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From: sonnens@mprvaxa
Newsgroups: net.math
Subject: Re: Mersenne primes and perfect numbers
Message-ID: <335@mprvaxa.UUCP>
Date: Wed, 19-Oct-83 18:25:46 EDT
Article-I.D.: mprvaxa.335
Posted: Wed Oct 19 18:25:46 1983
Date-Received: Sat, 22-Oct-83 03:24:24 EDT
Organization: Microtel Pacific Research, Burnaby BC
Lines: 44

Roger Noe gave the definition of a Mersenne prime (a prime of the form
2^n - 1), and correctly said: 

  "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.)"

halle1@houxz disputes this, saying:

  "Not all Mersenne primes produce perfect numbers.  (This should be obvious
  just from counting the known number of each.)  However, you have to go
  a ways before you come to the counter example, so he should be forgiven."

Obvious?!!  In fact, it would be extremely interesting if halle1@houxz could
produce "the counter example", since there is indeed a simple proof of Roger's
statement that has been universally accepted since the time of Euclid.

In modern notation, it uses the sigma "sum-of-divisors" function and the fact
that sigma is multiplicative.  A number m is `perfect' iff sigma(m) = 2m
(since sigma counts the number itself).  Here's the proof:

sigma([2^(n-1)]*[2^n - 1]) = sigma(2^[n-1])*sigma(2^n - 1)

= (1 + 2 + 2^2 + 2^3 + ... + 2^[n-1]) * 2^n       \*  since 2^n - 1 is prime

= (2^n - 1) * 2^n  =  2 * [2^(n-1)] * (2^n - 1).     Q.E.D.


Roger's statement leaves a few loose ends though.  One is
"whether or not n MUST be prime for 2^n-1 to be prime."  In fact it must,
since:

2^ab - 1 = (2^a - 1) * (2^(a[b-1]) + 2^(a[b-2]) + ... + 1).


Another thing is that Euler (I believe) proved the more difficult converse on
the form of even perfect numbers, namely that every even perfect number is of
the above form.  This fact was noted by halle@houxz, who also stated that it
is still unknown if all perfect numbers are even.  Results on this problem
usually end up in the form:  if it exists, an odd perfect number must be
greater than some value (where the value is some enormous number).

Dan Sonnenschein,
Microtel Pacific Research,
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