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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://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.

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  • $\begingroup$ You can still do $2y^3 = x^2-9=(x-3)(x+3)$, and the gcd of the factors $x-3$ and $x+3$ is a divisor of $6$. But now, instead of their product being exactly a cube, their product is two times a cube. So you can still conclude that $x-3=am^3$ and $x+3=bn^3$ with $a,b$ dividing $6^2$; but now, instead of $ab$ being a cube, $ab$ must be twice a cube. Otherwise the solution method is the same. $\endgroup$ – Greg Martin Sep 14 '16 at 18:28
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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.

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