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27.4.24

(Undeleted from MathOverflow) - If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a$, where $a < m$.


(Preamble: Andy Putman asserts, in the comments, that MO policy prohibits "requests to check completeness of proofs". I have therefore trimmed down my original question to the bare essentials. I hope this would already be OK.)

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The following is a proof that $m^2 - p^k$ is not a square, if $p^k m^2$ is an odd perfect number with special prime $p$.

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Assume that the estimate $p < m$ holds.  We want to show that the quantity $m^2 - p^k$ is not a square. Suppose to the contrary that $m^2 - p^k = s^2$. This is true if and only if
$$2s + 1 = p^k$$
and
$$2m - 1 = p^k.$$
This implies that $p < m < p^k$, from which we obtain $k > 1$.  Since $k \equiv 1 \pmod 4$, then we know that $k \geq 5$.  We can now use a proof by anonymous MSE user FredH to show that $m^2 - p^k$ is not a square (under the assumption $p < m$), as follows:

Since $N = p^k m^2$ is (odd) perfect, then we have the defining equation
$$\sigma(N) = 2N,$$
from which it follows that
$$\sigma(p^k)\sigma(m^2) = 2p^k m^2.$$

We know that $\sigma(p^k) = (p^{k+1} - 1)/(p - 1)$.  Since we have shown that $m = (p^k + 1)/2$, then we have the equation
$$2(p^{k+1} - 1)\sigma(m^2) = p^k (p - 1)(p^k + 1)^2. \hspace{0.76in} (*)$$

FredH considered the $GCD$ of $p^{k+1} - 1$ with the right-hand side of Equation $(*)$:
$$p^{k+1} - 1 = \gcd\left(p^{k+1} - 1, p^k (p - 1)(p^k + 1)^2\right) \leq (p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2$$
where FredH used the fact that $\left(p^{k+1} - 1\right) \mid RHS$ and the property that
$$\gcd(x,yz) \leq \gcd(x,y)\gcd(x,z).$$
But FredH also noticed that $p^{k+1} - 1 = p(p^k + 1) - (p + 1)$, whence FredH did also find
$$\gcd(p^{k+1} - 1, p^k + 1) = \gcd(p + 1, p^k + 1),$$
which is $p + 1$ because $k$ is odd. Thus,
$$(p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2 = (p - 1)(p + 1)^2.$$

Hence, the inequality
$$p^{k+1} - 1 \leq (p - 1)(p + 1)^2$$
holds.

Since $k \geq 5$, we obtain
$$p^5 < p^{k+1} - 1 \leq (p - 1)(p + 1)^2 < p^4,$$
which is a contradiction.

Hence, we now have the implication
$$p < m \Rightarrow m^2 - p^k \text{ is not a square}.$$

In other words, we have the contrapositive
$$m^2 - p^k \text{ is a square } \Rightarrow m < p.$$  
Now, suppose to the contrary that $m^2 - p^k$ is a square. This implies that $m < p$. Since $p^k < m^2$, we then have the implication $m < p \Rightarrow k = 1$.  Therefore, $k = 1$.  But we know (from the considerations above) that
$$m^2 - p^k \text{ is a square } \iff m = (p^k + 1)/2.$$  

Since $k = 1$, we infer that $m = (p + 1)/2$, or in other words, $p = 2m - 1$.  From Acquaah and Konyagin's results, we have the unconditional estimate $p < m \sqrt{3}$.  This implies that $2m - 1 = p < m \sqrt{3}$, from which we infer that
$$m(2 - \sqrt{3}) < 1$$
which contradicts the fact that $\omega(m) > 4$. (In fact, we do know that $m > {10}^{375}$, by using Ochem and Rao's lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number $N$, together with $p^k < m^2$.)

We conclude that $m^2 - p^k$ is not a square.

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Now, here goes the part where I am a bit unsure about its logical tightness, and is also my main question in this post:
> Does the following statement necessarily hold? "Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares."

