Timeline for Does there exist a rational polynomial $P(x)\in{\mathbb Q}[x]{}$ such that $P(\zeta(s))=\zeta(P(s))$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2020 at 12:42 | vote | accept | zeraoulia rafik | ||
| Mar 26, 2020 at 1:15 | answer | added | Fedor Petrov | timeline score: 15 | |
| S Mar 26, 2020 at 0:30 | history | suggested | RobPratt | CC BY-SA 4.0 |
Corrected grammar in title
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| Mar 26, 2020 at 0:29 | answer | added | Alexandre Eremenko | timeline score: 6 | |
| Mar 25, 2020 at 23:48 | review | Suggested edits | |||
| S Mar 26, 2020 at 0:30 | |||||
| Mar 25, 2020 at 23:42 | comment | added | YCor | The question doesn't match with the title. | |
| Mar 25, 2020 at 23:20 | history | edited | GH from MO |
edited tags
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| Mar 25, 2020 at 23:19 | answer | added | GH from MO | timeline score: 9 | |
| Mar 25, 2020 at 23:08 | comment | added | Sylvain JULIEN | Indeed, but the OP doesn't forbid degree $1$ polynomials. | |
| Mar 25, 2020 at 23:06 | comment | added | GH from MO | @SylvainJULIEN: Polynomials of degree at least $2$ are not injective on $\mathbb{C}$. | |
| Mar 25, 2020 at 23:03 | comment | added | Sylvain JULIEN | I have been working on a related problem, and my conclusion is that no injective $P$ fulfilling your requirements exists. | |
| Mar 25, 2020 at 23:00 | answer | added | Wojowu | timeline score: 9 | |
| Mar 25, 2020 at 22:52 | history | asked | zeraoulia rafik | CC BY-SA 4.0 |