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Oct 3, 2013 at 7:14 review First posts
Oct 3, 2013 at 7:16
Oct 3, 2013 at 7:06 comment added user40810 I should have thought of that. Thanks!
Oct 3, 2013 at 7:04 history edited user40810 CC BY-SA 3.0
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Oct 3, 2013 at 6:58 comment added Michael Zieve Substitute $z=yx$, then the rational points with $x\ne 0$ correspond to rational points of $z^2=xf(x)$.
Oct 3, 2013 at 6:55 history asked user40810 CC BY-SA 3.0