## 4.2.16

### If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?

(Note:  This blog post was pulled from this [MO link].)

The title says it all.

Question

If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?

Heuristic

From the Descartes spoof, with quasi-Euler prime $q_1$:
$$n_1 = 3003 < \sigma(n_1) = 5376 < q_1 = 22021$$

So it appears that it might be possible to prove that $q + 1 \neq \sigma(n)$.

Motivation

If $q + 1 \neq \sigma(n)$, then it follows that
$$I(q) + I(n) \neq \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
from which we obtain
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q},$$
since the reverse inequality
$$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$
will violate the inequality $I(q) < \sqrt[3]{2} < I(n)$ (see this [paper]).

But the inequality
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
implies that the biconditional
$$q < n \Longleftrightarrow \sigma(q) < \sigma(n)$$
holds.

This biconditional is then a key ingredient in the proof of the main result in this [arXiv preprint].

The methods in that preprint are only sufficient to rule out $\sigma(q) = q + 1 = \sigma(n)$ if $3$ does not divide $n$, since we obtain
$$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \left(\sqrt{3} + {10}^{-375}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
(where the smallest prime factor $u$ of $N$ satisfies $u \geq 5$), whence we arrive at a contradiction.

Further Considerations

If $\sigma(q) = q + 1 = \sigma(n)$ and $3 \mid n$, then the same method does not force a contradiction, because we then have
$$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \left(\sqrt{3} + {10}^{-375}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
where $u$ is the smallest prime factor of $N$.

## 30.1.16

### On even almost perfect numbers other than the powers of two, as compared to odd perfect numbers given in Eulerian form

(This question has been cross-posted from MSE to MO.)

Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) show that, if $M \neq 2^t$ is an even almost perfect number, then $M$ takes the form $M = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite.  We call $b^2$ the odd part of the even almost perfect number $M$.

Since $M$ is almost perfect, we have
$$\left(2^{r+1} - 1\right)\left(\sigma(b^2) - b^2\right) = \sigma(b^2) - 1$$
which further implies that

$$2^{r+1} = 1 + \frac{b^2 - 1}{\sigma(b^2) - b^2}.$$

Since $b$ is composite, then we have $\sigma(b^2) > b^2 + b + 1$, from which we obtain the inequality $2^{r + 1} < b$ using the last equation.  This last inequality implies that
$$2^r < 2^{r+1} < b < \sigma(b)$$
and
$$\sigma(2^r) = 2^{r+1} - 1 < b - 1 < b$$
so that we have
$$\frac{\sigma(2^r)}{b} < 1 < 2 < \frac{\sigma(b)}{2^r}.$$

Additionally, since $b^2$ is deficient, we can write $\sigma(b^2) = 2b^2 - c$, where we compute $c$ to be
$$c = b^2 - \frac{b^2 - 1}{\sigma(2^r)}$$
from which we obtain the upper bound
$$\frac{\sigma(b)}{b} < \frac{\sigma(b^2)}{b^2} < \frac{4}{3}.$$
Lastly, since $r \geq 1$ and $2 \mid 2^r$, then
$$\frac{3}{2} = \frac{\sigma(2)}{2} \leq \frac{\sigma(2^r)}{2^r},$$
so that
$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$

Compare the results we have obtained for even almost perfect numbers other than powers of two with the conjectured inequalities for the divisors of odd perfect numbers $N = {q^k}{n^2}$ given in Eulerian form (see this [link1] and [link2]):

$$\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(q^k)}{q^k} < \frac{5}{4} < \sqrt{\frac{8}{5}} < \frac{\sigma(n)}{n} < \frac{\sigma(n)}{q^k}$$

Lastly, observe that, for the lone spoof odd perfect number $D = m{n_1}^2 = 198585576189$ that we know of (see this [link3]), we actually have

$$\frac{\sigma(n_1)}{m} = \frac{5376}{22021} < 1 < \frac{22022}{22021} = \frac{\sigma(m)}{m} < \frac{5376}{3003} = \frac{\sigma(n_1)}{n_1} < \frac{\sigma(m)}{n_1} = \frac{22022}{3003},$$

where $m$ denotes the quasi-Euler prime of $D$.

