Timeline for Find all rational solutions of $x^2(x+1)(x^2+1)(x-1)=2(y+1)(y-1)$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2022 at 11:29 | vote | accept | math110 | ||
| Dec 1, 2016 at 15:32 | history | edited | GH from MO |
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| Dec 1, 2016 at 15:13 | answer | added | Michael Stoll | timeline score: 17 | |
| Dec 1, 2016 at 14:28 | comment | added | Jeremy Rouse | The Jacobian is isogenous to the product of two elliptic curves ($y^2 = x^3 - 4x + 16$ and $y^2 = x^3 +x^2 - x + 31$), and this shows that the rank of the Jacobian over $\mathbb{Q}$ is $3$. This will make applying Chabauty's method difficult. | |
| Dec 1, 2016 at 11:16 | comment | added | Daniel Loughran | Sorry I made the typo. The genus of the curve is $2$! | |
| Dec 1, 2016 at 10:56 | comment | added | Yaakov Baruch | Another quick search for small (|num|, |dem| <= 320) rationals adds $(\pm 4/7, \pm 317/343)$. | |
| Dec 1, 2016 at 10:41 | comment | added | coudy | A quick computer search gives $(-3,-19), (3,-19), (-1,-1), (0,-1), (1,-1), (-1,1), (0,1), (1,1), (-3,19), (3,19)$ for integer solutions between -100000 and 100000. | |
| Dec 1, 2016 at 10:19 | comment | added | Ilya Bogdanov | Rewriting in the form $(x^2-1)x^2(x^2+1)=8(y/2-1/2)(y/2+1/2)$, we find two more solutions $x^2=9=y/2-1/2$, i.e. $(\pm3,19)$. | |
| Dec 1, 2016 at 10:18 | comment | added | Daniel Loughran | This is a hyperelliptic curve of genus $3$. Normally either one of Michael Stoll or Noam Elkies comes along and solves such problems... | |
| Dec 1, 2016 at 9:44 | comment | added | Lior Bary-Soroker | By the theorem of Faltings there are only finitely many (since the genus is bigger than 1) | |
| Dec 1, 2016 at 9:34 | history | edited | HeinrichD | CC BY-SA 3.0 |
added 1 character in body; edited title
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| Dec 1, 2016 at 9:28 | history | asked | math110 | CC BY-SA 3.0 |