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Thread: gcd, lcm

  1. #1
    Sea
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    gcd, lcm

    $\displaystyle a\in\mathbb{Z^{+}}$ (a,a+1)=? , [a,a+1]=? (proof)
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  2. #2
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    #1: Let $\displaystyle d = (a, a+1) \geq 1$. This means that $\displaystyle d \mid a$ and $\displaystyle d \mid (a+1)$.

    Fact: If $\displaystyle d \mid x$ and $\displaystyle d \mid y$, then $\displaystyle d \mid (x+y)$.

    This means that $\displaystyle d \mid \left[(a+1) - a\right] \ \Leftrightarrow \ d \mid 1$.

    So $\displaystyle d = \cdots$

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    #2: Use the fact that if $\displaystyle (a,b) = 1$ then $\displaystyle [a,b] = ab$
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