Search This Blog

6.2.17

On a conjectured biconditional for odd perfect numbers - Part 2

Can this heuristic on odd perfect numbers be made rigorous?

Preliminaries

Euler showed that an odd perfect number, if one exists, must take the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  (It follows that $q \geq 5$.)

Let $\sigma(x)$ be the sum of the divisors of $x \in \mathbb{N}$.  Then by definition,
$$\sigma(q^k)\sigma(n^2) = \sigma(q^k n^2) = \sigma(N) = 2N = 2 q^k n^2$$
from which it follows that
$$I(q^k)I(n^2) = \frac{\sigma(q^k)}{q^k}\cdot\frac{\sigma(n^2)}{n^2} = 2$$
where $I(x)=\sigma(x)/x$ is the *abundancy index* of $x$.

Numerics

By a well-known formula for the sum of divisors of a prime power, we have
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{{q^k}(q - 1)}.$$

This is bounded by
$$\frac{q+1}{q} = I(q) \leq I(q^k) < \frac{q^{k+1}}{{q^k}(q - 1)} = \frac{q}{q - 1}.$$

Now, note that, when $k=1$, we have the bounds
$$1 < I(q^k) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq 1 + \frac{1}{5} = \frac{6}{5},$$
from which it follows that
$$\frac{5}{3} \leq I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} < 2.$$

Additionally, when $k>1$, we have the bounds
$$1 < I(q^k) < \frac{q}{q - 1}.$$
Since
$$\frac{q}{q - 1} = \frac{1}{1 - (1/q)}$$
we obtain
$$q \geq 5 \iff 1/q \leq 1/5 \iff 1 - (1/q) \geq 4/5 \iff \frac{1}{1 - (1/q)} \leq 5/4$$
so that
$$1 < I(q^k) < \frac{q}{q - 1} \leq \frac{5}{4}.$$

This then implies that
$$\frac{8}{5} < I(n^2) = \frac{2}{I(q^k)} < 2.$$

Implications

Note that we have proved the following implications:

(1) If $N=q^k n^2$ is an odd perfect number with $k=1$, then $I(n^2) \geq 5/3$.

(2) If $N=q^k n^2$ is an odd perfect number with $k>1$, then $I(n^2) > 8/5$.

Contrapositives

(3) Contrapositive of (1) If $N=q^k n^2$ is an odd perfect number with $I(n^2) < 5/3$, then $k>1$.

(4) Contrapositive of (2) If $N=q^k n^2$ is an odd perfect number with $I(n^2) \leq 8/5$, then $k=1$.

Analysis

By (3) and (2), we have
$$I(n^2) < 5/3 \Longrightarrow k>1 \Longrightarrow I(n^2) > 8/5 \Longrightarrow 8/5 < I(n^2) < 5/3$$
resulting in no contradiction, since ${8}\cdot{3} = 24 < 25 = {5}\cdot{5}$.

By (4) and (1), we get
$$I(n^2) \leq 8/5 \Longrightarrow k=1 \Longrightarrow I(n^2) \geq 5/3 \Longrightarrow 5/3 \leq I(n^2) \leq 8/5$$
resulting in a contradiction, since ${5}\cdot{5}= 25 \leq 24 = {8}\cdot{3}$ is false.

This implies that $I(n^2) > 8/5$ is unconditionally true.

Now here is the heuristic:

Heuristic

If we identify the lower bound $I(n^2) \geq 5/3$ with $k=1$ and similarly identify the lower bound $I(n^2) > 8/5$ with $k>1$, and since assuming the negation of the latter lower bound contradicts (4) and (1) above, does this mean that we can predict $k>1$ to be true?

That is, can we "assume" or prove that the following biconditionals hold?
$$k=1 \iff I(n^2) \geq 5/3$$
$$k>1 \iff I(n^2) > 8/5$$

Response by MSE user Noah Schweber
(Link to MSE question)

You can do nothing of the sort, and the axiom of choice plays no role here.

Let's ignore the specific math leading up to the "Analysis" part, and just look at the logic of what you're trying to do. (In particular, I make no claim about whether said math is correct.)

You have two numbers, "$k$" and "$I=I(n^2)$," that you care about; and you know $$k\ge 1,\quad k=1\Longrightarrow I\ge{5/3},\quad\mbox{ and }\quad k>1\Longrightarrow I>{8/5}.$$ All this lets you conclude is that $I<{5/3}$ implies $k>1$. If $I\ge{5/3}$, you can't conclude anything - both "$k=17$, $I=17$" and "$k=1, I=17$" satisfy the three facts above. In particular, from what you know so far there is no way to identify $k=1$ with $I\ge{5/3}$. (It's always a good idea to think through a few specific examples before trying to jump to a conclusion.)

