Timeline for Is there a formula I can use to count the number of k-potent elements over gaussian ring? [closed]
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2022 at 3:42 | history | closed |
Alex M. M.G. David E Speyer Max Alekseyev paul garrett |
Not suitable for this site | |
| Feb 19, 2022 at 23:07 | answer | added | Antoine Labelle | timeline score: 1 | |
| Feb 19, 2022 at 14:25 | comment | added | David E Speyer | I've voted to move this to math stackexchange, since it looks like you are missing a bunch of basic concepts. That said, the relevant approach here is (1) solve the problem for $n$ a power of a Gaussian prime and (2) use the Chinese Remainder Theorem. | |
| Feb 19, 2022 at 13:58 | comment | added | Max Alekseyev | For $k=2$, see oeis.org/A332476 | |
| Feb 19, 2022 at 12:03 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
MathJax: \pmod
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| Feb 18, 2022 at 23:49 | history | edited | user477306 | CC BY-SA 4.0 |
edited title
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| Feb 18, 2022 at 10:41 | comment | added | user477306 | Yes, Sir! I was looking for a formula to determine the numbers or the elements | |
| Feb 18, 2022 at 10:30 | comment | added | Alex M. | Are you looking for a formula giving the number of $k$-potent elements, or for an algorithm allowing you to count them? Are you interested in the computational complexity of such an algorithm, or not? It is not clear what exactly you are looking for. | |
| Feb 18, 2022 at 10:10 | review | Close votes | |||
| Feb 20, 2022 at 3:51 | |||||
| Feb 18, 2022 at 9:54 | comment | added | user477306 | Like a technique in finding the number of elements? | |
| Feb 18, 2022 at 9:51 | comment | added | Alex M. | More efficient than what? | |
| Feb 18, 2022 at 9:08 | history | edited | user477306 | CC BY-SA 4.0 |
added 145 characters in body
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| S Feb 18, 2022 at 8:40 | review | First questions | |||
| Feb 18, 2022 at 9:51 | |||||
| S Feb 18, 2022 at 8:40 | history | asked | user477306 | CC BY-SA 4.0 |