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Math Help - [SOLVED] Prove: For all integers a and b, if a|b then a^2|b^2?

  1. #1
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    Question [SOLVED] Prove: For all integers a and b, if a|b then a^2|b^2?

    I dont really know where to get start on this, all I can really figure out is:

    a|b <=> b = a (d)

    a^2|b^2 <=> b^2 = a^2 (d)

    therefore there is some common d that is in a|b and a^2|b^2

    I kinda doubt this is enough for a proof or if it really even proves anything really.

    Can someone help me out here?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Kitizhi View Post
    I dont really know where to get start on this, all I can really figure out is:

    a|b <=> b = a (d)

    a^2|b^2 <=> b^2 = a^2 (d)

    therefore there is some common d that is in a|b and a^2|b^2

    I kinda doubt this is enough for a proof or if it really even proves anything really.

    Can someone help me out here?
    from a|b implies b=ad for some integer d we have b^2=(ad)^2 = a^2d^2 and d^2 is also an integer.. thus a^2|b^2..
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