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Given $a,b,n \in \mathbb N$, prove that $a-b|a^n-b^n$. I think about induction. The assertion is obviously true for $n=1$. If I assume that assertive is true for a given $k \in \mathbb N$, i.e.: $a-b|a^k-b^k$, I should be able to find that $a-b|a^{k+1}-b^{k+1}$, but I can't do it. Any help is welcome. Thanks!

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marked as duplicate by leo, Lord_Farin, Bruno Joyal, Rick Decker, Matthew Pressland Nov 25 '13 at 18:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Rankeya's answer addresses what is asked here. – leo Nov 25 '13 at 17:16
The assertion is even moreobviously true for $n=0$, by the way :) – Hagen von Eitzen Nov 25 '13 at 17:32
up vote 4 down vote accepted

To complete the induction, note that

$a^{k + 1} - b^{k + 1} = a^{k + 1} - a^kb + a^kb - b^{k + 1} = a^k(a - b) + b(a^k - b^k), \tag{1}$

then simply observe that

$(a - b) \mid a^k(a - b), \tag{2}$

which is obvious, and that

$(a - b) \mid (a^k -b^k) \tag{3}$

by the induction hypothesis

$(a - b) \mid (a^k - b^k). \tag{4}$

Since $a - b$ divides both summands, it divides their sum.QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Thank you very much, Robert! – Walter r Nov 25 '13 at 17:57
@ Walter: my pleasure, glad to be of service. And thanks for your "acceptance"! I'm sorry to see your question was closed (not my doing, by the way), but I guess it has been asked before on MSE. Best Regards, Robert Kenneth Lewis – Robert Lewis Nov 25 '13 at 18:17

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