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(a,b)



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|am1 and d|an1 and then if c|am1 and c|an1, then cd.



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



I can't seem to be able to get dK=am1.




Any help would be beautiful!

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