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

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.)

In what follows, 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 $I(z)=\sigma(z)/z$.

We will also denote the *deficiency* of $z$ by $D(z)=2z-\sigma(z)$, and the *aliquot sum* of $z$ by $s(z)=\sigma(z)-z$.

I learned five (5) very important math lessons two days ago:**(1)** From "The Abundancy Index of Divisors of Odd Perfect Numbers" (JIS, 09/2012),
we have the estimate $p^k < m^2$. We infer that the implication $m < p \Rightarrow k = 1$ holds, which is equivalent to $k > 1 \Rightarrow p < m$.

From Acquaah and Konyagin's result in "On Prime Factors of Odd Perfect Numbers", we have that the implication $k = 1 \Rightarrow p < m\sqrt{3}$ is true, which implies (together with the implication $k > 1 \Rightarrow p < m$) that $p < m \sqrt{3}$ holds in general.

**(2)** Assume that the estimate $p < m$ holds. We want to show that the quantity $m^2 - p^k$ is not a square. (We follow the proof from pages $22$ to $23$ of this paper.) 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 $\gcd(s,p)=1$.**

**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 $\gcd(p,m)=1$.**

__Proof:__*special prime*, then $p \equiv 1 \pmod 4$ implies that $\gcd(2,p)=1$. Consequently, from the two simultaneous equations

and

$$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1)$$

we obtain that $u = 0$. This implies that

$$2m = p^k + 1$$

which is equivalent to

$$2m - 1 = p^k$$

or, expressed differently, as

$$p^{k+1} - 1 = \gcd\left(p^{k+1} - 1, p^k (p - 1)(p^k + 1)^2\right) \leq (p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2$$

$$\gcd(x,yz) \leq \gcd(x,y)\gcd(x,z).$$

Now, suppose to the contrary that $m^2 - p^k$ is a square. This implies that $m < p$. From **(1)**, we 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, $p = 2m - 1$. Again from **(1)**, 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 $m$ is odd. (It suffices to take $m > 6$, since $m$ must be composite and $\omega(m) \geq 2$. 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 $p^k < m^2$ from **(1)**.)

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

$$2^{p - 1} - (2^p - 1) = 1 - 2^{p - 1}$$

is likewise **not a square**.)**(3)** Notice that, from **(1)**, the implication

$$m < p \Rightarrow k = 1$$

means that we have the conditional statement

$$m < p^k \Rightarrow \left((m < p) \iff (k = 1)\right).$$**(4)** From **(2)**, we have concluded that $m^2 - p^k$ is not a square. In particular, $m^2 - p^k$ is not equal to $(m - 1)^2$. It follows that the conjunction

$$\left(m^2 - p^k \leq (m - 1)^2\right) \land \left((m - 1)^2 \leq m^2 - p^k\right)$$

is false. Thus, we have that exactly one of

$$\left(m^2 - p^k \leq (m - 1)^2\right) \oplus \left((m - 1)^2 \leq m^2 - p^k\right)$$

holds. Since $m^2 - p^k$ is not a square, and hence $m^2 - p^k \neq (m - 1)^2$, then this means that exactly one of the following logical statements is true:

$$\left(2m - 1 < p^k\right) \oplus \left(p^k < 2m - 1\right).$$

If $2m - 1 < p^k$ holds, then $m < p^k$ is true.

If $p^k < 2m - 1$ holds, then $p^k < 2m$ is true.

In particular, note that if the equation $\sigma(p^k)=2m$ is true, then the chain of inequalities

$$m < p^k < 2m$$

holds. (Moreover, the divisibility constraint $\sigma(p^k) \mid 2m$ holds, for reasons that will soon become crystal clear. This is equivalent to the condition $m \mid \sigma(m^2)$.)

**(5)** Recall that the main idea behind the findings in **(4)** is the following logical statement:

$$(2m - 1 < p^k) \oplus (p^k < 2m - 1).$$

We claim that, in fact, the inequality $p^k < 2m - 1$ holds.

Define the GCDs

$$G=\gcd\left(\sigma(p^k),\sigma(m^2)\right)=\gcd\left(\sigma(p^k)/2,\sigma(m^2)/p^k\right),$$

$$H=\gcd\left(m,\sigma(m^2)\right)=\gcd\left(m,\sigma(m^2)/p^k\right),$$

where

$$I=\frac{\sigma(m^2)}{p^k}=\frac{m^2}{\sigma(p^k)/2},$$

as proved in [Dris (NNTDM, August - 2017)]. Note that

$$G \times I = H^2.$$

(See [Dris (OMM, October - 2022)].) We give a quick proof of this fact here:

Since

$$G=\gcd\left(\sigma(p^k)/2,\sigma(m^2)/p^k\right)=\gcd\left(\sigma(p^k)/2,I\right),$$

$$H=\gcd\left(m,\sigma(m^2)/p^k\right)=\gcd(m,I),$$

and

$$I=\gcd\left(m^2,\sigma(m^2)/p^k\right)=\gcd(m^2,I),$$

then

$$G\times{I}=\gcd(\sigma(p^k)/2,I)\times{I}= \gcd\left(I\cdot{\sigma(p^k)/2},I^2\right)=(*)$$

