Re: lcm (p-1,p^n-1)
- From: rusin@xxxxxxxxxxxxxxxxxxxxx (Dave Rusin)
- Date: 13 Dec 2005 21:53:16 GMT
In article <dnn46b$msg$1@xxxxxxxxxxxxxxxxxxx>,
Keith A. Lewis <klewis@xxxxxxxxxxxxxxx> wrote:
>rusin@xxxxxxxxxxxxxxxxxxxxx (Dave Rusin) writes in article <dnmqts$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.
Right -- when a,b are interpreted the way we all read it. I meant that
if we thought the person was asking about the case a = p-1, b = p^(n-1),
then at least the proposed value of the lcm would be correct, but
who would imagine there would be a smaller common multiple of these a & b ?
You need all the factors of a and all n-1 factors of the prime p,
so what's left to show (assuming the Fundamental Theorem of Arithmetic,
of course) ?
dave
.
- References:
- lcm (p-1,p^n-1)
- From: maarte
- Re: lcm (p-1,p^n-1)
- From: José Carlos Santos
- Re: lcm (p-1,p^n-1)
- From: Dave Rusin
- Re: lcm (p-1,p^n-1)
- From: Keith A. Lewis
- lcm (p-1,p^n-1)
- Prev by Date: Re: what "REALLY" is derivative?
- Next by Date: Re: analytic prolongation
- Previous by thread: Re: lcm (p-1,p^n-1)
- Next by thread: Re: lcm (p-1,p^n-1)
- Index(es):
Relevant Pages
|
