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](http://www.wolframalpha.com/input/?i=Table+of[sigma_11%28n%29+mod+sigma_1%28n%29]%2C+for+n%3D2+to+1000)). Could it be that the solutions are always of the form $y=11x$?



For which $x$ and $y$ does $\sigma_x(n) $ divide $\sigma_y(n)$ for all $n$?
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.