You are trying to invoke the axiom of choice to get around a logical gap - that you can't conclude the converse of a statement from the statement. But this isn't a thing the axiom of choice does. 

****


EDIT: Note that by Shoenfield absoluteness (actually that's massive overkill but oh well), the truth of Sorli's conjecture can't possibly depend on the axiom of choice: if there is a proof using AC, there is also a proof not using AC. (Incidentally, this doesn't mean that the proof via AC might be easier to find, just that AC won't ultimately be necessary.) I strongly suspect that the same goes for each of the Millennium problems - indeed $P=NP$, Birch Swinnerton-Dyer, and Navier-Stokes are all not hard to cast in arithmetic terms - but I am worried about the Hodge Conjecture on the general principle that it looks scary to me (but it also does look $\Pi^1_2$).

14.10.16

On a conjectured biconditional for odd perfect numbers - Part 1

If $q^k n^2$ is an odd perfect number with Euler prime $q$, then

$$k = 1 \Longleftrightarrow D(n^2) \mid n^2,$$
where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of $n^2$.

In particular, this shows that the Descartes-Frenicle-Sorli conjecture for odd perfect numbers is true if and only if the non-Euler part $n^2$ is deficient-perfect.

(I have a new paper out there that discusses the proof for the biconditional
$$k = 1 \Longleftrightarrow D(n^2) \mid n^2,$$
if $q^k n^2$ is an odd perfect number with Euler prime $q$.)

25.9.16

Some Recent Improvements on Results Contained in the Paper Titled "New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II"

In this blog post, we present some recent improvements on the results contained in the paper titled "New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II".

We refer the interested reader to the following arXiv preprint for the latest version of this paper.

Preliminaries

Denote the sum of the divisors of the positive integer $x$ by $\sigma(x)$.  For example, $\sigma(6)=1+2+3+6=12$. A number $M$ is called perfect when $\sigma(M)=2M$.  Euler showed that an odd perfect number must have the form $q^k n^2$ where $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.

Denote the abundancy index I by
$$I(x)=\sigma(x)/x.$$
Note that, if $M$ is perfect, then
$$I(M)=\sigma(M)/M=2M/M=2.$$

Recall that $\sigma$ is weakly multiplicative, that is, it satisfies
$$\sigma(yz)=\sigma(y)\sigma(z)$$
if $\gcd(y,z)=1$.  In particular, note that the abundancy index $I$ is also weakly multiplicative, so that if $q^k n^2$ is an odd perfect number, then we have the equation
$$I(q^k)I(n^2) = I({q^k}{n^2}) = 2.$$

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

Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.  Notice that $\gcd(q, n) = 1$ implies that $q \neq n$ and $q^k \neq n$.  Also, we have the bounds $1 < I(qn) \leq I({q^k}n) < 2$, which follows from the fact that ${q^k}n$ is deficient, being a proper divisor of the perfect number $N = q^k n^2$.  It follows that we have the following inequations:
$$\sigma(q)/n \neq \sigma(n)/q$$
and
$$\sigma(q^k)/n \neq \sigma(n)/q^k.$$
We will be using the following ("general") bounds from this publication:
$$1 < 1 + \frac{1}{q} = I(q) \leq I(q^k) < \frac{5}{4} < \sqrt[3]{2} < \sqrt{\frac{8}{5}} < I(n) < I(n^2) < 2,$$
or when $k=1$ is known to hold, we will utilize the ("slightly") stronger bounds
$$1 < I(q) = 1 + \frac{1}{q} \leq \frac{6}{5} < \sqrt[3]{2} < \sqrt{\frac{5}{3}} < I(n) < I(n^2) < 2.$$
(Hereinafter, both sets of bounds will be abbreviated as $I(q^k) < \sqrt[3]{2} < I(n)$.)

Improved Lower Bound for $I(n)$

The following result was communicated to Arnie Dris (the author of this blog post) by Pascal Ochem via e-mail, on April 17 2013:

THEOREM P
$$I(n) > \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}.$$

In particular, note that
$$\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} > \sqrt{2}.$$
A proof of this claimed lower bound for $I(n)$ was likewise communicated by Ochem to Dris (in the same e-mail), and the details of the proof are presented in this preprint.

