Timeline for Minimal $n$ such that $(a-1)^m | a^n - 1$ for a given $a,m > 1$
Current License: CC BY-SA 3.0
10 events
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Jan 25, 2018 at 15:12 | vote | accept | Bryan Bush | ||
Jan 25, 2018 at 15:04 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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Jan 25, 2018 at 15:02 | comment | added | Max Alekseyev | @FedorPetrov: Yes, I forgot to enforce $n$ be even. Thanks! | |
Jan 25, 2018 at 14:45 | comment | added | Fedor Petrov | Your final expression still may be non integral, it should be $(a-1)^{m-1}2^{\max(2-m,-k)}$. | |
Jan 25, 2018 at 14:35 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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Jan 25, 2018 at 14:26 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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Jan 25, 2018 at 14:25 | comment | added | Max Alekseyev | @FedorPetrov: Indeed, this was my oversight. $2n$ was considered to enable the application of LTE. | |
Jan 25, 2018 at 14:20 | comment | added | Fedor Petrov | If $a$ is odd and $(a-1)^2$ divides $a^n-1=(a-1)(1+a+\dots+a^{n-1})$, then $n$ is of course even and moreover divisible by $a-1$. And why do you care on $a^{2n}-1$ at all? | |
Jan 25, 2018 at 13:26 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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Jan 25, 2018 at 13:19 | history | answered | Max Alekseyev | CC BY-SA 3.0 |