A number N is perfect if an only if sigma(N)=2N, where the sigma function is
the sum of alldivisors of N, including 1 and N.
The sigma function verify:
i) sigma(p)=p+1, if p is prime
ii) sigma(p^n)=1+p+p^2+...+p^n=(p^(n+1)-p)/(p-1), if p is prime
iii) sigma(a�b)=sigma(a)�sigma(b), if gcd(a,b)=1 (it's a multiplicative
function)
Then, if N=2^(p-1)(2^p-1), with 2^p-1 prime 8a Mersenne prime), we have
sigma(N)=sigma(2^(p-1)(2^p-1))=sigma(2^(p-1))sigma(2^p-1)=((2^p-1)/(2-1))(2^
p-1+1)=
(2^p-1)2^p=2(2^(p-1))(2^p-1))=2N.
There is a partial converse: If N is perfect AND EVEN, then
N=2^(p-1)(2^p-1), with 2^p-1 prime.
It is not proved the inexistence of perfect odd numbers, althought the
minimum cote is very high.
Un saludo,
Ignacio Larrosa Ca�estro
A Coru�a (Espa�a)
[EMAIL PROTECTED]
----- Original Message -----
From: Simon J Rubinstein-Salzedo <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Saturday, February 26, 2000 12:27 AM
Subject: Mersenne: Perfect numbers
> Can someone please outline a proof as to why (2^p-1)(2(p-1)) is a perfect
> number if 2^p-1 is prime?
>
> 2^6972593 - 1 is prime.
> e^(i*pi) + 1 = 0.
> This is the e-mail address of Simon Rubinstein-Salzedo.
> When you read this e-mail, Simon will probably be at a math contest.
> Don't forget to check Simon's website at
http://www.albanyconsort.com/simon
> Thanks
> SJRS
>
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