We will utilize Theorem P in getting improvements on the results contained in this preprint.

Lemmas

In light of this more recent preprint, we have the following unconditional result:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, then $q^k < n$ is true if and only if the biconditional
$$q^k < n \Longleftrightarrow \left(q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \sigma(q^k)/n < \sigma(n)/q^k\right)$$
holds.

From Remark 2.2 (page 2) of this preprint, we have the biconditional
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longleftrightarrow \left(\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right).$$

Here is a brief outline of the proof (taken from this preprint, to appear in JANTA):

First, we show that
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longrightarrow \left(\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right).$$

Since we have the inequality
$$\frac{\sigma(q^k)}{q^k} = I(q^k) < \sqrt[3]{2} < I(n) = \frac{\sigma(n)}{n},$$
it follows that
$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}.$$
By assumption, we have
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k},$$
which is equivalent to
$$\frac{q^k}{n} < \frac{\sigma(n)}{\sigma(q^k)}.$$
Putting everything together, we obtain
$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < \frac{\sigma(n)}{\sigma(q^k)}.$$

Now consider the product
$$\left(\frac{\sigma(q^k)}{\sigma(n)} - \frac{q^k}{n}\right)\cdot\left(\frac{\sigma(n)}{\sigma(q^k)} - \frac{q^k}{n}\right),$$
which should be negative.  Thus, we have
$$0 > \left(\frac{\sigma(q^k)}{\sigma(n)} - \frac{q^k}{n}\right)\cdot\left(\frac{\sigma(n)}{\sigma(q^k)} - \frac{q^k}{n}\right) = 1 + \left(\frac{q^k}{n}\right)^2 - {\frac{q^k}{n}}\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right),$$
which implies that
$${\frac{q^k}{n}}\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right) > 1 + \left(\frac{q^k}{n}\right)^2.$$

Finally, we obtain
$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}.$$

The proof for the other direction
$$\left(\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right) \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
is very similar, and is left as an exercise for the interested reader.

Main Results

By considerations in the preceding paragraphs, we know that $q^k \neq n$.  We therefore need to cover two cases:

Case A  $q^k < n$


This means that the biconditionals
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longrightarrow \sigma(q^k)/n < \sigma(n)/q^k$$
and
$$\sigma(q^k)/n < \sigma(n)/q^k \Longleftrightarrow {q^k}/n + n/{q^k} < \sigma(q^k)/\sigma(n) + \sigma(n)/\sigma(q^k)$$
both hold.

Since $q^k < n$ is true by assumption, all of the inequalities
$$\sigma(q^k) < \sigma(n),$$
$$\sigma(q^k)/n < \sigma(n)/q^k,$$
and
$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}$$
simultaneously hold.  Note further that we also have
$$\frac{\sigma(q^k)}{n}=\frac{q^k}{n}I(q^k)<\frac{n}{q^k}I(n)=\frac{\sigma(n)}{q^k}.$$
It follows that
$$\left(\frac{q^k}{n}\right)^2 < \frac{I(n)}{I(q^k)} < 2.$$
Now, note that
$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < \sqrt{2}$$
and
$$\frac{1}{\sqrt{2}} < \frac{n}{q^k} < \frac{\sigma(n)}{\sigma(q^k)}.$$
Taking square roots, we get
$$\sqrt{\frac{\sigma(q^k)}{\sigma(n)}} < \sqrt[4]{2}$$
and
$$\frac{1}{\sqrt[4]{2}} < \sqrt{\frac{\sigma(n)}{\sigma(q^k)}}.$$
Subtracting, we obtain
$$\sqrt{\frac{\sigma(q^k)}{\sigma(n)}} - \sqrt{\frac{\sigma(n)}{\sigma(q^k)}} < \sqrt[4]{2} - \frac{1}{\sqrt[4]{2}}.$$
Squaring both sides, we now have
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} - 2 = \left(\sqrt{\frac{\sigma(q^k)}{\sigma(n)}} - \sqrt{\frac{\sigma(n)}{\sigma(q^k)}}\right)^2 < \left(\sqrt[4]{2} - \frac{1}{\sqrt[4]{2}}\right)^2 = \sqrt{2} + \frac{1}{\sqrt{2}} - 2,$$
so that
$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}.$$
Note the rational approximation
$$\frac{3}{\sqrt{2}} \approx 2.12132.$$

