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- January 23rd 2008, 12:58 PM #1

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- January 23rd 2008, 02:34 PM #2

- January 30th 2008, 06:29 AM #3

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- January 30th 2008, 06:57 PM #4
A sketch would go something like this:

Call P(a) be the prime factorization of a. So any element of P(a) appears n more times in P(a^n). (The point being that taking a^n introduces no new prime factors to the factorization.) Now, if GCD(a, b) = 1 then no element of P(b) is the same as any as any element of P(a). Thus no element of P(b^n) is the same as any element of P(a^n). Thus GCD(a^n, b^n) = 1.

-Dan