Prove that $\left \{ (x,y)\in\mathbb{Z}^2:x^2+2=y^3 \right \}\subseteq \left \{ (-5,3),(5,3) \right \}$.
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This is a classic tricky Diophantine equation. I forget how to solve it. My first inclination is to try to use unique factorization in $\mathbb Z[\sqrt {-2}]$ – Thomas Andrews Aug 25 '15 at 08:37
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This is actually an elliptic curve and I believe the standard methods will work here for this question although I'm not posting this as an answer as it may assume too much knowledge on the OP's part – Matt B Aug 25 '15 at 09:52
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It's due to Pierre de Fermat (early 1600's). – DanielWainfleet Aug 26 '15 at 00:12
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See also Solve $x^2+2=y^3$ using infinite descent? – Sil Jan 30 '20 at 13:22
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First note that $x$ must be odd.
Then show that in $R=\mathbb Z[\sqrt {-2}]$ that $x+\sqrt{-2}$ and $x-\sqrt{-2}$ are relatively prime. Since $R$ is a unique factorization domain, and all units are cubes, this means the $x+\sqrt{-2}$ is a cube in $R$ (Since their product is a cube.)
Then solve:
$$x+\sqrt{-2}=(a+b\sqrt{-2})^3.$$
Thomas Andrews
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So $x$ has to be odd. If $x$ was even if implies $y$ is even by congruences mod $2$. Then you can let $x=2x_1$ and $y=2y_1$. But then we can cancel a factor of two, and taking the equation mod two again gives a contradiction? – snulty Aug 25 '15 at 09:17
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