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  1. #1
    Junior Member
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    n

    Prove that:

    if a and b are positive integers with (a, b) = 1, and if ab is a perfect square,

    then a and b are perfect squares.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let ab=k^2

    Let k=p_1p_2\ldots p_n. Then ab=p_1^2p_2^2\ldots p_n^2

    Then p_1^2|ab. But (a,b)=1\Rightarrow p_1^2|a or p_1^2|b. Suppose that p_1^2|a

    In the same way, suppose p_2^2|a,\ldots ,p_i^2|a.

    Then p_{i+1}^2,\ldots ,p_n^2|b

    So, a=p_1^2p_2^2\ldots p_i^2a_1, \ b=p_{i+1}^2p_{i+2}^2\ldots p_n^2b_1

    Then ab=p_1^2p_2^2\ldots p_n^2a_1b_1\Rightarrow a_1=b_1=1\Rightarrow a=(p_1p_2\ldots p_i)^2, \ b=(p_{i+1}p_{i+2}\ldots p_n)^2
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