## 5.11.22

### On the equation $s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$, if $q^k n^2$ is an odd perfect number with special prime $q$

(Note: The contents of this blog post were taken verbatim from this Mathematics Stack Exchange question.  In this regard, kindly see the rebuttals from Professor Dujella in this chat room.)

Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

---

In what follows, let us keep in mind the following lemma:

LEMMA:
If $q^k n^2$ is an odd perfect number given in Eulerian form, then $k = 1$ holds if and only if
$$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)},$$
where $s(x)=\sigma(x)-x$ is the aliquot sum of the positive integer $x$, $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.

---

$$\gcd(n^2,\sigma(n^2)) = \left(q^k t - 2(q - 1)\right)n^2 + \left((-\sigma(q^k)/2)\cdot{t} + q\right)\sigma(n^2),$$

like Tony Kuria Kimani did in this recent ResearchGate preprint.

But the equation
$$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \frac{s(n^2)}{D(q^k)/2} = \frac{D(n^2)}{s(q^k)}$$
holds.  Consequently, we obtain the simultaneous equations
$$s(n^2) = (D(q^k)/2)\cdot\gcd(n^2,\sigma(n^2))$$
$$= \left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$
$$D(n^2) = {s(q^k)}\cdot\gcd(n^2,\sigma(n^2)) = \left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2).$$

We now test whether the equation
$$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$$
holds; that is, whether the equation
$$\left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$
$$= \left(\frac{q-1}{2}\right)\cdot\Bigg(\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2)\Bigg) \tag{1}$$
is true.

For simpler algebra, let
$$u_1 = q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)$$
$$u_2 = \left(\frac{q-1}{2}\right)\cdot\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)$$
$$v_1 = (-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)$$
and
$$v_2 = \left(\frac{q-1}{2}\right)\cdot\left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right).$$

Then Equation $(1)$ becomes
$$\left(u_1 - u_2\right)\cdot{n^2} = \left(v_2 - v_1\right)\cdot{\sigma(n^2)}. \tag{2}$$

We now attempt to get simplified expressions for $u_1 - u_2$ and $v_2 - v_1$ using WolframAlpha.  We obtain the following:
$$u_1 - u_2 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(tq^k - 2q + 2\right)$$
$$v_2 - v_1 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$

Dividing both sides of Equation $(2)$ by $d = \gcd(n^2,\sigma(n^2))$, we obtain
$$\left(u_1 - u_2\right)\cdot\Bigg({\frac{n^2}{d}}\Bigg) = \left(v_2 - v_1\right)\cdot\Bigg(\frac{\sigma(n^2)}{d}\Bigg).$$
Recall that
$$d = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$
Consequently, we have
$$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = \left(v_2 - v_1\right)\cdot{q^k}. \tag{4}$$

Since $\gcd(q^k,\sigma(q^k)/2)=1$, then there exists an integer $G$ such that
$$v_2 - v_1 = G\cdot\left(\sigma(q^k)/2\right).$$
Substituting into Equation $(4)$, we get
$$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = G\cdot\left(\sigma(q^k)/2\right)\cdot{q^k},$$
from which we finally obtain
$$u_1 - u_2 = G\cdot{q^k}.$$

We therefore finally get
$$\gcd(u_1 - u_2, v_2 - v_1) = \gcd\Bigg(G\cdot{q^k}, G\cdot\left(\sigma(q^k)/2\right)\Bigg) = G\cdot\gcd(q^k, \sigma(q^k)/2) = G. \tag{5}$$

---

To recap, we have obtained
$$G = \gcd(u_1 - u_2, v_2 - v_1).$$

However, from the equations in $(3)$, we also get
$$\gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$

Hence, we finally have
$$G = \gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$

Since GCDs are always nonnegative, then $G \geq 0$, which means that we have
$$q^k - q \leq 0.$$
This implies that $k \leq 1$.  Since $k \geq 1$ ought to hold (because $k$ is a positive integer satisfying $k \equiv 1 \pmod 4$), then we now know that $k=1$.

Quite apparently, this implies that $G = 0$, whereupon we obtain $u_1 = u_2$ and $v_1 = v_2$.

---

Here is my question:

Can I define GCDs to be always positive (so that $G > 0$ in the previous section) and thereby get a proof for $k \neq 1$?

## 25.7.22

### Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms? | A proposed proof (?) for the nonexistence of odd perfect numbers

(Note 1: The following blog post is taken verbatim from this MSE question.)

(Note 2: This was cross-posted from MO, because it was not well-received there.  Will delete the MO post in a few.)

