## 15.3.16

## 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.

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

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 $q + 1 \neq \sigma(n)$.

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

$$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$

$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$

(where the smallest prime factor $u$ of $N$ satisfies $u \geq 5$), whence we arrive at a contradiction.

If $\sigma(q) = q + 1 = \sigma(n)$ and $3 \mid n$, then the same method does not force a contradiction, because we then have

$$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$

$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$

where $u$ is the smallest prime factor of $N$.

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

$$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$

$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$

(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

$$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$

$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$

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

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

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

Lastly, observe that, for the lone

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

where $m$ denotes the

My question is: Could there be a simple logical explanation for the

Antalan and Tagle (in a 2004 preprint titled

**) 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***Revisiting forms of almost perfect numbers***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 $k = 1 \Rightarrow q < n$. (

__Update (Feb 6 2016)__: Dr. Brown's preprint has appeared in the arXiv.)
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

$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = \gcd\left(n^2, \sigma(n^2)\right).$$

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

$$q^k.$$

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 Numbers*, https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html
(2) J. A. B. Dris and F. Luca,

*A note on odd perfect numbers*, http://arxiv.org/pdf/1103.1437v3.pdf
(3) F. J. Chen and Y. G. Chen,

*On Odd Perfect Numbers*, http://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 number*, http://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 with $\sigma(q^k) \neq \sigma(n)$, 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 $1 < I(q^k)I(n) < 2$.

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):

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

Hence, in general, the inequality

$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

holds.

Under this case, we have

Since $q^k \neq n$, this means that

Hence, in general, the inequality

$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

holds.

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

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

$$\sigma(q^k) = \sigma(n)$$

which is not true by assumption.

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

Under this case, we have

$$\left(\sigma(q^k) - \sigma(n)\right)\left(n - q^k\right) < 0.$$

Consequently, we have the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n).$$

This then implies that we have the biconditionals

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

under the assumption $\sigma(q^k) \neq \sigma(n)$.

This then implies that we have the biconditionals

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

under the assumption $\sigma(q^k) \neq \sigma(n)$.

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 $q < n$. (

__Update (Feb 6 2016)__: Patrick's preprint containing a proof for $q < n$ (for all odd perfect numbers) as well as for $q^k < n$ (under some mild conditions) has appeared in the arXiv.)
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 $q < n$.