I'm supposed to prove the following,
$n$ is an even perfect number if and only if $n=2^{p-1}(2^p-1)$ where $(2^p-1)$ and $p$ are prime.
My proof:
Suppose $n=2^{p-1}(2^p-1)$. Let $\sigma(n)$ denote the sum of the divisors of $n$. Then, $$\sigma(n) = \sigma(2^{p-1})\sigma(2^p-1) = (2^p-1)2^p = 2n.$$ For the other direction, suppose $n$ is an even perfect number. Let $n= 2^sj$ where $j$ is an odd integer. We know that $$\frac{\sigma(n)}{2} = \frac{(2^{s+1}-1)\sigma(j)}{2}= n.$$ So, $$(2^{s+1}-1)\frac{\sigma(j)}{2} = 2^sj.$$ By inspecting the parity of the terms, we see that $j = (2^{s+1}-1)$ and $\sigma(j) = 2^{s+1} = j+1$. Since $j$ has only two divisors, it is prime. Moreover, since $(2^{s+1}-1)$ is prime, clearly $s+1=p$ where $p$ is some prime number ( if not, we would be able to factor $(2^{s+1}-1)$ but that is impossible because it is prime.) Hence, $$n=2^ej = 2^{p-1}(2^p-1).\square$$
Is my proof correct?