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How to prove that the equation x^2+2=y^3 admits a unique solution in positive integer?

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closed as too localized by Felipe Voloch, GH from MO, Cam McLeman, Fernando Muro, Todd Trimble Jun 8 '13 at 19:18

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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
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...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

1 Answer 1

up vote 11 down vote accepted

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.

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