# Sum of divisors is power of 2, is n square-free?

We were given this question on one of our recent exams, and we can't seem to generate a proof, nor find a counter-example.

If we let $\sigma(n)$ be the sum of the divisors of $n$, the question is, if

$$\sigma(n)=2^k$$

for some $k$, does this imply $n$ is square free?

This is clearly true if we consider just the number of divisors being a power of two, but we cannot extend it to the sum of the divisors. We have tried numerous $n$ that are not square-free, and none have given a counter-example. We made some progress in noting that if

$$n=p_1^{k_1}p_2^{k_2}...p_j^{k_j}$$

is the prime factorization of $n$, then

$$2^k=\sigma(n)=(1+p_1+p_1^2+...+p_1^{k_1})...(1+p_j+p_j^2+...+p_j^{k_j})$$

This then implies that each factor must be a power of two itself, and thus for each $p_i$,

$$p_i+p_i^2+...+p_i^{k_i}=2^t-1$$

for some $t$. Even with this, we are unable to find a counter-example. There does not seem to be a non-trivial solution.

Is there a proof we can't see, or is there a counter-example we can't find?

EDIT: From a comment by @lulu $\sigma(n)=2^k$ when $n$ is a product of distinct Mersenne primes, and is thus square-free. Since we have not covered these in our class, I'm still open to other solutions.

• If we take n=$2^n$ then its sum of divisor is $2^{n+1}$ but $2^n$ isn't square free for $n\geq 2$. Perhaps problem is when n is odd – arberavdullahu Nov 3 '16 at 16:59
• @arberavdullahu You forgot 1 as a divisor. The sum is squarefree. – Jacob Wakem Nov 3 '16 at 17:01
• @Alephnull yes you're right, thank you. – arberavdullahu Nov 3 '16 at 17:02
• this question is relevant. the only such $n$ are products of distinct Mersenne primes (hence square free). – lulu Nov 3 '16 at 17:03
• Thanks @lulu. That clears it up. – superckl Nov 3 '16 at 17:15