Find all positive integral solutions to the equation
$ x^3+2y^3=4z^3$
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1A start: $x$ is even, say $2s$. We get $4s^3+y^3=2z^3$. Go two more rounds. – André Nicolas Mar 17 '16 at 04:14
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Yeah it's like infinite descent – Mar 17 '16 at 04:20
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Yes, that's right, a descent argument. – André Nicolas Mar 17 '16 at 04:26
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Duplicate: http://math.stackexchange.com/questions/1663985/show-that-there-exist-no-a-b-c-in-mathbb-z-such-that-a3-2b3-4c3 – TheRandomGuy Mar 17 '16 at 05:51
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$x$ is even, so if $x'=x/2$ then $$x'^3=\frac{2z^3-y^3}{4},$$ Now $y$ is even, so $$x'^3=\frac{z^3}{2}-2y'^3$$ Now $z$ is even and $$x'^3+2y'^3=4z'^3$$ By the same manner we can conclude that $(x,y,z)=(0,0,0)$
Sergey Zaitsev
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