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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?

Ethan Bolker
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user459947
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  • There are a couple of solutions with $x=-1$ though then $y$ would be positive – Henry Sep 26 '18 at 00:09
  • What do you mean by "negative integral"? If you mean $x$ and $y$ are both negative integers, then $(x,y)=(0,0)$ would not work since $0$ is not negative. – Eric Wofsey Sep 26 '18 at 00:27

2 Answers2

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For fixed $x$

$$ y = \frac{-6x \pm \sqrt{36x^2+32x}}{2}\\ = -3 x \pm \sqrt{9x^2+8x} $$

If $x$ is negative, then to make $y$ negative as well we would need to choose the - branch and have

$$ 9x^2+8x > 9x^2 $$

but that doesn't happen in this case. So no integral negative solutions.

AHusain
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$y^2+6xy-8x=0$ 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: $$ \begin{array}{r|rrr} y & 2 & 0 & 4 \\ x & -1 & 0 & -1 \end{array} $$

lhf
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