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Math Help - abstract algebra #2

  1. #1
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    abstract algebra #2

    if p is prime and (a,b)=p, then (a^2,b^2)=?
    Last edited by mr fantastic; April 15th 2010 at 03:36 PM. Reason: Edited post title
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  2. #2
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    Dear Deepu,

    (a,b)=p

    Take (a^2,b^2)=k

    Now since, k\mid{a^2} and k\mid{b^2}

    Therefore, k\mid{a} and k\mid{b}

    Then since the gcd of "a" and "b" is "p" ; k\leq{p} ---------------A

    Also since p=(a,b)\Rightarrow{p\mid{a}} and p\mid{b}\Rightarrow{p\mid{a^2}} and p\mid{b^2}

    Then since the gcd of a^2 and b^2 is "k" ; p\leq{k} ----------------B

    From A and B ; p=k

    Therefore (a^2,b^2)=p

    Hope this helps.
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  3. #3
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    Thank you very much.
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  4. #4
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    Quote Originally Posted by Sudharaka View Post
    ...
    Now since, k\mid{a^2} and k\mid{b^2}

    Therefore, k\mid{a} and k\mid{b}
    This implication is incorrect.

    Consider, for example, a=6,b=9 \Rightarrow a^2=36, ~b^2=81 \Rightarrow 9 | 36, ~ 9 | 81 but 9 obviously does not divide 6.

    ---

    Let (a,b)=p \Rightarrow p|a, ~ p|b \Rightarrow p^2|a^2,~p^2|b^2

    So, we now know that p^2 is a common divisor of a^2,~b^2. I'll leave it to you to show that it is the greatest common divisor.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Suppose  a=p_1^{\alpha_1} p_2^{\alpha_2} \cdot\cdot\cdot p_n^{\alpha_n} and
     b=p_1^{\beta_1} p_2^{\beta_2} \cdot\cdot\cdot p_n^{\beta_n} where  \alpha_i,\beta_j \geq 0 .

    Set  k=(a,b) . Then  k=p_1^{\min{\alpha_1,\beta_1}} p_2^{\min{\alpha_2,\beta_2}} \cdot\cdot\cdot p_n^{\min{\alpha_n,\beta_n}} .

    Now  a^2=p_1^{2\alpha_1} p_2^{2\alpha_2} \cdot\cdot\cdot p_n^{2\alpha_n} and  b^2=p_1^{2\beta_1} p_2^{2\beta_2} \cdot\cdot\cdot p_n^{2\beta_n} .

    So  (a^2,b^2) = p_1^{\min{2\alpha_1,2\beta_1}} p_2^{\min{2\alpha_2,2\beta_2}} \cdot\cdot\cdot p_n^{\min{2\alpha_n,2\beta_n}} = p_1^{2\min{\alpha_1,\beta_1}} p_2^{2\min{\alpha_2,\beta_2}} \cdot\cdot\cdot p_n^{2\min{\alpha_n,\beta_n}} = k^2 .

    Hence  (a^2,b^2) = (a,b)^2 .
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  6. #6
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    Quote Originally Posted by Defunkt View Post
    This implication is incorrect.

    Consider, for example, a=6,b=9 \Rightarrow a^2=36, ~b^2=81 \Rightarrow 9 | 36, ~ 9 | 81 but 9 obviously does not divide 6.

    ---

    Let (a,b)=p \Rightarrow p|a, ~ p|b \Rightarrow p^2|a^2,~p^2|b^2

    So, we now know that p^2 is a common divisor of a^2,~b^2. I'll leave it to you to show that it is the greatest common divisor.
    Dear Defunkt,

    Thank you for showing my mistake. I tried to solve the problem again using the suggestion you had given, but still haven't had any luck.
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