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Find positive integers a,b,n that satisfy this expression

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Are there any positive integers n, a and b such that

    [tex]96n+88=a^2+b^2[/tex]


    2. Relevant equations
    3. The attempt at a solution
    It resembles the Pythagorean theorem but I'm not sure how that would help me solve it. I factored the LHS

    [tex]2^3((2^2)(3)n+11)=a^2+b^2[/tex]

    How do I proceed? Thanks.
     
  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    What can you say about perfect squares modulo 4?
     
  4. Nov 6, 2012 #3
    I'm very bad with the whole modulus concept but I can see that a perfect square divided by 4 will always have a remainder of either 0 or 1.
     
  5. Nov 6, 2012 #4

    haruspex

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    Right, so what can the sum of two squares be mod 4?
     
  6. Nov 6, 2012 #5
    0,1 or 2, right?
     
  7. Nov 6, 2012 #6

    haruspex

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    Right. Now which of those could match the LHS (mod 4)?
     
  8. Nov 6, 2012 #7
    96 is divisible by 4 (remainder = 0). 88 is also divisible by 4 (remainder = 0). So sum of remainders (0 + 0) = 0.
     
  9. Nov 6, 2012 #8

    SammyS

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    To get the remainder of zero, means that both a and b are even. Correct ?
     
  10. Nov 6, 2012 #9
    Yes, they'd both have to be even. Should I substitute, a = 2u, b = 2v ?
     
  11. Nov 6, 2012 #10

    haruspex

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    Yes, which allows some cancellation. See if you can then repeat the logic.
     
  12. Nov 7, 2012 #11
    96n + 88 = 4u^2 + 4v^2
    24n + 22 = u^2 + v^2
    (LHS= 2 mod 4 --> both u and v are odd)

    24n + 22 = (2w+1)^2 + (2x+1)^2
    6n + 5 = w^2 + w + x^2 + x
    (LHS = 3 mod 4 --> ??)

    If k^2 = 0 mod 4 then either k = 0 mod 4 or k = 2 mod 4
    If k^2 = 1 mod 4 then either k = 1 mod 4 or k = 3 mod 4

    Is this correct? And please correct me if my notation is wrong. Thanks.
     
  13. Nov 7, 2012 #12

    haruspex

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    w2+w = w(w+1). What does that tell you about w2+w mod 2?
     
  14. Nov 7, 2012 #13
    w(w+1)=odd*even=even --> w mod 2 = 0. same goes for x^2 + x. So the RHS is even but the LHS is odd. therefore it's not possible. Is that it?
     
  15. Nov 7, 2012 #14

    haruspex

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    That's it.
     
  16. Nov 7, 2012 #15
    Thank you so much haruspex and SammyS for breaking it down and guiding me through the problem. It helped me a lot :)
     
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