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Solution needed 4 a very good problem!

  1. Sep 29, 2005 #1
    Solve x^3 - y^2 = 2006 in integers.
     
  2. jcsd
  3. Sep 29, 2005 #2

    Zurtex

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    What have you tried so far? What sort of mathematics have you learnt to solve this problem?
     
  4. Sep 29, 2005 #3
    number theory problem? do u knowsolution r not?
     
  5. Sep 29, 2005 #4
    zurtex? do u hve ny idea?
     
  6. Sep 29, 2005 #5

    Zurtex

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    Well yes I do know a solution and I would have a guess it's the only solution, but I only know this solution by running the problem through mathematica, I was wondering what sort of number theory you know so I might be able to help you rather than just give you an answer.
     
  7. Sep 29, 2005 #6
    [tex]x^3 - y^2 = 2006[/tex]

    [tex]y^2 = x^3 - 2006[/tex]

    [tex]y = \sqrt{x^3 - 2006}[/tex]

    So then, using substitution...

    [tex]x^3 - \sqrt{x^3 - 2006} = 2006[/tex]

    I'm sure this isn't nearly sophisticated enough, and I've probably gotten into a rut of some sort (I know I don't know how to continue ><)... It's what I tried, though-- and now I'm curious as to how you would continue, or if I'm completely wrong... If this is digressing from the topic, I apologize-- just ignore me if that is the case...
     
    Last edited: Sep 29, 2005
  8. Sep 29, 2005 #7
    You can't do that. You have a single equation with two variables which you substituted back into your equation. It will give you something in along the lines of 0 = 0 or x = x. And it does:

    [tex]x^3 - x^3 + 2006 = 2006[/tex]
     
  9. Sep 29, 2005 #8
    I don't know the theory behind it, but it only took a few minutes of guess and check to solve.
     
  10. Sep 29, 2005 #9

    David

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    Is this a math olympiad problem by any chance?
     
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