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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 2, 2015 at 12:05 answer added castor timeline score: 1
Jul 22, 2015 at 6:13 comment added joro I edited my answer for the case $n=4$ giving map to the original curve.
Jul 21, 2015 at 20:44 answer added Jeremy Rouse timeline score: 5
Jul 21, 2015 at 20:28 comment added Jeremy Rouse For a hyperelliptic curve $y^{2} = f(x)$, where $f(x)$ has even degree, there are two points at infinity on the non-singular model. In addition, there are the points $(-5/2,3/2)$ and $(-5/2,-3/2)$. This explains the four points other than those with $y = 0$.
Jul 21, 2015 at 18:47 comment added Vesselin Dimitrov Well, one of them is just the point $[0:1:0]$ at infinity, which is the neutral element. So there are three more.
Jul 21, 2015 at 18:45 comment added Igor Rivin @VesselinDimitrov Yes, exactly. But what about those four other mysterious points.
Jul 21, 2015 at 18:44 comment added Vesselin Dimitrov I do not know if I can read the conclusion of no other integral solutions without actually computing all seven non-trivial points of the transformed equation, as well as the change of coordinates map. In any case, we know that four of these seven non-identity rational points were given as $y = 0$ and $x \in \{-1,\ldots,-4\}$ in the original coordinates.
Jul 21, 2015 at 18:24 comment added Igor Rivin @VesselinDimitrov Oops, forgot that $4$ was a square :), thanks! But still, can you parse what the sage result actually means?
Jul 21, 2015 at 18:19 comment added Vesselin Dimitrov Regarding the addendum: For $n = 4$, there are no non-trivial ($y \neq 0$) integer solutions, by the Erdos-Selfridge result that I quoted.
Jul 21, 2015 at 18:14 history edited Igor Rivin CC BY-SA 3.0
added sage results, added tag
Jul 21, 2015 at 7:04 answer added Vesselin Dimitrov timeline score: 7
Jul 21, 2015 at 6:33 answer added joro timeline score: 2
Jul 21, 2015 at 3:45 history asked Igor Rivin CC BY-SA 3.0