- Bill,

You seem to have missed Jack's point -- he wasn't talking

about 22021 as being the odd-prime "candidate". It was the

number N defined as

> N = 3^2*7^2*11^2*13^2*22021

which your alleged "proof" might get fooled by. There are

two main sources of confusion. Before getting to the "hard"

one which Jack had in mind; would you care to explain how

your proof deals with number which are not square-free (e.g. 9,

25, 45, ...)?

Following your line of reasoning, when looking at number 9,

one could either use f1 = 1 and f2 = 9, or f1 = f2 = 3.

The problem is that in neither case, the LHS (1 + f1 + f2)

manages to express the sum of divisors of number 9 -- in

the first case, it equals 1+1+9 = 11, in the other, it's

1+3+3 = 7. So yes, the proof can correctly conclude that

LHS cannot be equal to RHS... but it since LHS wasn't equal

to the sum of divisors of the product f1*f2 in the first

place, this inequality says nothing about the perfect-ness

of the number f1*f2.

Peter - Thx, I will rework it.

________________________________

From: Tom Hadley <kctom99@...>

To: Mathieu Therrien <mathieu344@...>

Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>

Sent: Thursday, July 14, 2011 4:41:26 PM

Subject: Re: [PrimeNumbers] Post-Cartesian Puzzle

Mathieu Therrien <mathieu344@...> wrote:

>

I think you have misunderstood the sigma() function. In Pari-GP, sigma(x) is the sum of the divisors of x. So sigma(9) = 1+3+9 = 13.

>If I understood the purpose of Sigma(N/M) = N/P_1 = M*P_2 for P = P_1 * P_2 correctly,

>

>then Many solution are possibles as long as (M+1) is divided by 2 only once

>

>for example u have m=5 ; N = P_2 * m * 3 = 15*P_2 and as long that P_2 is odd

>

>So m=5 and N=45 is 1 solution

>

>

The puzzle is: Find a pair of odd integers (N,m) with m|N,

sigma(N/m)*(1+m) = 2*N, and bigomega(m) = 1.

The proposed solution, N=45, m=5 doesn't work, since

sigma(45/5)*(1+5) = sigma(9)*6 = 13*6 = 78, which is not 2*N=2*45=90.

Tom Hadley

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