## 10.10.18

### Arnie Dris's Publications - 3rd Quarter, 2018

Conditions equivalent to the Descartes–Frenicle–Sorli Conjecture on odd perfect numbers – Part II (co-authored with Doli-Jane Uvales Tejada)

## 2.10.18

### On the abundancy index/outlaw status of the fraction $\frac{p+2}{p}$, where $p$ is an odd prime

As before, we consider the equation

$$\frac{\sigma(x)}{x} = \frac{p+2}{p}$$
where $p$ is an odd prime.

We consider several cases.  (Note that the list of cases presented here is not exhaustive.)

Case 1:  $3 \mid x$

Since
$$\frac{\sigma(x)}{x} = \frac{p+2}{p}$$
implies that $x$ is an odd square, and since $3 \mid x$, it follows that $9 = 3^2 \mid x$, so that
$$\frac{13}{9} = \frac{1 + 3 + 9}{9} = \frac{\sigma(3^2)}{3^2} \leq \frac{\sigma(x)}{x} = \frac{p+2}{p}$$
from which we obtain
$$13p \leq 9(p + 2)$$
$$4p \leq 18$$
$$p \leq \frac{9}{2}.$$

Since $p$ is an odd prime, it follows that $p = 3$.  (Since $p = 3$ also implies that $3 \mid x$, this means that we know that the biconditional
$$\bigg(3 \mid x\bigg) \Longleftrightarrow \bigg(p = 3\bigg)$$
must hold.)

Case 2:  $5 \mid x$

Similar to the proof for Case 1, we get
$$\frac{31}{25} = \frac{1 + 5 + 25}{25} = \frac{\sigma(5^2)}{5^2} \leq \frac{\sigma(x)}{x} = \frac{p+2}{p}$$
from which we have
$$31p \leq 25(p + 2)$$
$$6p \leq 50$$
$$p \leq \frac{25}{3}.$$

Since $p$ is an odd prime, it follows that either $p = 3$, $p = 5$, or $p = 7$.

Case 3:  $p \mid x$

Note that this case holds in general (since $\gcd(p, p+2)=1$ follows from $p$ is an odd prime).

As in Case 1, it follows that
$$\frac{1 + p + p^2}{p^2} = \frac{\sigma(p^2)}{p^2} \leq \frac{\sigma(x)}{x} = \frac{p+2}{p}$$
from which we obtain
$$\frac{1}{p^2} + \frac{1 + p}{p} \leq \frac{2 + p}{p}.$$
This last inequality implies that
$$\frac{1}{p^2} \leq \frac{1}{p}$$
from which we get
$$p \leq p^2$$

## 1.7.18

### Arnie Dris's Publications - 2nd Quarter, 2018

The Non-Euler Part of a Spoof Odd Perfect Number is not Almost Perfect (co-authored with Doli-Jane T. Lugatiman)

## 9.6.18

### Would like to get numerical (lower [and upper?]) bounds for $p$

(This post is copied verbatim from this MSE question.)

This question is an offshoot of this earlier MSE question.

Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$, the set of positive integers.  Denote the abundancy index of $z$ by $I(z) := \sigma(z)/z$.

If $N={p^k}{m^2}$ is an odd perfect number with Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then it is somewhat trivial to prove that
$$3 - \frac{p - 2}{p(p-1)} = \frac{3p^2 - 4p + 2}{p(p-1)} < I(p^k) + I(m^2)$$
and
$$I(p^k) + I(m^2) \leq \frac{3p^2 + 2p + 1}{p(p+1)} = 3 - \frac{p - 1}{p(p+1)}.$$
Now, setting $x := 3 - \bigg(I(p^k) + I(m^2)\bigg)$, we have the simultaneous inequalities
$$\frac{p-1}{p(p+1)} \leq x < \frac{p - 2}{p(p - 1)}$$
resulting in the inequalities
$$\bigg((p - 2) > xp(p-1)\bigg) \land \bigg((p - 1) \leq xp(p+1)\bigg).$$

Notice that it is known that
$$\frac{57}{20} < I(p^k)+I(m^2) < 3$$
so that we know that
$$0 < x < \frac{3}{20}.$$

We now solve the inequalities one by one.

Solution is
$$\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < p < \frac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}$$

Solution is
$$p \in \bigg(-\infty, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x}\bigg] \bigcup \bigg[\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x},\infty\bigg)$$

Now, this is where the computations start to get messy.  Can I ask for some help?

Basically, I would like to get numerical (lower [and upper?]) bounds for $p$.

It is easily seen that
$$\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < 2, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x} < 2$$
$$\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x} \leq p < \frac{1+x + \sqrt{x^2 - 6x + 1}}{2x}$$
then, one can see that the lower bound and upper bound differs by $1$ and both are unbounded in the given range of $x$. That is, there can be only one prime $p$ for any given value $x$.

Related Paper:  The abundancy index of divisors of odd perfect numbers - Part III, http://nntdm.net/volume-23-2017/number-3/53-59/

## 5.5.18

### Can the following argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?

(The following post is extracted verbatim from this MSE question.)

This is related to this earlier MSE question.

Let $\sigma(x)$ be the sum of the divisors of $x$, and denote the abundancy index of $x$ by $I(x):=\sigma(x)/x$.

