If $p=2^n-1$ is prime, therefore $2^{n-1}·(2^n-1)$ is perfect

Question

If $p=2^n-1$ is prime, therefore $2^{n-1}·(2^n-1)$ is perfect.

Well I would want to prove that but I don't really know where to start. Any ideas?

Answer 1

So you want to know how this formula is resulted!

This is based on assumption that the required perfect number N is a multiple like $a^{\alpha}$ of a Mersenne prime $2^n-1$:

$N=a^{\alpha}(2^n-1)$

$\sigma_N=S_{a^{\alpha}}\times S_{2^n-1}$

Where S is the sum of divisors, we have:

$S_{a^{\alpha}}=\frac{a^{\alpha +1}-1}{a-1}$

And $S_{2^n-1}=2^n-1+1=2^n$

Because $2^n-1$is prime and it's divisors is itself and the number 1. Therefore we may write:

$S_N=2^n \times \frac{a^{\alpha+1}-1}{a-1}$

Now due to the definition of perfect number we must have:

$S_N=2 N$ ⇒ $2^n \times \frac{a^{\alpha+1}-1}{a-1}=2 a^{\alpha}(2^n-1)$

It can easily been seen that if $a=2$ we will have:

$S_N=2^n \times \frac{2^{\alpha+1}-1}{2-1}=2 \times2^{\alpha}(2^n-1)$

Or

$2^n(2^{\alpha+1}-1)=2^{\alpha +1}(2^n-1)$

Which results in $n=\alpha+1$ and gives the general form of perfect numbers as:

$N=2^{n-1}(2^n -1)$