When for given $n$ and $m$ we get $p^n-1=k(p^m-1)$ provided that $k$ is an even number Consider $p$ is an odd prime  number.
Assume that  $n$ and $m$ are two positive integer numbers
provided that $m \mid n$ which results in $p^m-1 \mid p^n-1$.
Therefore, we get $p^n-1=k(p^m-1)$ where $k$ is a positive integer number. 
My question: Suppose that $n$ is fixed. 
With which condition over $m$, we get $k$ is an even number 
($k$ may be odd or even number).
Thanks for any suggestions 

Since $m|n$ we find $n=tm$ for some integer $t$. Then $p^n-1=(p^m-1)(1+p^m+p^{2m}+...+p^{(t-1)m})$, so $k=1+p^m+p^{2m}+...+p^{(t-1)m}$. Since all the terms $p^{im}$ are odd, $k$ is even if and only if $t$ is even.
