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

(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.

No, if an odd perfect number exists, then $n$ must contain a square factor. This is a 1937 result of Steuerwald:
• It is absurd that someone chose to vote down this answer! (It took me a little effort to parse your answer -- maybe it would be clearer if the first line said something like: No: if an odd perfect number exists, then $n$ must necessarily contain a square factor.") – Lucia Dec 18 '15 at 22:33
• @so-calledfriendDon, mind if you check out this related MSE question, where I ask if $n$ is a square? Let me know if you need me to ask a separate MO question. – Arnie Bebita-Dris Dec 24 '15 at 14:10