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Hi everybody. I need to know if the cubic Thue equation $x^3 + x^2y + 3xy^2 - y^3 = \pm 1$ is completely solved. I know that there are effective algorithms to solve any cubic Thue equation and that some of them are implemented in computer programs. However, I think that since the coefficients of that equation are small, it may have already been discussed in the literature. Thank you.

edited Dec 1 at 13:21

J. H. S.
2,415●12●31

asked Dec 1 at 10:15

Richard Bonne
205●5

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What context gave rise to this equation for you? The equation can be rewritten as $|(x-y)^3 + 4x^2y| = 1$, and then changing variables by $u=x-y$ and $v = y$, it becomes $|u^3 + 4u^2v + 8uv^2 + 4v^3| = 1$. The left side in integers $u$ and $v$ (necessarily not both 0) is the index $[{\mathbf Z}[\alpha]:{\mathbf Z}[u\alpha+v\alpha^2]]$, where $\alpha$ is a root of $t^3 - 2t^2 + 4t - 4$. Therefore the question asks for all possible ring generators of ${\mathbf Z}[\alpha]$ up to addition by an integer, e.g., $(x,y) = (1,0)$ corresponds to $\alpha$ and $(x,y)=(1,-1)$ corr. to $\alpha-\alpha^2$. – KConrad Dec 22 at 0:08

John Cremona · Answer 1 · 2012-12-21 17:49:06Z UTC

5

For a cheaper solution, use Magma. For a free solution, use pari/gp:

(17:47) gp > thue(thueinit(x^3+x^2+3*x-1,1),1)

%2 = [[1, 0], [0, -1]]

answered Dec 21 at 17:49

John Cremona
711●3●6

Thanks, I didn't know about this function! I just checked that thue(thueinit(x^3+x^2+3*x-1,1),4) works too, in a few milliseconds even using the "certify unconditionally" flag (as opposed to assuming GRH). – Noam D. Elkies Dec 21 at 19:24

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...and come to think of it, this means gp also solves the second part of problem B-1 on the 1982 Math Olympiad: thue(thueinit(x^3-3*x+1,1),2891) returns []. (Does this gp routine thue know to look for shortcuts such as the intended obstruction mod 9, or the possibly unintended obstruction mod 7 which is what I used?) – Noam D. Elkies Dec 21 at 22:07

Carlo Beenakker · Answer 2 · 2012-12-22 15:15:50Z UTC

4

Dec. 1, 2012: your equation has only two integer solutions:

$x=0,y=-1$ and $x=1,y=0$ if the right-hand-side equals $+1$,

or $x=-1,y=0$ and $x=0,y=1$ if the right-hand-side equals $-1$.

if you have access to Mathematica, to find these solutions is a simple one-liner:

Reduce[x^3 + x^2*y + 3*x*y^2 - y^3 == 1, {x, y}, Integers]

Dec. 22, 2012: the question has evolved a bit into the direction of the (monetary -- not computational) cost of the software used to solve the Thue equation; in that connection it might be of some interest to note that the algorithm implemented by Mathematica can actually be called free of charge through the Wolfram Alpha interface:

solve for integer x,y: x^3+x^2*y+3*x*y^2−y^3=1

edited Dec 22 at 15:15

answered Dec 1 at 16:24

Carlo Beenakker
5,731●10●20

$(47,159)$ is close, though... (Giving $-4$.) – Noam D. Elkies Dec 1 at 16:39

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en.wikipedia.org/wiki/Thue_equation Thue's equation has only a finite number of integer solutions, and there exists a bound on the solutions, so by simply trying all integers within the bound you are guaranteed to find all solutions. The calculation can be lengthy, I enlisted the help of Mathematica. – Carlo Beenakker Dec 1 at 17:11

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@Richard: You should convince people who read your article that solving that equation is a routine matter - given the bounds established by Baker, say. Indeed, all that Mathematica uses is the existence of these bounds on $x,y$, which are part of the (by now) standard theory. The fact that you learned about this only recently doesn't force you to belabor the point; on the contrary, if you spend too many words on this your readers might get the false impression that there's something unusual going on here (which there ain't). – René Pannekoek Dec 2 at 3:51

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more precisely, the algorithm implemented by Mathematica is described in Journal of Symbolic Computation, 38 (2004) 1145 ---finanz.math.tu-graz.ac.at/~ziegler/Papers/… – Carlo Beenakker Dec 2 at 13:34

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@John Cremona: Richard Pinch once disproved the GRH using Mathematica (in the 1990s). In the manual, Mathematica claimed "our primality test is the following [blah] and this test is proved to be correct assuming GRH". Pinch then found a non-prime number that Mathematica's primality test asserted was prime, and got in touch with them. He managed to persuade them to let him see part of the source code, and told me that the code in the actual program bore no resemblance to the claims made in the manual! Hopefully this story answers your question, at least in a weak sense. – wccanard Dec 22 at 17:01

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Solved cubic Thue equation

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