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Thread: Prime GCD

  1. #1
    Super Member
    Mar 2006

    Prime GCD

    Suppose that gcd(m,n)=p, p prime, find gcd(m^2,n^2) .

    I'm guessing that p^2 is the answer to this thing, and surely it divides both m^2 and n^2. But I'm having trouble afterward, thanks!
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  2. #2
    MHF Contributor

    Aug 2008
    Paris, France
    You can first look at prime divisors. If q is a prime number dividing both m^2 and n^2, then q divides both m and n (because of primality), hence q divides p and q=p (again because of primality of p).

    So \gcd(m^2,n^2) is a power of p. It is at least p^2, as you said, so it suffices to rule out the case \gcd(m^2,n^2)=p^3. However, in the prime decomposition of m^2, the exponents are even (twice what they are for m), so that p^3|m^2 implies p^2|m (you could also simply say that p|\left({m\over p}\right)^2 and p is prime, so that p|{m\over p}, which means p^2|m). And the same holds with n, whereas p^2 does not divide both m and n (it is greater than their gcd). This concludes that p^3 can't be the gcd, hence it is p^2.

    Last edited by Laurent; Sep 17th 2008 at 07:40 AM.
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