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December 16th, 2007, 02:49 PM   #1
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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.
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December 16th, 2007, 04:57 PM   #2
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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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