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Is $N - \varphi(N)$ a square, if $N = q^k m^2$ is an odd perfect number with special prime $q$?

This question was inspired by this MSE question. In MSE, it is shown that $$n - \varphi(n) = (2^{p-1})^2$$ if $n = {2^{p-1}}(2^p - 1)$ is an even perfect number. Here is my question in this post: Is $...
Jose Arnaldo Bebita's user avatar
-4 votes
2 answers
173 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$ [closed]

(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 ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
137 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then must $m^2 - p^k = s^2 - t^2$ hold for some $s$ and $t$?

My present question is as is in the title: If $p^k m^2$ is an odd perfect number with special prime $p$, then must $m^2 - p^k = s^2 - t^2$ hold for some $s$ and $t$? It is known that $m^2 - p^k$ is ...
Jose Arnaldo Bebita's user avatar
0 votes
0 answers
55 views

If $p^k m^2$ is an odd perfect number with special prime $p$, is it possible to have $p = k$?

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. My question is as is in the title: If $p^k m^2$ is an odd perfect number with special prime $p$, is it ...
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
420 views

On a GCD approach to odd perfect numbers

Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Let $\sigma(z)$ denote the classical sum of divisors of the positive ...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
167 views

On "Euclidean" odd perfect numbers

In what follows, we let $N = r^s u^2$ be an odd perfect number given in Eulerian form, i.e. $r$ is the special prime satisfying $r \equiv s \equiv 1 \pmod 4$ and $\gcd(r,u)=1$. In this preprint, ...
Jose Arnaldo Bebita's user avatar
4 votes
0 answers
87 views

On Carmichael function and aliquot parts of odd perfect numbers

I've asked nine months ago this question on Mathematics Stack Exchange with identifier 4430381 and same title. There is not answer for this question on Mathematics Stack Exchange, I wondered if this ...
user142929's user avatar
0 votes
1 answer
114 views

Given that $H = \frac{n^2}{\sigma(q^k)/2} = G \times J^2$, where $q^k n^2$ is an odd perfect number, then what is the value of $\gcd(G, J)$?

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\...
Jose Arnaldo Bebita's user avatar
4 votes
1 answer
263 views

Divisibility relation with a specific sum of divisors

Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n\mid\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$ I have checked this up to $n=100$, and I ...
JoshuaZ's user avatar
  • 7,059
1 vote
1 answer
154 views

Number of distinct near-squares primes dividing an odd perfect number

I'm curious about if the following question is in the literature or what work can be done about it. Denote the number of distinct primes dividing an odd perfect number $N$ with the arithmetic function ...
user142929's user avatar
0 votes
1 answer
203 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditions on $\sigma(p^k)/2$ does $k=1$ follow?

Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Descartes (1638), Frenicle (1657), and subsequently [Sorli (2003) - ...
Jose Arnaldo Bebita's user avatar
3 votes
0 answers
180 views

Does this Theorem 2 from Dandapat et al. imply that $\gcd(\sigma(p^k),\sigma(a^2)) > 1$?

Write the odd perfect number $m=p^k a^2$ as a product of primes $$m = p^k {p_1}^{2a_1} \cdots {p_v}^{2a_v}.$$ (Note that it is known that $v \geq 9$ by work of Nielsen.) Let $N(m)$ be the number of ...
Jose Arnaldo Bebita's user avatar
3 votes
2 answers
234 views

Help with R. Ryan's "A simpler dense proof regarding the abundancy Index."

I'm reading Richard Ryan's article "A simpler dense proof regarding the abundancy index" and got stuck in his proof for Theorem 2. The Theorem is stated as follows: Suppose we have a ...
jvkloc's user avatar
  • 133
2 votes
2 answers
484 views

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

(Preamble: We have asked this same question in MSE two weeks ago, without getting any answers. We have therefore cross-posted it to MO, hoping that it gets answered here.) The topic of odd perfect ...
Jose Arnaldo Bebita's user avatar
0 votes
0 answers
107 views

On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$

(Preamble: This question is an offshoot of this answer to an MSE question with the same title.) Denote the classical sum of the divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$ and the ...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
751 views

Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. It is known that $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
321 views

On odd perfect numbers and a GCD - Part III

Let $m = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. It is known that $$\gcd(\sigma(q^k),\sigma(n^2)) = \frac{(\gcd(n,\sigma(n^...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
345 views

On odd perfect numbers $q^k n^2$ satisfying $n^2 - q^k = 2^r t$

Let $N = q^k n^2$ be an odd perfect number with special prime $q$, satisfying $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2,t)=1$. We could prove that: (1) $2^r t > 2n$. (We can modestly ...
Jose Arnaldo Bebita's user avatar
1 vote
2 answers
387 views

Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Define the abundancy index $$I(x)=\frac{\sigma(x)}{x}$$ where $\sigma(x)$ ...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
342 views

On odd perfect numbers and a GCD - Part II

(Note: A detailed version of this question was posted in MSE last April 15, 2020. It has not received any responses there as of yet. I have therefore cross-posted it here, hoping that it is ...
Jose Arnaldo Bebita's user avatar
2 votes
2 answers
643 views

On the nearest-square function and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number

This question has been cross-posted from this MSE question and is an offshoot of this other MSE question. (Note that MSE user mathlove has posted an answer in MSE, which I could not completely ...
Jose Arnaldo Bebita's user avatar
8 votes
2 answers
722 views

Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?

This question is related to the last question about van der Pol's identity for the sum of divisors. In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following ...
mathoverflowUser's user avatar
9 votes
0 answers
695 views

Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following recurrence relation ($n>1$): $$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
Perfect Number's user avatar
5 votes
0 answers
171 views

Touchard / van der Pol's identity for the sum of divisors and an elliptic curve for perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$, satisifies the following recurrence relation ($n>1$): $$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
Perfect Number's user avatar
2 votes
0 answers
117 views

If $n$ is a multiperfect number, then necessarily does one of its prime factors $p$ satisfy $p \parallel n$?

My question is as in the title: If $n$ is a multiperfect number, then necessarily does one of its prime factors $p$ satisfy $p \parallel n$? I quote from an answer by Varun Vejalla to a closely ...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
76 views

Least number of factors $\sigma(p^e)$ of representation of $\sigma(N)$ to get the least multiple of $\operatorname{rad}(N)$, for odd perfect numbers

I've cross-posted this from the post of Mathematics Stack Exchange that I've asked (Apr, 2nd 2020) with title On the least number of factors $\sigma(q^{e_q})$ to get the least multiple of $\...
user142929's user avatar
2 votes
1 answer
198 views

Bounds for two arithmetic functions, when one assumes that $n$ are odd perfect numbers

For an integer $n>1$ in this post we denote the Dedekind psi function as $\psi(n)=n\prod_{\substack{p\mid n\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$ and the product of distinct primes dividing ...
user142929's user avatar
1 vote
1 answer
186 views

Around a characterization for even perfect numbers, similar than Euclides-Euler theorem, in terms of totatives

In this post we denote the sum of divisors function as $$\sigma(n)=\sum_{1\leq d\mid n}d,$$ then an even perfect number is a positive integer $n\equiv 0\text{ mod }2$ for which $\sigma(n)=2n.$ As ...
user142929's user avatar
2 votes
0 answers
68 views

Is it possible to deduce statements for odd perfect numbers from the convolution sums involving divisor functions or other arithmetic functions?

Dividing and using some identities of [1] I've deduced the following facts, see also the remarks below. After these introductory paragraphs, to motivate our question, I am asking if we can deduce some ...
user142929's user avatar
1 vote
0 answers
56 views

Equations involving quasiperfect numbers: a first search of odd solutions for this type of equations or well succinct reasonings about these

In this post we study the following equations that involve quasiperfect numbers, denoted as $x$, that are integers such that the sum of all its positive divisors is equals to $2x+1$, and certain ...
user142929's user avatar
2 votes
0 answers
57 views

On $\sum_{\substack{1\leq d\mid n\\d<f(n)}}d$ and odd perfect numbers, for $f(n)$ the greatest prime factor or $\operatorname{rad}(n)$, respectively

First, in this paragraph we remember the definitions/notations for two number theoretic functions, for an integer $m>1$, we denote its greatest prime factor as $\operatorname{gpf}(m)$, and the ...
user142929's user avatar
1 vote
0 answers
222 views

Attempt of exploit the equation $1/\operatorname{rad}(n)=1/2-2\varphi(n)/\sigma(n)$ in the context of even perfect numbers, and a related conjecture

It is well known that the problem concerning even perfect numbers is to prove or refute if there are infinitely many of them. Few weeks ago I wrote the following conjecture, where $\varphi(n)$ denotes ...
user142929's user avatar
3 votes
4 answers
1k views

A conjecture regarding odd perfect numbers

(Note: I asked this question in MSE this June 2018 but did not receive any responses there. I have therefore cross-posted it here, hoping that it gets answered.) Let $\sigma(z)$ denote the sum of ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
482 views

If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?

STATEMENT OF THE PROBLEM If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above? MOTIVATION Let $\sigma=\sigma_{1}$ denote the classical ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
141 views

Is there an integer $r \neq q$ (with $r>1$) such that $N = q^k n^2 = \frac{r(r+1)}{2}\cdot{d}$ is an odd perfect number with $d>1$?

Slowak showed in 1999 that every odd perfect number $N = q^k n^2$ can be written in the form $$N = \dfrac{{q^k}\sigma(q^k)}{2}\cdot{D}$$ where $D>1$. From this result, it follows that every odd ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
259 views

On attempting a proof for $r > 1$, if $M = {2^r}{b^2}$ is an even almost perfect number which is not a power of two

(Preamble: I first thought that this question might be more appropriate for MSE. However, I posted it here nonetheless in the hope that someone with that brilliant idea can help with answering my ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
348 views

Is it possible to have an even superperfect number and an odd superperfect number whose product is an almost perfect number?

A number $n \in \mathbb{N}$ is said to be superperfect if $$\sigma(\sigma(n)) = 2n.$$ A number $m \in \mathbb{N}$ is said to be almost perfect if $$\sigma(m) = 2m - 1.$$ Here is my question: Is ...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
256 views

On even almost perfect numbers other than powers of two

(Note: This question is an improved version of and has been cross-posted from this MSE post.) Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an ...
Jose Arnaldo Bebita's user avatar
13 votes
2 answers
2k views

Has it been proved that odd perfect numbers cannot be triangular?

(Note: This question has been cross-posted from MSE.) Euclid and Euler proved that every even perfect number is of the form $m = \frac{{M_p}\left(M_p + 1\right)}{2}$ where $M_p = 2^p - 1$ is a prime ...
Jose Arnaldo Bebita's user avatar
6 votes
1 answer
2k views

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

The title says it all. Question If $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$? Heuristic From the Descartes spoof, with ...
Jose Arnaldo Bebita's user avatar
-1 votes
1 answer
291 views

On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$, Part II

I posted this question on MSE two days ago, but did not receive any responses. I have cross-posted it on MO, hoping it gets more attention here and that it is appropriate for this site. A positive ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
326 views

Does there exist an integer that is both solitary and almost perfect?

This question is an offshoot from the following MSE post. I hope that it is appropriate for this site. Let $\sigma(x)$ be the sum of the divisors of $x$. An integer $a$ is said to be solitary if ...
Jose Arnaldo Bebita's user avatar
5 votes
1 answer
607 views

Who is attributed with the conjecture that every multiply-perfect number greater than $1$ is even?

I know that Descartes is considered to be the first to ask whether or not odd perfect numbers exist ($n$ such that $\sigma(n)=2n$, where $\sigma(n)$ is the sum of divisors of $n$), and he also ...
Jaycob Coleman's user avatar
2 votes
0 answers
221 views

Determine whether if $n$ is a primitive pseudoperfect (semiperfect) number, then $\sigma(n)<2^{\sigma_0(n)}$

Determine whether if $n$ is a primitive pseudoperfect (semiperfect) number, then $\sigma(n)<2^{\sigma_0(n)}$. $\sigma_k(n)$ is the division function and $\sigma(n)=\sigma_1(n)$. A number is ...
Jaycob Coleman's user avatar
0 votes
1 answer
314 views

Are all known $k$-multiperfect numbers (for $k > 2$) not squarefree?

I asked the following question in MSE four ($4$) days ago, but so far nobody has posted an answer. The gist of the question is as follows: Are all known $k$-multiperfect numbers (for $k > 2$...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
286 views

What are the divisors of $2n^2 - \sigma_{1}(n^2)$ for composite $n$?

What are the divisors of $2n^2 - \sigma_{1}(n^2)$ for composite $n$? Here, $\sigma_{1}$ is the classical sum-of-divisors function. For example, $\sigma_{1}(3^2) = 1 + 3 + {3^2} = 13$. (The function ...
Jose Arnaldo Bebita's user avatar