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From: rjnoe@ihlts.UUCP
Newsgroups: net.math
Subject: Re: Mersenne primes and perfect numbers
Message-ID: <230@ihlts.UUCP>
Date: Tue, 18-Oct-83 10:32:22 EDT
Article-I.D.: ihlts.230
Posted: Tue Oct 18 10:32:22 1983
Date-Received: Wed, 19-Oct-83 22:38:37 EDT
References: <221@ihlts.UUCP> <463@nsc.uucp>, <712@ihuxr.UUCP>
Organization: AT&T Bell Labs, Naperville, Il
Lines: 22

Briefly, a Mersenne prime is a prime number of the form (2^n - 1) for
some integer n.  They are named after some monk who wrote about them
a long time ago.  (Do I look like a history book?)  I don't think the
definition of Mersenne primes requires n to be prime, yet I believe
all known examples follow that rule.  I haven't begun to investigate
whether or not n MUST be prime for 2^n-1 to be prime.  I don't know if
it's been proven one way or another, either.  (Actually, I should let
nsc!chongo respond to this question--he's the expert.)  Clearly not
all prime numbers are Mersenne primes.  Now, a perfect number is a
positive integer equal to the sum of its divisors, excluding itself.
The first perfect number is 6=1+2+3.  The next is 28=1+2+4+7+14.  From
there we go to 496, then up to 8128 for the next perfect number.  As you
can see, these are all generated from Mersenne primes:  6=3*2, 28=7*4,
496=31*16, and 8128=127*64.  Given any Mersenne prime we can always
generate a perfect number of the form [2^(n-1)]*(2^n - 1).  (The proof
is quite simple.)  The question remains whether or not all perfect
numbers can be generated this way.  I don't know because I haven't
followed number theory very closely at all for a few years.  If there is
a proof either way (counterexample is simplest for the one way) I would
love to be pointed to it.
-- 
		Roger Noe		...ihnp4!ihlts!rjnoe