Sum of divisors and indices: $\sigma (n)= 2^k$ Are there infinitely many positive integers (say $n$ is one of them), sum of whose divisors are powers of two, i.e, $ \sigma (n)= 2^k $ ? 
 A: The only possibilities here are products of distinct Mersenne primes (thus it is unknown whether or not there are infinitely many). (Note:  it's clear that any product of distinct Mersenne primes does indeed have the desired property.)
To see this, consider $n$ such that $\sigma(n)$ is a power of two, and write its prime factorization as $n=\prod p_i^{a_i}$.  
Since the divisor sum function, $\sigma$, is weakly multiplicative it suffices to work prime by prime.  Thus, if $p|n$, with $v_p(n)=a$, we know that $\sigma (p^a)=2^k$ for some $k$. We want to argue that $p$ is a Mersenne Prime and that $a=1$.
We have:  $$2^k=\sigma(p^a)=1+p+p^2+\cdots+p^a$$
Whence both $p$ and $a$ must be odd.  Writing $a=2b+1$ we can factor to get $$2^k=(1+p)(1+p^2+p^4+\cdots+p^{2b})$$
This shows that $1+p$ is a power of $2$, whence $p$ is a Mersenne prime. We write $p=2^r-1$.
Now, to see that $a=1$ we remark that the factoring we gave also implies that $1+p^2+\cdots+p^{2b}$ is a power of $2$ (write it as $2^l$).  Thus, if  $b>0$ we must have $b$ odd, so we can write $b=2c+1$.  Then, as before we can factor to see that $$2^l=1+p^2+\cdots+p^{2b}=(1+p^2)(1+p^4+\cdots+p^{4c})$$
Therefore $1+p^2$ is also a power of $2$.  As $p=2^r-1$ we get $$1+p^2=1+(2^r-1)(2^r-1)=1+2^{2r}-2^{r+1}+1=2+2^{2r}-2^{r+1}$$  Which is clearly not divisible by $4$, a contradiction.
