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Math Help - GCD = p (prime)

  1. #1
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    GCD = p (prime)

    Suppose (a, b) = p, which is a prime, then determine EVERY possible value of:

    1.) (a^2, b^2)

    2.) (a^2, b)

    3.) (a^3, b^2)

    Note: (a, b) means gcd(a, b).
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  2. #2
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    Hello, Ideasman!

    I don't know what they mean by "EVERY possible value".
    . . For a given set of numbers, there is exactly one GCD.


    Suppose GCD(a, b) = p, a prime,
    then determine EVERY possible value of:

    1) .GCD(a, b)
    Let: .a = pm .and .b = pn
    . . where m and n are relatively prime (they have no common factors).

    Then: .a .= .pm
    . and: .b .= .pn

    Therefore: .GCD(a, b) .= .p



    2) .GCD(a, b)
    We have: .a .= .pm
    . . . .and: . b . = .pn

    Therefore: .GCD(a, b) .= .p



    3) .GCD(a, b)

    We have: .a .= .pm
    . . . and: . b .= .pn

    Therefore: .GCD(a, b) .= .p

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  3. #3
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    Are you sure that these are unique?
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  4. #4
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    Quote Originally Posted by Ideasman View Post
    Are you sure that these are unique?
    Why would you think it could be otherwise?
    Do you understand the unique prime factorization theorem?
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  5. #5
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    You cannot define gcd(a,b) without making it unique, otherwise the function gcd: N x N --> N is not well-defined.
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