Thursday, 18 April 2019

divisibility - Gcd number theory proof: (an1,am1)=a(m,n)1

Prove that if a>1 then (an1,am1)=a(m,n)1




where (a,b)=gcd



I've seen one proof using the Euclidean algorithm, but I didn't fully understand it because it wasn't very well written.
I was thinking something along the lines of have d= a^{(m,n)} - 1 and then showing
d|a^m-1 and d|a^n-1 and then if c|a^m-1 and c|a^n-1, then c\le d.



I don't really know how to show this though...



I can't seem to be able to get d* \mathbb{K} = a^m-1.




Any help would be beautiful!

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