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Diophantine equations of the form x^3-dy^3=1

  1. Dec 5, 2011 #1
    I was recently thinking about these diophantine equations. I'm looking for solutions (x,y) in integers with d as an integer also. I have found that whenever d is a perfect cube, the equation has finitely many solutions. However, I can't seem to figure anything else out regarding equations of this type. Any ideas?



    Sincerely,

    Mathguy


    EDIT:eek:h, and i don't know any advanced mathematics except for some linear algebra and bits and pieces of abstract algebra. So, please don't overwhelm this mere high schooler. Thanks
     
    Last edited: Dec 5, 2011
  2. jcsd
  3. Dec 5, 2011 #2
    Oh, and I've also figured out that if d is a cube-free integer such that the number of solutions to the equation is finite, then all integers of the equation x^3-d(n^3)(y^3)=1 also has a finite number of solutions.
     
  4. Dec 6, 2011 #3
    Well, we know it has no solutions x>1 (There's a special case x=0, d=1, and y=-1, though) when d is a cubed integer, as then d^3*y^3 would also be a cubed integer. And since there is no difference between two cubed integers that is only 1 (they grow further and further apart from each other), there is a finite number of solutions.

    And there is an infinite number of solutions when y=1, as then x>1 would be also have d=x^3-1 to make the solution true. But on the specifics of x=/=1 and y=/=1, I do not know.
     
  5. Dec 6, 2011 #4
    Look up Delaunay-Nagell theorem (or equation). However, this result uses advanced algebraic techniques, so you might not be ready to understand the proof (yet!).
     
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