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I would like to know if there are infinitely many $k$ for which $$\sigma(k)/k=n^p$$ such that $m=k{n}^{p-1}$ with $m,n>0$ and $p$ is an odd prime?

Note: $\sigma(\frac{m}{{n}^{p-1}})$ is the sum of divisors function of $\frac{m}{ {n}^{p-1}}$ such that $m=k{n}^{p-1}$.

Edit 2: I edited the question as it has the same meaning for the precedent.

Thank you for any help.

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closed as off-topic by GH from MO, Boris Bukh, Marco Golla, Alex Degtyarev, Yoav Kallus Jul 29 '15 at 17:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – GH from MO, Boris Bukh, Marco Golla, Alex Degtyarev, Yoav Kallus
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ It is trivial that $\sigma(k)>k$ for any $k>1$, since $1$ and $k$ are two distinct divisors of $k$. Hence your equation is only satified when $m=n=1$. At any rate, this question is not of research level, voting to close. $\endgroup$ – GH from MO Jul 29 '15 at 1:35
  • $\begingroup$ @GHfromMO, thank you for your comments and i didn't get attention for no solution existed for sigma(k)=k for all k\geq 1 $\endgroup$ – zeraoulia rafik Jul 29 '15 at 2:16
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    $\begingroup$ With the notation $m=kn$ your new equation becomes $\sigma(k)/k=n^2$. Hence it would be more natural to ask if there are infinitely many $k$'s for which $\sigma(k)/k$ is a square. $\endgroup$ – GH from MO Jul 29 '15 at 2:22
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    $\begingroup$ BTW there should be infinitely many $k$, such as $51590^2$, for which both $k$ and $\sigma(k)$ are square, and thus $\sigma(k)/k$ is a rational square. Perhaps this already follows from known results. $\endgroup$ – Noam D. Elkies Jul 29 '15 at 4:37
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    $\begingroup$ @Noam, "Numbers $n$ such that sum of divisors of $n^2$ is a square" is oeis.org/A008847 which has a reference to a preprint of Kaplansky. I can't find the Kaplansky work anywhere, so I don't know whether he proved anything in it. $\endgroup$ – Gerry Myerson Jul 29 '15 at 23:53
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The question can be reformulated as:

Question. Are there infinitely many $k$'s such that $\sigma(k)/k$ is a square?

If the answer is yes, then (trivially) there are infinitely many multiperfect numbers. This consequence is not known unconditionally, so at the moment we cannot prove that the answer is yes.

If the answer is no, then it implies that each perfect number is divisible by a prime from a fixed finite set of primes, because the product of two coprime perfect numbers satisfies $\sigma(k)/k=4$. This consequence is not known unconditionally, so at the moment we cannot prove that the answer is no.

In short, it seems that above question is out of reach at the moment.

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  • $\begingroup$ thank you for your answer, if it was asked before ,could you give me a paper or .. ? $\endgroup$ – zeraoulia rafik Jul 29 '15 at 3:09
  • $\begingroup$ @zeraouliarafik: I don't think it was asked before, but I am not familiar with the literature of multiperfect numbers. $\endgroup$ – GH from MO Jul 29 '15 at 3:35
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    $\begingroup$ @zeraouliarafik, hard, yes; interesting, that's in the eye of the beholder. In two short paragraphs, GH has reduced it to some well-known hopeless problems. There is no difficulty in Number Theory in tweaking well-known hard problems to come up with any number of further hard problems. $\endgroup$ – Gerry Myerson Jul 29 '15 at 6:47
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    $\begingroup$ @zeraouliarafik: My answer applies almost verbatim for your new question: if $k$ is the product of $p$ pairwise coprime perfect numbers, then $\sigma(k)/k=2^p$. Also, please do not edit questions for which there is an accepted answer. Instead, open a new question. $\endgroup$ – GH from MO Jul 30 '15 at 9:23
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    $\begingroup$ @WillJagy: I don't have an opinion as I haven't thought about it. Maybe later, but at the moment I am busy with other things. Also, I am not too familiar with this type of problems. $\endgroup$ – GH from MO Jul 30 '15 at 19:30

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