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Math Help - Pell's equation

  1. #1
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    Pell's equation

    I have an equation of the form

    2(2x-1)^2 -(2n-1)^2 = 1

    I know that this is a pell's equation, but I don't know what to do with it, since its not in the form.

    x^2 - dn^2 = 1

    Thanks for your help!
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  2. #2
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    Quote Originally Posted by davismj View Post
    I have an equation of the form

    2(2x-1)^2 -(2n-1)^2 = 1

    I know that this is a pell's equation, but I don't know what to do with it, since its not in the form.

    x^2 - dn^2 = 1

    Thanks for your help!
    You have u^2 - D v^2 = 1 where u = \sqrt{2} (2x - 1), v = (2n-1) and D = 1.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You have u^2 - D v^2 = 1 where u = \sqrt{2} (2x - 1), v = (2n-1) and D = 1.
    Thanks for the response.

    Hmm. Still confused cause if I apply the algorithm, I get:

    (\sqrt{2}-1)^2 = 3 - 2\sqrt{2}

    which implies u = -2\sqrt{2}. But that is not a solution since u is positive. What am I doing wrong?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by davismj View Post
    I have an equation of the form

    2(2x-1)^2 -(2n-1)^2 = 1

    I know that this is a pell's equation, but I don't know what to do with it, since its not in the form.

    x^2 - dn^2 = 1

    Thanks for your help!
    Let  a=2n-1 and  b=2x-1 . We then have  a^2-2b^2=-1

    Thus  (a^2-2b^2)^2=(-1)^2=1 \implies (a^2-2b^2)^2-2(2ab)^2=1 .

    But  2b^2=a^2+1 , so we get  (2a^2+1)^2-2(2ab)^2=1 .

    Summarizing, we see our original equation was transformed into Pell's equation.
    Last edited by chiph588@; April 25th 2010 at 09:38 AM.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by mr fantastic View Post
    You have u^2 - D v^2 = 1 where u = \sqrt{2} (2x - 1), v = (2n-1) and D = 1.
    The  \sqrt2 kind of scares me. Aren't we dealing with integers here?
    Last edited by chiph588@; April 25th 2010 at 12:20 AM.
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    Quote Originally Posted by chiph588@ View Post
    Let  a=2n-1 and  b=2x-1 . We then have  a^2-2b^2=-1

    Thus  (a^2-2b^2)^2=(-1)^2=1 \implies (a^2-2b^2)^2-2(2ab)^2=1 .
    I don't think I agree with this. This seems to be true iff 2(2ab)^2 = 0, yes?
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by davismj View Post
    I don't think I agree with this. This seems to be true iff 2(2ab)^2 = 0, yes?
    Sorry, I had a typo there. That line should read

     (a^2-2b^2)^2=(-1)^2=1 \implies (a^2+2b^2)^2-2(2ab)^2=1
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  8. #8
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    Quote Originally Posted by chiph588@ View Post
    Sorry, I had a typo there. That line should read

     (a^2-2b^2)^2=(-1)^2=1 \implies (a^2+2b^2)^2-2(2ab)^2=1
    Gotcha. You'll have to forgive me because our class isn't a number theory class nor does it have a textbook, so I have a wikipedia understanding of Pell's equation. So now I need to come up with a solution of a and b and then raise it to a subsequent power?
    Last edited by davismj; April 25th 2010 at 10:47 PM.
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