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Does anyone have any ideas towards solving these four equations one at a time?

  • $a^3 - 3a^2b + b^3 = \pm 1$
  • $a^3 + 3a^2b - 6 ab^2 + b^3 = \pm 1$

I am guessing that the $1$ might mean we can use units in some algebraic number field to solve these but I have no idea which one or how to find it. Maybe I am wrong entirely.


These are both Thue equations. Mordell shows how to solve $a^3 - 3a^2b + b^3 = 1$ in his book using $p$-adic methods, it is found the solutions are (x,y) = (1,0),(0,-1),(-1,1),(1,-3),(-3,2),(2,1).

These equations can be solved by pari/gp

? p = thueinit(x^3 - 3*x^2 + 1); ? thue(p,1) % = [[-1, 2], [0, 1], [-1, -1], [-2, -3], [3, 1], [1, 0]] ? thue(p,-1) % = [[1, -2], [0, -1], [1, 1], [2, 3], [-3, -1], [-1, 0]] ? p = thueinit(x^3 + 3*x^2 - 6*x + 1); ? thue(p,1) % = [[0, 1], [-1, -1], [1, 0]] ? thue(p,-1) % = [[0, -1], [1, 1], [-1, 0]] 

but it is not clear how they are being solved.

  • 0
    When you say solve these at one time, what exactly do you mean? Do you mean "assume all 4 hold simultaneously, what is a solution?" or do you mean "find a way that will solve each of these seperately."2011-04-03
  • 0
    solved Separately not simultaneously.2011-04-03
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    It seems one can make a little bit of progress in two of the cases as $1-b^3=(1-b)(1+b+b^2)$, which then determines some small solutions and helps begin a case analysis. But I have not been able to push this through to determine that there are no non-trivial solutions.2011-04-03
  • 0
    Seems to have been solved in this paper http://dx.doi.org/10.1016/0022-314X(84)90053-2 but we cannot access it.2011-05-06

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