When for given $n$ and $m$ we get $p^n-1=k(p^m-1)$ provided that $k$ is an even number

Question

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

Answer 1

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.

Answer 2

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.