# Strange Cubic Diophantine Equations

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.

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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." – Eric Naslund Apr 3 '11 at 14:09
solved Separately not simultaneously. – quanta Apr 3 '11 at 14:11
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. – András Salamon Apr 3 '11 at 16:37
Seems to have been solved in this paper dx.doi.org/10.1016/0022-314X(84)90053-2 but we cannot access it. – quanta May 6 '11 at 0:19

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.

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This seems like significant progress. Now $Y$ has to be a cube of the form $1 \pmod{3}$, so $Y=(3n+1)^3$. For any given $Y$ there is a cubic for $v$ which can be checked to see if it is integral. – Ross Millikan Apr 3 '11 at 17:30
@Ross, the cubic formula of Cardano?! – quanta Apr 3 '11 at 18:30
@quanta: exactly. Or you have $v^3+3uv^2-u^2v+5u^3+1=0$ You can "just" factor $5u^3+1$ and use the rational root theorem. – Ross Millikan Apr 3 '11 at 18:38
If we take $n=1$ then $Y=64, u=4$. The cubic becomes $v^3+48v^2-16v+321=0$. By the rational root theorem, the only possible integer roots are $v=\pm 1, \pm3, \pm107, \pm 321$. You can just try those, or submit the equation to Alpha and learn there are no integral roots. – Ross Millikan Apr 4 '11 at 14:46
The only way I could deal with all the numbers would be to find a modulus that proves there is no solution. The obvious try of 24 doesn't help. Others have more tools available to them. – Ross Millikan Apr 4 '11 at 15:05