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From: rjnoe@ihlts.UUCP (Roger Noe)
Newsgroups: net.math
Subject: Re: Mersenne primes and perfect numbers
Message-ID: <236@ihlts.UUCP>
Date: Wed, 19-Oct-83 14:04:23 EDT
Article-I.D.: ihlts.236
Posted: Wed Oct 19 14:04:23 1983
Date-Received: Thu, 20-Oct-83 08:42:53 EDT
References: <230@ihlts.UUCP>, <550@houxz.UUCP>
Organization: AT&T Bell Labs, Naperville, Il
Lines: 30

I must contend that I was correct in stating that any prime number of the
form (2^n-1) generates a perfect number of the form (2^n-1)*2^(n-1).  Note
I did NOT say that every perfect number is so generated.  I am aware that
there may or may not be odd perfect numbers.  The proof for my statement
follows:

Given a prime number p=2^n-1 for some integer n, p*2^(n-1) is a perfect number.

PROOF
    The divisors of p*2^(n-1) must each either contain p as a factor or not,
    in which case they would just be powers of two.  The divisors of the
    number in question which are not multiples of p are:
	1, 2, 4, . . ., 2^(n-1)		i.e.  2^m for 0<=m<=n-1
    and the divisors which are multiples of p are:
	p, 2p, 4p, . . ., p*2^(n-2)	i.e.  p*2^q for 0<=q<=n-2.
    Excluding p*2^(n-1) itself but including 1, this is a complete list of
    the divisors of p*2^(n-1) and there can be no duplications because p is
    a Mersenne prime.

    The sum of these divisors is:
	 n-1           n-2
	SIGMA 2^m  +  SIGMA p*2^q  =  (2^n - 1) + p*(2^(n-1) - 1)
	 m=0           q=0
                                   =  p*[1 + 2^(n-1) - 1]  =  p*2^(n-1)

    which of course was the original number.  Therefore p*2^(n-1) is perfect.

quod erat demonstrandum
-- 
		Roger Noe		...ihnp4!ihlts!rjnoe