Timeline for Can repunits be perfect cubes?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 11, 2013 at 0:21 | review | Late answers | |||
| Sep 11, 2013 at 0:32 | |||||
| S Sep 11, 2013 at 0:21 | review | First posts | |||
| Sep 11, 2013 at 3:21 | |||||
| Sep 10, 2013 at 14:29 | comment | added | David E Speyer | More generally, for any modulus $M$ relatively prime to $3$, if $n \equiv 1 \bmod \phi(M)$, then $(10^n-1)/9 \equiv (10-1)/9 \equiv 1 \bmod M$, so there is always an arithmetic progression where $M$ cannot be used to prove $(10^n-1)/9$ is prime. | |
| Sep 10, 2013 at 14:26 | comment | added | David E Speyer | If $n=18m+1$ then $(10^n-1)/9$ is $1 \bmod 7$, which is a cube. | |
| Sep 10, 2013 at 13:23 | history | answered | Luara | CC BY-SA 3.0 |