Now, because of the fact that $\sigma(q^k) \neq n$ (since $\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4$ and $n$ is odd), we have two further sub-cases to analyze:

Sub-case A-1   $q^k < \sigma(q^k) < n < \sigma(n)$

Under this Sub-case A-1, we have
$$\sigma(q^k)/n < 1 < \sigma(n)/q^k$$
from which we obtain the lower bound
$$I(q^k)I(n) + 1 < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k},$$
which we get by considering the negative product
$$\left(\sigma(q^k)/n - 1\right)\left(\sigma(n)/q^k - 1\right).$$
Thus, we have
$$1 + \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(q^k)I(n) + 1 < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}.$$
This validates the (trivial!) inequality
$$\frac{q^k}{n}+\frac{n}{q^k} < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}.$$

We now compute a lower bound for the quantity
$$\frac{\sigma(q^k)}{\sigma(n)}+\frac{\sigma(n)}{\sigma(q^k)}.$$
Note that a trivial lower bound is given by the Arithmetic Mean-Geometric Mean Inequality, as follows:
$$2 < \frac{\sigma(q^k)}{\sigma(n)}+\frac{\sigma(n)}{\sigma(q^k)}$$
since $\sigma(q^k) < \sigma(n)$.  Note further that $\sigma(n)/\sigma(q^k) > 1$.

We now attempt to get an improved upper bound for $\sigma(q^k)/\sigma(n)$:
$$\sigma(q^k)/n < 1 < \sigma(q^k)/q^k < 5/4 < (8/5)^{\ln(4/3)/\ln(13/9)} < \sigma(n)/n < \sigma(n)/q^k$$
$$\frac{\sigma(q^k)}{\sigma(n)}=\frac{\sigma(q^k)/n + \sigma(q^k)/q^k}{\sigma(n)/n + \sigma(n)/q^k} < \frac{1 + (5/4)}{2\cdot{(8/5)^{\ln(4/3)/\ln(13/9)}}} = \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 0.778867$$
This translates to the improved lower bound
$$\frac{\sigma(n)}{\sigma(q^k)} > \frac{8}{9}\cdot{\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}} \approx 1.283916$$
for $\sigma(n)/\sigma(q^k)$.
From another perspective, since
$$2 < \frac{\sigma(q^k)}{\sigma(n)}+\frac{\sigma(n)}{\sigma(q^k)}$$
and because
$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 0.778867,$$
then we get
$$\frac{\sigma(n)}{\sigma(q^k)} > 2 - \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 1.2211.$$
(Notice that, in general, multiplicative estimates are better/tighter than additive estimates.)

Let
$$\theta_1 := \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 0.778867.$$
Then
$$\frac{1}{\theta_1} = \frac{8}{9}\cdot{\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}} \approx 1.283916.$$
Note that the product
$$\bigg(\frac{\sigma(q^k)}{\sigma(n)} - \theta_1\bigg)\bigg(\frac{\sigma(n)}{\sigma(q^k)} - \theta_1\bigg)$$
is negative since $\sigma(q^k)/\sigma(n) < \theta_1 < \sigma(n)/\sigma(q^k)$. In particular, we finally get the (improved) lower bound
$$1 + {\theta_1}^2 < {\theta_1}\cdot\bigg(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\bigg),$$
which implies that
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} > \theta_1 + \frac{1}{\theta_1} \approx 2.062783179.$$

We now try to derive an improved estimate for ${q^k}/n$.
$$\frac{q^k}{n}=\frac{{q^k}/\sigma(q^k) + {q^k}/\sigma(n)}{n/\sigma(q^k) + n/\sigma(n)}<\frac{1+(5/8)^{\ln(4/3)/\ln(13/9)}}{1+(1/2)}=\frac{2}{3}\cdot\bigg(1 + \left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}\bigg) \approx 1.1282,$$
which is trivial as compared to $q^k < n$.

Lastly, since we have
$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{3}{\sqrt{2}},$$
and because $n/{q^k} > 1$, we obtain
$$\frac{q^k}{n} < \frac{3}{\sqrt{2}} - 1 = \frac{3\sqrt{2} - 2}{2} \approx 1.12132,$$
which again is trivial in comparison to $q^k < n$.  Hence, it appears that it will not be possible to improve on the upper bound
$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{3}{\sqrt{2}}$$
using our current method.

