Timeline for Transforming the Kondo quintic $5T2$ into the Lehmer quintic $5T1$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2023 at 8:15 | comment | added | Tito Piezas III | You may like this MSE question on elliptic curves for $a^4+b^4=c^4+d^4$. | |
| Jan 3, 2023 at 13:27 | comment | added | Jeremy Rouse | The "method" mentioned in an earlier comment doesn't apply to a question like this. I did play a bit with this and found a rational function expression (specifically $a = t^{2}$, $b = \frac{9}{7t^{2}}$, $y = 14t^{3} + \frac{54}{7t^{3}}$). | |
| Jan 2, 2023 at 19:25 | comment | added | Tito Piezas III | Does your method work for other two-parameter elliptic curves? I need to make the discriminant of a special cubic into a square, $$-27 + 196 a^3 + 126 a b + 49 a^2 b^2 + 28 b^3 = y^2$$ I can easily find integer points, but I'm having a devil of a time finding polynomial expressions for $a,b$. | |
| Dec 31, 2022 at 6:11 | comment | added | Tito Piezas III | I knew it would involve $a = -(n^2+7)$. But, based on the Lehmer quintic, didn't expect $b,c$ would have denominators. That book "Generic Polynomials" is a gold mine! I see the Kondo-Brumer quintic is there, so now I'm not sure who found it earlier. I'm glad I found the transformation on my own, though that book certainly will give me plenty of material to play with. | |
| Dec 31, 2022 at 6:02 | history | bounty ended | Tito Piezas III | ||
| Dec 31, 2022 at 6:02 | vote | accept | Tito Piezas III | ||
| Dec 31, 2022 at 0:12 | comment | added | Jeremy Rouse | Applying that method to your second family of cyclic quintics gives $a = -n^{2} - 7$, $b = (-8n^{4} - 145n^{2} - 625)/n^{2}$. | |
| Dec 31, 2022 at 0:11 | comment | added | Jeremy Rouse | Yes, standard methods in the theory of elliptic curves will yield infinitely many solutions with a $c^{2}$. It seems that only one of those has a $c^{4}$. By the way, the Kondo-Brumer quintic is a generic polynomial for the Galois group $5T2$ and any $5T2$ extension (or even a $5T1$ extension) over any field can be gotten from it. There's a description of how to do this on page 45 of the book "Generic Polynomials" by Jensen, Ledet and Yui. | |
| Dec 30, 2022 at 13:18 | comment | added | Tito Piezas III | By the way, using these initial rational points, the standard methods for elliptic curves will yield only a $c^2$, and not a $c^4$, correct? Because I know a second family of cyclic quintics here but I can't transform the Kondo-Brumer into that so far. | |
| Dec 30, 2022 at 12:58 | comment | added | Tito Piezas III | Thanks, the first question I was 99% sure. The second question, I actually found the parameterization rather serendipitously. I asked Mathematica to find $a,b$ such that the cubic would be a 2nd power. To my surprise, they were 4th powers. And yes, I did try the positive case of $a = \pm (7+10n+5n^2+n^3)$ first and it didn't work. I was about to give up, figured trying the negative case, and the $b$ factored out. | |
| Dec 30, 2022 at 2:11 | history | edited | Jeremy Rouse | CC BY-SA 4.0 |
Cleaned up reasoning for first answer.
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| Dec 30, 2022 at 2:06 | history | answered | Jeremy Rouse | CC BY-SA 4.0 |