This blog post is an elucidation of some of the recent discoveries/advances in the preprint titled "On a Curious Biconditional Involving the Divisors of Odd Perfect Numbers" as applied to the case of spoof odd perfect numbers, with details in the older preprint titled "The Abundancy Index of Divisors of Spoof Odd Perfect Numbers".
Recall that we call $n$ a spoof odd perfect number if $n$ is odd and $n=km$ for two integers $k, m > 1$, such that $\sigma(k)(m+1)=2n=2km$.
We begin with the following very useful lemmas. (In what follows, we take $\sigma(x)$ to be the sum of the divisors of $x$, and denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$, where $\mathbb{N}$ is the set of natural numbers or positive integers.)
Lemma 1. Let $n = km$ be a spoof odd perfect number. Then
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} \text{ is bounded } \iff \frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m} \text{ is bounded }.$$
Remark 1.1. Note that, in Lemma 1, we define
$$\sigma(m) := m+1.$$
Proof of Lemma 1. The claimed result follows from the inequalities
$$m < \sigma(m) = m + 1 < 2m$$
$$\sqrt{k} < \sigma(\sqrt{k}) < 2\sqrt{k}$$
(since $\sqrt{k}$ is a proper divisor of the deficient number $k$). Consequently,
$$\frac{m}{\sqrt{k}} < \frac{\sigma(m)}{\sqrt{k}} < 2\cdot\frac{m}{\sqrt{k}}$$
$$\frac{\sqrt{k}}{m} < \frac{\sigma(\sqrt{k})}{m} < 2\cdot\frac{\sqrt{k}}{m}$$
from which it follows that
$$\frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m} < \frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} < 2\cdot\bigg(\frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m}\bigg).$$
We therefore conclude that
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} \text{ is bounded } \iff \frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m} \text{ is bounded },$$
as desired.
Remark 1.2. In general, the function
$$f(z) := z + \frac{1}{z}$$
is not bounded from above. (It suffices to consider the cases $z \to 0^{+}$ and $z \to +\infty$.) This means that we do not expect the sum
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m}$$
to be bounded from above.
Corollary 1.1. Let $n = km$ be a spoof odd perfect number. Then the following conditions hold:
(a) $\sigma(\sqrt{k}) \neq m+1=\sigma(m)$
(b) $\sigma(\sqrt{k}) \neq m$
Proof of Corollary 1.1.
(a) Suppose that $n=km$ is a spoof satisfying the condition
$$\sigma(\sqrt{k}) = \sigma(m).$$
It follows that
$$\frac{\sigma(\sqrt{k})}{\sqrt{k}} = \frac{\sigma(m)}{\sqrt{k}}$$
and
$$\frac{\sigma(\sqrt{k})}{m} = \frac{\sigma(m)}{m}$$
from which we obtain
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m}=\frac{\sigma(\sqrt{k})}{\sqrt{k}}+\frac{\sigma(m)}{m}.$$
But we also have
$$\frac{\sigma(\sqrt{k})}{\sqrt{k}}+\frac{\sigma(m)}{m} < \frac{\sigma(k)}{k}+\frac{\sigma(m)}{m} = I(k) + \frac{m+1}{m} < 2 + \frac{10}{9} = \frac{28}{9}.$$
Finally, we get
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} < \frac{28}{9},$$
which contradicts Lemma 1 and Remark 1.2. We conclude that
$$\sigma(\sqrt{k}) \neq m+1=\sigma(m).$$
(b) Suppose that $n=km$ is a spoof satisfying the condition
$$\sigma(\sqrt{k}) = m.$$
It follows that
$$\frac{\sigma(\sqrt{k})}{m}=1.$$
But
$$1 < \frac{\sigma(\sqrt{k})}{\sqrt{k}}\cdot\frac{\sigma(m)}{m} = \frac{\sigma(\sqrt{k})}{m}\cdot\frac{\sigma(m)}{\sqrt{k}} < 2.$$
This implies that
$$1 < \frac{\sigma(m)}{\sqrt{k}} < 2.$$
In particular,
$$\frac{\sigma(\sqrt{k})}{m}+\frac{\sigma(m)}{\sqrt{k}}=1+\frac{\sigma(m)}{\sqrt{k}} < 1+2 = 3.$$
This contradicts Lemma 1 and Remark 1.2. We conclude that
$$\sigma(\sqrt{k}) \neq m.$$
Lemma 2. Let $a, b \in \mathbb{N}$.
(a) If $I(a) + I(b) < \sigma(a)/b + \sigma(b)/a$, then $a < b \iff \sigma(a) < \sigma(b)$ holds.
(b) If $\sigma(a)/b + \sigma(b)/a < I(a) + I(b)$, then $a < b \iff \sigma(b) < \sigma(a)$ holds.
