## 22.5.23

### Some New Results on Odd Perfect Numbers - Part III

Let $N = p^k m^2$ be an odd perfect number with special prime $p$. Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x) = \sigma_1(x)$ is the classical sum of divisors of $x$. Additionally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$. I claim that the estimate $m < p$ holds if and only if $D(m) = 1$.

Recall the following material facts:
First, we prove the following claims:

LEMMA 1. $I(m^2) \neq 2m/(m+1)$

Proof: Suppose to the contrary that $I(m^2) = 2m/(m+1)$. Since
$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1},$$
then we have $$I(m^2) = \frac{2m}{m+1} \leq \frac{2p}{p + 1}.$$
This then implies that $m < p$ (since $\gcd(p,m) = 1$). But then $m < p$ implies $k = 1$.

Consequently, we obtain
$$\frac{2m}{m+1} = I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(p)} = \frac{2p}{p+1},$$
which then gives $m = p$. This contradicts $\gcd(p,m)=1$.

LEMMA 2. $I(m^2) < 2m/(m+1)$ if and only if $p < m$.

Proof. By Lemma 1, we know that $I(m^2) \neq 2m/(m+1)$. It is obvious that $2m/(m+1) < I(m^2)$ implies $m < p$, since
$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$
We want to show that $I(m^2) < 2m/(m+1)$ implies $p < m$. To this end, suppose that the inequality $I(m^2) < 2m/(m+1)$ is true. Assume that the estimate $m < p$ holds. This implies that $k = 1$, whereupon we obtain
$$\frac{2p}{p+1} = \frac{2}{I(p)} = \frac{2}{I(p^k)} = I(m^2) < \frac{2m}{m+1}.$$
Thus, we have $2p/(p+1) < 2m/(m+1)$, which implies that $p < m$, contradicting our earlier assumption that $m < p$. It follows that $I(m^2) < 2m/(m+1) \iff p < m$, and we are done.

It is known that $D(m) = 1 \Rightarrow m < p$, since $m > 1$ is deficient, whence by a criterion for deficient numbers by Dris:
$$D(m) = 1 \iff \frac{2m}{m+1} = \frac{2m}{m+D(m)} < I(m) < \frac{2m+D(m)}{m+D(m)} = \frac{2m+1}{m+1}$$
(Use the $LHS$ inequality for $I(m)$.)
$$\bigg(\frac{2m}{m+1} < I(m) < I(m^2) \leq \frac{2p}{p+1}\bigg) \Rightarrow m < p$$
However, it is also known (by Dris [JIS (09/2012)]) that $m < p$ implies $k = 1$ (since $p^k < m^2$). Consequently, $$\left(D(m) = 1\right) \Rightarrow \left(m < p\right) \Rightarrow \left(k = 1\right).$$ We want to determine whether the implication $m < p \Rightarrow D(m) = 1$ is true.

Assume that $m < p$. Suppose to the contrary that $D(m) \neq 1$. (For brevity, in what follows we let $r$ denote the fraction $2m/(m+1)$.)

We consider three cases:

(1) $$\left(r < I(m) < I(m^2)\right) \Rightarrow \left((D(m) = 1) \land (m < p)\right) \Rightarrow m \text{ is almost perfect.}$$
(2) $$\left(I(m) < r < I(m^2)\right) \Rightarrow \left((D(m) \neq 1) \land (m < p)\right) \Rightarrow m \text{ is not almost perfect.}$$
(3) $$\left(I(m) < I(m^2) < r\right) \Rightarrow \left((D(m) \neq 1) \land (p < m)\right) \Rightarrow m \text{ is not almost perfect.}$$ Note that Case (3) is ruled out under the assumption $m < p$.

We consider the remaining problematic case:

Case (2): Under this case, $I(m) < r < I(m^2)$. Since $m < p$ is true under Case (2), then we infer that (mainly because $k = 1$), $$I(m^2) \leq \left(I(m)\right)^{\ln(13/9)/\ln(4/3)}$$ $$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\ln(13/12)/\ln(4/3)}$$ where $\ln(13/12)/\ln(4/3)$ is approximately $0.27823321415675831$.

Consequently, we obtain $$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\ln(13/12)/\ln(4/3)} < \left(2\right)^{\ln(13/12)/\ln(4/3)}$$ where $$\left(2\right)^{\ln(13/12)/\ln(4/3)} \approx 1.212708839706.$$ But since $k = 1$, then $p$ is the special prime implies that $p \geq 5$, so that $$I(p^k) = I(p) = \frac{p+1}{p} = 1+\frac{1}{p} \leq \frac{6}{5}$$ $$I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(p)} = \frac{2p}{p+1} \geq \frac{5}{3}$$ Now consider the quantity $Q = \left(I(m^2) - r\right)\left(I(m) - r\right)$. This quantity is negative under Case (2). Thus, we have $$I(m^2)I(m) + r^2 < r\left(I(m^2) + I(m)\right).$$

Dividing both sides of the last inequality by $\left(I(m^2)\right)^2$, we get $$\frac{I(m)}{I(m^2)} + \left(\frac{r}{I(m^2)}\right)^2 < \left(\frac{r}{I(m^2)}\right)\cdot\left(\frac{I(m)}{I(m^2)} + 1\right).$$

