TITLE: Equation: $(x^2-9y^2)^2=33y+16$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 $y$.