(c) If $I(a) + I(b) = \sigma(a)/b + \sigma(b)/a$ holds, then either $a = b$ or $\sigma(a) = \sigma(b)$ is true.
Proof of Lemma 2. We refer the interested reader to a proof of (b) in page 6 of this preprint. The proofs for parts (a) and (c) are very similar.
Remark 2.1. Note that if we let $a = m$ and $b = \sqrt{k}$ in Lemma 2, and if we make the additional assumption that $\gcd(a, b) = \gcd(m, k) = 1$, then case (c) is immediately ruled out, as $\gcd(m, \sqrt{k}) = 1$ implies that $m \neq \sqrt{k}$. Additionally, note that $\sigma(m) \neq \sigma(\sqrt{k})$ per Corollary 1.1 (a).
Likewise, note that Lemma 1 and Remark 1.2 rules out case (b), as it implies that
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} < I(m) + I(\sqrt{k}) < \frac{m+1}{m} + I(k) < \frac{10}{9}+2=\frac{28}{9},$$
a contradiction.
Hence, we are left with the scenario under part (a):
$$I(m)+I(\sqrt{k}) = \frac{m+1}{m}+\frac{\sigma(\sqrt{k})}{\sqrt{k}} < \frac{m+1}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m},$$
which per Lemma 2 implies that
$$m < \sqrt{k} \iff m+1<\sigma(\sqrt{k}).$$
The considerations in Remark 2.1 prove the following proposition.
Theorem 3. Let $n = km$ be a spoof odd perfect number. Then the series of biconditionals
$$m < \sqrt{k} \iff m+1<\sigma(\sqrt{k}) \iff \frac{m+1}{\sqrt{k}}<\frac{\sigma(\sqrt{k})}{m}$$
hold.
Proof of Theorem 3. Trivial.
Remark 3.1. Note that
$$\frac{m+1}{\sqrt{k}} \neq \frac{\sigma(\sqrt{k})}{m}$$
is in general true under the assumption $\gcd(m, \sqrt{k})=1$, since it follows from the fact that
$$1 < \frac{m+1}{\sqrt{k}}\cdot\frac{\sigma(\sqrt{k})}{m}=\frac{m+1}{m}\cdot\frac{\sigma(\sqrt{k})}{\sqrt{k}} < 2.$$
Also, note that since $m \neq \sqrt{k}$ (because $\gcd(m, \sqrt{k})=1$), and $m+1 = \sigma(m) \neq \sigma(\sqrt{k})$ (by Corollary 1.1 (a)), then equivalently, we have the series of biconditionals
$$\sqrt{k} < m \iff \sigma(\sqrt{k}) < m+1 \iff \frac{\sigma(\sqrt{k})}{m} < \frac{m+1}{\sqrt{k}}.$$
Remark 3.2. Let us double-check the findings of Theorem 3 (and Remark 3.1) using the only spoof that we know of, as a test case.
In the Descartes spoof,
$$m = 22021 = {{19}^2}\cdot{61}$$
and
$$\sqrt{k} = {3}\cdot{7}\cdot{11}\cdot{13} = 3003$$
so that we obtain
$$m + 1 = 22022 = 2\cdot{11011}$$
and
$$\sigma(\sqrt{k}) = (3+1)\cdot(7+1)\cdot({11}+1)\cdot({13}+1) = 5376 = {2^8}\cdot{3}\cdot{7}.$$
Notice that we then have
$$\sqrt{k} < m,$$
$$\sigma(\sqrt{k}) < m+1,$$
and
$$\frac{\sigma(\sqrt{k})}{m} = \frac{5376}{22021} < 1 < \frac{22022}{3003} = \frac{m+1}{\sqrt{k}},$$
in perfect agreement with the results in Theorem 3.
Remark 3.3. Using Theorem 3, we list down all allowable permutations of the set
$$\left\{m, m+1, \sqrt{k}, \sigma(\sqrt{k})\right\}$$
(following the usual ordering on $\mathbb{N}$) subject to the biconditional
$$m < \sqrt{k} \iff m+1<\sigma(\sqrt{k}) \iff \frac{m+1}{\sqrt{k}}<\frac{\sigma(\sqrt{k})}{m}$$
and the constraints
$$m < m + 1, \text{ } \sqrt{k} < \sigma(\sqrt{k}).$$
We have the following cases to consider:
Case A: $m < m+1 := \sigma(m) < \sqrt{k} < \sigma(\sqrt{k})$
Under this case, $m < \sqrt{k} < k$ which is true if and only if $k$ is not an odd almost perfect number (see page 6 of this preprint). That is,
$$D(k) := 2k - \sigma(k) = \frac{\sigma(k)}{m} > \frac{\sigma(\sqrt{k})}{m} > 1,$$
whence there is no contradiction.
