Re: lcm (p-1,p^n-1)
- From: Phil Carmody <thefatphil_demunged@xxxxxxxxxxx>
- Date: 13 Dec 2005 23:21:17 +0200
rusin@xxxxxxxxxxxxxxxxxxxxx (Dave Rusin) writes:
> In article <407jdcF18hg8nU1@xxxxxxxxxxxxxx>,
> =?ISO-8859-1?Q?Jos=E9_Carlos_Santos?= <jcsantos@xxxxxxxx> wrote:
> >On 12-12-2005 18:01, Dave Rusin wrote:
> >
> >>>I tried to answer the following question, but I didn't came far;
> >>>lcm(p-1,p^n-1)=(p-1)(p^n-1) where p is an (odd) prime. Anyone?
> >>
> >> Is it possible you meant the n-1 to be in the exponent?
> >
> >I thought about that possibility, but:
>
> Hey, I just finished grading a tall stack of student papers.
> I get really good at trying to invent ways of interpreting
> what is written so that the written becomes true!
>
> (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...)
Bugger, I'm a no-one, then.
Phil
--
What is it: is man only a blunder of God, or God only a blunder of man?
-- Friedrich Nietzsche (1844-1900), The Twilight of the Gods
.
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