Here is a summary of the results that we have obtained for
Sub-case A-1:
$$2 < \frac{q^k}{n} + \frac{n}{q^k} < \frac{3}{\sqrt{2}} \approx 2.12132$$
$$2.062783179 \approx \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} + \frac{8}{9}\cdot{\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{3}{\sqrt{2}}.$$

Sub-case A-2   $q^k < n < \sigma(q^k) < \sigma(n)$

Note that Sub-case A-2 implies that $k>1$.


Under this Sub-case A-2, we have

$$1 <\sigma(q^k)/n < \sigma(q^k)/q^k < 5/4 < (8/5)^{\ln(4/3)/\ln(13/9)} < \sigma(n)/n < \sigma(n)/q^k < 2,$$
from which we obtain
$$\frac{1}{2}=\frac{1+1}{2+2}<\frac{\sigma(q^k)}{\sigma(n)}=\frac{\sigma(q^k)/n + \sigma(q^k)/q^k}{\sigma(n)/n + \sigma(n)/q^k}<\frac{(5/4)+(5/4)}{2\cdot{(8/5)^{\ln(4/3)/\ln(13/9)}}}=\frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)} \approx 0.8654.$$
It follows that
$$2 > \frac{\sigma(n)}{\sigma(q^k)}>\frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)} \approx 1.1555.$$

Let
$$\theta_2 := \frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}$$
and
$$\frac{1}{\theta_2} = \frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}.$$
Since the product
$$\bigg(\frac{\sigma(q^k)}{\sigma(n)} - \theta_2\bigg)\cdot\bigg(\frac{\sigma(n)}{\sigma(q^k)} - \theta_2\bigg)$$
is negative, we have
$$1 + {\theta_2}^2 < {\theta_2}\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right),$$
which implies that
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} > \theta_2 + \frac{1}{\theta_2} = \frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)} + \frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)} \approx 2.02093.$$

Similarly, since the product
$$\bigg(\frac{\sigma(q^k)}{\sigma(n)} - 2\bigg)\cdot\bigg(\frac{\sigma(n)}{\sigma(q^k)} - 2\bigg)$$
is positive, we obtain
$$1 + 4 > 2\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right),$$
which implies that
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{5}{2}.$$
This last upper bound is trivial when compared to the earlier bound
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{3}{\sqrt{2}} \approx 2.12132.$$

Here is a summary of the results that we have obtained for
Sub-case A-2:
$$2 < \frac{q^k}{n}+\frac{n}{q^k} < \frac{3}{\sqrt{2}} \approx 2.12132$$
$$2.02093 \approx \frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)} + \frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}  < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{3}{\sqrt{2}} \approx 2.12132$$

Case B  $n < q^k$

It remains to consider three remaining sub-cases under Case B:

Sub-case B-1   $n < q^k < \sigma(q^k) \leq \sigma(n)$

Sub-case B-2   $n < q^k \leq \sigma(n) < \sigma(q^k)$

Sub-case B-3   $n < \sigma(n) < q^k < \sigma(q^k)$

We remark that work is underway to fill in the gaps in Brown's proof for $q^k < n$.  (The interested reader is hereby referred to this preprint for more details.)

THIS POST IS CURRENTLY A WORK IN PROGRESS.

8.9.16

An Elementary Proof (???) of the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers

EUREKA???

Introduction

If $N=q^k n^2$ is an odd perfect number with Euler prime $q$, then the Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

In this blog post, we will attempt to give an elementary proof for this conjecture. We will also derive the possible value(s) for the Euler prime $q$!  

We denote the abundancy index of the positive integer $x$ by
$$I(x) = \frac{\sigma(x)}{x}.$$

Preliminaries

First, we restate the following results from these earlier posts in this blog:



Proposition 1
Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$.  Then $k=1$ implies
$$I(n^2) \leq 2 - \frac{5}{3q}.$$

Proposition 2
Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$I(n^2) \geq 2 - \frac{5}{3q}$$
implies that $k=1$ and $q=5$.

Main Results

Theorem
Suppose that  $N=q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$I(n^2) \leq 2 - \frac{5}{3q}.$$
Additionally, $I(n^2) = 2 - \frac{5}{3q}$ if and only if $k=1$ and $q=5$.

Proof
Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.

By the contrapositive of Proposition 1,
$$I(n^2) > 2 - \frac{5}{3q} \Rightarrow k \neq 1.$$

By Proposition 2, we have
$$I(n^2) \geq 2 - \frac{5}{3q} \Rightarrow \left\{k = 1 \land q = 5\right\}.$$

Now, assume that
$$I(n^2) \geq 2 - \frac{5}{3q}$$
is true.  Then we have $k=1$ and $q=5$.  In particular, we are sure that $k=1$ must hold.