Observe that an even perfect number $M = (2^p - 1)\cdot{2^{p - 1}}$ and an odd perfect number $N = q^k n^2$ have similar multiplicative forms.  (Indeed, it is conjectured that $k=1$, and this prediction goes back to Descartes ($1638$).)

Here is my initial question:

INITIAL QUESTION
Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms?

I ask because of a related query here.

Indeed, if we could show that $$\left|2^r - t\right| = 1,$$
for $r$ and $t$ satisfying $n^2 - q^k = 2^r t$ and $\gcd(2, t)=1$, then we would have a proof for $n < q^k$, which together with Brown's estimate $q < n$ ($2016$) would yield a refutation of Descartes's Conjecture that $k = 1$.

Note that, for even perfect numbers, we actually have
$$(2^p - 1) - 2^{p-1} = 2^{p-1} - 1 = \bigg(2^{\frac{p-1}{2}} + 1\bigg)\bigg(2^{\frac{p-1}{2}} - 1\bigg) = ab,$$
where the factorization works whenever $M \neq 6$.  (Note that $\gcd(a, b) = 1$.)

We compute that
$$\left|a - b\right| = 2.$$

DISCUSSION

In the hyperlinked MO question, the following (summarized) cases were considered for odd perfect numbers, which we now examine for even perfect numbers:

Case 1:
$$2^{\frac{p-1}{2}} < \min(a,b) = 2^{\frac{p-1}{2}} - 1$$
Notice that Case 1 clearly does not hold.

Case 2:
$$\min(a,b) = 2^{\frac{p-1}{2}} - 1 < 2^{\frac{p-1}{2}} < \max(a,b) = 2^{\frac{p-1}{2}} + 1$$
Notice that Case 2 clearly holds.

It follows (from mimicking the resulting inequality $q^k < n\cdot{\left|2^r - t\right|}$ for odd perfect numbers) that
$$2^p - 1 < {2^{\frac{p-1}{2}}}\cdot{\left|a - b\right|} = {2^{\frac{p-1}{2}}}\cdot{2} = {2^{\frac{p+1}{2}}}.$$

This last inequality implies that
$$p < -\frac{2\bigg(\log(2) - \log(\sqrt{2} + \sqrt{6})\bigg)}{\log(2)} \approx 1.89997,$$
which is a contradiction to $p \geq 2$.

Case 3:
$$\max(a,b) = 2^{\frac{p-1}{2}} + 1 < 2^{\frac{p-1}{2}}$$
Notice that Case 3 clearly does not hold.

Note that for even perfect numbers, we do have
$$2^{\frac{p-1}{2}} < 2^p - 1$$
which "mimics" the conjecture $n < q^k$ for odd perfect numbers above.

Here is my final question for this post:

FINAL QUESTION
Does the exhaustion of all possible cases in the DISCUSSION section essentially disprove the existence of odd perfect numbers?

## 4.3.22

### My ResearchGate Profile

The link to my ResearchGate profile is here.

Here is a preview, as of March 4, 2022 - 9:09 PM (Manila time):

## 7.2.22

### A new approach to odd perfect numbers via GCDs

Update (February 17, 2022 - 9:39 AM [Manila time]):  This article has just been announced on math.NT in arXiv.

Here is the abstract:

Let $q^k n^2$ be an odd perfect number with special prime $q$.  Define the GCDs

$$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$

$$H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$$

and

$$I = \gcd\bigg(n,\sigma(n^2)\bigg).$$

We prove that $G \times H = I^2$.  (Note that it is trivial to show that $G \mid I$ and $I \mid H$ both hold.)  We then compute expressions for $G, H,$ and $I$ in terms of $\sigma(q^k)/2, n,$ and $\gcd\bigg(\sigma(q^k)/2,n\bigg)$.  Afterwards, we prove that if $G = H = I$, then $\sigma(q^k)/2$ is not squarefree.  Other natural and related results are derived further.  Lastly, we conjecture that the set

$$\mathscr{A} = \{m : \gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))\}$$

has asymptotic density zero.

## 2.2.22

### On factors of odd perfect numbers

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

In a previous blog post, I showed that the following equation must hold:

$$GH = I^2$$

where $G, H,$ and $I$ are defined as

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg)$$

$$I := \gcd\bigg(n,\sigma(n^2)\bigg).$$

Note that

$$H = i(q)$$

is the index of the odd perfect number $N$ at the prime $q$, and that

$$i(q) := \frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{s(q^k)} = \frac{2s(n^2)}{D(q^k)}. \tag{1}$$

Note further that

$$i(q) = q\sigma(n^2) - 2(q - 1)n^2$$

so that it follows from this blog post that

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

Since $q \equiv k \equiv 1 \pmod 4$ and $q$ is prime, then

$$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4,$$

so that $\sigma(q^k)/2$ is an integer.

