# Solutions of $rad(\sigma(m))=2rad(m)$

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?

• The equation is close to defining n as an odd k-multiperfect where k rational happens to be composed of prime factors n. An even weaker condition is P(rad(sigma(n))) <= P(n), where P is the largest prime factor of n > 1. As far as I know, even this weaker condition for n a prime power has not been investigated. Gerhard "Would Like To See Answers" Paseman, 2014.02.27 Feb 27, 2014 at 17:58
• It sounds like you will find the (very similar) problems considered in this paper of interest: math.uga.edu/~pollack/pperfs16.pdf Feb 27, 2014 at 18:36
• This MSE question might be related. Nov 12, 2015 at 20:34

The smallest solution to your equation ${\rm rad}(\sigma(n)) = 2{\rm rad}(n)$ is $n = 135$.
• @SylvainJULIEN: Why would you expect this to be true? -- By the way, the further solutions $n < 1000000$ are 891, 200655, 307125, 544635. Feb 27, 2014 at 18:16