$20x^3 - 14y^2 = 2014$
The solution has to be a pair of integers (x, y). I've been trying to solve this one for the past few hours. I've brute forced pairs of values where $x$ and $y$ are in the range of $[1, 10000]$ or where $x$ is in the range of $[1, 1000000]$ with no success.
Can you help me solve it? Is there a general solution for equations that have the form $ax^3-by^2=10a + b$?