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 

You can write $$\frac{p^n-1}{p^m-1}=\frac{(p^{m})^l-1}{p^m-1}=\sum_{i=0}^{l-1}(p^m)^i$$ There are $l$ terms on the RHS all being odd, so you need $l$ to be even, that is $\frac{n}{m}$ must be even.
