Mersenne primes and superperfect numbers

Question

Definition: Let $n\in\mathbb{Z}$ with $n>0$. Then $n$ is said to be superperfect if $\sigma(\sigma(n)) = 2n$. Where $\sigma$ is the sum of positive divisors arithmetic function. ($\sigma(n) = \sum_{d|n, d>0} d $)

I'm working on a problem which asks me to prove that if $2^p - 1$ is a Mersenne prime, then $2^{p-1}$ is superperfect.

So far all I've said in my proof is:

If $2^p - 1$ is a Mersenne prime, then p must be a prime number.

I'm not really sure how to proceed. Could someone nudge me in the right direction?

Answer 1

Observe the following:

• $n\in\mathbb{N}\implies\sigma(2^{n-1})=\color{red}{2^n-1}$
• $2^n-1\in\mathbb{P}\implies\sigma(2^n-1)=1+(2^n-1)=\color{green}{2^n}$
• $2^n-1\in\mathbb{P}\implies\sigma(\sigma(\color{blue}{2^{n-1}}))=\sigma(\color{red}{2^n-1})=\color{green}{2^n}=2\cdot\color{blue}{2^{n-1}}$