If $p^k m^2$ is an odd perfect number, then what is the optimal constant $C$ such that $\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$?

Question

The topic of odd perfect numbers likely needs no introduction.

The question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then what is the optimal constant $C$ such that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}?$$

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

MY OWN PROOF FOR $C = 1$

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$. This implies that $$\frac{\sigma(m^2)}{p^k} \neq m^2 - p^k$$ since $\sigma(m^2)/p^k$ is always odd.

Now, suppose to the contrary that $$\sigma(m^2) > p^k m^2 - p^{2k}.$$ $$p^{2k} + \sigma(m^2) > p^k m^2$$ $$2p^{2k} + 2\sigma(m^2) > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$ $$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 2\bigg)$$ $$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 2} \leq \frac{2p^{2k}}{p^k - 1}$$ where we have used the lower bound $\sigma(p^k) \geq p^k + 1$. But from the paper Dris (2012), we have the lower bound $$3p^k \leq \sigma(m^2)$$ while we also have the upper bound $$\frac{2p^{2k}}{p^k - 1} = 2(p^k + 1) + \frac{2}{p^k - 1}.$$

This implies that $$3p^k < 2(p^k + 1) + \frac{2}{p^k - 1}$$ $$p^k - 2 < \frac{2}{p^k - 1}$$ $$(p^k - 1)(p^k - 2) < 2.$$ This last inequality is a contradiction, as $p^k \geq 5$.

We therefore conclude that the inequality $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$$ holds when $C=1$.

Does the inequality still hold when the constant $C > 1$? If so, then what is the optimal value for $C$?

Answer 1

This is a partial answer.

Does the inequality still hold when the constant $C > 1$?

Yes, one can get $$\frac{\sigma(m^2)}{p^k} \le \frac{m^2 - p^k}{\color{red}{4/3}}\tag1$$

I don't know if this is the optimal value for $C$.

Proof for $(1)$ :

Note that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}\iff \frac{p^k(m^2 - p^k)}{\sigma(m^2)}\gt C$$

Let $F:=\dfrac{p^k(m^2 - p^k)}{\sigma(m^2)}$. We can write $F$ as

$$F=\frac{p^k(m^2 - p^k)\sigma(p^k)}{\sigma(m^2)\sigma(p^k)}=\frac{p^k(m^2 - p^k)\sigma(p^k)}{2p^km^2}=\frac{m^2\sigma(p^k)-p^{k}\sigma(p^k)}{2m^2}$$

$$=\frac{\sigma(p^k)}{2}-\frac{p^{k}\sigma(p^k)}{2m^2}=\frac{\sigma(p^k)}{2}-\frac{p^{2k}}{\sigma(m^2)}$$

Using $3p^k\le \sigma(m^2)$, we get

$$F=\frac{\sigma(p^k)}{2}-\frac{p^{2k}}{\sigma(m^2)}\ge \frac{\sigma(p^k)}{2}-\frac{p^{k}}{3}=\frac{p^{k+1}-1}{2(p-1)}-\frac{p^{k}}{3}=\frac{p^{k+1}+2p^k-3}{6(p-1)}$$

Let $g(k):=\dfrac{p^{k+1}+2p^k-3}{6(p-1)}$. Since $g(k)$ is increasing, we get $$F\ge g(k)\ge g(1)=\frac{p+3}{6}\ge \frac{5+3}{6}=\frac 43$$ to have $$F\ge \frac 43$$ which is equivalent to $(1)$.$\quad\blacksquare$

Answer 2

(Last updated on March 2, 2021 - 5:04 PM Manila time)

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Suppose to the contrary that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{C}$$ and that $C=3$.

Following in the footsteps of this answer, we have $$3\sigma(m^2) > p^k m^2 - p^{2k}$$ $$6\sigma(m^2) + 2p^{2k} > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$ $$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 6\bigg)$$ Assuming $\sigma(p^k) \neq 6$ (that is, assuming it is not the case that $p=5$ and $k=1$ holds), then $$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 6} \leq \frac{2p^{2k}}{p^k - 5} = \frac{2p^{2k} - 50}{p^k - 5} + \frac{50}{p^k - 5} = 2(p^k + 5) + \frac{50}{p^k - 5}.$$

But according to Broughan et al. (2013), we have the lower bound $315p^k \leq \sigma(m^2)$.

