> either you have a complete decomposition into factors or not.

Nope. Your alleged proof doesn't say anything about needing

> when Euler looked for�N = (4x+1)^(4y+1) * Q^2 he wrote

> out p1, p2, p3, p4, etc. in a general decomposition as the

> most basic assumption, so 22021 would have to be decom-

> posed. Does that answer your question?

the decomposition to be complete; it only required it to be

/a/ decomposition in which all the factors are odd. But well,

let's play your game of conjuring extra assumptions on the

way for a bit... and assume that the undescribed quantities

f1, f2, f3, ... in your proof are supposed to represent the

complete factorization of N.

Now, calling upon the powers of Euler was not the best move --

since he delivers two more fatal blows (for the price of one!)

to your proof :-)

Fatal blow #1: The LHS of equations you have on your website

assumes that all the f1, f2, ... are distinct. Otherwise, you'd

be including some factors more than once. That's what I asked

you about in my previous unanwered reply, regarding number 9;

if you write it down factored fully, f1 = f2 = 3, the expression

on the LHS would be equal to 1 + f1 + f2 = 1 + 3 + 3 = 7. Yet,

the sum 9's factors (apart from 9 itself) is just 4. Ka-pow!

Maybe you'd like to correct your proof to support higher

powers of primes? Looking at something simple, like

N = 5(P^2)(Q^2)(R^2) (where P, Q, R are distinct odd primes,

all different from 5) might be a good start...

Fatal blow #2: Quoting from your site:

- let an 'even' number of variables f1, f2 be 'odd'

- let an 'even' number of variables f1, f2, f3, f4 be 'odd'

- let an 'even' number of varibles f1, f2, f3,... f(2n) be 'odd'

Now, what is this obsession with 'even' number of variables?

The very form you quoted above and attributed to Euler shows

that /odd/ number of prime factors (counting multiplicities).

In other words, even if your proof was corrected to work

with multiplicities, it would still only cover numbers which

are already known to be non-candiates for odd-perfects. Bang!

Peter

[Non-text portions of this message have been removed]- 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

[Non-text portions of this message have been removed]