## 10.3.19

### Breaking the barriers at $q=5$ and $q=13$ for $q^k n^2$ an odd perfect number with special prime $q$

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

Let $\sigma(x)$ be the sum of divisors of the positive integer $x$.  If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.

Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $\sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.

So now suppose that $\sigma(n^2)/q^k$ is a square.  This implies that $k=1$, and also that $\sigma(n^2) \equiv 1 \pmod 4$, since $\sigma(n^2)/q^k$ is odd and $q \equiv k \equiv 1 \pmod 4$.

The congruence $\sigma(n^2) \equiv 1 \pmod 4$ then implies that $q \equiv k \pmod 8$.  (See this MO post for the details.)  Substituting $k=1$, we obtain
$$q \equiv 1 \pmod 8.$$

This implies that the lowest possible value for the special prime $q$ is $17$.  (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $\sigma(n^2)/q^k$ is a square.)  Note that, if $q=17$, then $(q+1)/2 = 3^2 \mid n^2$.

Here is my question:
Can we push the lowest possible value from $q \geq 17$, to say, $q \geq 41$ or even $q \geq 97$, using the ideas in this post, and possibly more?

Note that if
$$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}$$
is a square, then $k=1$ and $\sigma(q^k)/2 = (q+1)/2$ is also a square.

The possible values for the special prime satisfying $q < 100$ and $q \equiv 1 \pmod 8$ are $17$, $41$, $73$, $89$, and $97$.

For each of these values:
$$\frac{q_1 + 1}{2} = \frac{17 + 1}{2} = 9 = 3^2$$
$$\frac{q_2 + 1}{2} = \frac{41 + 1}{2} = 21 \text{ which is not a square.}$$
$$\frac{q_3 + 1}{2} = \frac{73 + 1}{2} = 37 \text{ which is not a square.}$$
$$\frac{q_4 + 1}{2} = \frac{89 + 1}{2} = 45 \text{ which is not a square.}$$
$$\frac{q_5 + 1}{2} = \frac{97 + 1}{2} = 49 = 7^2$$

Thus, if $\sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q \geq 97$.

## 23.2.19

### On the golden ratio and odd perfect numbers

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

On the golden ratio and odd perfect numbers

Here is my question:
Is $I(n^2) - 1 > 1/I(n^2)$ true when $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $k>1$?

My Attempt

If $k>1$, then since $q$ is the special prime, then $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  In particular, we know that $q \geq 5$ and $k \geq 5$.

We know that
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} \leq \frac{5}{4}.$$

It follows that
$$I(n^2) = \frac{2}{I(q^k)} > \frac{2(q - 1)}{q} \geq \frac{8}{5}.$$

Thus,
$$I(n^2) - 1 > \frac{2(q - 1)}{q} - 1 = \frac{(2q - 2) - q}{q} = \frac{q - 2}{q} > \frac{q}{2(q - 1)} > \frac{1}{I(n^2)}$$
where the inequality
$$\frac{q - 2}{q} > \frac{q}{2(q - 1)}$$
holds provided $q > 3+\sqrt{5} \approx 5.23607$.

However, the resulting inequality for $I(n^2)$ from
$$I(n^2) - 1 > \frac{1}{I(n^2)}$$
together with the following upper bound for $I(n^2)$ (which holds when $k>1$)
$$\frac{2q}{q+1} > I(n^2)$$
only yields
$$\frac{2q}{q+1} > I(n^2) > \frac{\sqrt{5}+1}{2}$$
thereby giving
$$q > \frac{1+\sqrt{5}}{3-\sqrt{5}} = 2+\sqrt{5} \approx 4.23607.$$

Note that there is no discrepancy when $k=1$, as then we have
$$I(n^2) - 1 \geq \frac{2}{3} > \frac{3}{5} \geq \frac{1}{I(n^2)}$$
yielding the lower bound $q > 2+\sqrt{5} \approx 4.23607$ from
$$\frac{2q}{q+1}=I(n^2) > \frac{\sqrt{5}+1}{2}.$$

When $k>1$, we get
$$I(n^2) - 1 > \frac{3}{5} \not\gt \frac{5}{8} > \frac{1}{I(n^2)}.$$

## 4.1.19

### On Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College

(Note:  This blog post was lifted verbatim from this MSE question and the answer contained therein.)

THE QUESTION

In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$.  Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)$.  Finally, let $s(X)=\sigma(X)-X$ denote the sum of the aliquot divisors of $X$.

This is a question about Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College, titled Characterization of the Positive Integers with Abundancy Index of the Form $(2x-1)/x$.  (A copy of the poster presentation is available via the following hyperlink.)

In the abstract of the paper, it is stated in the fourth sentence that
Rational numbers of the form $(2x-1)/x$ are important since both even and odd perfect numbers have a divisor with abundancy index of this form.

Let $M = 2^{p-1}(2^p - 1)$ be an even perfect number, and let $N = q^k n^2$ be an odd perfect number.

Clearly,
$$I(2^{p-1}) = \frac{2^p - 1}{2^{p-1}} = \frac{2x_1 - 1}{x_1}$$
where $x_1 = 2^{p-1}$.  (In other words, $2^{p-1}$ is an even almost perfect number, since it is a power of two.)

However,
$$I(p^k) = \frac{p^{k+1} - 1}{p^{k+1} - p^k}$$
and
$$I(n^2) = \frac{2}{I(p^k)} = \frac{2(p^{k+1} - p^k)}{p^{k+1} - 1}$$
so clearly $p^k$ is not almost perfect (since $p$ must be odd).

$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)} \geq 3$$
(see the paper [Dris, 2012]), then clearly $n^2$ is likewise not almost perfect.  (Similarly, it can be proved that $n$ and $q^k n$ are not almost perfect.)

So I think the trivial divisor $1$ of an odd perfect number has the required abundancy index
$$I(1) = 1 = \frac{2\cdot{1} - 1}{1} = \frac{2x_2 - 1}{x_2}$$
where $x_2 = 1$.

Here is my question:
Is there any other divisor $m > 1$ of an odd perfect number $N = q^k n^2$ such that
$$I(m) = \frac{2x - 1}{x}$$
for some positive integer $x$?

Let $N = q^k n^2$ be an odd perfect number.

Suppose that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.

Then
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = 2 - \frac{1}{(q+1)/2} = \frac{2((q+1)/2) - 1}{(q+1)/2},$$
where $(q+1)/2$ is an integer (since $q \equiv 1 \pmod 4$).

It remains to consider the case when $k>1$.  (Note that $k \equiv 1 \pmod 4$.)

## 24.12.18

### Arnie Dris's Publications - 4th Quarter, 2018

Revisiting some old results on odd perfect numbers (co-authored with Doli-Jane Uvales Tejada)

## 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
$$p \leq p^2$$