If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

Question

Let $\sigma(x)$ be the sum of the divisors of $x$. Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$.

Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$? $$D(q^k)D(n^2)=2s(q^k)s(n^2)$$

I only know that $k=1$ is true if and only if (one of) the following conditions hold:

(1) $\sigma(n^2)/q \mid n^2$

(2) $D(n^2) \mid n^2$

(3) $\gcd(n^2, \sigma(n^2)) = D(n^2)$

Source of Equation

From the fundamental equation $$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$ we obtain $$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \frac{D(n^2)}{s(q^k)} = \gcd(n^2, \sigma(n^2))$$ and $$\frac{\sigma(n^2) - n^2}{\frac{2q^k - \sigma(q^k)}{2}} = \frac{2s(n^2)}{D(q^k)} = \gcd(n^2, \sigma(n^2)),$$ using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}.$$

Reference

Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers

@user477343, I think there is a mistake in your second equation. It should be $$\frac{2s(n^2)}{D(q^k)}=\frac{D(n^2)}{s(q^k)}.$$ — Jose Arnaldo Bebita Dris, Mar 19 '18 at 09:28
Oh yes, woops. Typo. Sorry about that :) .... Oh, but now that changes everything.... — Mr Pie, Mar 19 '18 at 09:30
From the wikipedia page off perfect numbers: [https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers]. Apparently it still an open problem, but it is proven that k = 1 mod 4 — Dadedidododu, Mar 19 '18 at 09:30
@user477343, of course following your lead, $k=1$ if and only if $\sigma(q^k)-q^k=\sigma(q^{k-1})=s(q^k)=1$, which is trivial. — Jose Arnaldo Bebita Dris, Mar 19 '18 at 09:30
@Dadedidododu, thanks for pointing me to Wikipedia. I do am aware of the fact that $q \equiv k \equiv 1 \pmod 4$. — Jose Arnaldo Bebita Dris, Mar 19 '18 at 09:32
@user477343, that is fine. It is always good to see more and more people getting interested in this kind of maths stuff. ^_^ — Jose Arnaldo Bebita Dris, Mar 19 '18 at 09:33
@JoseArnaldoBebitaDris Oh yeah, my comment is useless, apparently you know a lot about this subject, sorry ^^. Did you already try to ask one of your questions on MathOverflow? — Dadedidododu, Mar 19 '18 at 09:39
@Dadedidododu, yes I have already asked some of my questions in MathOverflow. Check them out here. — Jose Arnaldo Bebita Dris, Mar 19 '18 at 09:41
In this question, I think you don't know that the equation is actually an identity, i.e. that it holds for every odd perfect number (of the form $q^kn^2$), so I've undeleted my answer. — mathlove, Oct 11 '18 at 14:32

score 1 · Accepted Answer · answered Oct 11 '18 at 12:54

The equation is actually an identity since we get

$$\begin{align}D(q^k)D(n^2)&=(2q^k-\sigma(q^k))(2n^2-\sigma(n^2)) \\\\&=(2q^k-\sigma(q^k))\left(2n^2-\frac{2n^2q^k}{\sigma(q^k)}\right) \\\\&=4n^2q^k-\frac{4n^2q^{2k}}{\sigma(q^k)}-2n^2\sigma(q^k)+2n^2q^k \\\\&=4n^2q^k-2n^2\sigma(q^k)-\frac{4n^2q^{2k}}{\sigma(q^k)}+2n^2q^k \\\\&=2\left(\sigma(q^k)-q^k\right)\left(\frac{2n^2q^k}{\sigma(q^k)}-n^2\right) \\\\&=2\left(\sigma(q^k)-q^k\right)\left(\sigma(n^2)-n^2\right) \\\\&=2s(q^k)s(n^2) \end{align}$$

score 0 · Answer 2 · answered Apr 25 '18 at 21:10

This is not an answer, just some comments that are too long to fit in the space beneath my original question.

For what it is worth, there is the simple relationship $$D(x) := x - s(x)$$ so that we obtain $$2s(q^k)s(n^2) = D(q^k)D(n^2) = \bigg(q^k - s(q^k)\bigg)\bigg(n^2 - s(n^2)\bigg)$$ $$= q^k n^2 - q^k s(n^2) - n^2 s(q^k) + s(q^k)s(n^2),$$ from which it follows that $$s(q^k)s(n^2) + q^k s(n^2) + n^2 s(q^k) - q^k n^2 = 0$$ $$s(q^k)s(n^2) + q^k s(n^2) + n^2 s(q^k) + q^k n^2 = 2q^k n^2$$ $$\bigg(s(q^k) + q^k\bigg)s(n^2) + \bigg(s(q^k) + q^k\bigg)n^2 = 2q^k n^2$$ $$\bigg(s(q^k) + q^k\bigg)\bigg(s(n^2) + n^2\bigg) = 2q^k n^2$$ But $s(x) + x = (\sigma(x) - x) + x = \sigma(x)$. Therefore, we have: $$\sigma(q^k)\sigma(n^2) = 2q^k n^2.$$

However, it is not clear to me whether the preceding argument can or cannot force $k=1$.

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

2 Answers2

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