How to prove that the equation x^2+2=y^3 admits a unique solution in positive integer?
closed as too localized by Felipe Voloch, GH from MO, Cam McLeman, Fernando Muro, Todd Trimble♦ Jun 8 '13 at 19:18This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. 


You don't need the general theory of Mordell's equation to handle this particular case. Since $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain whose only units are $ \pm 1,$ it follows that there are integers $a$ and $b$ such that $x + \sqrt{2} = (a+b\sqrt{2})^{3}.$ Then $1 = 3a^{2}b 2b^{3}.$ Hence $b = \pm 1.$ The case $b = 1$ leads to $a = \pm 1$ and $y=3$. The case $b = 1$ leads to a contradiction. 

