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This is my first Question here so i hope i am not doing that many mistakes. I am currently still at school and soon i would be able to study. Still i am interested in doing some basic proofs. I checked if there is a similiar Question asked yet but i couldn´t find any.

For the question - i struggle to show following divisibility:

$\forall a,b \in \mathbb{Z} :(a^m-b^m)|(a^n-b^n)$

with $n=k\cdot m$ and $k,m,n \in \mathbb{N} \setminus\{0\}$

What i wanted to shows is that $(a^n-b^n)=(a^m-b^m)\cdot j$

What i did so far was showing that:

$(a^m-b^m)|(a^n-b^n)$

$(a^m-b^m)|(a^{m \cdot k}-b^{m \cdot k})$

$(a^m-b^m)|[(a^m)^k-1 +1 -(b^m)^k]$

and with $(a^m-1)=(a-1)\cdot \sum_{q=0}^{m-1}(a^q)$

it goes to

$(a^m-b^m)|[(a^m-1)\cdot (\sum_{q=0}^{k-1}a^{m \cdot q}) -(b^m-1) \cdot (\sum_{q=0}^{k-1}b^{m \cdot q})]$

I feel like this might be close to get it done but since i am new to this kind of proof i am not sure if this goes to the right direction at all. Also i am stuck at this part. I would like to know if the proof untill now contains any severe mistakes and i would appreciate if anyone could give a small hint how to take further steps on this proof. Another question would be if the factorisation

$(a^m-1)=(a-1)\cdot \sum_{q=0}^{m-1}(a^q)$

is valid since i proofed it on my own through mathematical induction. Thanks in advance.

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  • $\begingroup$ $a^{mk}- b^{mk}=\left(a^m- b^m\right)\left(a^{mk-m}+ a^{mk-m-1}b+\cdots+ b^{mk-m}\right)$. $\endgroup$ – user236182 Oct 25 '15 at 19:16
  • $\begingroup$ $(a^m -1) = (a -1 ) \sum_{q-0}^{m-1} (a^q)$ this is true. I cannot see how this $(a^m-b^m)|[(a^m-1)\cdot (\sum_{q=0}^{k-1}a^{m \cdot q}) -(b^m-1) \cdot (\sum_{q=0}^{k-1}b^{m \cdot q})]$ can lead to the answer.To solve the problem you can think about this let $x = a^m , y = b^m$ why $x-y | x^k - y^k$ ? $\endgroup$ – Ameryr Oct 25 '15 at 19:32
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Hint:

$a^{mk}-b^{mk}=(a^m)^k-(b^m)^k $. You must know the factorisation of $A^k-B^k$.

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