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Math Help - cubic diophantine equation

  1. #1
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    cubic diophantine equation

    Please help me to find a source/reference for solving the equation (or similar) 4u^3 - v^2 = 3.

    I am guessing that the only solutions are (1,1), (1,-1), (7,37), and (7,-37), but my efforts have not been successful.

    Thanks,

    Wayne
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  2. #2
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    Quote Originally Posted by dioph View Post
    Please help me to find a source/reference for solving the equation (or similar) 4u^3 - v^2 = 3.

    I am guessing that the only solutions are (1,1), (1,-1), (7,37), and (7,-37), but my efforts have not been successful.
    There are infinitely many rational solutions, for example u=\frac{73}{36},\ v=\pm\frac{595}{108}. It seems unlikely that there are any more integer solutions, but I don't know how one might prove that.
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  3. #3
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    integer

    Sorry - I meant to say integer solutions. Do you know of a source/reference for solving specifically equations of this form?

    Thanks,

    Wayne
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  4. #4
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    Yes. This looks like a problem with Elliptic curves (hence my avatar). A book on Elliptic curves might mention something about this equation. I have a book at home higher number theory, it mentions some stuff, but nothing like this.

    Another idea:
    If you can reduce the equation to y^2 = x^3 - k where k is square-free, then it might work. The first step is to note (2u)^3 = 2v^2 - k. And so (2u)^3 = (\sqrt{2} v)^2 + 6. And we are left with an equation, x^3 = y^2 + 6, but here x,y are not integers. Perhaps you can take advantage of the fact that \mathbb{Z}[\sqrt{-2}] has unique factorization and use it to prove that x^3 = y^2 + 6 has not only solutions in \mathbb{Z} but more, no solutions in \mathbb{Z}[\sqrt{-2}].
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