Hence, $H = i(q) \mid n^2 \mid N$, so that $H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$ is a factor of the odd perfect number $N$.

Next, we show that $I \mid H$.

Since $a \mid b$ implies that $\gcd(a, c) \mid \gcd(b, c)$, then setting $a = n$, $b = n^2$, and $c = \sigma(n^2)$, we get that

$$n \mid n^2 \Rightarrow \gcd\bigg(n,\sigma(n^2)\bigg) = I \mid H = \gcd\bigg(n^2,\sigma(n^2)\bigg).$$

Since $n \mid n^2$ is true in general, we conclude that $I \mid H$.

Lastly, since $GH = I^2$ is true, then we obtain

$$\frac{H}{I} = \frac{I}{G}. \tag{2}$$

Because $I \mid H$, the LHS of Equation (2) is an integer.  It follows that the RHS of Equation (2) is also an integer.  Consequently, the divisibility constraint $G \mid I$ holds.

Hence, we have the following proposition:

THEOREM (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, and $G, H,$ and $I$ defined as follows

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$

and

$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$

then we have the equation

$$G \times H = I^2$$

and the divisibility chain

$$G \mid I \mid H.$$

In particular, we have the following corollary:

COROLLARY (02/02/2022):  If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, and $G, H,$ and $I$ defined as follows

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$

and

$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$

then $G, H,$ and $I$ are factors of $N$.

Remark:

(1)   Note that $G = H$ holds if and only if $H = I$.

(2)  Suppose that $\sigma(q^k)/2$ is squarefree.  We want to show that $G \neq H$.  Assume to the contrary that $G = H$.  The assumption that $\sigma(q^k)/2$ is squarefree implies that $\sigma(q^k)/2 \mid n$ (since $\sigma(q^k)/2 \mid n^2$ holds in general).  But the condition $\sigma(q^k)/2 \mid n$ is equivalent to $n \mid \sigma(n^2)$, because of the equation

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

Hence, since $n \mid \sigma(n^2)$, then it follows that

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

since we can compute

$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot\Bigg(\gcd\bigg(1,\frac{\sigma(n^2)}{n}\bigg)\Bigg)^2.$$

But we know from Equation (1) that

$$H = \frac{n^2}{\sigma(q^k)/2}.$$

It follows from the assumption $G = H$ that $\sigma(q^k)/2 = n$.  This contradicts Steuerwald (1937) who showed that $n$ must contain a square factor.

(Reference: R. Steuerwald, "Verschärfung einer notwendigen Bedingung für die Existenz einer ungeraden vollkommenen Zahl," S.-B. Math.-Nat. Abt. Bayer. Akad. Wiss., 1937, pp. 68-73. - See this MathOverflow answer for more information.)

Thus, we also have the following proposition:

THEOREM (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, $\sigma(q^k)/2$ is squarefree, and $G, H,$ and $I$ defined as follows

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$

and

$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$

then we have $G < I < H$.

## 24.12.21

### On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$

(The first part of this blog post is taken from this MSE question, which was asked on June 12, 2021, and the answer contained therein, which was last updated on June 20, 2021.)

QUESTION

The topic of odd perfect numbers likely needs no introduction.

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

If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an odd perfect number.  Euler proved that a hypothetical odd perfect number must necessarily have the form $n = p^k m^2$ where $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.  Dris conjectured that the inequality $p^k < m$ is true in his M. Sc. thesis, and Brown (2016) eventually produced a proof for the weaker inequality $p < m$.

Now, recent evidence suggests that $p^k < m$ may in fact be false.

THE ARGUMENT

Let $n = p^k m^2$ be an odd perfect number with special prime $p$.

Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$.  Moreover, $m^2 - p^k$ is not a square (Dris and San Diego (2020)).

This implies that we may write

$$m^2 - p^k = 2^r t$$

where $2^r \neq t$, $r \geq 2$, and $\gcd(2,t)=1$.

It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:

$$\text{Case (1): } m > t > 2^r$$

$$\text{Case (2): } m > 2^r > t$$

$$\text{Case (3): } t > m > 2^r$$

$$\text{Case (4): } 2^r > m > t$$

$$\text{Case (5): } t > 2^r > m$$

$$\text{Case (6): } 2^r > t > m$$

We can rule out Case (5) and Case (6), and under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds.

