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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? Tags arithmetic, gcd, modular, possibly, primes Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gelatine1 Number Theory 4 March 16th, 2014 08:00 AM BenFRayfield Physics 0 February 20th, 2014 09:39 AM HmBe Number Theory 0 November 27th, 2012 06:45 AM Joolz Number Theory 2 October 3rd, 2012 06:15 AM Hoempa Number Theory 2 September 22nd, 2010 02:23 AM

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