Timeline for On Choudhry's $x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+y_4^k$ for $k=7$?
Current License: CC BY-SA 4.0
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| S Feb 23, 2023 at 8:07 | history | bounty ended | CommunityBot | ||
| S Feb 23, 2023 at 8:07 | history | notice removed | CommunityBot | ||
| Feb 15, 2023 at 6:46 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
added 16 characters in body
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| S Feb 15, 2023 at 6:17 | history | bounty started | Tito Piezas III | ||
| S Feb 15, 2023 at 6:17 | history | notice added | Tito Piezas III | Draw attention | |
| Feb 15, 2023 at 6:14 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Major revision of post
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 4, 2015 at 21:46 | comment | added | Tito Piezas III | @Wolfgang: I updated the post with more details. See Questions. | |
| Jan 4, 2015 at 21:33 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
More details, notation.
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| Jan 4, 2015 at 20:24 | comment | added | Tito Piezas III | Oh, those $u_i$ are different. I keep gravitating to the same variables. My bad. :) P.S There are several (7.3.5) found by Wroblewski using Choudhry's cubic, but the others don't have the "accidental" constraint that $c=2f$. | |
| Jan 4, 2015 at 20:21 | comment | added | Wolfgang | Yes, I figured that. The problem was just that those $u_i$ are different from the $u_1,...,u_4$ in your question :) Interesting that there is just one of type (7,3,5). | |
| Jan 4, 2015 at 19:58 | comment | added | Tito Piezas III | @Wolfgang: The thing with odd powers is it is better to move them all to one side. Thus I prefer $\sum\limits^8 u_i^k = 0$ versus $\sum\limits^4 x_i^k = \sum\limits^4 y_i^k$. The latter can be mis-leading. For example, one of the seven in the table actually is a $\sum\limits^3 x_i^k = \sum\limits^5 y_i^k$ in positive terms. | |
| Jan 4, 2015 at 19:55 | comment | added | Wolfgang | You mean $x_i$ & $y_i$ (and up to sign), not $u_i$ ? | |
| Jan 4, 2015 at 19:34 | comment | added | Tito Piezas III | @Wolfgang: Yes, you are correct. I did not phrase it correctly. There are four constraints: that $k=1,3,7$ and 4th, $\sum\limits^4 x_i = \sum\limits^4 y_i = 0$. There in fact is a solution to $k=1,3,7$, namely $$u_i = 184, 443, 556, 698, -230, -353, -625, -673$$ found by Nuutti Kuosa in 1999 (before Choudhry's paper) that does not have the 4th constraint, namely there is no partition such that $\sum\limits^4 u_i = 0$. (I had checked it before.) | |
| Jan 4, 2015 at 18:28 | comment | added | Wolfgang | So if I get it right, does that mean that all solutions of Choudhry always have certain terms with equal sums, as e.g. $ 1741+ 3476= 1937+ 3280$ and $2435 + 3004 = 2111 + 3328$ in the first example? Which means he has added a heavy constraint. So he has not exactly "reduced" the original equation (2) [if true for k=1, 3, 7] to (3), but could we rather say: he was lucky that his additional constraint made a computer search more efficient, which allowed him to find a bunch of solutions? | |
| Jan 2, 2015 at 21:24 | comment | added | Tito Piezas III | @JesperPetersen: These solutions have the special property that $\sum_i^4 x_i = \sum_j^4 y_j = 0$. Thus, some of terms are necessarily negative. For aesthetics, Choudhry transposed negative terms to the other side of the equation, so labelling is not rigid. For example, $X_1+X_2+X_3=324+5439+893=2\cdot3328$ where one assumes $y_4 = 3328$. But $-3328$ is really one of the $x_i$ transposed to the other side. | |
| Jan 2, 2015 at 20:04 | comment | added | Jesper Petersen | This does not answer your questions, but upon reading the paper, I came across one minute detail. In the first solution mentioned, where $X_1 = 324, X_2 = 5439, X_3 = 893$, what is the "suitable transposition" involved? The $X_i$'s mentioned does not imply $x_1 = 1741$ without that transposition. | |
| Jan 2, 2015 at 3:59 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Better notation.
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| Jan 2, 2015 at 3:03 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Greek alphabet.
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| Jan 2, 2015 at 2:45 | history | asked | Tito Piezas III | CC BY-SA 3.0 |