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Math Help - GCD

  1. #1
    Junior Member
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    Mar 2012
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    Bogota D.D
    Posts
    30

    GCD

    Hi everyone.

    Let a,b,c\in{\mathbb{Z}}. If a|bc, show that a|(a,b)(a,c) where (,) is GCD



    Well, if a|bc\Rightarrow{bc=ak}, then I can multiply in the both sides by (a,b) and (a,c) hence

    bc(a,b)(a,c)=ak(a,b)(a,c) next, (a,b)(a,c)=ak(\frac{a}{b},1)(\frac{a}{c},1), but \frac{a}{b},\frac{a}{c}\in{\mathbb{Z}} by hypothesis.

    Therefore a|(a,b)(a,c).
    But the problem is that bc\neq{0}
    Correcto?
    Thanks.
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  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
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    420
    Thanks
    64

    Re: GCD

    \begin{cases}\gcd(a,b)=\frac{b}{k_1}\\\gcd(a,c)= \frac{c}{k_2} \end{cases}

    hence :

    \gcd(a,b) \cdot \gcd(a,c)=\frac{b\cdot c}{k_1 \cdot k_2}

    since :

    a \mid b\cdot c \Rightarrow a \mid \frac{b\cdot c }{k_1 \cdot k_2} \Rightarrow a \mid \gcd(a,b) \cdot \gcd(a,c)
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  3. #3
    Junior Member
    Joined
    Mar 2012
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    Bogota D.D
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    Re: GCD

    Hi, I have another idea.

    Let d=(a,b) and r=(a,c) be then d=ax+by and r=ax'+cy', now we multiplity d\cdot r hence

    d\cdot r=(ax+by)(ax'+cy')=a^2xx'+acxy'+abx'y+bcyy' but by hypothesis bc=ak therefore (a,b)(a,c)=a(axx'+cxy'+bx'y+kyy') so a|(a,b)(a,c)
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