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Thread: gcd

  1. #1
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    gcd

    (a,b)=1 and c divides a+b prove (a,c)=1 and (b,c)=1
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  2. #2
    o_O
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    Let $\displaystyle d_{1} = (a,c) \geq 1$ and $\displaystyle d_{2} = (b,c) \geq 1$.

    Since $\displaystyle c \mid (a+b)$, then that means $\displaystyle d_{1} \mid (a+b)$ and $\displaystyle d_{2} \mid (a+b)$ as both $\displaystyle d_{1}, d_{2}$ are factors of $\displaystyle c$.

    However, since $\displaystyle d_{1} \mid (a+b)$ and $\displaystyle d_{1} \mid a$ & $\displaystyle d_{2} \mid (a+b)$ and $\displaystyle d_{2} \mid b$, then $\displaystyle d_{1} \mid b$ and $\displaystyle d_{2} \mid a$.

    This means $\displaystyle d_{1} \mid (a,b) $ and $\displaystyle d_{2} \mid (a,b)$. But $\displaystyle (a,b) = 1$. Conclude?

    Can you conclude?
    Last edited by o_O; Oct 5th 2008 at 07:57 PM.
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