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From: bill@milford.UUCP (bill)
Newsgroups: net.math
Subject: Re: Can you prove/disprove this?
Message-ID: <112@milford.UUCP>
Date: Mon, 2-Dec-85 10:39:38 EST
Article-I.D.: milford.112
Posted: Mon Dec  2 10:39:38 1985
Date-Received: Thu, 5-Dec-85 05:38:59 EST
References: <34080@lanl.ARPA> <526@klipper.UUCP>
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Organization: Telecomp,Inc. , Milford Ct.
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> In article <34080@lanl.ARPA> dxm@lanl.ARPA writes:
> I believe, it is much harder to proof that all even perfect numbers
> are of this form, which result is due to Euler if I'm not mistaking.
> 

Not >that< much harder: let s(n) = the sum of the divisors of n, so
s(n) = 2n if and only if n is 'perfect'.

We are given that 2n = s(n) and n = k*2^a;

so k*2^(a+1) = s(n) = ((2^(a+1)) - 1)*s(k) or 


s(k) =           (k*2^(a+1))
                 -----------
		 (2^(a+1)-1)

from (k*(2^(a+1)) = (k*(2^(a+1)) - k + k gives:

s(k) = k +     k
          -----------
         (2^(a+1) - 1)  


and both parts of the right side divide k (*)

but s(k) is the sum of the divisors of k (including k); so there are only
two divisors and

1 = (k/(2^(a+1) -1)) or 
			k = 2^(a+1) - 1 is prime.


To push on to bigger and better things beyond elementary algebra: 

a)The proof breaks for n odd at the line marked (*) because then the two
parts of the right side represent the same number (k), and the question
of odd perfect numbers is still up in the air.

b)Anyone want to investigate values n for which the equation s(x) - x = n
has no solutions? n=5 apparently is the only odd 'untouchable' number.

c)Generalize perfect numbers from s(n) = 2*n to s(n) = i*n; is there any
similar formulae for these i-ply perfect numbers? Is there an i for which
it can be proven there are an infinite number of i-ply perfect numbers?
Is there an upper bound to i?