My question is:  Could there be a simple logical explanation for the discrepancies in the inequalities relating the divisors of even almost perfect numbers other than powers of two, odd perfect numbers, and spoof odd perfect numbers

## 28.1.16

### On odd perfect numbers given in Eulerian form - Part 3

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

Dr. Patrick Brown (PatrickBrown496 AT gmail.com) has communicated to me an attempt to prove my 2008 conjecture that $q^k < n$.  In particular, Dr. Brown appears to have completed a proof for the inequality $q < n$.  He accomplished this by proving the implication $k = 1 \Rightarrow q < n$.

In this post, we will investigate the implications of a proof for $k = 1$, in addition to Dr. Brown's claim that $q < n$.

Hereinafter, we will assume the Descartes-Frenicle-Sorli conjecture that $k = 1$.

We then have
$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = \gcd(n^2, \sigma(n^2).$$

If
$$\frac{\sigma(n^2)}{q} \leq q,$$
then $n^2 < \sigma(n^2) \leq q^2$, which would contradict $q < n$.

Hence
$$\frac{\sigma(n^2)}{q} > q.$$

We want to show that
$$\frac{\sigma(n^2)}{q} \neq n^2.$$

Assume that
$$\frac{\sigma(n^2)}{q} = n^2.$$

Then
$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = n^2$$
which implies that
$$\sigma(n^2) = n^2.$$
This contradicts the fact that $n > \sqrt[3]{N} > {10}^{500}$.

Observe that the inequality
$$\frac{\sigma(n^2)}{q} > n^2$$
cannot hold because this, together with $q \geq 5$, will imply that $n^2$ is abundant, contradicting $I(n^2) < 2$.

Consequently, we have
$$\frac{\sigma(n^2)}{q} < n^2.$$

Claim:
$$\frac{\sigma(n^2)}{q} > n$$

Suppose that $\sigma(n^2)/q \leq n$.  Then $\sigma(n^2) \leq qn < n^2$, a contradiction.

## 18.1.16

### On odd perfect numbers given in Eulerian form - Part 2

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

The following equations can be easily derived:

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

$$\sigma(n^2) = {q^k}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$

$$n^2 = {\frac{\sigma(q^k)}{2}}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$

From the last two equations, it can be proved that
$$\gcd(n^2, \sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})},$$
and
$$\sigma(n^2) - n^2 = \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(q^k - \frac{\sigma(q^k)}{2}\right)$$
$$= \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right).$$
This last equation expresses a relationship among the quantities
$$\sigma(n^2) - n^2,$$
$$2n^2 - \sigma(n^2),$$
$$\sigma(q^{k-1}),$$
and
$$q^k.$$

In particular, we know that
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
and
$$\frac{\sigma(q^k)}{n^2} = 2\cdot\left(\frac{\sigma(q^{k-1})}{2n^2 - \sigma(n^2)}\right).$$

Notice that, since $\sigma(q^k)\sigma(n^2) = \sigma(N) = 2N = 2{q^k}{n^2}$ and $\gcd(q^k,\sigma(q^k)) = 1$, then$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
is an odd integer.

The following papers obtain (ever-increasing) lower bounds for $\sigma(n^2)/q^k$:

(1)  J. A. B. Dris, The Abundancy Index of Divisors of Odd Perfect Numbershttps://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html
(2)  J. A. B. Dris and F. Luca, A note on odd perfect numbershttp://arxiv.org/pdf/1103.1437v3.pdf
(3)  F. J. Chen and Y. G. Chen, On Odd Perfect Numbershttp://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=8738171
(4)  K. A. Broughan, D. Delbourgo, and Q. Zhou, Improving the Chen and Chen result for odd perfect numbers, http://www.emis.de/journals/INTEGERS/papers/n39/n39.pdf
(5)  F. J. Chen and Y. G. Chen, On the index of an odd perfect numberhttp://zmath.sub.uni-goettingen.de/ZMATH/serials/en/search/zmath/?q=an:1301.11003&format=complete

From this M. Sc. thesis, we know that
$$\frac{11}{3} \leq \frac{\sigma(q^k)}{n^2} + \frac{\sigma(n^2)}{q^k}.$$

## 3.1.16

### On odd perfect numbers given in Eulerian form - Part 1

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.

That is, $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.

