"Vincent J. Mooney Jr." wrote:
>
> 5 is an odd prime. 2^5 = 32 and minus one, is 31. 31 is not divisible by 3.
>
>
> At 07:50 PM 10/26/99 +0200, you wrote:
> >Hello all,
> >
> >a simple Number Theory question. Is always
> >
> >(2^p-1) / p ,odd prime p, divisible by 3 ?
> >
Well, I managed to mess up the equation so badly that noone could
possibly make any sense of it. It should have been: (2^(p-1) -1) /p, and
by now I figured out why these numbers are divisible by 3 for odd (and
not just for prime) p>3 (if you kick out the /p, its for p>2).
a^k == 1 (mod a^k -1)
a^n == a^(n-k) (mod a^k -1), n>=k
a^(nk) == 1 (mod a^k -1)
(or a^(nk)-1 == 0 (mod a^k-1) ,which shows that Mersenne primes must
have prime exponents)
Since 2^2-1=3, 2^(2k)-1 == 0 (mod 3) for all integer k and, in the above
formula,
2^(p-1)-1 == 0 (mod 3) for odd p.
Thats what happens when you ask a question before thinking it to the
end, sorry for the confusion.
Ciao,
Alex.
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