TITLE: Can $\sigma(p^k)=2^n$ for some $k>1$?
QUESTION [5 upvotes]: If $k=1$ then the only solutions to
$$
\sigma(p^k)=2^n
$$
are when $p$ is a Mersenne prime (of course $p$ is restricted to the primes). Is there a solution for larger $k$? It doesn't seem so but I can't find a proof. It's easy enough to exclude some congruence classes but that's a long way from showing that there are no examples.

REPLY [8 votes]: Clearly $p=2$ is out of the reckoning.
I use the
Fact. If $\ell+1\mid k+1$, then $\sigma(p^{\ell})\mid \sigma(p^k)$. 
This follows by cancelling a factor $p-1$ from the relation $p^{\ell+1}-1\mid p^{k+1}-1$.
Let $q$ be a prime factor of $k+1$. Because
$\sigma(p^{q-1})\mid\sigma(p^k)$
we can deduce that $\sigma(p^{q-1})$ is also a power of two. But $p$ is odd, so $$p^i\equiv1\pmod2$$ for all $i$. Hence $$\sigma(p^{q-1})\equiv q\pmod2.$$ Therefore $q=2$, and we can conclude that $k+1$ must be a power of two as it has no odd prime factors.
But, if $k+1$ is a multiple of four, we get the divisibility relation
$
\sigma(p^3)\mid \sigma(p^k)
$
implying that
$$
\sigma(p^3)=1+p+p^2+p^3=(1+p)(1+p^2)
$$
must also be a power of two. Consequently both $1+p$ and $1+p^2$ are powers of two as well.
But $1+p^2\equiv2\pmod4$ and $p>2$, so this is impossible.