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.

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.

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 $k = 1$.

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.

(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:

$$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$

holds in general.

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:

$$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$

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?

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.

(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 $q$.

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).

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.