This implies that $$315p^k < 2(p^k + 5) + \frac{50}{p^k - 5}$$ $$(313p^k - 10)(p^k - 5) < 50$$ which is clearly a contradiction if $p \neq 5$ or $k \neq 1$.

We therefore have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $\sigma(p^k) \neq 6$, then $$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{3}.$$

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SECOND ATTEMPT

Using mathlove's approach in his/her answer below, together with Broughan et al. (2013)'s result, we have the lower bound $\frac{\sigma(m^2)}{p^k} \geq 315$. As in mathlove's answer, let the quantity $F$ be given by $$F = \frac{\sigma(p^k)}{2} - \frac{p^{2k}}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{315} = \frac{p^{k+1} - 1}{2(p-1)} - \frac{p^k}{315} := f_1(k).$$ Since $f_1(k)$ is increasing, we get $$F \geq f_1(k) \geq f_1(1) = \frac{p+1}{2} - \frac{p}{315} \geq \frac{188}{63} \approx 2.98412698412698$$ whence we have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then $$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{\frac{188}{63}}.$$

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THIRD ATTEMPT

Using mathlove's approach in his/her answer below, together with mathlove's answer to a closely related question, we have the lower bound $\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375$. As in mathlove's answer, let the quantity $F$ be given by $$F = \frac{\sigma(p^k)}{2} - \frac{p^{2k}}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p-1)} - \frac{p^k}{3375} := f_2(k).$$ Since $f_2(k)$ is increasing, we get $$F \geq f_2(k) \geq f_2(1) = \frac{p+1}{2} - \frac{p}{3375} \geq \frac{2024}{675} = 2.99\overline{851}$$ whence we have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then $$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{\frac{2024}{675}}.$$

Answer 3

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Suppose to the contrary that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{C}$$ and that $C=2$.

(Note that $$\frac{\sigma(m^2)}{p^k} \neq \frac{m^2 - p^k}{2}$$ since the LHS is odd while the RHS is even, as $m^2 - p^k \equiv 0 \pmod 4$.)

Following in the footsteps of the proof in the OP, we have $$2\sigma(m^2) > p^k m^2 - p^{2k}$$ $$4\sigma(m^2) + 2p^{2k} > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$ $$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 4\bigg)$$ (Note that, since $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$, then $p^k \geq 5$, so that $\sigma(p^k) \geq p^k + 1 \geq 6$. This shows that the factor $\bigg(\sigma(p^k) - 4\bigg)$ is always positive.) This implies that $$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 4} \leq \frac{2p^{2k}}{p^k - 3}$$ where we have used the lower bound $\sigma(p^k) \geq p^k + 1$. But the upper bound can be rewritten as $$\frac{2p^{2k}}{p^k - 3} = \frac{2p^{2k} - 18}{p^k - 3} + \frac{18}{p^k - 3} = 2(p^k + 3) + \frac{18}{p^k - 3}.$$ However, as before, we have the lower bound $3p^k \leq \sigma(m^2)$. This implies that $$3p^k < 2(p^k + 3) + \frac{18}{p^k - 3}$$ which means that $$(p^k - 6)(p^k - 3) < 18,$$ an inequality that is satisfied when $p=5$ and $k=1$.

We infer that we need to use a larger lower bound for $\sigma(m^2)/p^k$. We use the one by Dris and Luca (2016): $$7p^k \leq \sigma(m^2) < 2(p^k + 3) + \frac{18}{p^k - 3}$$ $$(5p^k - 6)(p^k - 3) < 18.$$ This is a contradiction as $p^k \geq 5$ implies that $$(5p^k - 6)(p^k - 3) \geq 38.$$

We therefore have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{2}.$$