Consequently,
$$I(n^2) \geq 2 - \frac{5}{3q} \Rightarrow k = 1$$
holds.

Note that
$$I(n^2) > 2 - \frac{5}{3q} \Rightarrow k = 1$$
also holds.

Putting it all together,
$$\left\{I(n^2) > 2 - \frac{5}{3q} \Rightarrow k \neq 1\right\} \land \left\{I(n^2) > 2 - \frac{5}{3q} \Rightarrow k = 1\right\} \Rightarrow I(n^2) \leq 2 - \frac{5}{3q}.$$

Notice that
$$I(n^2) = 2 - \frac{5}{3q} \Rightarrow \left\{k=1 \land q=5\right\}$$
holds.

Thus, it remains to rule out the case
$$I(n^2) < 2 - \frac{5}{3q}.$$

It is easy to show that if
$$I(n^2) < 2 - \frac{5}{3q}$$
is true, then it cannot happen that both $k=1$ and $q=5$ are true.  Thus, either $k \neq 1$ or $q \neq 5$ is true.  This means that the implication
$$k = 1 \Rightarrow q \neq 5$$
is true.

However, Iannucci (1999) [Lemma 12, page 873] proved that 
$$q=5$$
implies
$$k=1.$$

Consequently, we have that $q \neq 5$.  (Note that this implies that $q \geq 13$.)

(THIS BLOG POST IS CURRENTLY A WORK IN PROGRESS.)

27.7.16

When $p$ is an odd prime, is $\frac{p+2}{p}$ an outlaw or an index?

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the abundancy index $I$ of $x$ by
$$I(x)=\frac{\sigma(x)}{x}$$
and the deficiency $D$ of $y$ by
$$D(y)=2y - \sigma(y).$$

We first show some preliminary lemmata.

Lemma 1.  If $p$ is an odd prime, and $\frac{p+2}{p}$ is the abundancy index of some integer $n$, then $n$ is deficient.
Proof.  Since $p$ is an odd prime, $p > 2$.  Furthermore, since

$$I(n) = \frac{p+2}{p} = 1 + \frac{2}{p} < 1 + 1 = 2,$$
then $n$ is deficient.          QED

Lemma 2.  If $p$ is odd, then $\gcd(p, p+2)=1$.
Proof.
$$\gcd(p,p+2)=\gcd(p,(p+2)-p)=\gcd(p,2)=1$$
where the last equality uses the fact that $p$ is odd.          QED

Lemma 3.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $D(n) = 2n - \sigma(n) \neq 1$.
Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Now, assume to the contrary that
$$D(n) = 2n - \sigma(n) = 1.$$
Note that this implies
$$\gcd(n,\sigma(n))=\gcd(n,2n-1)=1.$$
Additionally, since $p$ is odd, by Lemma 2 we obtain
$$\gcd(p,p+2)=1.$$
Consequently, since $I(n)=\frac{p+2}{p}$, we get
$$p\sigma(n)=(p+2)n.$$
But $\gcd(p,p+2)=1$ implies that $p \mid n$, and $\gcd(n,\sigma(n))=1$ implies that $n \mid p$.  Hence, it follows that $p=n$.

But since $p$ is an odd prime, $I(n)=I(p)=\frac{p+1}{p}$, which contradicts $I(n)=\frac{p+2}{p}$.

We conclude that $D(n) = 2n - \sigma(n) \neq 1$.          QED

The following lemmas show that $p < n$, if $I(n)=\frac{p+2}{p}$ is true.  In particular, $p < n$ follows from Lemma 4 and $p \mid n$.

Lemma 4.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $I(p) < I(n)$.
Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.

We have
$$I(p)=\frac{p+1}{p}<\frac{p+2}{p}=I(n).$$

In fact, we have
$$I(n) - I(p) > I(n) - \frac{p}{p-1} = \left(\frac{1}{p-1}\right)\cdot\left(\frac{D(n)}{n}\right).$$
QED

Lemma 5.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $\gcd(n,\sigma(n)) \neq 1$.
Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Suppose to the contrary that we have $\gcd(n,\sigma(n))=1$.  Following the proof of Lemma 3, since $\gcd(p,p+2)=1$ (by Lemma 2), we have
$$n = p$$
and
$$\sigma(n) = p+2.$$
Thus,
$$p+1=\sigma(p)=\sigma(n)=p+2.$$
This is a contradiction.                        QED

