Equation: $(x^2-9y^2)^2=33y+16$

Question

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.

Answer 1

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 $y$.

Answer 2

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