Timeline for Mod n, are all higher powers also lower powers?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2023 at 13:43 | vote | accept | Charles | ||
| Jan 6, 2023 at 8:30 | answer | added | Aphelli | timeline score: 1 | |
| Dec 27, 2022 at 19:01 | history | edited | Charles | CC BY-SA 4.0 |
added 48 characters in body
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| Dec 21, 2022 at 3:08 | review | Close votes | |||
| Dec 29, 2022 at 3:02 | |||||
| Dec 20, 2022 at 14:52 | comment | added | Charles | @ChristopheLeuridan I'm only looking at powers mod $n$ in this question. For integer powers there's no $n$ for the $H(n)$, and the analogue with fixed $H$ fails for integer powers because, e.g., $2^p$ for $p>H$ is a $p-th$ power but not any lower power. | |
| Dec 20, 2022 at 14:45 | comment | added | Christophe Leuridan | I do not understand the question: every $e$-th power is an $e$-th power modulo every integer. Do you ask for some converse? | |
| Dec 20, 2022 at 13:58 | history | asked | Charles | CC BY-SA 4.0 |