Since prime powers are not perfect and $\gcd(q, n) = 1$, then $q \neq n$ and $q^k \neq n$ both hold.

We want to prove the following proposition.

PROPOSITION 1:  If $N = {q^k}{n^2}$ is an odd perfect number given in
Eulerian form, then the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

holds.

PROOF:

Note that the inequation
$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k}$$
is trivial, and follows from the fact that $\gcd(q, n) = 1$ and $1 < I(q^k)I(n) < 2$.

The direction
$$q^k < n \Longrightarrow \sigma(q^k) < \sigma(n) \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
holds because $I(q^k) < \sqrt[3]{2} < I(n)$, where $I(x) = \frac{\sigma(x)}{x}$ is the abundancy index of $x$.

The implication
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longrightarrow \sigma(q^k) < \sigma(n)$$
also holds for the same reason (i.e., $I(q^k) < I(n)$).

We now prove that $\sigma(q^k) < \sigma(n) \Longrightarrow q^k < n$.

We consider three cases:

Case 1.  $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n)$

Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(q^k - n\right) < 0.$$

Consequently, we have the biconditional
$$q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k).$$

However, this biconditional contradicts $I(q^k) < I(n)$ (which can be clearly seen when written in the following form):
$$1 < \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < 1.$$

Case 2.  $I(q^k) + I(n) < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$

Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(n - q^k\right) < 0.$$

Consequently, we have the biconditional
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n).$$

In particular, the implication $\sigma(q^k) < \sigma(n) \Longrightarrow q^k < n$ holds.

Case 3.  $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} = I(q^k) + I(n)$

Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(q^k - n\right) = 0.$$

Since $q^k \neq n$, this means that
$$1 = \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}.$$

Thus, we obtain
$$\sigma(q^k) = \sigma(n) \land n < q^k.$$

This then implies that the implication
$$n < q^k \Longrightarrow \sigma(n) \leq \sigma(q^k)$$
is true.

By the contrapositive, the implication
$$\sigma(q^k) < \sigma(n) \Longrightarrow q^k < n$$
is also true.

QED.

## 24.12.15

### From MO post http://mathoverflow.net/questions/226841

The title says it all.

What is wrong with this proof that $q < n$, if $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q, n) = 1$?

Acquaah and Konyagin showed that $q < (3N)^{1/3}$.  The following proof (communicated to me by Dr. Patrick Brown) is a modification of theirs to strengthen the result to show $q < n$.

For the three cases of the proof we write
$$N = q{p^{2b}}{{r_1}^{2\beta_1}}{{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$$
where $p$ is the unique prime whereby $q \mid \sigma(p^{2b})$.  When convenient we will let $w^2 = {{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$, and write $N = q{p^{2b}}{{r_1}^{2\beta_1}}{w^2}$.

Case 1
$$q = \sigma(p^{2b})$$

Note that the assumption $q = \sigma(p^{2b})$ means $p \not{\mid} \sigma(q)$ since $q + 1 \equiv 2 \pmod p$.  So we let $p^{c_i} || \sigma({r_i}^{2\beta_i})$ for $1 \leq i \leq k$.  It is possible that $p^{c_i} = \sigma({r_i}^{2\beta_i})$ for any particular $i$, but since we know $N$ has at least ten components, at least one of the $\sigma({r_i}^{2\beta_i})$ has to have factors other than $p$.  Thus we may rewrite subscripts and assume:
$$\sigma({r_1}^{2\beta_1}) = p^{c_1}{r_2}v$$
for $c_1 \geq 0$, where $v$ is equal to any other primes dividing $\sigma({r_1}^{2\beta_1})$, including multiplicities of $r_2$ should they appear.

We now have what we need to prove this case.  Observe,
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma({r_1}^{2\beta_1})\sigma(w^2).$$
Since $p \not{\mid} \sigma(q)$, then $p^{2b - c_1} \mid \sigma(w^2)$, thus
$$2N > (q + 1)q({p^{c_1}}{r_2})(p^{2b - c_1})$$
$$2N > {q^2}{r_2}{p^{2b}}$$
Now, $r_2$ being an odd prime means $r_2 \geq 3$.  We also note that $p^{2b} > (2/3)\sigma(p^{2b})$.
Consequently,
$$2N > {q^2}(3)\frac{2}{3}\sigma(p^{2b})$$
$$N > q^3$$
from which it easily follows that $q < n$.

