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31.7.21

A failed attempt at a proof for the nonexistence of odd perfect numbers

The following hyperlink points to a failed attempt of the author of this blog post at a proof for the nonexistence of odd perfect numbers.

The ideas contained therein may prove useful to somebody else in the future.

13.7.21

Interesting identity involving divisors of odd perfect numbers that I discovered today (July 13, 2021)

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.  Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.

I discovered an interesting identity involving divisors of odd perfect numbers given in the Eulerian form $N = q^k n^2$ today (July 13, 2021).

The identity is:

Proposition:  If $N = q^k n^2$ is an odd perfect number with special prime $q$, then

$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$


Proof:

Our starting point is the following blog post, where it is proved that

$$\gcd(n^2, \sigma(n^2)) = 2(1 - q)n^2 + q\sigma(n^2).$$

However, note that we have

$$\frac{\sigma(n^2)}{q^k} = \gcd(n^2, \sigma(n^2)).$$

These equations are equivalent to

$$2(1 - q)n^2 + q\sigma(n^2) = \frac{\sigma(n^2)}{q^k}.$$

Factoring out $qn^2$ on the LHS, we obtain

$$qn^2 \Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q^k}.$$

Multiplying both sides of the last equation by $q^{k-1}$, we get

$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$

(Note that the RHS of the last equation is an odd integer.)

This concludes our proof.

QED.

In particular, we have

$$\frac{N}{\sigma(n^2)/q} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)}.$$

But we also know from the following MSE post that

$$I(n^2) - \frac{2(q - 1)}{q} = \frac{2(q - 1)}{q\bigg(q^{k+1} - 1\bigg)}.$$

This means that we obtain

$$\frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\bigg(q^{k+1} - 1\bigg)}{2(q - 1)} = \frac{q\sigma(q^k)}{2}.$$

But $\sigma(q^k) \equiv k + 1 \pmod 4$, since $q \equiv 1 \pmod 4$, and since $k \equiv 1 \pmod 4$, then $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$

This finding implies that $\sigma(n^2)/q$ divides $N = q^k n^2$, which is almost a proof of the desired divisibility constraint $\sigma(n^2)/q \mid n^2$, as the latter constraint would imply $k = 1$.

Got published with Keneth Adrian Precillas Dagal in NNTDM 27/2

My joint paper with Keneth Adrian Precillas Dagal was published in NNTDM 27/2 last June 12, 2021.  Here is the abstract:

In this note, we show that if $N=q^k n^2$ is an odd perfect number with special prime $q$, and $N$ is not divisible by $3$, then the inequality $q < n$ holds. We then give another unconditional proof for the inequality $q < n$ which is independent of the results of Brown and Starni.

25.3.21

On (the series of inequalities) $\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$, where $q^k n^2$ is an odd perfect number

(Preamble: The contents of this blog post were pulled verbatim from this MSE question.)

Let $N=q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.

Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, the aliquot sum of $x$ by $s(x)=\sigma(x)-x$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

Since $N=q^k n^2$ is perfect and $\gcd(q,n)=1$, we obtain
$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2.$$
Now using the fact that $\gcd(q^k,\sigma(q^k))=1$, we see that $q^k$ must divide $\sigma(n^2)$, so that
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$
Now, using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B},$$
where $B \neq 0$, $D \neq 0$, and $D-B \neq 0$, then we obtain
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}.$$

We now prove the following:

CLAIM:  If $N=q^k n^2$ is an odd perfect number with special prime $q$, then the series of inequalities
$$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$
holds in general.

PROOF:  We prove each inequality one by one, left to right.

First, assume to the contrary that
$$2n^2 - \sigma(n^2)=D(n^2)=\frac{D(n^2)}{s(q)} < \frac{2n^2}{\sigma(q)} = \frac{2n^2}{q+1}.$$
This inequality is equivalent to
$$\frac{2q}{q+1}=2 - \frac{2}{q+1} < \frac{\sigma(n^2)}{n^2} = I(n^2),$$
contradicting $I(n^2) \leq 2q/(q+1)$.

Next, suppose to the contrary that
$$\frac{2n^2}{q+1}=\frac{2n^2}{\sigma(q)} < \frac{\sigma(n^2)}{q}.$$
This inequality is equivalent to
$$\frac{2q}{q+1} < \frac{\sigma(n^2)}{n^2} = I(n^2),$$
contradicting $I(n^2) \leq 2q/(q+1)$.

Lastly, assume to the contrary that
$$\frac{\sigma(n^2)}{q} < \frac{2s(n^2)}{D(q)} = \frac{2(\sigma(n^2) - n^2)}{q - 1}.$$
This inequality is equivalent to
$$(q - 1)\sigma(n^2) < 2q\sigma(n^2) - 2qn^2 \iff 2qn^2 < (q+1)\sigma(n^2) \iff \frac{2q}{q+1} < I(n^2),$$
contradicting $I(n^2) \leq 2q/(q+1)$.

Now here is my:

QUESTION:  Note that equality holds in
$$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$
if and only if $k=1$.
So do we have
$$k=1 \iff \bigg(\frac{D(n^2)}{s(q)}=\frac{2n^2}{\sigma(q)}=\frac{\sigma(n^2)}{q}=\frac{2s(n^2)}{D(q)}\bigg) \iff \bigg(\frac{D(n^2)}{s(q^k)}=\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2s(n^2)}{D(q^k)}\bigg),$$
by treating $k$ as a placeholder for $1$?
If so, do we then have a proof for $k=1$, since
$$\bigg(\frac{D(n^2)}{s(q^k)}=\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2s(n^2)}{D(q^k)}\bigg)$$
holds unconditionally?

