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December 16th, 2007, 02:49 PM  #1 
Newbie Joined: Dec 2007 Posts: 1 Thanks: 0  GCD's Primes possibly modular arithmetic
Can anyone provide me with a rough outline of the proof to the following statement? If a, b are positive integers such that gcd(a, b) = p for some prime p, then gcd(a^2, b^2) = p^2. 
December 16th, 2007, 04:57 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
If any prime q divides a^2, then q divides a because all primes dividing a^2 divide a. Since gcd(a, b) = p, gcd(a^2, b^2) = p^k for some integer k. Can you finish this?


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arithmetic, gcd, modular, possibly, primes 
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