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.
The change of variables $a = u+v$ and $b = u-v$ gives $$a^3 - 3a^2b + b^3=-(u^3 + 3vu^2 - 9v^2u - 3v^3) = 3(u-v)(u+v)(3u+v) - 8u^3 = 3X - 8Y = 1.$$
So we could have $(X,Y) = \ldots(-13,-5),(-5,-2),(3,1),(11,4),(19,7),\ldots$, each case could be dealt with one by one but to deal with all (infinity) of them at once is not practical..
The change of variables $a = u-v$, $b = v$ gives $$a^3 + 3a^2b - 6 ab^2 + b^3 = u^3 - 9 v^2 (u-1) = X - 9 Y = \pm 1.$$ this puts us in a similar situation as above.