Re: Nearing breakthrough...need magic :-)
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 04 Nov 2005 02:19:35 GMT
In article <1131066657.345014.30590@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Erik" <Erik99@xxxxxxxxx> wrote:
> Does anyone know if there's a mathmatical (as opposed to computational)
> way to find an integer value for m and e, such that for a given odd
> integer n,
>
> mn = (2^e)-1
>
> OR
>
> mn = (2^e)+1
>
> in other words, find a multiple of n that is equal to a power of two,
> minus (or plus) one?
>
> (i'm guessing no).
If you know the factors of n, it's easy; just calculate r = phi(n),
where phi is the Euler phi-function, then 2^r - 1 is a multiple of n.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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