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Homework Help: Larson 4.1.10

  1. Jan 7, 2008 #1
    [SOLVED] Larson 4.1.10

    1. The problem statement, all variables and given/known data
    Determine all triplets of integers (x,y,x) satisfying the equation


    2. Relevant equations

    3. The attempt at a solution

    I think the only solutions are x=y=0, x=z=0, z=y=0, z=-y and x=0, x=-z and y=0, x=-y and z=0. If I rewrite the equation as (x+y)(x+z)(y+z)=-xyz, several things are obvious:
    1)x,y,z are not all odd
    2)x,y,z are not all positive and not all negative
    3)if for example, z is negative and x,y are positive, then abs(z)< x,y or abs(z)>x,y

    But how can I show there are no other solutions...
  2. jcsd
  3. Jan 7, 2008 #2
    certainly either all are 0 or none
    by fermat last theorm(diff. from earliar one)

    i will work it out some other time
  4. Jan 7, 2008 #3
    I am sure there is a way to do this without using Fermat's Last Theorem. None of Larson's problems require something that advanced.
  5. Jan 7, 2008 #4


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    I don't even see how Fermat is relevant here.

    I have one comment on your attempted solution:
    That doesn't look correct to me. Shouldn't the right side be simply 0?
  6. Jan 7, 2008 #5
    No. Multiply it out. The LHS only gives 2xyz.
  7. Jan 7, 2008 #6


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    Maybe it's the sleepiness, but it seems to me that (x+y+z)^3 = x^3 + y^3 + z^3 + 3(x+y)(x+z)(y+z).
  8. Jan 7, 2008 #7
    You're right. It does. The solutions to (x+y)(x+z)(y+z)=0 (and to the the original equation) are x=-y,y=-z, or z=-x.
    Last edited: Jan 7, 2008
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