Timeline for Rational points on the elliptic curve $y^2 = x^{3} - t^{2}z^3$
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2020 at 14:48 | comment | added | Sam | @user44191. Thanks for catching my typo. You have a typo too. In your value's for (x,y,z) the variable 't ' need's to have the power two. There is a difference in your method & my method. I took the sum of two cubes [ (m^2-1)^3+1^3] . Hence we get it= m^2(m^4-3m^2+3). Thus we just need to multiply the LHS by (m^4-3m^2+3) to make the LHS a square. | |
| Aug 16, 2020 at 19:25 | comment | added | user44191 | I think you have mixed up $x$ and $y$ in the last line; I also suspect that the method can be generalized rather easily to a 3-parameter family: choose $t, x_0, z_0$; then let $(x, y, z, t) = (x_0 (x_0^3 - t z_0^3), (x_0^3 - t z_0^3)^2, z_0 (x_0^3 - t z_0^3), t)$. | |
| Aug 16, 2020 at 14:11 | review | First posts | |||
| Aug 16, 2020 at 19:25 | |||||
| Aug 16, 2020 at 14:07 | history | answered | Sam | CC BY-SA 4.0 |