(I have asked a similar question in MSE around a week ago, but did not receive any responses. I have therefore cross-posted it to this site, hoping to get some answers.)

An odd perfect number $N$ is said to be given in **Eulerian form** if $N = {q^k}{n^2}$ where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$, and $\gcd(q, n) = 1$.

Here is my question:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, is $n$ squarefree?

I would certainly appreciate it if someone could point me to existing papers in the literature where this particular question is addressed.