Case 2
$$q{r_1} \mid \sigma(p^{2b}), p \not{\mid} \sigma(q)$$

In this case, since $q < \sigma(p^{2b})$, then there is another prime dividing $\sigma(p^{2b})$.  We assume without loss of generality that $r_1 \mid \sigma(p^{2b})$.  This time, we set $w^2 = {{r_1}^{2\beta_1}}\cdots{{r_k}^{2\beta_k}}$.  Observe that $p \not{\mid} \sigma(q)$ implies $p^{2b} || \sigma(w^2)$.  This is all the machinery we need to prove $q < n$ for this case.
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma(w^2)$$
$$2N > (q + 1)q(r_1)(p^{2b})$$
$$2N > {q^2}(r_1)\frac{2}{3}\sigma(p^{2b})$$
$$2N > {q^2}(r_1)\frac{2}{3}{r_1}q$$
$$2N > \frac{2}{3}{r_1}^2{q^3}$$
As above, $r_1 \geq 3$.  Therefore,
$$N > 3q^3$$
and again, $q < n$ easily follows.

Case 3
$$q{r_1} \mid \sigma(p^{2b}), p \mid \sigma(q)$$

We borrow the same proof method Acquaah and Konyagin borrowed from Luca and Pomerance.  We also would not really utilize the hypothesis that $r_1 \mid \sigma(p^{2b})$ as it would not buy us the extra factor we need.  For that, we look back to case 1, and let $p^{c_i} || \sigma({r_1}^{2\beta_i})$, for $1 \leq i \leq k$ and $p^{c_q} || \sigma(q)$.  Again, we assume without loss of generality that $\sigma({r_1}^{2\beta_1}) = p^{c_1}{r_2}v$ for $c_1 \geq 0$ as we did in case 1.

Let $u = \sigma(p^{2b})/q$.  Since
$$\sigma(p^{2b}) \equiv 1 \pmod p, q \equiv -1 \pmod p$$
we know $u \equiv -1 \pmod p$.  Since $u$ is odd we know $u \neq p - 1$ and thus $u \geq 2p - 1$.

By assumption, $c_q \geq 1$.  For $w^2 = {{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$, we have $p^{2b - c_q - c_1} || \sigma(w^2)$, which implies
$$\sigma(w^2) \geq p^{2b - c_q - c_1}.$$
Observe now,
$$p^{2b+1} - 1 = (p - 1)\sigma(p^{2b}) = (p - 1)uq = (p - 1)u\sigma(q) - (p - 1)u.$$
Therefore, $(p - 1)u \equiv 1 \pmod{p^{c_q}}$, which implies that $(p - 1)u > p^{c_q}$.

Combining inequalities yields,
$$\sigma(w^2)(p - 1)u > p^{2b - c_1} \Longrightarrow \sigma(w^2)u > \frac{p^{2b - c_1}}{p - 1}.$$
This should be all we need (for $w$ as defined in case 1):
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma({r_1}^{2\beta_1})\sigma(w^2)$$
$$2N = (q + 1)uq(p^{c_1}{r_2}v)\sigma(w^2)$$
$$2N > {q^2}\frac{p^{2b - c_1}}{p - 1}{p^{c_1}{r_2}}$$
$$2N > {q^2}{r_2}\frac{p^{2b}}{p - 1}$$
$$2N > {q^2}{r_2}\frac{2\sigma(p^{2b})}{3(p - 1)}$$
$$2N > {q^2}{r_2}\frac{2uq}{3(p - 1)}$$
Recall that $u \geq 2p - 1$ and again $r_2$ being an odd prime means $r_2 \geq 3$.
$$2N > {q^3}(3)\frac{2}{3}\frac{2p - 1}{p - 1}$$
$$2N > {q^3}(3)\frac{2}{3}(2)$$
$$N > 2q^3$$
And again, we get $q < n$.

## 6.12.15

### EAPN Research (October 2015) - Some updates on even almost perfect numbers other than the powers of two

Let $\sigma(x)$ be the sum of the divisors of $x$, and call the function

$$D(x) = 2x - \sigma(x)$$

as the deficiency of $x$.

