Timeline for If $p_1$ and $p_2$ are prime numbers, then either $p_1$ divides $\sum_{i=1}^{p_1-1} i^{p_1p_2-1}$ or $p_2$ divides $\sum_{i=1}^{p_2-1} i^{p_1p_2-1}$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2023 at 12:10 | vote | accept | Raj Pratap Singh | ||
| Nov 1, 2023 at 12:10 | vote | accept | Raj Pratap Singh | ||
| Nov 1, 2023 at 12:10 | |||||
| Nov 1, 2023 at 9:34 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
added 260 characters in body
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| Nov 1, 2023 at 0:31 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
edited body
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| Oct 31, 2023 at 20:16 | comment | added | GH from MO | The counterexamples to the original conjecture are precisely the Carmichael numbers (and they have at least 3 prime factors). | |
| Oct 31, 2023 at 19:15 | comment | added | GH from MO | @SalvoTringali Yes, I figured this out while preparing my dinner. See my comment before yours. | |
| Oct 31, 2023 at 19:15 | comment | added | Salvo Tringali | @GHfromMO For a reference, see e.g. Problem 8.2.12 in the 7th edition of Burton's Elementary Number Theory (it's a standard exercise on primitive roots). | |
| Oct 31, 2023 at 19:14 | comment | added | GH from MO | I see, if $p-1$ does not divide $k$, then picking a primitive root $g\bmod p$, we have $g^k\not\equiv 1\pmod{p}$, while the sum multiplied by $g^k$ is congruent to itself mod $p$, whence the sum is in fact divisible by $p$. | |
| Oct 31, 2023 at 19:09 | comment | added | GH from MO | Can you give a reference for the first sentence? | |
| Oct 31, 2023 at 19:02 | history | answered | Fedor Petrov | CC BY-SA 4.0 |