Timeline for On Euler's elliptic curve for $A^4+B^4 = C^4+D^4$?
Current License: CC BY-SA 4.0
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| Feb 19 at 16:24 | vote | accept | Tito Piezas III | ||
| Jan 28 at 6:24 | comment | added | Tito Piezas III | @DeyiChen You may like this new MSE post. It is about elliptic curves and $a^4+b^4+c^4=d^4$. | |
| Oct 11, 2023 at 20:03 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 11, 2023 at 19:53 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 9, 2023 at 4:07 | comment | added | Tito Piezas III | @DavidRoberts Answer has now been trimmed more. I looked at his profile, and maybe he is just not familiar with the norms of MO. But I do appreciate the great work and observations he's done. | |
| Oct 9, 2023 at 4:04 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Improved formatting, and trimmed to reduce byte size.
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| Oct 9, 2023 at 3:32 | comment | added | David Roberts♦ | @TitoPiezasIII well, up to you. (I appreciate the cleaning up). I just wanted to point out to Deyi that continuing to make incremental edits is going to keep bumping this to the top of the front page, and that is not seen as playing fair by MO standards. So if you two could communicate by email or whatever to continue this discussion/line of research that might be helpful. Or even in a chat room here on MO. | |
| Oct 9, 2023 at 2:24 | comment | added | Tito Piezas III | @DavidRoberts I also noticed how MathJax struggled. In fact, I already trimmed away some of Devi's longest formulas before. Do you want me to trim some more? (P.S. I also did it to some of my own formulas in the question.) | |
| Oct 8, 2023 at 6:52 | comment | added | David Roberts♦ | @Deyi the software has flagged this for excessive edits. It might be better to assemble further edits into a document hosted offside, and link to it here. The MathJax really struggles with such big formulas. | |
| Oct 8, 2023 at 6:03 | comment | added | Tito Piezas III | @DeyiChen I also modified the $b_1$ and table in my post and credited it to you. It was useful to consider both families at the same time, as Ghoshal and you noticed a pattern for the two, and you found an additional relation between the degrees of the denominators of the two. Seems a lot of mysteries for the simple equation $x^4+y^4=z^4+w^4$! | |
| Oct 8, 2023 at 5:54 | comment | added | Deyi Chen | In fact, I calculated the sum of the coefficients for the second family first, and found that there was only one difference from the first family's without modifying $b_1$, so I modified $b_1$. | |
| Oct 8, 2023 at 5:47 | comment | added | Tito Piezas III | @DeyiChen Thanks for adding the info. So the sums of the coefficients of the first and second family are the SAME. That is certainly interesting! | |
| Oct 8, 2023 at 5:01 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 8, 2023 at 4:54 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 8, 2023 at 4:52 | comment | added | Tito Piezas III | @DeyiChen I have trimmed the post as its file size was getting too large and the page was loading too slow. Also, you don't have the sums of the coefficients of the second family. Kindly check. | |
| Oct 8, 2023 at 4:48 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Improved formatting
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| Oct 8, 2023 at 4:41 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
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| Oct 8, 2023 at 4:28 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 8, 2023 at 2:59 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 8, 2023 at 2:34 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 15:39 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 15:17 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 15:09 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 14:40 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 13:56 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 13:26 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 13:19 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 12:50 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 12:41 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 12:24 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 12:18 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 10:18 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Minor clarifications.
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| Oct 7, 2023 at 9:24 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 8:54 | comment | added | Deyi Chen | Transform the point $(b_2,y) \in E_1$ to $P\in E_2$. | |
| Oct 7, 2023 at 8:47 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 8:33 | comment | added | Tomita | @Deyi Chen, How did you find $P=\left(3 n^{2}+1, -2 n^{6}-\frac{1}{2} n^{4}-5 n^{2}-\frac{1}{2}\right)$? | |
| Oct 7, 2023 at 7:55 | comment | added | Tito Piezas III | The same elliptic curve, $$(a^3 - b) (b^3 - a) = y^2$$ also has a second family of solutions when $a=n^3,$ namely, $$b_1 =n\,\frac{(1 - 2n^2 + n^4 + n^6)}{(1 + n^2 - 2n^4 + n^6)}$$ We may have found the first three points. Can you verify if we didn't skip one? See this post. | |
| Oct 7, 2023 at 7:41 | comment | added | Tito Piezas III | Ah, thanks for the explanation. And I was just going to suggest to change $b_n \to b_m$ since you already used the variable $n$ in the elliptic curve $E_1$. | |
| Oct 7, 2023 at 7:38 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 7:32 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 7:13 | comment | added | Tito Piezas III | +1 Thanks. I've verified that for $b_{10}$, the the sum of its coefficients is $2^{100}$ while for $b_{11}$ it is $2^{120}$, consistent with the results for previous $b_m$. If you state it can be done for ALL $m$, then that answers part of my question. How did you do it though? Did you use $2P, 3P, 4P, 5P$, etc, so a one-to-one correspondence to the $b_m$? | |
| Oct 7, 2023 at 6:58 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 6:39 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Oct 7, 2023 at 5:50 | history | answered | Deyi Chen | CC BY-SA 4.0 |