# The equation x^2+2=y^3 admits a unique solution in positive integer [closed]

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 19:18

This 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.

The first step is to switch the roles of $x$ and $y$, since $y^2 = x^3 - 2$ is the usual way it is written nowadays. Then read about Mordell's equation, which can be found in many references on Diophantine equations or elliptic curves. Your question is not a research-level question, so it's more suitable for math.stackexchange if you need further assistance. – KConrad Jun 8 '13 at 13:23
...though $x^2 + 2 = y^3$ is the equivalent form that starts the usual proof via unique factorization in ${\bf Z}[\sqrt{-2}]$ (as in Geoff Robinson's answer), which is more down-to-earth than elliptic curves. But yes, this is too well-known and elementary for MO. – Noam D. Elkies Jun 8 '13 at 18:41

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.