If the equation $I(a) = b/c$ has no solution $a \in \mathbb{N}$, then the rational number $b/c$ is said to be an abundancy outlaw.

In the earlier question, it is shown that:

(a) If $I(n) = (p+2)/p$, then $n$ is an odd square.

(b) If $I(n) = (p+2)/p$, then $p \mid n$.

So now suppose to the contrary that $I(n) = (p+2)/p$.  From (a) and (b), since $p$ is a prime number, then $p^2 \mid n$.  It follows that
$$\frac{p^2 + p + 1}{p^2} = I(p^2) \leq I(n) = \frac{p+2}{p},$$
whence there is still no contradiction.

From the divisibility constraint $p^2 \mid n$, we have that $p^2 \leq n$.  We claim that $p^2 \neq n$.  Suppose to the contrary that $p^2 = n$.

Then we have
$$\frac{\sigma(n)}{n} = \frac{p+2}{p} = \frac{p(p+2)}{p^2}$$
so that
$$p^2 + p + 1 = \sigma(p^2) = \sigma(n) = p^2 + 2p,$$
which contradicts the fact that $p$ is an odd prime.

This implies that $p < \sqrt{n}$, which further means that
$$\gcd(n, \sigma(n)) = \frac{n}{p} > \sqrt{n} > {10}^8$$
where the lower bound $n > {10}^{16}$ is due to Richard F. Ryan, "Results concerning uniqueness for $\sigma(x)/x = \sigma(p^n q^m)/(p^n q^m)$ and related topics, International Math. J., 2002 , V2#5pp497-514".

Here is my question:

Can the preceding argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?

## 19.3.18

### If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

(Note:  This post was copied verbatim from this MSE question.)

Let $\sigma(x)$ be the sum of the divisors of $x$.  Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$.

Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?
$$D(q^k)D(n^2)=2s(q^k)s(n^2)$$

I only know that $k=1$ is true if and only if (one of) the following conditions hold:

(1) $\sigma(n^2)/q \mid n^2$

(2) $D(n^2) \mid n^2$

(3) $\gcd(n^2, \sigma(n^2)) = D(n^2)$

Source of Equation

From the fundamental equation
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$
we obtain
$$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \frac{D(n^2)}{s(q^k)} = \gcd(n^2, \sigma(n^2))$$
and
$$\frac{\sigma(n^2) - n^2}{\frac{2q^k - \sigma(q^k)}{2}} = \frac{2s(n^2)}{D(q^k)} = \gcd(n^2, \sigma(n^2)),$$
using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}.$$

Reference

## 5.3.18

### If $\frac{σ(x)}{x}=\frac{p+2}p$ where $p$ is an odd prime, does it follow that $x$ is an odd square?

(Note:  The following proof was copied verbatim from the answer of MSE user Alex Francisco.)

First, note that for any coprime $a, b \in \mathbb{N}_+$, there is$$I(ab) = I(a) I(b).$$

Suppose there is an even number $n = 2^k \cdot l$, where $k \geq 1$ and $l$ odd, such that$$I(n) = \frac{p + 2}{p}.$$

Case 1: $k \geq 2$. Then$$\frac{p + 2}{p} = I(2^k \cdot l) = I(2^k) I(l) \geq I(2^k) = \frac{2^{k + 1} - 1}{2^k} = 2 - \frac{1}{2^k} \geq \frac{7}{4},$$
which implies $4(p + 2) \geq 7p$, contradictory to $p \geq 3$.

Case 2: $k = 1$ and $p \geq 5$. Then, analogously,$$\frac{p + 2}{p} \geq 2 - \frac{1}{2^k} = \frac{3}{2},$$
which implies $2(p + 2) \geq 3p$, contradictory to $p \geq 5$.

Case 3: $k = 1$ and $p = 3$. Then$$\frac{5}{3} = I(2l) = \frac{3}{2} I(l) \Longrightarrow I(l) = \frac{10}{9}.$$

If $l$ has an odd prime factor $q < 10$, suppose $q^m \mathbin{\|} l$, then$$\frac{10}{9} = I(l) = I(q^m) I\left( \frac{l}{q^m} \right) \geq I(q^m) = \frac{q^{m + 1} - 1}{q^m (q - 1)}\\ = 1 + \frac{1}{q - 1}\frac{q^m - 1}{q^m} \geq 1 + \frac{1}{q - 1}\frac{q - 1}{q} = 1 + \frac{1}{q} > \frac{10}{9},$$
a contradiction. Now suppose the prime factorization of $l$ is$$l = \prod_{i = 1}^s p_i^{a_i},$$
then $p_i \geq 11$. Because$$\frac{10}{9} = I(l) = \prod_{i = 1}^s \frac{1 + p_i + \cdots + p_i^{a_i}}{p_i^{a_i}},$$
then$$9 \prod_{i = 1}^s \sum_{j = 0}^{a_i} p_i^j = 10 \prod_{i = 1}^s p_i^{a_i} \Longrightarrow \left. 9 \,\middle|\, \prod_{i = 1}^s p_i^{a_i} \right.,$$
contradictory to $p_i \geq 11$.

Therefore, there does not exist an even positive integer $n$ such that$$I(n) = \frac{p + 2}{p}.$$