<s>[INST] What are all of the negative integral solutions of $y^2+6xy-8x=0$? I got the answer as $(0,0)$ by making $D\ge 0$ (quadratic in $y$).
However, how do I know this is the only possible answer?
 [/INST] $y^2+6xy-8x=0$ iff $y^2=2x(4-3y)$. Therefore, $\dfrac{y^2}{4-3y}=2x$ is an integer.
Computing the gcd of $y^2$ and $4-3y$ gives
$$
\frac{16}{4-3y} = 9 \frac{y^2}{4-3y} + (3y+4) = 18x + (3y+4)
$$
Therefore, $4-3y$ divides $16$ and so $4-3y \in \{ \pm 1, \pm 2, \pm 4, \pm 8,\pm 1 6 \}$. The only integer solutions $y$ for which $x$ is also an integer are:
$$
\begin{array}{r|rrr}
y & 2 & 0 & 4
\\
x & -1 & 0 & -1
\end{array}
$$
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