17.3.21

Improving the Upper Bound for $I(q^k)$

It is known that, for $q$ prime and $k$ a positive integer,

$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{1 + q + \ldots + q^k}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)},$$

where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.


This then gives the (trivial) upper bound

$$I(q^k) < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$


We prove the following theorem here.


THEOREM

If $q$ is a prime number and $k$ is a positive integer, then

$$I(q^k) < \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$

Moreover, we have

$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) < \frac{q}{q - 1}.$$


PROOF


Let $q$ be a prime number, and let $k$ be a positive integer.


First, we want to show that

$$I(q^k) < \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$

Assume to the contrary that

$$\frac{q^{k+1} - 1}{q^k (q - 1)} = I(q^k) \geq \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$

This implies that

$$q^{2k+2} - 1 = (q^{k+1} - 1)(q^{k+1} + 1) \geq {q^k}\cdot{q}\cdot{q^{k+1}} = q^{2k+2},$$

which is a contradiction.  This proves the first part of the proposition.


Next, we have to prove that

$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) < \frac{q}{q - 1}.$$

Suppose to the contrary that

$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) \geq \frac{q}{q - 1}.$$

It follows that

$$q^{k+1} \geq q^{k+1} + 1,$$

which is a contradiction.  This proves the second part of the proposition, and we are done.

26.1.21

Improving on the lower bound $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number

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


Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.


Define the abundancy index

$$I(x)=\frac{\sigma(x)}{x}$$

where $\sigma(x)$ is the classical sum of divisors of $x$.


Since $q$ is prime, we have the bounds

$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$

which implies, since $N$ is perfect, that

$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$


By considering the negative product

$$\bigg(I(q^k) - \frac{2(q-1)}{q}\bigg)\bigg(I(n^2) - \frac{2(q-1)}{q}\bigg) < 0,$$

since we obviously have

$$\frac{q}{q-1} < \frac{2(q-1)}{q},$$

then after some routine algebraic manipulations, we arrive at the lower bound

$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$


Now, a recent MO post improves on the lower bound for $I(n^2)$, as follows:

$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$

Repeating the same procedure as above, we have the negative product

$$\Bigg(I(q^k) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg)\Bigg(I(n^2) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg) < 0.$$


This implies, after some algebraic manipulations, that

$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}.$$


But WolframAlpha says that the partial fraction decomposition of the new lower bound is given by

$$\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}} = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q^{k+2}} - \frac{q}{(q - 1)(q^{k+1} + 1)}.$$


So essentially, my question boils down to:


QUESTION: Is it possible to produce an unconditional proof (that is, for all $k \geq 1$ and for all special primes $q \geq 5$) for the following inequality?

$$\frac{2(q - 1)}{q^{k+2}} > \frac{q}{(q - 1)(q^{k+1} + 1)}$$


MY ATTEMPT


I tried to ask WolframAlpha for a plot of the above inequality, it gave me the following GIF image:


WolframAlpha Inequality Plot



So it does appear that the inequality is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old.  Is it possible to prove this analytically?


And lastly:  Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?


Posted Answer


Suppose to the contrary that there exists an integer $k \geq 1$ and a (special) prime $q \geq 5$ such that

$$\frac{q}{(q-1)(q^{k+1}+1)} \geq \frac{2(q-1)}{q^{k+2}}.$$

This inequality is equivalent to

$$q^{k+3} \geq 2(q-1)^2 (q^{k+1}+1) = 2q^{k+1} - 4q^{k+2} + 2q^{k+3} + 2q^2 - 4q + 2,$$

which in turn is equivalent to

$$0 \geq q^{k+3} - 4q^{k+2} + 2q^{k+1} + 2q^2 - 4q + 2 = q^{k+2} (q - 4) + 2q^{k+1} + 2q(q - 2) + 2 > 0,$$

a contradiction.


We therefore conclude that

$$\frac{2(q-1)}{q^{k+2}} > \frac{q}{(q-1)(q^{k+1}+1)}$$

for all integers $k \geq 1$ and all (special) primes $q \geq 5$.


Hence, the new lower bound

$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$

does indeed improve on the old lower bound

$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q - 1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$


It remains to be seen whether this implies that there does exist an integer $a$ such that $k \leq a$, if $q^k n^2$ is an odd perfect number with special prime $q$.

11.12.20

On the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number (Published in NNTDM 26/4, November 2020)

My third joint paper with Immanuel T. San Diego, titled "On the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number", has been published in Notes on Number Theory and Discrete Mathematics (NNTDM 26/4, November 2020).

21.11.20

A disproof of the Dris and Descartes-Frenicle-Sorli Conjectures

On behalf of my co-author (Keneth Adrian P. Dagal), we would like to invite you to check out the following arXiv preprint on odd perfect numbers, which claims a disproof of the Dris and Descartes-Frenicle-Sorli Conjectures.  (This paper has already been submitted to a journal.)

In case you are interested, we also encourage you to join the closely related discussion in MathOverflow.