Definition: Let $n \in \Bbb Z$ with $n>0$. Then $n$ is said to be superperfect if $σ(σ(n))=2n$. Where $σ$ is the sum of positive divisors arithmetic function.
I am trying to solve a proof that asks: Prove that if $2^a$ is superperfect, then $2^{a+1} - 1$ is a Mersenne prime.
So far, this is what I got:
Assume $2^a$ is superperfect. Then, $σ(σ(2^a)) = 2(2^a) = 2^{a+1}$. Can someone point me in the right direction?
I don't understand how I could show $2^{a+1} - 1$ is a Mersenne prime from what I have so far. Thanks!
The key was stated in comments, you need do the substitution that you've computed as $$\sigma(2^a)=2^{a+1}-1$$ in your condition deduced from the assumption that $2^a$ is superperfect. Then do a comparison between the result and the implication (that is a proposition) $$\sigma(N)=N+1\implies N\text{ is prime}.$$ Notice that you've deduced before $\sigma(N)=N+1$, where the substitution is saying to you what is your $N$. Then since you have $\sigma(N)=N+1$ for a $N$ you can conclude that $N$ is prime (that is, the only difficulty here is if you know how show for any $m$ that: if $\sigma(m)=m+1$ then $m$ is prime).