I want to know the solution of the equation $(x^2-9y^2)^2=33y+16$ in positive integers. I know it has solution $(\pm2;0)$ but I can't prove that it doesn't have other solutions.

Please help.


2 Answers 2


You can prove a limit on the number of solutions like so:

Rewrite $(x^2-9y^2)^2=33y+16$ as $x^2 = (3y)^2 \pm \sqrt {33y+16}$.

For sufficiently large $y$,

$$(3y-1)^2 \lt (3y)^2 - \sqrt {33y+16} \lt (3y)^2 \lt (3y)^2 + \sqrt {33y+16} \lt (3y+1)^2$$

hence $\sqrt {(3y)^2 \pm \sqrt {33y+16}}$ cannot be an integer for sufficiently large $y$.

  • 2
    $\begingroup$ May I add on this? "Sufficiently large" $y$ is when $$\begin{align*}(3y)^2+\sqrt{33y+16}<&(3y)^2+6y+1=(3y)^2+\sqrt{36y^2+12y+1}\\ 33y+16<&36y^2+12y+1\\ 36y^2-21y-15>&0\\ y>&\frac{21+\sqrt{21^2-4\times36\times(-15)}}{2\times36} = 1\end{align*}$$So the other answer had found out all the solutions. $\endgroup$
    – peterwhy
    Jan 23, 2014 at 12:08
  • 1
    $\begingroup$ Ah, that large? I guess knowing what we know we can start from: For $y > 2$,$$0 < 36y^2 - 45y - 15$$ $$33y+16 < 36y^2 - 12y+1$$ $$33y+16 < (6y-1)^2 < (6y+1)^2$$ $$\sqrt {33y+16} < 6y-1 < 6y+1$$ $$-(6y-1) < -\sqrt {33y+16} < 0 < \sqrt {33y+16} < 6y+1$$ etc. $\endgroup$
    – Neil W
    Jan 23, 2014 at 13:31

It does have other solutions. (4, 1) for example

  • $\begingroup$ the solution of the equation should be in positive integers $\endgroup$
    – nadia-liza
    Jan 23, 2014 at 10:53
  • $\begingroup$ I guess (4,1) was the only valid one in the list I had mentioned. My bad. $\endgroup$
    – Dhanesh
    Jan 23, 2014 at 11:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .