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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions

1 vote
0 answers
57 views

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
-3 votes

If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some posi...

(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 …
Jose Arnaldo Bebita's user avatar
-4 votes
2 answers
167 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some posi... [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 essentials. I hope …
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
134 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then must $m^2 - p^k = s^2 - 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 pos …
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
410 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 integ …
Jose Arnaldo Bebita's user avatar
0 votes

A conjecture regarding odd perfect numbers

This answer is a direct response to Professor Pace Nielsen's suggestion to perform a "sanity check" and complements this other answer by providing a proof of the following biconditionals: If $p^k m^2 …
Jose Arnaldo Bebita's user avatar
0 votes

A conjecture regarding odd perfect numbers

Further to this recent corrigendum in NNTDM, which corrects an oversight in Some modular considerations regarding odd perfect numbers – Part II, we realized that we do in fact have the following bicon …
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
163 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, Brow …
Jose Arnaldo Bebita's user avatar
0 votes
Accepted

Given that $H = \frac{n^2}{\sigma(q^k)/2} = G \times J^2$, where $q^k n^2$ is an odd perfect...

This is a partial answer, which uses the ideas in my earlier comments. We compute $$\gcd(G,J)={p_1}^{\min\left(\min(b_1,2a_1 - b_1),2a_1 - b_1 - \min(a_1,2a_1 - b_1)\right)} \cdots {p_m}^{\min\left(\ …
Jose Arnaldo Bebita'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...

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 $\sig …
Jose Arnaldo Bebita's user avatar
0 votes
Accepted

If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditio...

This is a partial response, as it does not directly answer the original question that was asked. Additionally, what follows are actually some remarks that would be too long to fit in the Comments sec …
Jose Arnaldo Bebita's user avatar
0 votes
Accepted

On odd perfect numbers and a GCD - Part III

Let $p^s Q^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv s \equiv 1 \pmod 4$ and $\gcd(p,Q)=1$. I did some more digging on when the equations $$\gcd(Q^2, \sigma(Q^2)) = \gcd( …
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
200 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditio...

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) - Conjec …
Jose Arnaldo Bebita's user avatar
2 votes

Resources where I can find open problems in number theory along with their level of difficulty

Do check out the Handbook of Number Theory (Volumes I and II), if you need to refer to a compendium of the latest results on number theory. (I forget which Volume has open problems.)

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