Re: lcm (p-1,p^n-1)

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...)

dave
.

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