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Sum of squares equation problem

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove only solution in integers of the equation

    x2 + y 2 + z2 = 2xyz

    is x = y = z =0


    2. The attempt at a solution

    Well, using common sense got the idea but dont exactly know how to prove it!
    Can anyone plz help as how to start ...?

    Thanks!
     
  2. jcsd
  3. Jul 12, 2010 #2

    HallsofIvy

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    Re: Equation

    Well, a proof is often in the details of common sense. What "common sense" gave you the idea?
     
  4. Jul 13, 2010 #3
    Re: Equation

    Suppose we take in negative integers.
    then the sum of squares is positive and the product beside is negative.(eliminated possibility)

    Then again taking 0 the answer is satisfied.

    Then taking positive ints by a bit of trial none (That I found ) satisfy the eqn
    (though this trial is the main problem that i faced)

    So the thing is I need the exact proof and not trial.
     
  5. Jul 13, 2010 #4

    Dick

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    Re: Equation

    The general idea is to show that if x,y,z is a solution then x/2,y/2,z/2 is also a solution. It's an infinite descent thing. One of x, y, and z must be even, right? Now consider divisibility by 4. You won't get the exact proof here. You only get hints. You'll have to work the proof out for yourself.
     
  6. Jul 14, 2010 #5
    Re: Equation

    Ok .. heres what i did after your suggestion( thanks!!)

    since it must keep on working for x/2 , y/2 , z/2 and then by 4 and so on finally it must come to x/x , y/y , z/z (that is 1 ,1 ,1)

    So the equation comes as this

    1^2 + 1^2 + 1^2 = 2 * 1 * 1 *1

    But since the equality doesnt hold true... well the only answer remaining is x = y = z =0.

    Is this right??
     
  7. Jul 14, 2010 #6

    Dick

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    Re: Equation

    Ooops. Actually my 'hint' isn't even right. The equation isn't homogeneous in x, y, z. Sorry. Here's what I meant to say. Suppose 2^k is the largest power of two dividing all of the numbers x, y and z. Then x=2^k*x', y=2^k*y', z=2^k*z', where at least one of x', y', z' is odd. Write out the equation in terms of x', y' and z' and show you have a problem mod 4.
     
  8. Jul 14, 2010 #7
    Re: Equation

    Got it!

    writing in terms of x', y', z'

    (2kx')2 + (2ky')2 + (2kz')2 = 2 * (2kx') * (2ky') * (2kz')

    x' + y' + z' = 23k+1-2k * x'* y'* z'

    x' + y' + z' = 2k+1 * x'* y'* z'

    But this contradicts the main problem statement..in terms of x'* y'* z'(for all values except 0 = x = y = z)

    Right?

    And also I wanna know what was that mod 4 cause I dont know how to use it in a sum. Could you please tell me how to do this sum by that method?
     
  9. Jul 14, 2010 #8

    Dick

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    Re: Equation

    What happened the squares on x' etc? And how is that a contradiction? If you square an odd number what's its value mod 4? How about squaring an even number?
     
  10. Jul 15, 2010 #9
    Re: Equation

    Since I dont know much about this mod 4(nothing at all actually!) I will try this out when I learn about it. Can you suggest from which website can i learn about this mod?? (Its not in my syllabus so no school teacher gonna teach me nor do I have the books havin that )
     
    Last edited: Jul 15, 2010
  11. Jul 15, 2010 #10

    Dick

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    Re: Equation

    A value of x 'mod 4' is just the remainder you get if you divide x by 4. If x is even what's the remainder if you divide x^2 by 4? Answer the same question in the case x is odd. You can also try looking at http://en.wikipedia.org/wiki/Modular_arithmetic or there must be a lot of other references.

    Examples:

    2 mod 4=2
    3 mod 4=3
    4 mod 4=0
    5 mod 4=1
    6 mod 4=2
    7 mod 4=3
    8 mod 4=0
    etc etc
     
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