Timeline for Show $p\nmid k!+1$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2022 at 8:27 | comment | added | Geoff Robinson | Exact wording of footnote f in DA: "Manifestly $M$ cannot be of the form $3z$ because otherwise $4n$ would be divisible by $3$. With regard to the ambiguity as to whether $b-c$ must $=N$ or $=-N$,it is unnecessary to consider the question here, and by the nature of the case it cannot be determined because it depends on the selection of the primitive root $g$. For some primitive roots the difference $b-c$ will be positive, for others negative". | |
| Sep 27, 2022 at 13:56 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
added 14 characters in body
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| Sep 27, 2022 at 13:50 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
Cited Lemmermeyer
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| Sep 27, 2022 at 13:49 | comment | added | Ilya Bogdanov | @math110 I've added a PS about Gauss's theorem; sorry, I have no other information, maybe the communoty would help... | |
| Sep 27, 2022 at 13:46 | comment | added | Ilya Bogdanov | @GeoffRobinson Thanks for the clarification --- I do not have access to the book. I've finished that case as well, and the modulo 6 congruence magically works! | |
| Sep 27, 2022 at 13:44 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
Cited Lemmermeyer
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| Sep 27, 2022 at 10:55 | comment | added | math110 | Nice, can you link the Gauss's theorem proof,I searched for a while and found no proof of this theorem | |
| Sep 27, 2022 at 10:54 | comment | added | Geoff Robinson | Actually, footnote f to Article 358 of Disquisitiones Arithmeticae seems to say that the sign can't be determined, because it can be either according to the choice of a certain primitive root, so I am not sure you are justified in assuming $L$ positive. | |
| Sep 27, 2022 at 9:56 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |