Timeline for Rational zeros of a polynomial
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2013 at 22:17 | comment | added | Felipe Voloch | Stoll has code on his webpage for doing this kind of calculation: faculty.jacobs-university.de/mstoll/magma | |
| Jul 3, 2013 at 18:11 | comment | added | Noam D. Elkies | Also two pairs of rational points at $x = \pm 1$; and no other $x = m/n$ with $|m|,|n| \leq 10^6$ according to Stoll's ratpoints ("ratpoints '0 -1 0 2 2 1' 1000000" takes less than a second). The curve has good reduction away from $2$, and an automorphism $(x,y) \mapsto (-1/x, x/y^3)$ whose square is the hyperelliptic involution $(x,y) \leftrightarrow (x,-y)$. But neither of these observations seems to make the problem much easier. | |
| Jul 3, 2013 at 17:30 | history | edited | Remke Kloosterman | CC BY-SA 3.0 |
added 9 characters in body
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| Jul 3, 2013 at 13:23 | comment | added | Abhinav Kumar | it has two rational Weierstrass points. Also, the substitution x <- x -1 transforms it to the simpler looking curve $y^2 = x^5 + 2x^4 + 2x^2 - x = x(x^2 + 1)(x^2 + 2x - 1)$. | |
| Jul 3, 2013 at 12:35 | history | answered | Remke Kloosterman | CC BY-SA 3.0 |