Case B: $\sqrt{k} < \sigma(\sqrt{k}) < m < m + 1$
Noting that, for the Descartes spoof, we have
$$D(k) = D({3003}^2) = 2{3003}^2 - \sigma({3003}^2) = 819 > 1,$$
so that $k$ is not almost perfect, it would seem prudent to try to establish the inequality $m < k$ for Case B.
To do so, one would need to repeat the methodology starting in Lemma 2, but this time with $a = m$ and $b = k$. Again, we are left with the scenario under part (a):
$$I(m)+I(k) = \frac{m+1}{m}+\frac{\sigma(k)}{k} < \frac{m+1}{k}+\frac{\sigma(k)}{m},$$
which per Lemma 2 implies that
$$m < k \iff m+1<\sigma(k).$$
Thus we see that we have the following corollary to Theorem 3.
Corollary 3.1. Let $n = km$ be a spoof odd perfect number. Then the series of biconditionals
$$m < k \iff m+1<\sigma(k) \iff \frac{m+1}{k}<\frac{\sigma(k)}{m}$$
hold.
Remark 3.4. Note that
$$\frac{m+1}{k} \neq \frac{\sigma(k)}{m}$$
is in general true under the assumption $\gcd(m, k)=1$, since it follows from the fact that
$$1 < \frac{m+1}{k}\cdot\frac{\sigma(k)}{m}=\frac{m+1}{m}\cdot\frac{\sigma(k)}{k} = 2.$$
Otherwise, we have $2 = (m+1)/k$ (which implies that $m=2k-1$) and $\sigma(k)/m=1$, so that $D(k) = 2k - \sigma(k) = 1$, which is equivalent to $k < m$. (This is not true under Case A above, so it suffices to show that $m < k$ directly under Case B above.)
Also, note that since $m \neq k$ (because $\gcd(m, k)=1$), and $m+1 = \sigma(m) \neq \sigma(k)$ (by a similar result as proved in Corollary 1.1 (a)), then equivalently, we have the series of biconditionals
$$k < m \iff \sigma(k) < m+1 \iff \frac{\sigma(k)}{m} < \frac{m+1}{k}.$$
Lastly, note that $\sigma(k) \neq m$ (by a similar result as proved in Corollary 1.1 (b)).
Lastly, note that $\sigma(k) \neq m$ (by a similar result as proved in Corollary 1.1 (b)).
Remark 3.5. Using Corollary 3.1 (and the considerations in Remark 3.4), we list down all allowable permutations of the set
$$\left\{m, m+1, k, \sigma(k)\right\}$$
(following the usual ordering on $\mathbb{N}$) subject to the biconditional
$$m < k \iff m+1<\sigma(k) \iff \frac{m+1}{k}<\frac{\sigma(k)}{m}$$
and the constraints
$$m < m + 1, \text{ } k < \sigma(k).$$
Since $m < k$ already holds under Case A above, we consider the following subcases under Case B above:
Case B.1: $\sqrt{k} < \sigma(\sqrt{k}) < m < m + 1 < k < \sigma(k)$
Under this subcase,
$$m < k \iff \frac{\sigma(k)}{m} > 1 \iff D(k) > 1 \iff k \text{ is not an odd almost perfect number }.$$
Since we ultimately want to prove $m < k$, this case is OK.
Case B.2: $\sqrt{k} < \sigma(\sqrt{k}) < k < \sigma(k) < m < m + 1$
Note that, under this subcase,
$$2k - \sigma(k) = D(k) = \frac{\sigma(k)}{m} < 1,$$
forcing $k$ perfect. This contradicts the fact that $k$ is a square.
Hence, we now conclude that:
CONCLUSION
MAIN THEOREM. Let $n = km$ be a spoof odd perfect number. Then $m < k$, which is true if and only if $k$ is not an odd almost perfect number.
Remark 3.6. Note that the MAIN THEOREM is an analogue of a similar result proved by Dris for the case of odd perfect numbers, i.e. the Euler factor $Q^K$ of an odd perfect number $Q^K N^2$ is less than the non-Euler part $N^2$.
Remark 3.7. Virtually everything written in this blog post (for spoofs) also apply to the case of the usual odd perfect numbers, except for the fact that in the case of spoofs, the quasi-Euler prime $m$ is tacitly assumed to have exponent $1$. (In the language of the usual odd perfect numbers, the Descartes-Frenicle-Sorli conjecture is a given for spoofs.)
ACKNOWLEDGMENTS. Arnie Dris thanks Professor Pace Nielsen for hinting over MathOverflow that most non-computational results on odd perfect numbers carry over to the case of spoofs, and vice-versa. (Non-computational results means OPN results that do not depend on numerical computations involving prime factorizations.)
