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May 29, 2014 at 19:18 comment added Peter Mueller @Jeremy Rouse: Thanks for remarking that $y=1$ isn't covered by the original answer. I fixed it now.
May 29, 2014 at 19:16 history edited Peter Mueller CC BY-SA 3.0
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May 29, 2014 at 16:06 comment added Jeremy Rouse It takes a little bit of work to handle the case that $y = 1$.
May 29, 2014 at 15:55 comment added Joe Silverman Jeremy and I were answering a more general question, but you're right, the specific question is easier (although Nagell-Lutz is only marginally easier than the height argument). More generally, you're argument should work for $y^2=x^3+D$ with $D$ square-free. But it's not clear (to me) if it can be made to work for 6'th power free $D$.
May 29, 2014 at 15:51 history answered Peter Mueller CC BY-SA 3.0