## 27.4.24

### (Deleted from MathOverflow) - If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a$, where $a < m$.

The following is a complete proof that $m^2 - p^k$ is not a square, if $p^k m^2$ is an odd perfect number with special prime $p$. (I apologize in advance for the somewhat lengthy post, I merely had to combine several posts together into one for ease of reference later on.)

Assume that the estimate $p < m$ holds.  We want to show that the quantity $m^2 - p^k$ is not a square. Suppose to the contrary that $m^2 - p^k = s^2$.  It follows that
$$(m + s)(m - s) = p^k.$$
Since $p$ is prime, we infer that we have the simultaneous equations
$$m + s = p^{k-u}$$
and
$$m - s = p^u,$$
where $u$ is an integer satisfying $0 \leq u \leq (k-1)/2$.  It follows that we have the system
$$2s = p^{k-u} - p^u = p^u (p^{k-2u} - 1)$$
and
$$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1).$$

We claim that $\gcd(s,p)=1$.

Proof: Suppose otherwise. Then $\gcd(s,p) > 1$. By the definition of GCD, we have both $\gcd(s,p) \mid p$ and $\gcd(s,p) \mid s$. Hence, either $\gcd(s,p) = 1$ (which contradicts our assumption) or $\gcd(s,p) = p$, since the only possible factors of the prime $p$ are $1$ and itself.  We infer that $p \mid s$.  This implies that $p \mid s^2 = m^2 - p^k$, from which we conclude that $p \mid m$. But this contradicts $\gcd(p,m)=1$.

Since $p$ is the special prime, then $p \equiv 1 \pmod 4$ implies that $\gcd(2,p)=1$.  Consequently, from the two simultaneous equations
$$2s = p^{k-u} - p^u = p^u (p^{k-2u} - 1)$$
and
$$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1)$$
we obtain that $u = 0$.  This implies that
$$2s = p^k - 1$$
and
$$2m = p^k + 1$$
which is equivalent to
$$2s + 1 = p^k$$
and
$$2m - 1 = p^k$$
or, expressed differently, as
$$s = \frac{p^k - 1}{2}$$
and
$$m = \frac{p^k + 1}{2}.$$

Notice that the resulting equation $m = (p^k + 1)/2$ implies that $m < p^k$.  Under the assumption $p < m$, then we obtain $k > 1$.  Since $k \equiv 1 \pmod 4$, then we know that $k \geq 5$.  We can now use a proof by anonymous MSE user FredH to show that $m^2 - p^k$ is not a square (under the assumption $p < m$), as follows:

Since $N = p^k m^2$ is (odd) perfect, then we have the defining equation
$$\sigma(N) = 2N,$$
from which it follows that
$$\sigma(p^k)\sigma(m^2) = 2p^k m^2,$$
since the divisor sum $\sigma$ is a multiplicative function.

We know that $\sigma(p^k) = (p^{k+1} - 1)/(p - 1)$.  Since we have shown that $m = (p^k + 1)/2$, then we have the equation
$$2(p^{k+1} - 1)\sigma(m^2) = p^k (p - 1)(p^k + 1)^2. \hspace{0.76in} (*)$$

FredH considered the $GCD$ of $p^{k+1} - 1$ with the right-hand side of Equation $(*)$:
$$p^{k+1} - 1 = \gcd\left(p^{k+1} - 1, p^k (p - 1)(p^k + 1)^2\right) \leq (p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2$$
where FredH used the fact that $\left(p^{k+1} - 1\right) \mid RHS$ and the property that
$$\gcd(x,yz) \leq \gcd(x,y)\gcd(x,z).$$
But FredH also noticed that $p^{k+1} - 1 = p(p^k + 1) - (p + 1)$, whence FredH did also find
$$\gcd(p^{k+1} - 1, p^k + 1) = \gcd(p + 1, p^k + 1),$$
which is $p + 1$ because $k$ is odd. Thus,
$$(p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2 = (p - 1)(p + 1)^2.$$

Hence, the inequality
$$p^{k+1} - 1 \leq (p - 1)(p + 1)^2$$
holds.

Since $k \geq 5$, we obtain
$$p^5 < p^{k+1} - 1 \leq (p - 1)(p + 1)^2 < p^4,$$

Hence, we now have the implication
$$p < m \Rightarrow m^2 - p^k \text{ is not a square.}$$

In other words, we have the contrapositive
$$m^2 - p^k \text{ is a square} \Rightarrow m < p.$$
Now, suppose to the contrary that $m^2 - p^k$ is a square. This implies that $m < p$. Since $p^k < m^2$, we then have the implication $m < p \Rightarrow k = 1$.  Therefore, $k = 1$.  But we know (from the considerations above) that
$$m^2 - p^k \text{ is a square } \iff m = (p^k + 1)/2.$$

Since $k = 1$, we infer that $m = (p + 1)/2$, or in other words, $p = 2m - 1$.  From Acquaah and Konyagin's results, we have the unconditional estimate $p < m \sqrt{3}$.  This implies that $2m - 1 = p < m \sqrt{3}$, from which we infer that
$$m(2 - \sqrt{3}) < 1$$
which contradicts the fact that $\omega(m) > 4$. (In fact, we do know that $m > {10}^{375}$, by using Ochem and Rao's lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number $N$, together with $p^k < m^2$.)

