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.
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$\begingroup$ The equation is close to defining n as an odd kmultiperfect 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 $\endgroup$– Gerhard PasemanFeb 27, 2014 at 17:58

1$\begingroup$ It sounds like you will find the (very similar) problems considered in this paper of interest: math.uga.edu/~pollack/pperfs16.pdf $\endgroup$– socalled friend DonFeb 27, 2014 at 18:36

$\begingroup$ This MSE question might be related. $\endgroup$– ARNIE BEBITADRISNov 12, 2015 at 20:34
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1 Answer
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The smallest solution to your equation ${\rm rad}(\sigma(n)) = 2{\rm rad}(n)$ is $n = 135$.

$\begingroup$ Thanks! Can it be proven that there are only finitely many solutions? $\endgroup$ Feb 27, 2014 at 18:11

$\begingroup$ @SylvainJULIEN: Why would you expect this to be true?  By the way, the further solutions $n < 1000000$ are 891, 200655, 307125, 544635. $\endgroup$– Stefan Kohl ♦Feb 27, 2014 at 18:16

$\begingroup$ I don't really expect it to be true. It would just be interesting to know whether there are finitely many solutions or not, so one has to pick up one of both possibilities. $\endgroup$ Feb 27, 2014 at 18:27

$\begingroup$ And if it ever turns out that there are finitely many solutions, this will entail that there are finitely many OPNs. $\endgroup$ Feb 27, 2014 at 18:30

1$\begingroup$ The sequence has now been entered into the online encyclopedia, oeis.org/A238330, with a link back here. $\endgroup$ Feb 27, 2014 at 22:01