## Re: [PrimeNumbers] Re: odd-perfect number don't exist

Expand Messages
• ... Nope. Your alleged proof doesn t say anything about needing the decomposition to be complete; it only required it to be /a/ decomposition in which all the
Message 1 of 33 , Jul 13, 2011
• 0 Attachment
> either you have a complete decomposition into factors or not.
> 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-

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!)

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
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 To: Mathieu Therrien Cc:
Message 33 of 33 , Jul 14, 2011
• 0 Attachment
Thx, I will rework it.

________________________________
To: Mathieu Therrien <mathieu344@...>
Sent: Thursday, July 14, 2011 4:41:26 PM

Mathieu Therrien <mathieu344@...> wrote:

>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
>
>
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.

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.