Re: lcm (p-1,p^n-1)
- From: klewis@xxxxxxxxxxxxxxx (Keith A. Lewis)
- Date: Tue, 13 Dec 2005 18:33:47 +0000 (UTC)
rusin@xxxxxxxxxxxxxxxxxxxxx (Dave Rusin) writes in article <dnmqts$qbf$1@xxxxxxxxxxxxxxxxx> dated 13 Dec 2005 15:55:40 GMT:
>(Mind you, it's quite a stretch to believe someone would not be
>able to show lcm(a,b) = a b under the present circumstances...)
Except in this case lcm(a,b) = b, because (p-1) is always a factor of
(p^n-1) unless there's something I missed.
Anyone? :^)
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
.
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