How can we solve in integers the following equation:
$$2y^3+9=x^2$$
? I found that the solutions are $(x,y)\in\{(-3,0),(3,0),(-5,2), (5,2), (-21,6), (21,6)\}$. I have tried to apply the coprime factors trick (http://alpha.math.uga.edu/~pete/4400MordellEquation.pdf) but this one works fine just in Mordell's diophantine equations: $y^3+k=x^2$.
Any ideas are very welcomed.
If you multiply the equation by $4$, you get $(2y)^3 + 36 = (2x)^2$, so you can substitute variables and work with $z^3 + 36 = w^2$, if you like, and then look up the solutions in a Mordell table. Divide the even ones by two and you'll have all your solutions. http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+ I did this and you have found all the solutions.