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S Jul 3, 2021 at 6:00 history bounty ended CommunityBot
S Jul 3, 2021 at 6:00 history notice removed CommunityBot
Jul 2, 2021 at 23:42 history edited Alex-Github-Programmer CC BY-SA 4.0
Corrected LaTeX
Jul 2, 2021 at 23:35 history edited Alex-Github-Programmer CC BY-SA 4.0
Corrected theorem formulation
Jul 2, 2021 at 17:35 comment added mathlove $2+\frac{(n-2)(n-1)}2$ is equal to $\frac{n^2-3n+6}{2}$, not $\frac{n^2-3n-6}{2}$.
Jul 2, 2021 at 11:58 comment added mathlove A nice work, but I think you have a typo. What you got is $S^{S_n}\le\dfrac{n^2-3n\color{red}+6}2$ from which I got $x_1=x_2=\cdots =x_{\lfloor 2n/3\rfloor}=0$ which is slightly better than claim 3 in my answer.
Jul 2, 2021 at 7:12 answer added mathlove timeline score: 1
Jul 2, 2021 at 6:19 answer added Alex-Github-Programmer timeline score: 0
Jul 2, 2021 at 6:06 history edited Alex-Github-Programmer CC BY-SA 4.0
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Jul 2, 2021 at 1:54 comment added Alex-Github-Programmer @mathlove Could you post your comment as an answer? Also show your process.
Jun 26, 2021 at 16:38 answer added Peter Taylor timeline score: 4
Jun 26, 2021 at 14:19 history rollback Alex-Github-Programmer
Rollback to Revision 5
Jun 26, 2021 at 14:19 history rollback Alex-Github-Programmer
Rollback to Revision 6
Jun 26, 2021 at 14:11 comment added Alex-Github-Programmer @PeterTaylor I see.
Jun 26, 2021 at 12:42 comment added mathlove If I'm not mistaken, $x_1=x_2=\cdots =x_{\lfloor (2n-1)/3\rfloor}=0$.
Jun 26, 2021 at 11:13 history edited Alex-Github-Programmer CC BY-SA 4.0
edited body
Jun 26, 2021 at 10:42 comment added Peter Taylor (Also, by exhaustive checking, $S=2$ giving the known non-trivial example or $S > 95$).
Jun 26, 2021 at 10:12 comment added mathlove A slightly better claim can be obtained. One can prove that for every non-trivial solution to $A_n$, $S^{S_n}\le n(n-1)/2$. I've already got the following small results, but don't know how to do for larger $n$. (1) $x_1=0$. (2) If $n\ge 4$, then $x_2=0$. (3) If $n\ge 6$, then $x_3=0$. (4) $A_2$ has no non-trivial solution. (5) $A_3$ has only one non-trivial solution. (6) Each of $A_4,A_5,A_6,A_7$ has no non-trivial solution.
Jun 26, 2021 at 9:39 history edited Alex-Github-Programmer CC BY-SA 4.0
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Jun 26, 2021 at 8:45 history edited Alex-Github-Programmer CC BY-SA 4.0
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Jun 25, 2021 at 4:07 history edited Alex-Github-Programmer CC BY-SA 4.0
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S Jun 25, 2021 at 4:01 history bounty started Alex-Github-Programmer
S Jun 25, 2021 at 4:01 history notice added Alex-Github-Programmer Draw attention
Jun 20, 2021 at 11:57 comment added Max Lonysa Muller @Alex-Github-Programmer interesting family of equations. Do you have an application in mind for this problem?
Jun 20, 2021 at 9:34 history edited YCor CC BY-SA 4.0
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Jun 19, 2021 at 23:39 comment added Alex-Github-Programmer @BrendanMcKay And for odd square $t$ it is also close to the square $(2t^{t/2})^2$.
Jun 19, 2021 at 23:28 history edited Alex-Github-Programmer CC BY-SA 4.0
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Jun 19, 2021 at 14:19 comment added Brendan McKay $4t^t+(t-1)(3t+1)$ is a square for $t=2$. If it is ever a square for larger integer $t$, that will give another 0-1 solution. It doesn't happen up to $t=10000$. For even $t$, it is too close to the square $(2t^{t/2})^2$; I'm not sure about odd $t$.
Jun 19, 2021 at 13:49 comment added Gerald Edgar I agree with @BrendanMcKay ... as given I would inerpret this in real numbers. Well, positive real numbers so that the powers are also real. In integers, you could allow negative values.
Jun 19, 2021 at 13:28 comment added Brendan McKay You want integers or real numbers?
Jun 19, 2021 at 7:25 review First posts
Jun 19, 2021 at 9:39
Jun 19, 2021 at 7:19 history asked Alex-Github-Programmer CC BY-SA 4.0