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I would like to know more about divisibility among power-divisor functions. Put $\sigma_k(n) = \sum_{d \mid n} d^k$ for all positive integers $k$ and $n$.

My question here is : for which positive integers $x$ and $y$ do we have that $\sigma_x(n) $ divides $\sigma_y(n) $ for all $n$?

EDIT01 : For instance, this is true for $y = 11$ and $x = 1$, at least for $n$ from $2$ to $1000$ (see this Wolfram|Alpha calculation). Could it be that the solutions are always of the form $y=11x$?

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    $\begingroup$ See my answer to mathoverflow.net/questions/227458 , and consider what happens there if you replace $p$ by $p^x$. If $n$ does not have a special form, I doubt you will improve much on that answer. Gerhard "Welcomes You To Try, Naturally" Paseman, 2016.01.22. $\endgroup$
    – Gerhard Paseman
    Commented Jan 23, 2016 at 0:58
  • $\begingroup$ @GerhardPaseman, thank you for this , look the result in Edit :01 $\endgroup$
    – zeraoulia rafik
    Commented Jan 24, 2016 at 2:28
  • $\begingroup$ @L SPICE ,thank you very much for your edit $\endgroup$
    – zeraoulia rafik
    Commented Jan 24, 2016 at 3:44
  • $\begingroup$ The linked result in the comment above will show that for prime powers of the form n=p^10 that sigma_11(n) mod sigma(n) is nonzero. See if you can find other powers e such that sigma_11(p^e) mod sigma(p^e) is nonzero. Gerhard "Pondering What One Read Helps" Paseman, 2016.01.23 $\endgroup$
    – Gerhard Paseman
    Commented Jan 24, 2016 at 4:30
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    $\begingroup$ The most recent edit to the problem makes the problem clear and easier to solve (and, in my opinion, less appropriate for this forum). Another version, which is likely to be nore interesting and appropriate and belongs in a separate question: Given some predicate P(n,x,y) of mathematical interest, for which triples of positive integers n,x, and y do we have [P(n,x,y) and sigma_x(n) divides sigma_y(n)] ? Even if P is always true, that question would be of interest and likely hard. Gerhard "Always Looking For Quality Improvements" Paseman, 2016.01.24. $\endgroup$
    – Gerhard Paseman
    Commented Jan 24, 2016 at 20:02

1 Answer 1

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The answer is when $x=y$. Using primes $p$ for $n$, one sees that $y$ has to be an odd multiple of $x$, say $y=kx$ for $k$ odd. Now if $k>1$, use $n= p^{k-1}$, set $q=p^x$, and note that $\sigma_x(n)$ divides $q^k - 1$, so $\sigma_y(n)$ is $k$ mod $\sigma_x(n)$. So we do not have (for $k > 1$) $\sigma_x(n)$ dividing $\sigma_{kx}(n)$ for all $n$.

Gerhard "Has Likely Revealed Some Folklore" Paseman, 2016.01.23.

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  • $\begingroup$ Thank you for your answer, I think for k=p >7 holds (k must be a prime number ) $\endgroup$
    – zeraoulia rafik
    Commented Jan 24, 2016 at 16:28
  • $\begingroup$ Ok. Try n=59049, x=1, and k=y=11. Gerhard "Opinion Often Resistant To Fact" Paseman, 2016.01.24 $\endgroup$
    – Gerhard Paseman
    Commented Jan 24, 2016 at 16:51
  • $\begingroup$ nice counter example , then it fails for n >1000 $\endgroup$
    – zeraoulia rafik
    Commented Jan 24, 2016 at 17:01
  • $\begingroup$ do you meant ,it can be holds for all n execept the form wich gives k mod sigma_x(n) $\endgroup$
    – zeraoulia rafik
    Commented Jan 24, 2016 at 17:39
  • $\begingroup$ I mean that the linked relation gives you some idea of, given n, when sigma(n) divides sigma_y(n). Although you can extend this to sigma_x dividing sigma_y somewhat, it highly depends on n and not just on the exponents of its prime factorization. Gerhard "Leaving Good Parts For You" Paseman, 2016.01.24. $\endgroup$
    – Gerhard Paseman
    Commented Jan 24, 2016 at 18:12

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