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From: sonnens@mprvaxa.UUCP
Newsgroups: net.math
Subject: A proof of Euler's theorem on perfect numbers
Message-ID: <347@mprvaxa.UUCP>
Date: Thu, 27-Oct-83 11:59:51 EST
Article-I.D.: mprvaxa.347
Posted: Thu Oct 27 11:59:51 1983
Date-Received: Mon, 31-Oct-83 09:58:43 EST
Organization: Microtel Pacific Research, Burnaby BC
Lines: 29

The following proof is by Dickson (1911), from the book
"First Course in Theory of Numbers" by Harry N. Wright.

Note.  An arithmetic function f is said to be multiplicative if, when a and b
are relatively prime, f(a*b) = f(a) * f(b).  It is fairly easy to prove that
the sum-of-divisors function sigma (called s here), is multiplicative.


Let m be an even perfect number, so that s(m) = 2m, and write m in the form

                  n-1
                 2   * q, where q is odd and n > 1.

                n-1          n-1             n                     n
Then s(m) =  s(2   * q) = s(2   ) * s(q) = (2  - 1) * s(q) = 2m = 2  * q.

                                                n                   n
Let d = s(q) - q, so that s(q) = q + d.  Then (2  - 1) * (q + d) = 2  * q
can be solved to give:  q = d(2^n - 1).

This shows d to be a proper divisor of q, so that if d > 1,

                s(q) >= q + d + 1 > q + d = s(q), -- a contradiction.

Therefore d = 1, and s(q) = q + d = q + 1, which is only true when q is a prime.
We conclude that q = d(2^n - 1) = 2^n - 1, so the perfect number is of the form:

                            n-1     n
                           2    * (2  - 1).