In reference to this question, is anyone could deduce that if $x^2+2=y^3$ and $x,y \in \mathbb{N}$, then $x=5$ and $y=3$. I already prove that the only natural number $x$ for which $x+\sqrt{-2}$ is a cube of $A$ is $x=5$, so I think it's related. I think we have to use the coprimality of $x + \sqrt{-2}$ and $x + \sqrt{-2}$ of the other question, but it is unclear.
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For starters, if $m$ and $n$ are relatively prime (ideal/number) and $mn$ is a $k$th power, then what do you infer about $m$ and $n$? (Hint: $m$ and $n$ will have to be individually $k$th powers; why?) – knsam Apr 07 '16 at 04:53
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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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With $(x+\sqrt{-2})$, $(x-\sqrt{-2})$ coprime, $x^2+2=(x+\sqrt{-2})(x-\sqrt{-2})=y^3$ implies $x+\sqrt{-2}$ and $x-\sqrt{-2}$ are both cubes. Having already shown that $x+\sqrt{-2}$ is only a cube for $x=5$, we are left with $y^3=(5+\sqrt{-2})(5-\sqrt{-2})=27$.
Quinn Greicius
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