Timeline for Non-trivial solution to $\sum^{n}_{i=1}\sum^{n}_{j=1,j\ne i}(x_{i})^{(x_j)}=(\sum^{n}_{i=1}x_i)^{(\sum^{n}_{i=1}x_i)}$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2021 at 6:00 | history | bounty ended | CommunityBot | ||
| Jun 28, 2021 at 8:26 | comment | added | rgvalenciaalbornoz | Sorry, it was upper, my typo. Thank you for the answer, I agree completely. It was just because I remembered Ellison's numdam.org/article/STNB_1970-1971____A9_0.pdf and maybe something similar could be applied in this case, but I don't know really. As you said, I hope someone else finds the next clue too. | |
| Jun 27, 2021 at 18:07 | comment | added | Peter Taylor | @rgvalenciaalbornoz, I think we want an upper bound on that sum rather than a lower bound, and since all of the quantities involved are integers I'm not sure how to apply Baker's theorem. Of course, maybe someone else can see something I can't. | |
| Jun 27, 2021 at 9:39 | comment | added | rgvalenciaalbornoz | Because of the shape of the sum involving the $\lambda_{i}^{\lambda_{j}}$, it is possible to reformulate the problem as a linear form in logarithms and then using a quantitative version of Baker's theorem to obtain a lower bound for $n$? Perhaps, it could lead to a contradiction, showing no non-trivial solutions for large $n$, or give a finite list of possible $n$ to verify. | |
| Jun 26, 2021 at 23:31 | comment | added | Peter Taylor | In fact, choosing partitions into two parts to maximise the LHS, I find that the LHS readily exceeds the RHS for odd non-square $n > 61$, so that this approach in itself isn't going to rule out the case of odd non-square $n$. | |
| Jun 26, 2021 at 16:38 | history | answered | Peter Taylor | CC BY-SA 4.0 |