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This question already has an answer here:

If $(a, b)=1$, prove that$$\gcd\left(\frac{a^n-b^n}{a-b}, a-b\right) = \gcd(n, a-b).$$

So since $a$ and $b$ are relatively prime that means $\gcd(a,b)=1$. Now, my question is, what theorems would I use to prove this statement? I was thinking something along the lines of using Euclidean algorithm but I wasn't quite sure how I would use it to prove this.

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marked as duplicate by Namaste, lulu, Community Mar 2 '18 at 3:30

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Hint $$a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\\=\left(a^{n-1}-a^{n-2}b \right)+2\left(a^{n-2}b-a^{n-3}b^2\right)+...+(n-1)\left(ab^{n-2}-b^{n-1}\right)+nb^{n-1}$$

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    $\begingroup$ Clever! I hadn't seen this before. $\endgroup$ – Steven Stadnicki Mar 2 '18 at 1:33
  • $\begingroup$ I'm sorry, but I'm still not grasping it. I'm not looking for a straight up answer, but could you explain why we're using this sequence of numbers and having the n decrement? I'm trying to teach myself the basics of number theory but it just won't stick in my head unless I drill it in... $\endgroup$ – CBsmith90 Mar 2 '18 at 1:42
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    $\begingroup$ @CBsmith90 Go step by step. The first term is $a^{n-1}$. We want to make $a-b$ a common factor, since this is the second element. To do this we subtract and then add back $a^{n-2}b$. The sum is now $$a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\\=\left(a^{n-1}-a^{n-2}b \right)+2a^{n-2}b+...+ab^{n-2}+b^{n-1}$$ Now, move to the next term $2a^{n-2}b$ and again try to make $a-b$ a factor. Repeat. $\endgroup$ – N. S. Mar 2 '18 at 1:46

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