Timeline for Solutions to diophantine equation related to an interpolation problem on hypercubes
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2022 at 21:24 | comment | added | Sidharth Ghoshal | strange, i did some experimental lookup and a lot of numbers to the power 8 are 1 mod 255. I wonder why that is: [1, 2, 4, 8, 13, 16, 19, 26, 32, 38, 43, 47, 49, 52, 53, 59, 64, 67, 76, 77, 83, 86, 89, 94, 98, 101, 103, 104, 106, 118, 121, 127, 128, 134, 137, 149, 151, 152, 154, 157, 161, 166, 169, 172, 178, 179, 188, 191, 196, 202, 203, 206, 208, 212, 217, 223, 229, 236, 239, 242, 247, 251, 253, 254]. Edit: its exactly 64 values | |
| Dec 20, 2022 at 21:22 | comment | added | Sidharth Ghoshal | ah clever in general the powers of $2$ will be good candidates in general | |
| Dec 20, 2022 at 21:05 | comment | added | Gerry Myerson | For $n=3$, we want $k^3\equiv1\bmod7$, which holds precisely for $k\equiv1,2,4\bmod7$. | |
| Dec 20, 2022 at 21:04 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
deleted 19 characters in body
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| Dec 20, 2022 at 21:04 | comment | added | Sidharth Ghoshal | yes let me edit that. | |
| Dec 20, 2022 at 21:00 | comment | added | Gerry Myerson | What does it mean to be "a carmichael number of $2^n-1$"? Do you mean, when $n$ doesn't equal $\lambda(2^n-1)$? | |
| Dec 20, 2022 at 19:40 | vote | accept | Manfred Weis | ||
| Dec 20, 2022 at 19:39 | comment | added | Manfred Weis | Very nice and extensive answer; I would of course appreciate if you could provide additional code. | |
| Dec 20, 2022 at 19:14 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
code
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| Dec 20, 2022 at 18:54 | history | answered | Sidharth Ghoshal | CC BY-SA 4.0 |