But we know that

$$I\cdot{\sigma(p^k)/2}=m^2.$$

Hence,

$$(*)=\gcd(m^2,I^2)=\left(\gcd(m,I)\right)^2=H^2,$$

so we can conclude that

$$G \times I = H^2. \hspace{1.50in} (**)$$

Compare Equation (**) with

$$\left(\frac{\sigma(p^k)}{2}\right)\cdot\left(\frac{\sigma(m^2)}{p^k}\right) = m^2.$$

Since $I=\sigma(m^2)/p^k$, and the cofactor of $\sigma(m^2)/p^k$ with respect to $m^2$ is $\sigma(p^k)/2$, then we obtain (at once) that

$$G=\gcd(\sigma(p^k),\sigma(m^2))=\sigma(p^k)/2$$

and

$$H=\gcd(m,\sigma(m^2))=m.$$

In particular, $m \mid \sigma(m^2)$, which is equivalent to $\sigma(p^k) \mid 2m$, mainly because of the equation

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

and the fact that $\gcd(p^k,\sigma(p^k)/2)=1$.

The divisibility condition $\sigma(p^k) \mid 2m$ then implies that

$$p^k + 1 \leq \sigma(p^k) < 2m,$$

so we obtain

$$p^k < 2m - 1.$$

Compare with what we obtained in **(4)**.

(Observe that $G$ is a square if and only if $I$ is a square.)

Lastly, note that the equation

$$\gcd(m,\sigma(m^2))=H=m=\sqrt{G \times I}=\sqrt{G} \times \sqrt{I} \hspace{0.76in} (***)$$

$$=\sqrt{\sigma(p^k)/2} \times \sqrt{\sigma(m^2)/p^k}$$

appears to force $G$ and $I$ to be squares, because the $LHS$ of Equation $(***)$ is a positive integer, and since any natural number $O$ can be written as the product of its square part $O_\rho$ and its squarefree part $O_\mu$.

In other words, if we suppose to the contrary that $I$ is not a square, then either $I$ is squarefree or $I$ is neither squarefree nor a square; in both cases, it then seems easy to show that the $RHS$ of Equation $(***)$ is irrational. Unfortunately, this problem turns out to be difficult. We shall explain why as we go along.

The following is a "proof" that $G=\sigma(p^k)/2$ is not squarefree. Suppose to the contrary that $\sigma(p^k)/2$ is squarefree. (This is currently a work in progress.)

First, we claim that $I=\sigma(m^2)/p^k={m^2}/{\sigma(p^k)/2}$ is not squarefree. Suppose that $I_{\rho} = 1$ holds. Then $I$ is squarefree. It follows that, since the identity $G \times I = H^2$ holds, then we obtain $I \mid H$ since $I$ is squarefree, by assumption. But the divisibility condition $\gcd(m,I)= H \mid I = \gcd(m^2,I)$ also holds via the GCD property

$$(u \mid v) \Rightarrow \left(\gcd(u,w) \mid \gcd(v,w)\right).$$

(Notice that we then have the chain of divisibility conditions $G \mid H \mid I$.) Since $H$ and $I$ are both positive, then $H=I$ (under the assumption that $I$ is squarefree.) This implies that $G=H=I$, so that

$$\sigma(p^k)/2 = G = H = m.$$

But $m$ must contain a square factor, as proved by Steuerwald \cite{Steuerwald}. Thus, $m$ is not squarefree, which contradicts $\sigma(p^k)/2 = m$. Consequently, we conclude that $I$ is not squarefree. (Notice that, along the way, we did also get a proof for the implication

$$I \text{ is squarefree} \Rightarrow G \text{ is not squarefree}.$$

This implication essentially gives that the conjunction

$$\left(I \text{ is squarefree}\right) \land \left(G \text{ is squarefree}\right)$$

is false.)

Hence, under the assumption that $I$ is not a square, since we have shown that $I_{\rho} = 1$ leads to a contradiction, then the remaining case is when $I$ is neither squarefree nor a square. This means that the square part $I_{\rho}$ and the squarefree part $I_{\mu}$ of $I$ are both greater than unity.

Since $G$ is not a square (because $I$ is not a square by assumption and

$$I \text{ is a square } \iff G \text{ is a square }$$

holds), we then consider whether $G=\sigma(p^k)/2$ is squarefree. (That is, whether the square part $G_{\rho}$ equals one.) Under this scenario, it follows that $G_{\mu} > 1$.

Otherwise, if $G=\sigma(p^k)/2$ is ** not squarefree**, then since $G$ is not a square by assumption, then it follows that $G_{\rho} > 1$ and $G_{\mu} > 1$ (whence $G$ is neither a square nor squarefree).

