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25.3.21

On (the series of inequalities) $\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$, where $q^k n^2$ is an odd perfect number

(Preamble: The contents of this blog post were pulled verbatim from this MSE question.)

Let $N=q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.

Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, the aliquot sum of $x$ by $s(x)=\sigma(x)-x$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

Since $N=q^k n^2$ is perfect and $\gcd(q,n)=1$, we obtain
$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2.$$
Now using the fact that $\gcd(q^k,\sigma(q^k))=1$, we see that $q^k$ must divide $\sigma(n^2)$, so that
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$
Now, using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B},$$
where $B \neq 0$, $D \neq 0$, and $D-B \neq 0$, then we obtain
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}.$$

We now prove the following:

CLAIM:  If $N=q^k n^2$ is an odd perfect number with special prime $q$, then the series of inequalities
$$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$
holds in general.

PROOF:  We prove each inequality one by one, left to right.

First, assume to the contrary that
$$2n^2 - \sigma(n^2)=D(n^2)=\frac{D(n^2)}{s(q)} < \frac{2n^2}{\sigma(q)} = \frac{2n^2}{q+1}.$$
This inequality is equivalent to
$$\frac{2q}{q+1}=2 - \frac{2}{q+1} < \frac{\sigma(n^2)}{n^2} = I(n^2),$$
contradicting $I(n^2) \leq 2q/(q+1)$.

Next, suppose to the contrary that
$$\frac{2n^2}{q+1}=\frac{2n^2}{\sigma(q)} < \frac{\sigma(n^2)}{q}.$$
This inequality is equivalent to
$$\frac{2q}{q+1} < \frac{\sigma(n^2)}{n^2} = I(n^2),$$
contradicting $I(n^2) \leq 2q/(q+1)$.

Lastly, assume to the contrary that
$$\frac{\sigma(n^2)}{q} < \frac{2s(n^2)}{D(q)} = \frac{2(\sigma(n^2) - n^2)}{q - 1}.$$
This inequality is equivalent to
$$(q - 1)\sigma(n^2) < 2q\sigma(n^2) - 2qn^2 \iff 2qn^2 < (q+1)\sigma(n^2) \iff \frac{2q}{q+1} < I(n^2),$$
contradicting $I(n^2) \leq 2q/(q+1)$.

Now here is my:

QUESTION:  Note that equality holds in
$$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$
if and only if $k=1$.
So do we have
$$k=1 \iff \bigg(\frac{D(n^2)}{s(q)}=\frac{2n^2}{\sigma(q)}=\frac{\sigma(n^2)}{q}=\frac{2s(n^2)}{D(q)}\bigg) \iff \bigg(\frac{D(n^2)}{s(q^k)}=\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2s(n^2)}{D(q^k)}\bigg),$$
by treating $k$ as a placeholder for $1$?
If so, do we then have a proof for $k=1$, since
$$\bigg(\frac{D(n^2)}{s(q^k)}=\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2s(n^2)}{D(q^k)}\bigg)$$
holds unconditionally?