We conclude that $m^2 - p^k$ is not a square. (Note that, in the case of even perfect numbers $M = 2^{q-1}(2^q - 1)$, the quantity
$$2^{q-1} - (2^q - 1) = 1 - 2^{q-1}$$
is likewise not a square.)

---

Now, here goes the part where I am a bit unsure about its logical tightness:

Does the following statement necessarily hold? "Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares."

If so, WLOG we may assume that
$$(m - a)^2 < m^2 - p^k < (m - a + 1)^2$$
for some positive integer $a$. We may likewise assume that $m > a$.

If I am not mistaken, these assumptions will then yield a proof for the inequality
$$m < p^k < 2am$$
except for the (*problematic*) case $a=1$, where we can only derive
$$p^k < 2m - 1.$$

Either way, I think the inequalities can be summarized as
$$p^k < 2am$$
for some positive integer $a$.

Here then, is my question for this post (which, I hope, is acceptable to the MathOverflow community):

Does this "proof" for the inequality $p^k < 2am$ hold water? If not, where does the argument break?

POSTED ANSWER (April 27, 2024 - Before 18:00 PM Manila time)

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

This answer says something about the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers (i.e. the prediction $k=1$).

---

Assume that $p^k < 2m - 1$. We will show that $k \neq 1$.

Then
$$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} < \frac{2pm - p - 1}{2(p - 1)}$$
which is less than $m$ if $p > 2m - 1$. By assumption, we have
$$p \leq p^k < 2m - 1$$
which means that
$$m < \frac{\sigma(p^k)}{2}.$$
We thus obtain
$$p^k < 2m - 1 < \sigma(p^k) - 1 < \sigma(p^k)$$
which implies that $k \neq 1$ holds, under the assumption $p^k < 2m - 1$.

---

Assume that $p^k > 2m - 1$. We will show that $k \neq 1$.

Then
$$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} > \frac{2pm - p - 1}{2(p - 1)}$$
which is more than $m$ if $p < 2m - 1$.

Note that
$$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} > m$$
does hold, under the assumption $p^k > 2m - 1$.

By assumption, we have
$$2m - 1 < p^k$$
which means that
$$p < 2m - 1 < p^k,$$
or $k \neq 1$ holds, under the assumption $p^k > 2m - 1$.

## 5.6.23

### A Proof That There Are No Odd Perfect Numbers

$$I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}$$
which is equivalent to
$$I(m^2) \leq \left(I(m)\right)^{\ln(13/9)/\ln(4/3)},$$
from which we get
$$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\ln(13/12)/\ln(4/3)}.$$
This last inequality implies that
$$1 > \frac{I(m)}{I(m^2)} \geq \left(I(m)\right)^{\ln(4/3)/\ln(13/12)}$$
$$> \Bigg(\left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}\Bigg)^{\ln(4/3)/\ln(13/12)}$$
$$= \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}.$$
But in general, we know that $I(m^2) > 8/5$, so that
$$1 > \frac{I(m)}{I(m^2)} > \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}$$
$$> \left(\frac{8}{5}\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))} > 3$$

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

We will stop here for the time being.

## 30.5.23

### Last Presenter (From Taguig City) During the 2023 MSP-NCR Annual Convention, Held Via Zoom Last May 27 (Saturday)

I was the last presenter during the 2023 MSP-NCR Annual Convention, held via Zoom last May 27 (Saturday), and hosted by the University of Santo Tomas - Manila, Philippines.

I was at home when I presented.

Please do check out the Beamer slides in ResearchGate, if you are interested to know more about what ideas were communicated during my talk.

For the record, no questions were asked after I was done.  (IIRC, some of the participants were graduate students from Polytechnic University of the Philippines and one of my former professors from De La Salle University.)

I finished presenting in less than fifteen minutes.

## 26.5.23

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

Continuing from this earlier blog post: (In particular, if $m$ is almost perfect, then $m$ must be a square (since $m$ is odd). Consequently, if the equation $\sigma(p^k)=2m$ holds (which is true if and only if $G=H=I$), then $\sigma(p^k)/2$ is also a square. This implies that $k=1$, since $$p^k + 1 \leq \sigma(p^k)=2m<2p$$ holds under the assumption $m<p$.)