Remarks.
In fact, one can show that, if $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then
$$\sigma(n)=\left(\frac{n}{p}\right)\cdot(p+2)$$
and
$$n=\left(\frac{\sigma(n)}{p+2}\right)\cdot{p}.$$
(Note that
$$\frac{n}{p}$$
and
$$\frac{\sigma(n)}{p+2}$$
are integers because of Lemma 2.)  Consequently,
$$\gcd\left(n,\sigma(n)\right) = \frac{n}{p} = \frac{\sigma(n)}{p+2}.$$
More is actually true.  One can also show that
$$p\left(2n-\sigma(n)\right)=(p-2)n$$
so that
$$D(n)=\left(p-2\right)\cdot\left(\frac{n}{p}\right)=\left(p-2\right)\cdot\left(\frac{\sigma(n)}{p+2}\right) = \left(p-2\right)\cdot\gcd\left(n,\sigma(n)\right).$$
We therefore conclude that
$$\frac{D(n)}{n}=\frac{p-2}{p}=\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$
To facilitate easier reference later, we compute:
$$1+\frac{D(n)}{n}=2\cdot\left(\frac{p-1}{p}\right)=1+\left(\frac{p-2}{p+2}\right)\cdot{I(n)},$$
and
$$2+\frac{D(n)}{n}=\frac{3p-2}{p}=2+\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$

The following lemma is actually a theorem from this preprint.  (We omit the proof.)

Lemma 6.  If $n$ is a positive integer satisfying $D(n) = 2n - \sigma(n) > 1$, then we have the following bounds for the abundancy index of $n$ in terms of the deficiency of $n$:
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}.$$

We now attempt to prove the following "result":

"Conjecture".   If $p$ is an odd prime, then $\frac{p+2}{p}$ is an abundancy outlaw.
"Proof Attempt".
Let $p$ be an odd prime, and assume to the contrary that $I(n)=\frac{p+2}{p}$.

By Lemma 1, $n$ is deficient (so that $D(n) \neq 0$).

By Lemma 3, $D(n) \neq 1$.

Thus, by Lemma 6, we have the bounds
$$\frac{2}{1 + \frac{D(n)}{n}}= \frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}}.$$

By the Remarks, $I(n) = \frac{p+2}{p}$ implies that
$$1+\frac{D(n)}{n}=2\cdot\left(\frac{p-1}{p}\right)=1+\left(\frac{p-2}{p+2}\right)\cdot{I(n)},$$
and
$$2+\frac{D(n)}{n}=\frac{3p-2}{p}=2+\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$

We compute:
$$I(n) > \frac{2n}{n+D(n)} = \frac{2}{1 + \frac{D(n)}{n}} = \frac{p}{p-1} > I(p),$$
(no contradictions so far), and
$$I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}} = \frac{\left(\frac{3p-2}{p}\right)}{\left(\frac{2(p-1)}{p}\right)} = \frac{3p-2}{2p-2} = \frac{4p-4}{2p-2} - \frac{p-2}{2(p-1)} = 2 - {\frac{1}{2}}\cdot\left(\frac{p-2}{p-1}\right),$$
(still no contradictions, per this WolframAlpha computational verification).

It thus seems fruitful to try to improve on the (trivial?) bounds
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}$$
for $n$ satisfying $D(n)>1$.

20.7.16

If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$ implies that $k=1$ and $q=5$.

(Note:  This blog post is essentially an elucidation of the answer to this MSE question.)

Here, we prove the following proposition.

Theorem.  If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$$
implies that
$$k=1$$
and
$$q=5.$$


Proof.  Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

Since $N$ is perfect, we have
$$2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \left(\frac{q^{k+1} - 1}{q - 1}\right)\cdot\sigma(n^2).$$
It follows that
$$\frac{\sigma(n^2)}{n^2} = \frac{2{q^k}\left(q - 1\right)}{q^{k+1} - 1} = \frac{2q^{k+1} - 2q^k}{q^{k+1} - 1} = \frac{2q^{k+1} - 2}{q^{k+1} - 1} - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right)$$
$$= 2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right).$$

By assumption, we have
$$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}.$$

This inequality is equivalent to
$$2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right) \geq 2 - \frac{5}{3q},$$
which, in turn, is equivalent to
$$6q\left(q^k - 1\right) \leq 5(q^{k+1} - 1).$$
This last inequality simplifies to
$$q^{k+1} - 6q + 5 \leq 0,$$
which cannot be true when $k > 1$.  Thus, we know that $k=1$.

