If p is prime, then $n= 2^{p-1}(2^p -1)$ is perfect. True or False.
Please, help

Question

If p is prime, then $n= 2^{p-1}(2^p -1)$ is perfect. True or False.
Please, help

ganeshie8 · Answer

The statement is clearly false. Below is true however : 

If $2^n-1$ is prime, then $2^{n-1}(2^n-1)$ is a perfect number.

loser66 · Answer

We should prove it, right?

ganeshie8 · Answer

which statement you wanto prove ?

loser66 · Answer

Your statement.

ganeshie8 · Answer

simply evaluate the sum of divisors and see

ganeshie8 · Answer

Let $$m = 2^{n-1}(2^n-1)$$

since $2^n-1$ is prime, the numbers $2^{n-1}$ and $2^n-1$ are relatively prime, yes ?

loser66 · Answer

yes

ganeshie8 · Answer

use the fact that $\sigma $ is multiplicative

ganeshie8 · Answer

$m = 2^{n-1}(2^n-1)$

$\sigma(m) = \sigma(2^{n-1}(2^n-1)) = \sigma(2^{n-1})*\sigma(2^n-1)$

loser66 · Answer

hold on. I don't know what $\sigma$ is. I saw it many times when searching practice problem online , but I was not taught what it is yet.

ganeshie8 · Answer

Ohk, we don't really need to know about $\sigma $ here

ganeshie8 · Answer

but we need to find some way to find the sum of divisors of an integer

ganeshie8 · Answer

suppose 
$m = 2^3*3$

can you find the sum of divisors of $m$ ?

pawanyadav · Answer

Is it (1+2+3+4+6+8+12+24)

dan815 · Answer

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