I would like to know if there are infinitely many $k$ for which $$\sigma(k)/k=n^p$$ such that $m=k{n}^{p-1}$ with $m,n>0$ and $p$ is an odd prime?


Note: $\sigma(\frac{m}{{n}^{p-1}})$ is the sum of divisors function of $\frac{m}{ {n}^{p-1}}$ such that $m=k{n}^{p-1}$.


**Edit 2**: I edited the question as it has the same meaning for the precedent.


Thank you for any help.



Are there infinitely many $k$ for which $\frac{\sigma(k)}{k}=n^p$ and $p$ is an odd prime?
The question can be reformulated as:



> 
> **Question.** Are there infinitely many $k$'s such that $\sigma(k)/k$ is a square?
> 
> 
> 


If the answer is yes, then (trivially) there are infinitely many [multiperfect numbers](https://en.wikipedia.org/wiki/Multiply_perfect_number). This consequence is not known unconditionally, so at the moment we cannot prove that the answer is yes.


If the answer is no, then it implies that each perfect number is divisible by a prime from a fixed finite set of primes, because the product of two coprime perfect numbers satisfies $\sigma(k)/k=4$. This consequence is not known unconditionally, so at the moment we cannot prove that the answer is no.


In short, it seems that above question is out of reach at the moment.


