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Jose Arnaldo Bebita's user avatar
Jose Arnaldo Bebita's user avatar
Jose Arnaldo Bebita's user avatar
Jose Arnaldo Bebita
  • Member for 14 years, 1 month
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What is known about sum of divisors of $n$ equal to $kn$ with $k$ not being two?
A list of such numbers is tabulated here.
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Re: Mordell's equation $y^2 = x^3 + k$ and perfect numbers
@SylvainJULIEN: Apologies for the delayed revert. But my recent preliminary computations suggest that we can take $K=k+1$. I am still working out the details. Will let you know.
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Re: Mordell's equation $y^2 = x^3 + k$ and perfect numbers
Thank you for this comprehensive answer, @JoeSilverman. I will keep tab of your hints as I attempt to write a solution tonight.
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Arithmetic derivatives and non-commutative generalizations
I am a bit busy right now, I will check and get back to you with an update ASAP.
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Arithmetic derivatives and non-commutative generalizations
In that case, I would highly suggest that you check out Pace Nielsen's papers on ring theory. Some of his results may apply to what you need.
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Arithmetic derivatives and non-commutative generalizations
I do not have much background in higher calculus too, but I cannot help but say: A non-commutative generalization of the arithmetic derivative would be very nice! =) Do let me know what you find out.
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Arithmetic derivatives and non-commutative generalizations
It is fair to question: What were you hoping to achieve with the non-commutative generalizations?
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Re: Mordell's equation $y^2 = x^3 + k$ and perfect numbers
TeXified post, added minor contextual updates
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Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?
@MarkVSapir: If you do not have a mathematical proof, then I find it surprising that you (as a professional mathematician) can plainly state that there are no odd perfect numbers.
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If $N = qn^2$ is an odd perfect number with $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
edited body
accepted
If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
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If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
Ohh, yes! But then that does not require proving that $m^2 - p^k \neq \square$! Ohh my goodness! If you could just write that out as an answer, I will be more than happy to accept it (and award the bounty to you), @AlexRavsky =)
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If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
Thank you for your comment, @AlexRavsky! I can see that your claimed inequality $p^k < 2(m - 1)m$ is equivalent to $m^2 - p^k > 0 > 2m - m^2$, which is always true. (We do know, however, that $p^k < m^2$ always holds. So does that take care of your concern about the failure of my "finishing arguments"?)
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If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
I appreciate your comment, @ToddTrimble!
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If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
deleted 225 characters in body
revised
If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
trimmed down the original question, deleted the request to check completeness of proof to conform to MO policy
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If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
@AndyPutman: In that case, please do allow me to edit my question, to conform to MO policy, so that it does not look like a "request to check completeness of proofs".
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If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
added context and a complete proof for the estimate $p < 2m$
revised
If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$
added context and a complete proof for the estimate $p < 2m$
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Is $N - \varphi(N)$ a square, if $N = q^k m^2$ is an odd perfect number with special prime $q$?
@JoshuaZ: I appreciate your comment, Josh! I was hoping someone with more experience on the Euler totient function would be able to settle this one. (This question was mainly triggered by an inquiry that I received via e-mail.)
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