Re: lcm (p-1,p^n-1)
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:
1) Then the statement would be true for *every* prime. Why to add the
hypothesis that _p_ is odd?
2) It would be more natural to state it as lcm(p - 1,p^n) = (p - 1)p^n.
Would would anyone use n - 1 instead of _n_?
Best regards,
Jose Carlos Santos
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