If m divides n, prove that a^m-b^m divides a^n-b^n.

Question

To make a long story short, I have a two part homework in an elementary number theory course I'm currently doing at uni. First part is to prove that $(a-b)$ divides $(a^n-b^n)$ with $a,b \in\mathbb{Z}$ and $n \in\mathbb{N}$, which I already managed to prove via induction.

The second part is proving more difficult for me (I've tried induction of various forms but failed so far to produce anything useful), so I wanted to see if any of you could drop me a useful hint, that makes me see in which direction the proof goes, so here is the second part:

"Prove that if $m$ divides $n$, then $(a^m-b^m)$ divides $(a^n-b^n)$ with $a,b \in\mathbb{Z}$ and $m,n \in\mathbb{N}$."

I'm assuming that I will at some point have to use part 1's result but I haven't seen an option to do so yet. Any help is greatly appreciated.^^

More generic : http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 — lab bhattacharjee, Mar 15 '14 at 17:41

score 1 · Accepted Answer · answered Mar 15 '14 at 03:25

1

Rewrite as

$$a^n - b^n = a^{mk} - b^{mk} = (a^m)^k - (b^m)^k$$

for some integer $k$. This highly resembles the first part of the question.

Let $x=a^m,y=b^m$. Then $a^n - b^n = x^k - y^k$. But you shown earlier that $x -y$ divides this, i.e. $a^m -b^m$ divides $a^n -b^n$.

answered Mar 15 '14 at 03:25

Yiyuan Lee

14,435

Thank you! This was much simpler than I thought, I wouldn't have thought of substituting this way, nice. :) – MM8 Mar 15 '14 at 03:32
1

I was thinking of something more complex too, until this came into my mind :) – Yiyuan Lee Mar 15 '14 at 03:36

If m divides n, prove that a^m-b^m divides a^n-b^n.

1 Answers1