Solve this: $y^2+2=x^3$ or
Prove that $y^2+2=x^3 => (x,y)=(3,\pm 5)$
I know that it could be obvious to some of you by trial and error but I need a methodical approach.
Thanks in advance!
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One observation is that $y$ is even if and only if $x$ is even, but if $y=2m,x=2n$ we have $4m^2+2 = 8n^3$, which happens if and only if $2m^2+1 = 4n^3$, which cannot be since LHS is odd while RHS is even. Hence, neither $x$ nor $y$ can be even... – gt6989b Dec 08 '14 at 22:53
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@gt6989b Yeah thats a nice observation thanks – L887 Dec 08 '14 at 22:55
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Another one is just as easy: since $y^2 \geq 0$, we must have $x>0$, and the problem is symmetric in $y$, i.e. if $(x,y)$ is a solution, then $(x,-y)$ is a solution as well. Hence, let's look for $y \geq 0$ and it suffices to show that $(3,5)$ is a unique solution. – gt6989b Dec 08 '14 at 22:58
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@gt6989b Nice one again but how can you say that this suffices to show that (3,5) is a unique solution? – L887 Dec 08 '14 at 23:00
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i am not saying this suffices to show that (3,5) is unique; i said, showing (3.5) is unique suffices to solve the problem, since we now can assume $y>0$... – gt6989b Dec 08 '14 at 23:01
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Duplicate of this. See also Fermat's sandwich theorem. – Lucian Dec 08 '14 at 23:32