However $$\frac{I(m)}{I(m^2)} \geq \left(I(m)\right)^{\ln(4/3)/\ln(13/12)} \geq \Bigg(\left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}\Bigg)^{\ln(4/3)/\ln(13/12)}.$$
But recall from basic laws of exponents that
$$(A^{\alpha})^{\beta} = A^{\alpha\beta}.$$
Therefore, we obtain
$$\Bigg(\left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}\Bigg)^{\ln(4/3)/\ln(13/12)} = \left(I(m^2)\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}}$$
where
$$\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)} \approx 2.811777355162111571454751.$$
But our assumption $m < p$ implies that $k = 1$, which in turn is equivalent to $I(m^2)=2p/(p+1)$.

Furthermore, we compute
$$\left(\frac{r}{I(m^2)}\right)^2 = \left(\frac{(2m)/(m+1)}{(2p)/(p+1)}\right)^2 = \left(\frac{m(p+1)}{p(m + 1)}\right)^2.$$
Since $m < p$ implies $k = 1$, then the odd perfect number $\overline{N}$ takes the form $\overline{N} = pm^2$. This implies that $m < p < m^2$, whence we obtain
$$p^3 > pm^2 > {10}^{1500}$$
and
$$m^4 > pm^2 > {10}^{1500},$$
by utilizing a result of Ochem and Rao on a lower bound for the magnitude of an odd perfect number.

Thus, the estimates $p > {10}^{500}$ and $m > {10}^{375}$ both hold. Then notice that
$$\frac{r}{I(m^2)}=\frac{m(p+1)}{p(m + 1)}=\left(\frac{m}{m+1}\right)\cdot\left(\frac{p+1}{p}\right)$$
where $(p + 1)/p > 1$ (since $p$ is very large), while
$$\frac{m}{m+1}=\frac{(m+1)-1}{m+1}=1-\frac{1}{m+1}>1-\frac{1}{m}>1-{10}^{-375} > 0.999.$$

Consequently,
$$\frac{I(m)}{I(m^2)} + \left(\frac{r}{I(m^2)}\right)^2 > \left(\frac{5}{3}\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}} + (0.999)^2$$
$$> 4.205 + 0.998 > 5.203$$

But we also know that
$$\left(\frac{r}{I(m^2)}\right)\cdot\left(\frac{I(m)}{I(m^2)} + 1\right)=\left(\frac{m}{m+1}\right)\cdot\left(\frac{p+1}{p}\right)\cdot\left(\frac{I(m)}{I(m^2)} + 1\right),$$
where
$$\frac{m}{m+1} < 1$$
$$\frac{p+1}{p}=1+\frac{1}{p} < 1 + {10}^{-1500} < 1.001,$$
and of course,
$$\frac{I(m)}{I(m^2)} + 1 < 2$$
since $I(m) < I(m^2)$ holds, in general.

Consequently,
$$\left(\frac{r}{I(m^2)}\right)\cdot\left(\frac{I(m)}{I(m^2)} + 1\right) < {1}\cdot\left(1 + {10}^{-1500}\right)\cdot{2} < 2\cdot(1.001) = 2.002.$$

We arrive at the contradiction
$$5.203 < \frac{I(m)}{I(m^2)} + \left(\frac{r}{I(m^2)}\right)^2 < \left(\frac{r}{I(m^2)}\right)\cdot\left(\frac{I(m)}{I(m^2)} + 1\right) < 2.002.$$

We therefore conclude that Case (2) does not hold.

Consequently, only Case (1) remains: Case (1): $$r < I(m) < I(m^2) \Rightarrow \left((D(m) = 1) \land (m < p)\right) \Rightarrow m \text{ is almost perfect}.$$ In particular, $m$ is almost perfect if and only if $m < p$. (In other words, $m < p$ if and only if $D(m) = 1$.)

PostScript:  It seems that we can remove the reliance of the proof on the assumption $k = 1$ (and therefore $m < p$), to obtain
$$I(m) \geq \left(I(m^2)\right)^{\frac{\ln(4/3)}{\ln(13/9)}}$$
which is equivalent to
$$I(m^2) \leq \left(I(m)\right)^{\frac{\ln(13/9)}{\ln(4/3)}},$$
from which we get
$$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\frac{\ln(13/12)}{\ln(4/3)}}.$$
This last inequality implies that
$$1 > \frac{I(m)}{I(m^2)} \geq \left(I(m)\right)^{\frac{\ln(4/3)}{\ln(13/12)}}$$
$$> \Bigg(\left(I(m^2)\right)^{\frac{\ln(4/3)}{\ln(13/9)}}\Bigg)^{\frac{\ln(4/3)}{\ln(13/12)}}$$
$$= \left(I(m^2)\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}}.$$

But in general, we know that $I(m^2) > 8/5$, so that
$$1 > \frac{I(m)}{I(m^2)} > \left(I(m^2)\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}}> \left(\frac{8}{5}\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}} > 3$$
which is a contradiction.

This seems to prove that there are, in fact, no odd perfect numbers.

We will stop here for the time being.