Timeline for The smallest solution to $2^{2k}-1=\text{powerful}$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 3 at 19:37 | comment | added | Max Alekseyev | @joro: Ok, why are you asking then? Btw, the same argument applies to $2^{2k}-1$, where $k\equiv 1,5\pmod 6$, not necessarily prime. | |
| Sep 3 at 16:22 | comment | added | joro | It is proved in the question: $4^p-1=(2^p+1)(2^p-1)$ and $2^p+1$ is divisible by $3$ with exponent one. | |
| Sep 3 at 16:16 | comment | added | Max Alekseyev | @joro: Most likely, but I don't see how to prove that. | |
| Sep 3 at 11:40 | comment | added | joro | So if $p$ is prime, $4^p-1$ is never powerful, right? | |
| Aug 22 at 20:46 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Aug 22 at 20:41 | history | answered | Max Alekseyev | CC BY-SA 4.0 |