# The sum of all divisors of $N$ is a 2 potency iff $N$ is a product of different Mersenne primes

As far as I have controlled:

$$\sigma(a)=2^n$$, for some $$n\in\mathbb N \iff$$ $$a$$ is a product of different Mersenne primes.

The $$\Leftarrow$$-part is an immediate consequence of that $$\sigma$$ is multiplicative, but the other part seems more complicated, if it is true.

Maybe this is interesting in the view of that it is unknown if there are only a finite number of Mersenne primes or not.

Here is a solution for $$n$$ squarefree:
Assume the lhs is true. Then you can split $$a$$ up into its primes. By definition you have $$\sigma(p)=p+1$$ for each, thus $$\sigma(\prod_{i=1}^r p_i)=\prod_{i=1}^r (p_i+1)=2^n$$ Since $$2$$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $$2^m-1$$. If $$m$$ is a prime, we're done. If $$m=pq$$ ($$p$$ prime, $$q$$ integer) is computed then consider $$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$ Repeat the procedure with $$2^{n/p}$$.
The generalization could come from the fact that $$\sigma(p_n\#)=2^n$$ where $$p_n\#$$ is the primorial, which only consists of square free primes.
• I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $\sigma(p^k)=2^m$ for some prime $p$ with $k>1$. Commented Nov 20, 2018 at 15:21