Summarizing our results so far, we have the chain of implications:
$$m < p \Rightarrow \Bigg((m < p^k) \land (k = 1) \land (D(m) = 1)\Bigg).$$
But this blog post proves that the condition $D(m) = 1$ is equivalent to $m < p$.

Since $\sigma(m^2)/p^k$ is also a square if $\sigma(p^k)/2$ is a square, then under the assumption that $m < p$, we obtain $\sigma(m^2)/p^k = \sigma(p^k)/2 = m$ is a square. (Note that the difference $m^2 - p^k$ is also a square.)

$$\sigma(p^k) = p^k + 1$$
since $k=1$ follows from the assumption $m < p$.  Since $m^2 - p^k$ is a square if and only if
$$m^2 - p^k = (m - 1)^2,$$
it then follows from the assumption $m < p$ that $2m - 1 = p^k$, whence we obtain
$$2m = p^k + 1 = \sigma(p^k).$$

But note that
$$m < p \Rightarrow \Bigg((k = 1) \land (m < p^k) \land (\sigma(p^k) = 2m)\Bigg).$$

Since $m^2 - p^k$ is not a square, then we obtain
$$m^2 - p^k = 2^r t$$
where $r \geq 2$ and $2^r \neq t$.

Now, consider the following scenario in this MathOverflow question, under which the inequality $m < p^k$ is sure to hold:

$$\text{Case (A): } m > \max(2^r, t)$$

Case (A) implies that

$$\left(m - \max(2^r, t)\right)\left(m + \min(2^r, t)\right) > 0$$
$$p^k = m^2 - 2^r t = m^2 - \min(2^r, t)\max(2^r, t)$$
$$> m\left(\max(2^r, t) - \min(2^r, t)\right) = m\left|2^r - t\right|,$$

from which we get

$$p^k > m\left|2^r - t\right|.$$

Note that this implies that $m < p^k$ since $\left|2^r - t\right| \geq 1$ must necessarily hold.

Next, notice that the other (remaining) cases are

$$\text{Case (B): } \min(2^r, t) < m < \max(2^r, t)$$

$$\text{Case (C): } m < \min(2^r, t).$$

The disposal of Case (C) is very easy and is left as an exercise for the interested reader.

We then have the biconditional

$$\left(m < \max(2^r, t)\right) \iff \left(p^k < m\left|2^r - t\right|\right).$$

$$p^k < m \Rightarrow \text{ Case (B) } \Rightarrow \left|2^r - t\right| \neq 1.$$

By the contrapositive, we obtain
$$\left|2^r - t\right| = 1 \Rightarrow m < p^k.$$

However, under the assumption $m < p$, we know that $\sigma(p^k) = 2m$.  It follows that $p^k < 2m$.

Consequently, under Case (A), we derive
$$m \leq m\left|2^r - t\right| < p^k < 2m$$
from which it follows that
$$1 \leq \left|2^r - t\right| < 2$$
which forces $\left|2^r - t\right| = 1$.

We conclude that
$$m < p^k \iff \left|2^r - t\right| = 1.$$

Conclusion:  We now claim that the inequality $p < m$ indeed holds. Indeed, assume to the contrary that $m < p$ holds. Then we obtain both $k = 1$ and $\sigma(p^k)=2m$ as valid inferences from this assumption.  In particular,
$$2m=\sigma(p^k)=p+1$$
$$2m - 1 = p$$

But then again, as a by-product of Acquaah and Konyagin's results published in the International Journal of Number Theory in 2012, we have the upper bound
$$p < m\sqrt{3}.$$

Consequently,
$$2m - 1 = p < m\sqrt{3}$$
$$m(2 - \sqrt{3}) < 1.$$

This contradicts $m \geq 4$.

Hence, we infer that the estimate $p < m$ must be true.

## 25.5.23

### Check out ResearchGate.

Check out ResearchGate.

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

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

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
$$1 > \frac{I(m)}{I(m^2)} > \left(I(m^2)\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}}> \left(\frac{8}{5}\right)^{\frac{\left(\ln(4/3)\right)^2}{\ln(13/9)\cdot\ln(13/12)}} > 3$$

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

We will stop here for the time being.

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

In what follows, we will denote the classical sum of divisors of the positive integer $x$ by
$$\sigma(x)=\sigma_1(x).$$
We will also denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

Lemma 1: If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then $I(m^2) < \zeta(2)$ if and only if $p = 5$ and $k \neq 1$.

Proof:

Suppose that $I(m^2) < \zeta(2)$.  Since $2(p - 1)/p < I(m^2)$, then it follows that
$$\frac{2(p - 1)}{p} < \zeta(2).$$

Hence, we derive
$$p < \frac{12}{12 - {\pi}^2} < 6,$$
from which we infer that $p = 5$ (since $p$ is a prime satisfying $p \equiv 1 \pmod 4$).

