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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