Timeline for Raising positive integer to $c\in\mathbb{R}-\mathbb{N}$ rarely gives an integer!
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 19 at 21:48 | history | edited | LSpice | CC BY-SA 4.0 |
Removed spurious comma; deleted "thanks"
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| Dec 18 at 21:18 | answer | added | Joshua Stucky | timeline score: 7 | |
| Dec 6 at 18:19 | comment | added | Joshua Stucky | @Lucia Could you provide a reference or sketch of the kind of argument you're thinking of? | |
| Jun 4, 2019 at 3:05 | review | Close votes | |||
| Jun 4, 2019 at 20:25 | |||||
| May 27, 2019 at 7:56 | vote | accept | user141127 | ||
| May 27, 2019 at 1:20 | comment | added | Lucia | The six exponential theorem would show that if $n_1$, $n_2$, $n_3$ are three integers with $n_i^c \in {\Bbb N}$ then $\log n_i$ are linearly dependent over ${\Bbb Q}$. (I assume that $c$ is irrational.) You should be able to show from this that at most $\ll (\log N)^2$ integers up to $N$ can satisfy $n^c \in {\Bbb N}$. | |
| May 26, 2019 at 17:57 | answer | added | Fedor Petrov | timeline score: 9 | |
| May 26, 2019 at 17:47 | comment | added | Will Jagy | crosspost from math.stackexchange.com/questions/3240407/… | |
| May 26, 2019 at 17:22 | comment | added | Fedor Petrov | @Wojowu you may use other difference type operators | |
| May 26, 2019 at 17:20 | review | Close votes | |||
| May 27, 2019 at 21:52 | |||||
| May 26, 2019 at 17:16 | comment | added | Wojowu | @FedorPetrov Ah, shame on me, I have actually not read the problem carefully enough! While the problem might still be of interest to the OP, I don't think the proof I link generalizes - it crucially depends on taking differences, and if we only know some fraction of terms are integers, we are not guaranteed any first difference is an integer! However, much more sophisticated tools, like ones discussed here, should give a proof. Of course, OP asks for an elementary proof, which I cannot really provide... | |
| May 26, 2019 at 17:11 | comment | added | Fedor Petrov | @Wojowu Putnam problem was for 100 percents, not 1 percent. Although the proof may be modified. | |
| May 26, 2019 at 16:47 | comment | added | Wojowu | This was the problem A6 in 1971 Putnam. There is a solution here | |
| May 26, 2019 at 16:35 | review | First posts | |||
| May 26, 2019 at 16:56 | |||||
| May 26, 2019 at 16:33 | history | asked | user141127 | CC BY-SA 4.0 |