Timeline for For $q$ prime and $\gcd(q, n) = 1$, can $N = qn^2$ be almost perfect?

9 events

when toggle format	what		by	license	comment
Dec 5, 2015 at 16:29	history	edited	Jose Arnaldo Bebita	CC BY-SA 3.0	fixed the brpken hyperlink for paper2
Nov 12, 2015 at 10:42	history	undeleted	Jose Arnaldo Bebita
Aug 13, 2014 at 15:34	history	deleted	Jose Arnaldo Bebita		via Vote
Aug 13, 2014 at 10:59	comment	added	Jose Arnaldo Bebita		If $N = 2n^2$ is almost perfect with odd $n > 1$, then $I(n^2) < 4/3 = 1.\bar{3}$, where $I(x) = \sigma(x)/x$ is the abundancy index of $x$. However, I currently do not know how to push this through to a contradiction.
Aug 13, 2014 at 10:02	comment	added	Jose Arnaldo Bebita		Exactly @GHfromMO!
Aug 13, 2014 at 9:32	comment	added	GH from MO		To your first question (first line): if $qn^2$ is almost perfect, then $\sigma(qn^2)=\sigma(q)\sigma(n^2)$ is odd, hence $\sigma(q)=q+1$ is odd, i.e. $q=2$. So you are really asking if a number of the form $N=2n^2$ with $n$ odd can be almost perfect.
Aug 13, 2014 at 8:31	history	edited	Jose Arnaldo Bebita	CC BY-SA 3.0	added ", where $p$ is prime with $\gcd(p,m)=1$" in the second sentence of the third paragraph
Aug 13, 2014 at 8:26	answer	added	Jose Arnaldo Bebita		timeline score: 0
Aug 13, 2014 at 7:59	history	asked	Jose Arnaldo Bebita	CC BY-SA 3.0