Timeline for Can we find two positive integers $n$ and $m$ ($n,m>1$) such that $n^\pi = m$? [duplicate]
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2018 at 1:52 | vote | accept | Héctor | ||
| Oct 13, 2017 at 21:28 | comment | added | Gerry Myerson | If you're still here, Héctor, you might consider accepting one of the answers by clicking in the check mark next to it. | |
| S Oct 13, 2017 at 19:15 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Corrected some English typos.
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| Oct 13, 2017 at 17:35 | review | Suggested edits | |||
| S Oct 13, 2017 at 19:15 | |||||
| Jun 19, 2015 at 18:34 | history | closed |
Stefan Kohl♦ Joonas Ilmavirta András Bátkai Johannes Hahn Alex Degtyarev |
Duplicate of Is pi = log_a(b) for some integers a, b > 1? | |
| Jun 19, 2015 at 16:18 | review | Close votes | |||
| Jun 19, 2015 at 18:34 | |||||
| Jun 17, 2015 at 23:18 | answer | added | Matt Papanikolas | timeline score: 31 | |
| Jun 15, 2015 at 0:08 | answer | added | Felipe Voloch | timeline score: 34 | |
| Jun 15, 2015 at 0:02 | comment | added | Todd Trimble | @FelipeVoloch Please consider adding your comment as an answer. | |
| S Jun 14, 2015 at 23:49 | history | suggested | BigM |
Needed Retagging
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| Jun 14, 2015 at 23:46 | review | Suggested edits | |||
| S Jun 14, 2015 at 23:49 | |||||
| Jun 14, 2015 at 23:21 | comment | added | Héctor | @FelipeVoloch Thanks, very clear argument, i didn't see that. This satisfy me as an answer, but i'll leave this open in case someone has some great idea of how to show this without Schanuel. | |
| Jun 14, 2015 at 23:11 | comment | added | Felipe Voloch | Apply Schanuel to $2\pi i, \log n, \log m$. The last two numbers are linearly independent over $\mathbb{Q}$ because of your hypothesis and the fact that $\pi$ is irrational. Then all three numbers are linearly independent over $\mathbb{Q}$ since $2\pi i$ is not real. Finally, the exponentials of all three numbers are rational, so Schanuel implies that the three numbers are algebraically independent over $\mathbb{Q}$ which is stronger than what you want. | |
| Jun 14, 2015 at 23:06 | history | edited | Héctor | CC BY-SA 3.0 |
edited body
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| Jun 14, 2015 at 23:03 | comment | added | Héctor | @FelipeVoloch If is not too much to ask, could you outline the argument of why this should follow from Schanuel's conjeture? transcendence theory it's not my area. Thanks. | |
| Jun 14, 2015 at 22:57 | comment | added | Felipe Voloch | This follows from Schanuel's conjecture but it's probably hard to prove unconditionally. | |
| Jun 14, 2015 at 22:43 | history | asked | Héctor | CC BY-SA 3.0 |