(I have asked a similar question in [MSE](https://math.stackexchange.com/questions/1572152) 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.



If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, is $n$ squarefree?
No, if an odd perfect number exists, then $n$ must contain a square factor. This is a 1937 result of Steuerwald:


R. Steuerwald, "Verschärfung einer notwendigen Bedingung für die Existenz einer
ungeraden vollkommenen Zahl," S.-B. Math.-Nat. Abt. Bayer. Akad. Wiss., 1937, pp. 68-73.


A very nice recent paper on this theme is by Fletcher, Nielsen, and Ochem (Math. Comp.); the preprint version is freely available on Nielsen's website: <https://math.byu.edu/~pace/OPNSieves_web.pdf>


[Edited as per Lucia's suggestion.]


