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How many pairs of positive integer a, b

  1. Oct 9, 2004 #1
    How many pairs of positive integer a, b are such that [tex]a^2 + b^2 = 121?[/tex]
     
    Last edited: Oct 9, 2004
  2. jcsd
  3. Oct 9, 2004 #2

    Tide

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    Are there any positive integers for which

    [tex](11-b)(11+b)[/tex]

    are perfect squares?
     
  4. Oct 9, 2004 #3
    How did u come out with (11-b)(11+b)?
     
  5. Oct 9, 2004 #4

    Tide

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    I subtracted [itex]b^2[/itex] from both sides and factored the right side and the left side is a perfect square as you indicated.
     
  6. Oct 9, 2004 #5
    So I guess there is only one?
     
  7. Oct 9, 2004 #6

    Tide

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    What would that be?
     
  8. Oct 9, 2004 #7
    erm... 11?
     
    Last edited: Oct 9, 2004
  9. Oct 9, 2004 #8

    Tide

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    If one is 11 then the other has to be 0 but you specified positive integers.
     
  10. Oct 9, 2004 #9
    Ok so how do i get it?
    Btw i know that 0 isn't a positive integer, so what is it called then?
     
  11. Oct 9, 2004 #10
    if u cannot find b such that (11-b)(11+b) is a perfect square, what do u call such a situation?

    0 is simply called zero.
    However some literature do put zero in both positive and negative and indicate them separately as +0 and -0.+0 is positive zero and -0 in negative zero.

    -- AI
     
  12. Oct 9, 2004 #11
    I don't know what u call such a situation.
    Anyway back to the question. Do u know how many pairs of positive integer a, b are such that [tex]a^2+b^2=121[/tex]
     
  13. Oct 10, 2004 #12
    a = (11-b)(11+b)
    i cannot find positive integer b such that a is a perfect square
    therefore there is no solution for a and b.

    -- AI
     
  14. Oct 10, 2004 #13
    Ok. Thanks!
     
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