y,z$.
Brute force calculations confirm the naïve intuition that there are many [read: surely infinite!] primitive solutions; numerical observation suggests the solutions have determinable [perhaps even “descent”- or “recurrence”-based?] patterns. For example, one “stream” of solutions is
$$(x,y,z) \in \{(5,2,1),\ (49,20,1),\ (485,198,1),\ (4801,1960,1),\dots\},$$
where $z=1$ and the $(x_n,y_n)$  satisfy the recurrence $t_n = 10t_{n-1}-t_{n-2}$. Another such stream exists for $z=19$, another for $z=25$, etc.
Evidently, ($\star$) is related to [but different from] the Pellian equation
$$X^2-6Y^2=Z^2,$$
so I’m wondering:

Question #1: Is a complete solution already known, either for the equation or for one of the variables (e.g. characterization of $z$)?


Question #2: If not, what are the most promising ways to attack the problem?

 [/INST] We can always construct a linear relation between x and y such as $x=ay+b; a, b \in \mathbb z$ to convert pell  equations  $x^2-Dy^2=1$ to a single unknown equation. We can find numerous families of solutions by this method. For Pell like equations we use rational solutions. I try to show this bellow by an example:
Let $x=y+a$ we have:
$x^2-6y^2=1\Rightarrow 5y^2-2ay-a^2+1=0$
$\Delta'=6a^2-5$
which have number of solutions:
$(a, \Delta')=(1, 1), (3, 7^2), (7, 17^2), \cdot \cdot\cdot$
which gives:
$(x, y)=(\frac 25, 1), (\frac 4 5, 1), (8, 5), (\frac {11}5, -\frac 45)\cdot\cdot\cdot $
For example take $(x, y)=(\frac {11}5, -\frac 45)$ we have:
$\big(\frac {11}5\big)^2-6\big(\frac 45\big)^2=1\Rightarrow\big(\frac {11^2}{25}\big)-6\big(\frac{4^2} {25}\big)=1$
We can rewrit this relation as:
$(11\times 25)^2-6(4\times 25)^2=25^3$
So: $(x, y, z)=(675, 100, 25) $
Similarly we can find more solutions where $z=25$ as a family of solutions.
Generally there can alway be a solution for equation $x^2-Dy^2=z^{2k+1}$ . for example  for $x^2-6x^2=z^5$ we can write:
$(11\times625)^2-6(4\times 625)^2=25^5$
For D=19 we have:
$x=\frac{35} 3$ , $y=\frac 83$ and we have:
$(35\times 9)^2-19(6\times 9)^2=9^3$
$(35\times 81)^2-19(8\times 81)^2=9^5$
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