(I have asked a similar question in [MSE](https://math.stackexchange.com/questions/1585550) four days ago, but did not receive any answers. I have therefore cross-posted it to this site, hoping to get some responses.)


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 $n$ a square?
> 
> 
> 


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 $n$ a square?
There is no reason to believe that the non-Eulerian part is a 4th power. Moreover, Descartes spoof OPN shows us that such a guess is probably unmotivated and that there is no purely combinatorial way to prove that $n$ is a square. A proof of such a result would fundamentally require restriction to actual (rather than spoof) prime factors; and as far as I'm aware most results on OPN's (besides the finite computational searches) don't make such a restriction.


However, it may be possible to prove that $n$ cannot be a perfect square. It has been long known that it cannot happen that $p^2||n$ for each prime $p|n$. For a vast generalization of this result see my paper with S. Adam Fletcher and Pascal Ochem: [Sieve methods of odd perfect numbers](http://www.ams.org/journals/mcom/2012-81-279/S0025-5718-2011-02576-7/home.html).


