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Another small number theory proof

  1. May 9, 2010 #1
    Show that if x, y ∈ Z and 3|x2+z2 then 3|x and 3|z

    Solution:

    3|(x-z)(x+z)

    => 3|x+z or 3|x-z

    if 3|x+z

    then 3|(x+z)2 = x2+z2 +2xz

    => 3|x2+z2 +2xz - (x2+z2)

    => 3|2xz

    => 3|xz

    so 3|x or 3|z

    where to go from here?

    just the arguement that if 3|x then 3|x2

    and therefore 3 must divide z2 => 3|z?

    Thanks
     
    Last edited: May 9, 2010
  2. jcsd
  3. May 9, 2010 #2

    Office_Shredder

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    That's pretty much it
     
  4. May 9, 2010 #3
    Is the statement of the problem correct? We see a "y" in the hypothesis, but a "z" in the conclusion. More serious, instead of

    3|x2+z2

    did you mean

    3|x2-z2?

    I'm guessing that you meant the latter since your solution starts with

    3|(x-z)(x+z)

    and (x-z)(x+z) = x2-z2

    Finally, you had two cases (3|x+z or 3|x-z), but only addressed the first.

    Petek
     
  5. May 9, 2010 #4

    Dick

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    It can't be x^2-z^2. If x=4 and z=2, 3|12, but 3 doesn't divide 4 OR 2. Firepanda should try thinking about what x^2 mod 3 equals.
     
    Last edited: May 9, 2010
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