Timeline for Magic hourglass of squares hyperelliptic equation
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2021 at 23:11 | comment | added | Thomas | The elliptic fibration has rank 2, so there are a potentially infinite number of choices I could have made. | |
| Dec 6, 2021 at 3:20 | comment | added | Simpson17866 | "This can be parametrized, and taking one of the generators, you end up with:" Are there other generators that you could've used instead to generate other equations than just "$49 m^{12} - ... + 1$" ? | |
| Nov 28, 2021 at 7:30 | comment | added | Thomas | Looking at the papers, it seems that Bremner has been where I am. However, they haven't resolved many of the hyperelliptic curves. They did find some curves of degree 8, which might be easier to analyze... | |
| Nov 27, 2021 at 23:00 | comment | added | Gerry Myerson | I'd be worried, Thomas, that you might be reinventing the wheel. As noted in other comments, Andrew Bremner and others have studied this problem before. You might get some ideas from their work, or you might find they've already been where you are. | |
| Nov 27, 2021 at 11:33 | comment | added | Thomas | @Gerry Myerson, I followed my own ideas and reasoning. I didn't get any of the ideas from a paper so as far as I know, this is a unique method, although others could have done it. If you want I can write out in an extension to my question a more fleshed out working. Potentially that would allow for more of the other possible curves to be checked as well. | |
| Nov 26, 2021 at 7:57 | comment | added | castor | Over $\mathbb{Q}(\sqrt{3})$ one has for example $$\begin{matrix}(7+\sqrt{3})^2 & (7-\sqrt{3})^2 & 14^2\\ & 10^2 &\\ 2^2 & (1+7\sqrt{3})^2 & (-1+7\sqrt{3})^2\end{matrix}$$ provided by Michaud-Rodgers link. | |
| Nov 26, 2021 at 7:49 | comment | added | castor | In Bremner's Squares of squares II article you may find solutions of the "hour-glass" configuration over $\mathbb{Q}(\sqrt{3}).$ | |
| Nov 26, 2021 at 7:46 | comment | added | castor | Bremner considered the problem of magic square of squares (see link and link). It is mentioned that Duncan Buell handled the "hour-glass" configuration, however I only found this preprint: link. | |
| Nov 26, 2021 at 5:08 | comment | added | individ | The formulation of the problem is not quite clear to me. Maybe it should be formulated in a formal way? If one approach to solving it does not give a result, maybe it's worth going back to the beginning and solving it in another way? Formulate it formally. | |
| Nov 26, 2021 at 2:53 | comment | added | Jeremy Rouse | The hyperelliptic curve $y^{2} = 49x^{12} + \cdots$ has genus $5$ and it has an automorphism group of order $4$. One quotient of this is the genus $3$ curve $X : y^{2} = (x^{2}-2)(x^{6} - 6x^{5} + 3x^{4} + 16x^{3} - 9x^{2} - 10x + 1)$. An upper bound for the rank of $J(X)$ is $2$. If someone could find two independent points of infinite order on $J(X)$, Chabauty's method could be used to provably find all the rational points on $X$ and hence on the original curve. | |
| Nov 26, 2021 at 2:21 | comment | added | Gerry Myerson | It's not clear to me from the presentation, where what's already known stops, and what's your new work, if any, begins. If the $y^2=49x^{12}+\cdots$ equation is already in the literature, and no nontrivial solutions are known, that suggests that so far the problem is above everyone's pay grade, not just yours. | |
| Nov 25, 2021 at 6:38 | comment | added | Thomas | Every m,n combination in mod 3, mod 5, and mod 8 works to make e^2 a square, so that is out. | |
| Nov 25, 2021 at 5:34 | comment | added | Thomas | The existence of a few trivial integer solutions seems to rule out the modular approach, especially for small bases such as 3, 5, or 8. | |
| Nov 25, 2021 at 5:09 | comment | added | Steven Stadnicki | First thought: have you tried working modulo small numbers? Mod 3, 8 and 5 all come to mind as being worth a quick try to prove impossibility from. (Though this will admittedly only give integer solutions, not rational.) | |
| Nov 25, 2021 at 4:49 | history | asked | Thomas | CC BY-SA 4.0 |