For $m$ a positive integer greater than $1$, let $rad(m)$ be the product of all distinct primes dividing $m$. If $n$ is an odd perfect number (conjectured not to exist), one would have $\sigma(n)=2n$, hence $rad(\sigma(n))=2rad(n)$. Has this equation been considered so far? Are there any known solutions to it?  

Thanks in advance.



Solutions of $rad(\sigma(m))=2rad(m)$
The smallest solution to your equation ${\rm rad}(\sigma(n)) = 2{\rm rad}(n)$ is $n = 135$.


