I would like to proof or disproof this claim ,but i don't have enough information about divisor function structure .
Claim : for any positive integer $x, y ,n $ such that :$x\neq y$ and $n>1$,then we have : $$\sigma_y(n) \bmod \sigma_x(n)\neq 0 $$ if $$y\bmod x \neq 0$$ "which it does means no integer $n >1$ satisfies the titled claim"
Edit01: I edited the question as my first wrong typo and i meant in the case of y mod $x \neq 0$ .
I run some computation in WA about that from $n=2 $ to $1000$ with $y =17$ and $x=15$ I have got this and it's works with many values of $y \bmod x \neq 0$ , i would like to know if there some one give me any counter example or any idea for proving it correct
Note:$\sigma_y(n) , \sigma_x(n) =\sum d^y,\sum d^x,$ respectively are sum power divisor function
Thank you for any help
A possible counterexample is $y=1$, $x=2$ and $n=6$. Then $$2\bmod 1=0,$$ but $$\sigma_2(6) \bmod \sigma_1(6)=1^2+2^2+3^2+6^2\bmod 12\neq 0.$$ So we have $\sigma_y(n)\bmod \sigma_x(n)\neq 0$, although $y\bmod x=0$, i.e., although $y\bmod x\neq 0$ is not satisfied. Hence the "iff" is not true.