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

An almost perfect number $N$ is a positive integer satisfying $\sigma(N) = 2N - 1$, where $\sigma(x)$ is the sum of all the divisors of $x$. For example, $\sigma(2) = 1 + 2 = 2\cdot2 - 1$, and thus $2$ is almost perfect.

A number $M$ is perfect if $\sigma(M) = 2M$. Recall that the Eulerian form of an odd perfect number is $M={p^k}{m^2}$, where $p$ is prime with $\gcd(p,m)=1$. This means the following:

If $k = 1$, then $M=p{m^2}$ is perfect.

If $k > 1$, then $M=p{m^2}$ is deficient.

I was thinking of applying the criterion from this paper1, but alas this is where I get stuck.

To recap, my main (and more specific) question for this post would be:

If $k > 1$, can the divisor $p{m^2}$ of the odd perfect number $M={p^k}{m^2}$ be almost perfect?

I already know that $p^k$, $m$, $pm$ and $m^2$ are not almost perfect if $M={p^k}{m^2}$ is perfect (see paper2).

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    $\begingroup$ 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. $\endgroup$
    – GH from MO
    Commented Aug 13, 2014 at 9:32
  • $\begingroup$ Exactly @GHfromMO! $\endgroup$
    – Jose Arnaldo Bebita
    Commented Aug 13, 2014 at 10:02
  • $\begingroup$ 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. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Aug 13, 2014 at 10:59

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My bad, I just realized the following: If $N$ is an odd almost perfect number, then since $\sigma(N) = 2N - 1$ is also odd, then $N$ has to be a square.

Consequently, for my question that:

If $k > 1$, can the divisor $p{m^2}$ of the odd perfect number $M={p^k}{m^2}$ be almost perfect?

the answer is NO, since $\gcd(p,m) = 1$ implies that $p{m^2}$ cannot be a square.

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  • $\begingroup$ The more general question remains open for the case of even almost perfect numbers. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Aug 13, 2014 at 8:35

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