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Math Help - [SOLVED] On Divisibility by 19

  1. #1
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    [SOLVED] On Divisibility by 19

    Why is it that 19 must divide either x,y,or z in the equation x^6+y^6=Z^6? This problem is really starting to bug me :/
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  2. #2
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    Nevermind guys, I just solved it . srry
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  3. #3
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    Quote Originally Posted by Nives View Post
    Nevermind guys, I just solved it . srry
    Would you mind sharing how you did it, please. I am stuck on it for a while now. Thanks
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  4. #4
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    Um ... I thought that a^n+b^n=c^n has no positive integer solutions a,b,c when n is greater than 2 (Fermat's last theorem??). But in the case of even powers, it doesn't matter if the integers are positive or negative ... so in your case, the only solution would be x = y = z = 0 ... and any number divides zero ...
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  5. #5
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    Quote Originally Posted by Bingk View Post
    Um ... I thought that a^n+b^n=c^n has no positive integer solutions a,b,c when n is greater than 2 (Fermat's last theorem??). But in the case of even powers, it doesn't matter if the integers are positive or negative ... so in your case, the only solution would be x = y = z = 0 ... and any number divides zero ...

    You can have a solution of  x=0 and y=z for all y,z\in\mathbb{Z}....
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  6. #6
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    Ah, yeah ... ugh, I forgot that case also ... so basically, either x or y equals zero (or both equal zero) will have solutions ... and since 19 divides zero ....
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