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Integers questions

  1. Oct 14, 2008 #1
    i've come across a question thatr reads
    x^3 + 2y^3 + 4z^3 =0
    show that x y z are all even

    part 2 requires to show that there are no such intergers

    i have no idea at all how to show something is even
    can anyone help plz thanks
  2. jcsd
  3. Oct 14, 2008 #2


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    It helps if you know that the cube of any even number is even and the cube of any odd number is odd! Obviously, for any y and z, 2y^3+ 4z^3= 2(y^3+ 2z^3) is even. In order that x3 cancel that, x3 must be even.

    But you can say more. The cube of an even number is divisible by 8: (2n)3= 8n3. So if x is even, what about 2(y3+ 2z3)? And then what about (y3+ 2z3)?
  4. Oct 15, 2008 #3
    thanks for the hints !
  5. Oct 15, 2008 #4


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    Oh, and you show something is even by showing it satisfies the definition of "even": it is equal to 2n for some integer n.
  6. Oct 15, 2008 #5
    got it ! =)
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