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Any solution?

  1. Mar 19, 2006 #1
    are there any positive integers k,a, b such that this equation is satisfied:

    k^2-1=a^2+b^2
     
  2. jcsd
  3. Mar 19, 2006 #2

    0rthodontist

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    Yes, for example k = 9, a = 4, b = 8.
     
  4. Mar 21, 2006 #3
    thanks, how do i go about proving that infinite solutions exist?
     
  5. Mar 21, 2006 #4
    The solution

    Here's the solution.
    We have:
    k^2-a^2=b^2-1
    (k+a)(k-a)=b^2+1

    Then choose b such that b is even.
    This implies b^2+1 is odd. If b^2+1 is a prime, then put k+a=b^2+1 and k-a=1. You will get k&a. If b^2+1 is not a prime, then choose k+a and k-a as its two odd factors. Solving, you get k & a.
    As there is solution for all b>0 and b even, there are infinitely many solutions.
     
  6. Mar 21, 2006 #5
    You can see that Orthodontist's solution is also one of these.
     
  7. Mar 21, 2006 #6
    the case with 1 is a special case of
    k^2=a^2+b^2+c^2
    a good question that might arose from this is what number is greater: the number of pythogrean triplets or the above qudroplets?
     
  8. Mar 21, 2006 #7

    matt grime

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    neither, they are both the same (countably infinite); it is not a difficulct question, and has indeed already been answered in this thread where it is asserted that there are infintely many solutions to

    k^2-1=a^2+b^2
     
  9. Mar 30, 2006 #8
    why should [itex]b[/itex] be even?
     
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