Solve $$2x^2 y - 3x^2 + 4y = 2009$$ for $$x$$ and $$y$$ in positive integers.

PLACEHOLDER ANSWER - the answer is that there is no solution. Rearranging, $$2y(x^2 + 2) - 3(x^2 + 2) = 2009 - 6$$, i.e. $$(2y - 3)(x^2 + 2) = 2003$$. Since 2003 is prime and $$x^2 + 2 \ge 3$$, we need $$x^2 + 2 = 2003$$, i.e. $$x^2 = 2001$$, which is not a perfect square. So there are no positive integer solutions.