I asked the following [question](https://math.stackexchange.com/questions/512573) in MSE four ($4$) days ago, but so far nobody has posted an answer.



> 
> The gist of the question is as follows:
> 
> 
> Are all known $k$-multiperfect numbers (for $k > 2$) *not* squarefree?
> 
> 
> 


From my own computational verifications using WolframAlpha, it appears that a *closely related* question is:



> 
> Are all known (*even*) $k$-multiperfect numbers (for $k > 2$)
>  *divisible* by $4$?
> 
> 
> 


Thank you!



Are all known $k$-multiperfect numbers (for $k > 2$) not squarefree?
It's easy to prove that if $n$ is squarefree with largest prime factor $p \geq 5$, then $n$ is not multiply perfect. Indeed, in that case $p$ is larger than any of the primes dividing $\sigma(n)$, so that $n \nmid \sigma(n)$. So the only squarefree multiply perfect number is $n=6$ (or also $n=1$, if you allow $k=1$). 