If so, WLOG we may assume that
$$(m - a)^2 < m^2 - p^k < (m - a + 1)^2$$
for some positive integer $a$. We may likewise assume that $m > a$.

If I am not mistaken, these assumptions will then yield a proof for the inequality
$$m < p^k < 2am$$
except for the (problematic) case $a=1$, where we can only derive
$$p^k < 2m - 1.$$

Either way, I think the inequalities can be summarized as
$$p^k < 2am$$
for some positive integer $a < m$.

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Posted Answer

(This is not a direct answer to the original question, as posed, but rather are some thoughts that recently occurred to me, which would be too long to fit in the Comments section.)

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

Since $m^2 - p^k \text{ is not a square}$, then $p^k \neq 2m - 1$.

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If $p^k < 2m - 1$ holds, then $p < 2m - 1$ is true. (In particular, note that we get $p \leq p^k < 2m - 1 < 2m$.)

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Now assume that $2m - 1 < p^k$. We get
$$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $\sigma(p^k) > 2m$

Proof: $\sigma(p^k) \geq p^k + 1 > 2m$.  QED

Claim #2:
$$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $2pm - 2m = 2(p - 1)m = 2pm - p - 1$, which is equivalent to $p = 2m - 1 > m$. This implies that $k = 1$. But then $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2$ is a square, contradicting our result.  QED

It thus remains to rule out the case
$$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$
The RHS inequality yields $p > 2m - 1 > m$, which is equivalent to $k = 1$, since the assumption $2m - 1 < p^k$ implies $m < p^k$, which in turn implies that the biconditional $m < p \iff k = 1$ holds.
Substituting $k = 1$ into the LHS inequality yields
$$(p + 1)/2 > m.$$
From Acquaah and Konyagin's results, we have
$$p < m\sqrt{3}.$$
This implies that
$$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$
This is a contradiction.

The only way out of the contradictions is to have
$$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$
which implies, under the assumption $2m - 1 < p^k$, that
$$p < 2m - 1.$$
In particular, we obtain $k \neq 1$. Since the assumption $2m - 1 < p^k$ implies $m < p^k$, and because $m < p^k$ implies the biconditional $m < p \iff k = 1$ holds, then $k \neq 1$ is equivalent to $p < m$.

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Either way, we conclude that $p < 2m$. (Note that Acquaah and Konyagin has already proved that $p < m\sqrt{3}$, so all of this is but an academic exercise.)

30.5.23

Last Presenter (From Taguig City) During the 2023 MSP-NCR Annual Convention, Held Via Zoom Last May 27 (Saturday)

I was the last presenter during the 2023 MSP-NCR Annual Convention, held via Zoom last May 27 (Saturday), and hosted by the University of Santo Tomas - Manila, Philippines.


I was at home when I presented.

Please do check out the Beamer slides in ResearchGate, if you are interested to know more about what ideas were communicated during my talk.

For the record, no questions were asked after I was done.  (IIRC, some of the participants were graduate students from Polytechnic University of the Philippines and one of my former professors from De La Salle University.)

I finished presenting in less than fifteen minutes.

26.5.23

Some New Results on Odd Perfect Numbers - Part IV

Continuing from this earlier blog post: For the remainder of this post, assuming that $m$ is almost perfect, then $m$ must be a square (since $m$ is odd). Consequently, if the equation $\sigma(p^k)=2m$ holds (which is true if and only if $G=H=I$), then $\sigma(p^k)/2$ is also a square. This implies that $k=1$, since $$p^k + 1 \leq \sigma(p^k)=2m<2p$$ holds under the assumption $m<p$.

Summarizing our results so far, we have the chain of implications:
$$m < p \Rightarrow \Bigg((m < p^k) \land (k = 1) \land (D(m) = 1)\Bigg).$$ 

Since $\sigma(m^2)/p^k$ is also a square if $\sigma(p^k)/2$ is a square, then under the assumption that $m < p$, we obtain $\sigma(m^2)/p^k = \sigma(p^k)/2 = m$ is a square. (Note that the difference $m^2 - p^k$ is also a square.)

