How to show that if p is prime and m,n are natural numbers then $p^m -1 | p^n -1$

Question

How to show that if $p$ is prime and $m,n$ are natural numbers such that $m|n$

then

$p^m -1 | p^n -1$

I'm really stucked with this problem. Please help? Thanks!

Answer 1

I tend to like to think of division more in terms of modular arithmetic — somehow it feels more accessible to me. With that said:

Since $m|n$ there exists some $k\in\mathbb{Z}_+$ such that $n=mk$. To show that $(p^m - 1)|(p^n-1)$ we want to show that $p^n - 1 \equiv 0\pmod{p^m-1}$. Since $p^m\equiv1\pmod{p^m-1}$ we have $$p^n = p^{mk} = (p^m)^k\equiv 1^k = 1\pmod{p^m - 1}.$$ This is exactly saying that $p^n-1\equiv0\pmod{p^m-1}$!

Answer 2

Let

$n = qm; \tag 1$

then use the identity

$y^k - 1 = (y - 1) \displaystyle \sum_0^{k - 1} y^j, \tag 2$

setting

$y = p^m, \; k = q. \tag 3$

By the way, we don't need $p$ prime here; it works for any $\pm 1 \ne p \in \Bbb Z$, and it works for $p = -1$ provided $m$ odd, so that $(-1)^m - 1 \ne 0$.

Answer 3

It's always good get a feel for why a theorem is true, and to get an idea of how to prove it, by looking at examples: $n = 9$, $m = 3$

$$\dfrac {p^9-1}{p^3-1} = \dfrac {(p-1)(p^8+p^7+p^6+p^5+p^4+p^3+p^2+p+1)}{(p-1)(p^2+p+1)}$$

$$= \dfrac {p^8+p^7+p^6+p^5+p^4+p^3+p^2+p+1}{p^2+p+1}$$

$$= \dfrac {p^6(p^2+p+1)+p^3(p^2+p+1) + (p^2+p+1)}{p^2+p+1} = p^6+p^3+1$$