In all cases where $I$ is not a square (and therefore $G$ is not a square), note that we (will) arrive at a contradiction, since

$$\text{LHS}=m=H=\sqrt{G\cdot{I}}=\sqrt{G}\cdot\sqrt{I}=\sqrt{{G_{\mu}}\cdot{G_{\rho}}}\cdot\sqrt{{I_{\mu}}\cdot{I_{\rho}}}=\text{RHS} \hspace{0.55in} (****)$$

where the $LHS$ is an integer while the $RHS$ is irrational, under the following cases:

- Case (A): $I_{\rho} > 1$ and $I_{\mu} > 1 \Rightarrow I$ is neither squarefree nor a square.
- Case (B): $G_{\rho} = 1$ and $G_{\mu} > 1 \Rightarrow G$ is squarefree.
- Case (C): $G_{\rho} > 1$ and $G_{\mu} > 1 \Rightarrow G$ is neither squarefree nor a square.

Note that $\gcd(G,I)=G=\sigma(p^k)/2 \neq 1$ since the chain of divisibility conditions $G \mid H \mid I$ holds.

Since we still do not know the squarefree status for $G=\sigma(p^k)/2$, it follows that we cannot easily determine whether $G=\sigma(p^k)/2$ and $I=\sigma(m^2)/p^k$ are squares.

If $G=\sigma(p^k)/2$ is a square, then $k=1$ follows from work of Broughan, Delbourgo, and Zhou (Lemma 8, page 7). Also, $k=1$ would have followed from the truth of the condition $1=\gcd(G,I)=G=\sigma(p^k)/2$. Unfortunately,

$$\sigma(p^k)/2 \geq (p^k + 1)/2 \geq 3,$$

so that $\gcd(G,I) \neq 1$.

The author is well-aware that he can choose $G=\sigma(p^k)/2$ such that $G$ is squarefree; however, this assumption may not really be necessary. (For the ultimate reasons, see the discussion in the next section.)

=================

**Future Research**

Suppose that $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

**Slowak** proved in the year 1999 that $N$ must have the form

$$N = {\frac{p^k \sigma(p^k)}{2}}\cdot{d}$$

for some $d > 1$.

**Dris** showed further in 2017 that $d$ must have the form

$$d = \frac{D(m^2)}{s(p^k)}$$

where $d$ must necessarily be a divisor of $N$.

We leave the following problem for other researchers to solve.

**Conjecture:** If $N = p^k m^2$ is an odd perfect number given in Eulerian form, then unconditionally we have $\sigma(p^k)/2 = m$.

**Underlying heuristic: **If $N = p^k m^2$ is an odd perfect number given in Eulerian form, then suppose to the contrary that $m < p$. It follows that the following statements hold:

- Descartes's conjecture [i.e. $k = 1$] holds.
- Dris conjecture that $p^k < m$ is false.
- The square root of the non-Euler part $m^2$ of the odd perfect number $N$ is almost perfect.

**Proof.**Only the last assertion is not too obvious. Here is a rough sketch of our argument:

Since $m$ is deficient (being a proper factor of the (odd) perfect number $N = p^k m^2$), then using Dris's criterion for deficient numbers, we obtain

$$\frac{2m}{m + D(m)} \leq I(m) < \frac{2m + D(m)}{m + D(m)}.$$

In particular, $D(m)=1$ if and only if

$$\frac{2m}{m + 1} \leq I(m) < \frac{2m + 1}{m + 1},$$

where equality does not hold because $m \neq 1$.

Note that $I(m) < I(m^2)$, so that

$$\frac{2m}{m + D(m)} < I(m) < I(m^2)$$

$$I(m^2)=\frac{2}{I(p^k)}=\frac{2}{I(p)}=\frac{2p}{p + 1}.$$

from which it follows that

$$\frac{2m}{m + D(m)} < \frac{2p}{p + 1}$$

$$\iff \bigg(2m(p + 1) = 2pm + 2m < 2pm + 2pD(m) = 2p(m + D(m)\bigg) \iff (m < pD(m)).$$

Now, consider the location of the rational number $2m/(m+1)$ relative to $I(m^2)$.

We claim that $I(m^2) \neq 2m/(m+1)$. Suppose to the contrary that $I(m^2) = 2m/(m+1)$. Then we obtain

$$\frac{2m}{m+1}=I(m^2)=\frac{2}{I(p^k)}$$

$$\frac{2}{I(p^k)} \leq \frac{2}{I(p)}=\frac{2p}{p+1}$$

which implies that $m < p$. This implies that $k=1$, so that we get

$$\frac{2m}{m+1}=I(m^2)=\frac{2}{I(p^k)}$$

$$\frac{2}{I(p^k)}=\frac{2}{I(p)}=\frac{2p}{p+1}$$

which is equivalent to $m=p$. This contradicts $\gcd(p,m)=1$.

By trichotomy, we obtain

$$\left(I(m^2) < 2m/(m+1)\right) \oplus \left(I(m^2) > 2m/(m+1)\right).$$

It is easy to show that the biconditional

$$I(m^2) < 2m/(m+1) \iff p < m$$

holds.

By assumption, $m < p$ holds. Consequently, $2m/(m+1) < I(m^2)$ is true.

It remains to rule out the case

$$I(m) < 2m/(m+1) < I(m^2).$$

This particular case is ruled out by this blog post.

## Comments

## Post a Comment