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Math Help - Divisibility (gcd) 15

  1. #1
    Sea
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    Divisibility (gcd) 15

    Show that:


    n \in \mathbb{N},


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

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

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


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


    and



    I think...

    (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, 1 \leq j \leq m.
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  4. #4
    Super Member PaulRS's Avatar
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    Suppose (a^n,b^n)>1 then there is a prime p such that p|a^n and p|b^n

    You can check by Euclid's Lemma that we must have p|a and p|b thus (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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