Now, since we already have $p = 5$, then
$$I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(5^k)} = \frac{2\cdot{5^k}\cdot(5 - 1)}{5^{k+1} - 1} < \zeta(2).$$
Solving for $k$, we obtain
$$k > 2\log_{5}{\pi} - \log_{5}\left(5{\pi}^2 - 48\right) \approx 1.23 > 1.$$

For the other direction, assume that $p = 5$ and $k \neq 1$.

Suppose to the contrary that
$$I(m^2) > \zeta(2) = \frac{{\pi}^2}{6} \approx 1.6449.$$

Since $k \equiv 1 \pmod 4$, it follows that $k \geq 5$.  Hence, under the assumption that $p=5$, we have $k \geq 5$.

It follows that
$$I(m^2)=\frac{2}{I(p^k)}=\frac{2}{I(5^k)} \leq \frac{2}{I(5^5)} = \frac{3125}{1953}$$
where
$$\frac{3125}{1953} \approx 1.6001.$$
However, this contradicts our assumption that $I(m^2) > \zeta(2)$.

This concludes the proof of Lemma 1.

QED.

Remark 2:  Note that we then have (either of) the improved bounds

$$\Bigg(I(m^2) > \zeta(2) \iff \left((p \geq 13) \lor (k = 1)\right)\Bigg) \oplus \Bigg(I(m^2) < \zeta(2) \iff \left((p = 5) \land (k \neq 1)\right)\Bigg).$$

Note that we have the implications
$$k = 1 \Rightarrow I(m^2)=\frac{2}{I(p^k)}=\frac{2}{I(p)}=\frac{2p}{(p+1)} \geq \frac{5}{3} = 1.\overline{666}$$
and
$$p \geq 13 \Rightarrow I(m^2) > \frac{2(p-1)}{p} \geq \frac{24}{13} \approx 1.\overline{846153}.$$

We thus have
$$k = 1 \Rightarrow I(m^2) > \zeta(2)$$
and
$$p \geq 13 \Rightarrow I(m^2) > \zeta(2)$$

Hence, we have the following proposition.

Theorem 3.  Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Then the following implications hold:
• If $k=1$, then $I(m^2) > \zeta(2)$.
• If $p=5$, then we have the two cases:
• If $k = 1$, then $I(m^2) = 5/3 > \zeta(2)$.
• If $k \neq 1$, then $I(m^2) < \zeta(2)$.
(This blog post is currently a Work In Progress/Under Construction.)

## 7.5.23

### Some New Results on Odd Perfect Numbers - A Summary

Let $N = p^k m^2$ be a hypothetical odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$.  Note that, since $\gcd(p, m) = 1$ and $p$ is (the special) prime, then $p^k \neq m$.  (This also follows from the fact that prime powers are deficient, contradicting $N$ is perfect.)  By trichotomy, either $(p^k < m) \oplus (m < p^k)$ is true, where $\oplus$ denotes exclusive-OR.  (That is, $A \oplus B$ is true if and only if exactly one of $A$ or $B$ holds.)

We will denote the classical sum of divisors of the positive integer $z$ by $\sigma(z)=\sigma_1(z)$, and the abundancy index of $z$ by $I(z)=\sigma(z)/z$. Furthermore, we will denote the deficiency of $z$ by $D(z)=2z-\sigma(z)$, and the aliquot sum of $z$ by $s(z)=\sigma(z)-z$.

Here is a summary of some new results on odd perfect numbers, which were realized by the author on May 5, 2023:

• If $p < m$, then the quantity $m^2 - p^k$ is not a square. (Kindly note the contrapositive.)
• If $m < p$, then the following statements hold:
• Descartes's conjecture holds (i.e. $k = 1$).
• Dris conjecture (i.e. $p^k < m$) is false.
• The quantity $m^2 - p^k$ is a square.
• The square root of the non-Euler part $m^2$ is almost perfect.
• Define the following GCDs:
• $G=\gcd\left(\sigma(p^k),\sigma(m^2)\right)$
• $H=\gcd\left(m,\sigma(m^2)\right)$
• $I=\gcd\left(m^2,\sigma(m^2)\right)$
where we know that $I=\sigma(m^2)/p^k = {m^2}/(\sigma(p^k)/2)$.
• Dris proved in February 10, 2022 that the chain of divisibility conditions
• $G \mid H \mid I$
holds. Later on, he realized that we do in fact have

• $G=\gcd\left(\sigma(p^k)/2,\sigma(m^2)/p^k\right)=\sigma(p^k)/2=\gcd(G,I)$
• $H=\gcd\left(m,\sigma(m^2)/p^k\right)=m=\gcd(H,I)$.
• In particular, the divisibility constraint $\sigma(p^k)/2 \mid m$ holds.
• The divisibility condition $\sigma(p^k) \mid 2m$ is equivalent to $m \mid \sigma(m^2)$.
• ( To be continued$\ldots$ )