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Prove the sum of two even perfect squares is not a perfect square

  1. Oct 8, 2010 #1
    1. The problem statement, all variables and given/known data
    For all natural numbers, a and b, if a and b are both even, then (a^2+b^2) is not a perfect square. (prove this)

    2. Relevant equations



    3. The attempt at a solution
    I tried proving by contradiction and got (2s)^2 +(2t)^2 =k^2.
    which translates to 4s^2 +4t^2=k^2.
    I don't know how to form the contradiction from here. Is it even possible?
     
  2. jcsd
  3. Oct 8, 2010 #2
    [tex](a+b)^2=a^2+2ab+b^2[/tex]
     
  4. Oct 8, 2010 #3

    D H

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    Are you sure you have the problem correct as stated? As stated this is easily proven false by counterexample.
     
  5. Oct 8, 2010 #4
    so I set up the negation, then we assume a and b are even and that a^2 +b^2 is a perfect square. Then subbing 4 for a and 6 for b, we get a contradiction?
     
  6. Oct 8, 2010 #5

    D H

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    Obviously 52 is not a perfect square. The conjecture does not say that the sum of squares of some specific pair of even numbers is not a square number. The conjecture says that the sum of squares of every pair of even numbers is not a square number.
     
  7. Oct 8, 2010 #6
    true, but using four and six as counterexamples.....
     
  8. Oct 8, 2010 #7

    D H

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    42+62=16+36=52. and 52 is not a perfect square. 4 and 6 do not form a counterexample.
     
  9. Oct 8, 2010 #8
    a counterexample in the negation of the conjecture.
     
  10. Oct 8, 2010 #9

    D H

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    Correct, and 52 is not a perfect square. 4 and 6 are consistent with the conjecture.
     
  11. Oct 8, 2010 #10
    we assume the negation of the conjecture. which is for all natural numbers, a and b, if a and b are both even, then (a^2 +b^2) IS a perfect square.

    if we use 4 and 6 as counterexamples we do not get a perfect square, so we have a contradiction.....
     
  12. Oct 8, 2010 #11
    or are you saying we don't need the negation, just provide 4 and 6 as counterexamples and be finished....?
     
  13. Oct 8, 2010 #12
    There is an error.
     
  14. Oct 8, 2010 #13
    or maybe we're supposed to prove the conjecture false by counterexample...

    using 6 and 8 perhaps.
     
  15. Oct 8, 2010 #14

    Dick

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    Well, sure. It is false, isn't it?
     
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