Hence, we have
$$q^2 - 6q + 5 \leq 0.$$
But this inequality implies that
$$1 \leq q \leq 5,$$
which, together with $q \geq 5$, implies that $q=5$.

QED.

18.7.16

On conditions equivalent to the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

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

The Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

From the equation $\sigma(N)=2N$, we get that
$$\left(q^k + \sigma(q^{k-1})\right)\sigma(n^2) = \sigma(q^k)\sigma(n^2) = 2{q^k}{n^2}$$
so that we obtain
$$\frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})} = 2n^2 - \sigma(n^2).$$
(Note that $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of $n^2$, and that
$$I(x) = \frac{\sigma(x)}{x}$$
is the abundancy index of $x$.)

Now suppose that
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
Trivially, we know that
$$\frac{\sigma(n^2)}{q} \mid \sigma(n^2).$$
Thus, we have
$$\frac{\sigma(n^2)}{q} \mid \left(2n^2 - \sigma(n^2)\right) = \frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})}.$$
This implies that $I(q^{k-1})$ is an integer;  in other words, $k=1$.

The other direction
$$k=1 \Longrightarrow \frac{\sigma(n^2)}{q} \mid n^2$$
is trivial.

We therefore have the following lemma.

Lemma 1.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longleftrightarrow \frac{\sigma(n^2)}{q} \mid n^2.$$

Now assume that $k=1$.

By Lemma 1, we have
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
This implies that there exists an integer $d$ such that
$$n^2 = d \cdot \left(\frac{\sigma(n^2)}{q}\right).$$

Note that, from the equation $\sigma(N)=2N$, we obtain (upon setting $k=1$)
$$(q+1)\sigma(n^2) = \sigma(q)\sigma(n^2) = 2q{n^2}$$
from which we get
$$d = \frac{n^2}{\frac{\sigma(n^2)}{q}} = \frac{q + 1}{2}.$$

Notice that, when $k=1$, we can derive
$$\frac{5}{3} \leq I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} < 2$$
so that we have
$$\frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

We state this latest result as our second lemma.

Lemma 2.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longrightarrow \frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

Note that, when $k=1$, we have
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2q+2}{q+1} - \frac{2}{q+1} = 2 - \frac{1}{\frac{q+1}{2}} = 2 - \frac{1}{d}$$

However, notice that we know, by Lemma 2,
$$\frac{q}{2} < d \leq \frac{3q}{5} \Longrightarrow \frac{5}{3q} \leq \frac{1}{d} < \frac{2}{q} \Longrightarrow 2 - \frac{2}{q} < 2 - \frac{1}{d} = I(n^2) \leq 2 - \frac{5}{3q}.$$

Since we already have
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1}$$
this WolframAlpha computation validates that, indeed,
$$\frac{2q}{q+1} \leq 2 - \frac{5}{3q}$$
implies
$$q \geq 5.$$

Iterating the computations, we obtain
$$\frac{2 - \frac{2}{q}}{q} < \frac{1}{d} = \frac{I(n^2)}{q} \leq \frac{2 - \frac{5}{3q}}{q}$$
so that
$$2 - \frac{2 - \frac{5}{3q}}{q} \leq 2 - \frac{1}{d} = I(n^2) < 2 - \frac{2 - \frac{2}{q}}{q}.$$
But
$$2 - \frac{2 - \frac{5}{3q}}{q} = \frac{6q^2 - 6q + 5}{3q^2} \geq \frac{5}{3}$$
$$2 - \frac{2 - \frac{2}{q}}{q} = \frac{2q^2 - 2q + 2}{q^2} < 2.$$

Lastly, note that, after multiplying throughout
$$\frac{6q^2 - 6q + 5}{3q^2} \leq I(n^2) < \frac{2q^2 - 2q + 2}{q^2}$$
by $I(q) = \frac{q+1}{q}$, and asking WolframAlpha to solve the resulting inequality
$$\left(\frac{q+1}{q}\right)\cdot\left(\frac{6q^2 - 6q + 5}{3q^2}\right) \leq 2 = I(q)I(n^2) = I(qn^2) < \left(\frac{q+1}{q}\right)\cdot\left(\frac{2q^2 - 2q + 2}{q^2}\right)$$
we obtain
$$q \geq 5.$$

This blog post is currently a WORK IN PROGRESS.