So we are now left with Case (3) and Case (4):

Under both cases left under consideration, we have

$$(m - 2^r)(m - t) < 0$$

$$m^2 + 2^r t < m(2^r + t)$$

$$m^2 + (m^2 - p^k) < m(2^r + t)$$

$$2m^2 < m(2^r + t) + p^k.$$

Since we want to prove $m < p^k$, assume to the contrary that $p^k < m$.  We get

$$2m^2 < m(2^r + t) + p^k < m(2^r + t) + m < m(2^r + t + 1)$$

which implies, since $m > 0$, that

$$2m < 2^r + t + 1.$$

Here then is our question:

Will it be possible to derive a contradiction from the inequality

$$2m < 2^r + t + 1,$$

under Case (3) and Case (4) above, considering that $2m$ is large?  (In fact, it is known that $m > {10}^{375}$.)

On OP's request, I am converting my comment into an answer.

$p^k\lt m$ is equivalent to $$m\lt\frac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have

$$p^k\lt m \iff m^2-2^rt\lt m \iff m^2-m-2^rt\lt 0 \iff m\lt\frac{1+\sqrt{1+2^{r+2}t}}{2}$$

$(7)$ is better than $2m\lt 2^r+t+1$ since

$$\frac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$

holds. To see that $(8)$ holds, note that

$$(8) \iff \sqrt{1+2^{r+2}t}\lt 2^r+t \iff 1+2^{r+2}t\lt 2^{2r}+2^{r+1}t+t^2 \iff (2^r-t)^2\gt 1 \iff |2^r-t|\gt 1$$

which does hold.

We can say that $$\bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$

since

$$(9) \iff \bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\frac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\frac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0$$

$$\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0 \iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t}$$

$$\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t) \iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0 \iff (2^r-t)^2\gt 1 \iff |2^r-t|\gt 1$$

which does hold.

It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r).$$

(The second part of this blog post is taken from this MO question.)

THE ARGUMENT

Let $n = p^k m^2$ be an odd perfect number with special prime $p$.

Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$.  Moreover, $m^2 - p^k$ is not a square (Dris and San Diego (2020)).

This implies that we may write

$$m^2 - p^k = 2^r t$$

where $2^r \neq t$, $r \geq 2$, and $\gcd(2,t)=1$.

It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:

$$\text{Case (1): } m > t > 2^r$$

$$\text{Case (2): } m > 2^r > t$$

$$\text{Case (3): } t > m > 2^r$$

$$\text{Case (4): } 2^r > m > t$$

$$\text{Case (5): } t > 2^r > m$$

$$\text{Case (6): } 2^r > t > m$$

We can easily rule out Case (5) and Case (6), as follows:

Under Case (5), we have $m < t$ and $m < 2^r$, which implies that $m^2 < 2^r t$.  This gives

$$5 \leq p^k = m^2 - 2^r t < 0,$$

Under Case (6), we have $m < 2^r$ and $m < t$, which implies that $m^2 < 2^r t$.  This gives

$$5 \leq p^k = m^2 - 2^r t < 0,$$

Under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds, as follows:

Under Case (1), we have:

$$(m - t)(m + 2^r) > 0$$

$$p^k = m^2 - 2^r t > m(t - 2^r) = m\left|2^r - t\right|.$$

Under Case (2), we have:

$$(m - 2^r)(m + t) > 0$$

$$p^k = m^2 - 2^r t > m(2^r - t) = m\left|2^r - t\right|.$$

So we are now left with Case (3) and Case (4).

Under Case (3), we have:

$$(m + 2^r)(m - t) < 0$$

$$p^k = m^2 - 2^r t < m(t - 2^r) = m\left|2^r - t\right|.$$

Under Case (4), we have:

$$(m - 2^r)(m + t) < 0$$

$$p^k = m^2 - 2^r t < m(2^r - t) = m\left|2^r - t\right|.$$

Note that, under Case (3) and Case (4), we actually have

$$\min(2^r,t) < m < \max(2^r,t).$$

But the condition $\left|2^r - t\right|=1$ is sufficient for $p^k < m$ to hold.

Our inquiry is:

QUESTION: Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case (3) and Case (4)?

Note that the condition $\left|2^r - t\right|=1$ contradicts the inequality

$$\min(2^r,t) < m < \max(2^r,t),$$

under the remaining Case (3) and Case (4), and the fact that $m$ is an integer.