Divisibility of a power

Question

I'm being asked to prove that if $n|m$ and $a > 1$ then $\frac{a^m-1}{a^n-1}$ is an integer. I would like an algebraic prove of the problem.

More generic form : http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 — lab bhattacharjee, Sep 22 '13 at 03:35

score 2 · Accepted Answer · answered Sep 21 '13 at 20:04

2

Since $n \vert m$, we have $m = nk$ where $k \in \mathbb{Z}^+$. Hence, we want to show that $\dfrac{(a^n)^k-1}{a^n-1}$ is an integer. Let $a^n = b$. We then want to show that $\dfrac{b^k-1}{b-1}$ is an integer. Now make use of the fact that $$b^k-1 = (b-1)(b^{k-1} + b^{k-2} + \cdots + b + 1)$$ to conclude what you want.

answered Sep 21 '13 at 20:04

Exactly the kind of answer I was looking for. – Stalin Pimentel Sep 21 '13 at 21:00

score 1 · Answer 2 · answered Sep 21 '13 at 20:14

1

$n|m \implies m=nc$. Where $c \in \mathbb{N}$.

$$a^n-1 \equiv 0 \pmod{a^n-1} \implies a^n \equiv 1 \pmod {a^n-1}$$

$$(a^n)^c \equiv 1^c \pmod{a^n-1} \implies a^m-1 \equiv 0 \pmod {a^n-1}$$

answered Sep 21 '13 at 20:14

Inceptio

7,881

Divisibility of a power

2 Answers2