Super Perfect numbers A super-perfect number is a number with $\sigma(\sigma (n))=2n$.
How can I prove that every even super perfect number is from the form $n=2^k$ when $2^{k+1}-1$ is prime.
I tried every way please give me some guidance
 A: Let $p$ be a prime. We know that $\sigma (p^\alpha)= \dfrac{p^{\alpha+1} -1}{p-1}$.
Now, let $m=p^\alpha r$ be an integer with $p\not| r$.
 If $p \not| m$ then $\sigma(p m) = \sigma(p) \sigma(m) =(p+1) \sigma(m) > p \sigma(m)$.
If $p | m$ then we have $\sigma(pm) = \sigma (p^{\alpha +1})\sigma(r)= \dfrac{p^{\alpha +2} -1} {p-1} \sigma(r) >\dfrac{p^{\alpha +2} -p} {p-1} \sigma(r)= p\sigma(p^\alpha)\sigma(r) = p \sigma(m)$.

So, for any product $ab$ we have $\sigma(ab) \ge a \sigma(b)$ and equality holds exactly for $a=1$.


Now, we can conclude:
Let $n= 2^ku$ with $u$ odd and $k \ge 1$.
We have $\sigma(n)= (2^{k+1}-1) s$ where $s=\sigma(u) \ge u$.
Therefore $\sigma(\sigma(n))= \sigma((2^{k+1}-1)s) \ge \sigma((2^{k+1}-1) s  \ge 2^{k+1} u = 2n$.
So, for a super-perfect number, equality has to hold in this chain of inequalities.
The first inequality is true iff $s=1$ and the second inequality is true iff $2^{k+1}-1$ is prime (because the two trivial divisors already give $2^{k+1}$ and the number is clearly bigger than 1, this is where we use $k \ge 1$) and $s=u$.

So, in total $u=1$ and $2^{k+1}-1$ is prime and $n=2^k$.



Just to add a word of motivation:
Perfect numbers have the property $\sigma(n)=2n$, so with $\rho(n)=\sigma(n)-n$, this becomes $\rho(n)=n$.
The concatenation that is expected to stay the same size "on average" would be $\rho(\rho(n))=n$ which is related to the search for amicable numbers.
Now, instead we apply $\sigma$ twice, so we can expect that the number grows too much most of the time. This motivates the search for an inequality chain of the type $\sigma(\sigma(n))\ge 2n$ which then turns out to only work for even $n$. 