Additionally, we obtain
$$\sigma(p^k) = p^k + 1$$
since $k=1$ follows from the assumption $m < p$.  Since $m^2 - p^k$ is a square if and only if
$$m^2 - p^k = (m - 1)^2,$$
it then follows from the assumption $m < p$ that $\sigma(m) = 2m - 1 = p^k$, whence we obtain
$$2m = p^k + 1 = \sigma(p^k)$$
and
$$2m - 1 = \sigma(m).$$

Since $m^2 - p^k$ is not a square, then we obtain
$$m^2 - p^k = 2^r t$$
where $r \geq 2$ and $2^r \neq t$.

Now, consider the following scenario in this MathOverflow question, under which the inequality $m < p^k$ is sure to hold:

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

Case (A) implies that

$$\left(m - \max(2^r, t)\right)\left(m + \min(2^r, t)\right) > 0$$
$$p^k = m^2 - 2^r t = m^2 - \min(2^r, t)\max(2^r, t)$$
$$ > m\left(\max(2^r, t) - \min(2^r, t)\right) = m\left|2^r - t\right|,$$

from which we get

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

Note that this implies that $m < p^k$ since $\left|2^r - t\right| \geq 1$ must necessarily hold.

Next, notice that the other (remaining) cases are

$$\text{Case (B): } \min(2^r, t) < m < \max(2^r, t)$$

$$\text{Case (C): } m < \min(2^r, t).$$

The disposal of Case (C) is very easy and is left as an exercise for the interested reader.

We then have the biconditional

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

Furthermore, note mathlove's accepted answer:

$$p^k < m \Rightarrow \text{ Case (B) } \Rightarrow \left|2^r - t\right| \neq 1.$$

By the contrapositive, we obtain
$$\left|2^r - t\right| = 1 \Rightarrow m < p^k.$$

However, under the assumption $m < p$, we know that $k = 1$. Together with Acquaah and Konyagin's estimate (2012), we obtain
$$p^k < m\sqrt{3}$$
and this works under the assumption $k=1$. It follows that $m < p^k < 2m$ holds when $k=1$ is true.

Consequently, under Case (A), we derive
$$m \leq m\left|2^r - t\right| < p^k < 2m$$
from which it follows that
$$1 \leq \left|2^r - t\right| < 2$$
which forces $\left|2^r - t\right| = 1$.

We conclude that
$$m < p^k \iff \left|2^r - t\right| = 1.$$

Conclusion:  We now claim that the inequality $p < m$ indeed holds. In deed, assume to the contrary that $m < p$ holds. It follows that $k = 1$. If the equation $\sigma(p^k)=2m$ holds, then we obtain
$$2m=\sigma(p^k)=p+1$$
$$2m - 1 = p$$

But then again, as a by-product of Acquaah and Konyagin's results published in the International Journal of Number Theory in 2012, we have the upper bound
$$p < m\sqrt{3}.$$

Consequently,
$$2m - 1 = p < m\sqrt{3}$$
$$m(2 - \sqrt{3}) < 1.$$

This contradicts $m \geq 4$.

Hence, we infer that either the estimate 
$$p < m$$
or the inequation 
$$\sigma(p^k) \neq 2m$$
must be true.

26.7.22

Some Recent Updates to My Past Mathematics@StackExchange Posts

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

The recent answer to this MSE question titled "On odd perfect numbers and a GCD - Part V" appears to have successfully completed a proof for the inequation

$$\gcd(n,\sigma(n^2)) \neq \gcd(n^2,\sigma(n^2)).$$

If the proof holds water, then this shows that there cannot be an odd perfect number $N' = p^j m^2$ with special prime $p$ satisfying $p \equiv j \equiv 1 \pmod 4$ and $\gcd(p, m)=1$, of the form

$$N' = \frac{p^j \sigma(p^j)}{2}\cdot{m}.$$

You may refer to this MSE question (and the answer contained therein) for more information.


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.

 

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