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Thread: Divisibility (gcd) 15

  1. #1
    Sea
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    Divisibility (gcd) 15

    Show that:


    $\displaystyle n \in \mathbb{N},$


    $\displaystyle (a,b)=1 \Rightarrow (a^{n},b^{n})=1$
    Last edited by Sea; Dec 21st 2008 at 12:07 AM.
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  2. #2
    o_O
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    Let the prime factoirzations be: $\displaystyle a = p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}$ and $\displaystyle b = q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m}$ .

    Since $\displaystyle (a,b)=1$, there are no common factors between them, i.e. $\displaystyle p_i \neq q_j$ for any $\displaystyle 1 \leq i \leq n$, $\displaystyle 1 \leq j \leq m$.

    What can you say about $\displaystyle a^n$ and $\displaystyle b^n$?
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  3. #3
    Sea
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    $\displaystyle a^n = (p_1^{e_1})^{n}(p_2^{e_2})^{n}\cdots (p_n^{e_n})^{n}$


    $\displaystyle b ^n= (q_1^{f_1})^{n}(q_2^{f_2})^{n}\cdots (q_m^{f_m})^{n}$


    and



    I think...

    $\displaystyle (a,b)=(p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n},q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m})=1 \Rightarrow (p_i,q_j)=1, \hspace{0.3cm}p_i \neq q_j ,\hspace{0.3cm}1 \leq i \leq n$, $\displaystyle 1 \leq j \leq m$.
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  4. #4
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    Suppose $\displaystyle (a^n,b^n)>1$ then there is a prime $\displaystyle p$ such that $\displaystyle p|a^n$ and $\displaystyle p|b^n$

    You can check by Euclid's Lemma that we must have $\displaystyle p|a$ and $\displaystyle p|b$ thus $\displaystyle (a,b)\geq{p}>1$ CONTRADICTION!
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  5. #5
    Sea
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    I understand......


    Thanks a million......
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