If $D(n) = 1$, then $n$ is called an almost perfect number.

Antalan and Tagle (in a 2014 preprint titled "Revisiting forms of almost perfect numbers") show that, if $n \neq 2^k$ is an even almost perfect number, then $n$ takes the form

$$n = {2^r}{b^2}$$

where $b$ is an odd composite integer.

Using their result, we have

$${2^{r+1}}{b^2} - 1 = 2n - 1 = \sigma(n) = \sigma(2^r)\sigma(b^2) = (2^{r+1} - 1)\sigma(b^2)$$

from which we obtain

$${2^{r+1}}\left(\sigma(b^2) - b^2\right) = \sigma(b^2) - 1.$$

From this equation, we obtain the following results (summarized from this paper):

Claim 1
$$\sigma(b^2) = 2b^2 - c$$
where
$$c = b^2 - \frac{b^2 - 1}{2^{r + 1} - 1}$$

Claim 2
$$c \geq \frac{2b^2 + 1}{3}$$

Claim 3
If ${2^r}{b^2}$ is an almost perfect number with $\gcd(2, b) = 1$ and $b > 1$, then $b$ is solitary.

Claim 4
If ${2^r}{b^2}$ is an almost perfect number with $\gcd(2, b) = 1$ and $b > 1$, then $\left(\sigma(b^2) - b^2\right) \mid \left(b^2 - 1\right)$.

Claim 5
Suppose that there exist at least two distinct even almost perfect numbers
$$M_1 = {2^{r_1}}{b_1}^2$$
and
$$M_2 = {2^{r_2}}{b_2}^2$$
with $\gcd(2, b_1) = \gcd(2, b_2) = 1$, $b_1 > 1$, $b_2 > 1$, and $r_1 \neq r_2$.  Then $b_1 \neq b_2$.

Claim 6
If ${2^r}{b^2}$ is an almost perfect number with $\gcd(2, b) = 1$ and $b > 1$, then $r < \log_{2}{b} - 1$.

### An improvement to "On a Conjecture of Dris Regarding Odd Perfect Numbers"

Here, we outline an improvement to On a Conjecture of Dris Regarding Odd Perfect Numbers:

Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form. We want to show that
$$\frac{\sigma(n)}{q} \ne \frac{\sigma(q^k)}{n}.$$

Assume that
$$n\sigma(n) = q\sigma(q^k).$$

Since $\gcd(q,n) = 1$, this means that $n \mid \sigma(q^k)$ and $q \mid \sigma(n)$, so that
$$\frac{\sigma(q^k)}{n}$$
and
$$\frac{\sigma(n)}{q}$$
are equal positive integers.

But $\sigma(q^k)$ is even and $n$ is odd.  We therefore have:
$$2 \le \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}$$

The case
$$2 = \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}$$
is ruled out in this MSE post.

Therefore, we obtain
$$4 \le \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}$$

Multiplying both sides of the inequality and equation by $\frac{\sigma(n)}{q^k}$, we get

$$4\cdot{\frac{\sigma(n)}{q^k}} \le {\frac{\sigma(n)}{q}}\cdot{\frac{\sigma(n)}{q^k}} = {\frac{\sigma(q^k)}{n}}\cdot{\frac{\sigma(n)}{q^k}} < 2$$

Multiplying both sides of the inequality and equation by $\frac{\sigma(q)}{n}$, we obtain

$$4\cdot{\frac{\sigma(q)}{n}} \le {\frac{\sigma(q^k)}{n}}\cdot{\frac{\sigma(q)}{n}} = {\frac{\sigma(q)}{n}}\cdot{\frac{\sigma(n)}{q}} < 2$$

Consequently, we have
$$\frac{\sigma(n)}{q^k} < \frac{1}{2}$$
and

$$\frac{\sigma(q)}{n} < \frac{1}{2}$$
from which it follows that
$$\frac{\sigma(n)}{q^k} < \frac{1}{2} < 4 \le \frac{\sigma(q^k)}{n}$$
and
$$\frac{\sigma(q)}{n} < \frac{1}{2} < 4 \le \frac{\sigma(n)}{q}.$$
We conclude that we must have
$$q < n < q^k$$
so that $k > 1$.

THIS POST IS CURRENTLY A WORK IN PROGRESS.