If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$?

Question

(Note: This question is an offshoot of this closely related one.)

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

The topic of odd perfect numbers likely needs no introduction.

The initial question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then is there a constant $D$ such that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$

(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)

MY ATTEMPT

Since $\gcd(p^k,\sigma(p^k))=1$, we know that $$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}=\frac{D(m^2)}{s(p^k)},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.

Now, let $D > 0$ be a constant such that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}.$$ This is equivalent to $$D > \frac{p^k(m^2 - p^k)}{\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{\sigma(p^k)\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{2p^k m^2}$$ $$= \frac{\sigma(p^k)}{2} - \frac{p^k \sigma(p^k)}{2 m^2} = \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)}.$$

Using mathlove's answer to a closely related question, we have the lower bound $$\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375.$$

Hence, we have $$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p - 1)} - \frac{p^k}{3375} = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$ Let $$f(k) = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$ Then the first derivative $$\frac{\partial f}{\partial k} = \frac{(3373p + 2)p^k \log(p)}{6750(p - 1)}$$ is positive for $p \geq 5$. This implies that $f$ is an increasing function of $k$.

Therefore, since $k \geq 1$, we obtain $$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2}.$$

But we know that $p$ is at least $5$, since $p$ is the special prime satisfying $p \equiv 1 \pmod 4$. Therefore, we have $$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2} \geq \frac{2024}{675} = 2.9985\overline{185}.$$

Here are my final questions:

Does this mean that we can take $D=3$ in the inequality $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$ If $D=3$ does not work, then what value of $D$ works?

Answer 1

Using your idea, one has$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}\implies p^k\lt \frac{6750D}{3373}$$So, I think that for any given $D\gt 0$, $\dfrac{\sigma(m^2)}{p^k} > \dfrac{m^2 - p^k}{D}$ does not hold in general.

Answer 2

This is a partial answer.

If $D=3$, then $p=5$ and $k=1$ both hold.

Suppose that $D=3$.

$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{3}$$ $$2 > I(m^2)=\frac{\sigma(m^2)}{m^2} > \frac{p^k \bigg(1 - \dfrac{p^k}{m^2}\bigg)}{3}$$ $$\implies 6 > p^k \bigg(1 - \frac{p^k}{m^2}\bigg) \geq p^k \bigg(1 - \frac{2}{3375}\bigg) = \frac{3373p^k}{3375}$$ since $$\frac{\sigma(m^2)}{p^k} \geq 3375 \implies \frac{p^k}{m^2} \leq \frac{2}{3375}.$$

But the inequality $$\frac{3373p^k}{3375} < 6$$ only holds when $p = 5$ and $k = 1$. QED

Hence, we may not take $D=3$, as it will not make the inequality hold in general.