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Can anyone solve this?

  1. Sep 21, 2005 #1
    This question goes:

    a, b and c are positive integrers.

    What is the sum of all possible values of a and b between 0 and 100 if

    a^(-2) + b^(-2)=c^(-2)
     
  2. jcsd
  3. Sep 21, 2005 #2
    [tex]c = (\sqrt{a}+\sqrt{b})^2[/tex]

    [tex] a \in [0, 100] \rightarrow \sqrt{a} \in [0, 10] [/tex]

    so the sum of all values of a is:

    [tex] \sum_{x=1}^{10}{x^2}[/tex]

    the same for b.
     
  4. Sep 22, 2005 #3

    VietDao29

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    Homework Helper

    I don't understand...
    [tex]a ^ {-2} = \frac{1}{a ^ 2}[/tex], and that's not [tex]a ^ {-2} = \sqrt{a}[/tex]
    Viet Dao,
     
    Last edited: Sep 22, 2005
  5. Sep 22, 2005 #4
    I've solved it...

    1/c^2=1/b^2 + 1/a^2
    =(a^2+ b^2)/(ab)^2
    c^2=ab^2/(a^2+ b^2)
    c=ab/(a^2+ b^2)^1/2

    Since a^2+ b^2 is an integrer, it's root is either another integrer or irrational. Thus in order for c to be an integrer, (a^2+ b^2)^1/2 must be an integrer.
    Listing possible result;

    3^2 + 4^2 = 5^2
    5^2 + 12^2 = 13^2
    20^2 + 21^2 = 29^2
    ....

    Then considering the first possibility, a=4x and b=3x

    c=12x^2/5x
    =12x/5

    We conclude that x must be a factor of 5. Since a=4x, a is multiple of 20. Adding up all multiples of 20 between 0 and 100

    20 + 40 + 60 + 80 + 100=300

    Now for b, b=3x

    15 + 30 + 45 + 60 + 75 + 90=315

    Now the second possibility,

    c=60x/13

    a=5x, b=12x

    x must be a multiple of 13

    5(13)=65, and 12(13)>100.

    The next possibility, we knoe that a=20x and b=21x, and x has to be a factor of 29. Since 21(29)>20(29)>100, then there is no further solution.

    300 + 315 + 65=680.

    So the awnser is 680.
     
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