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Thread: GCD

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

    GCD

    Hi everyone.

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



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

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

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

    Re: GCD

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

    hence :

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

    since :

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

    Hi, I have another idea.

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

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