What kind of methods there are to find integer solutions of $x^3+y^3-xy^2=1$? I tried some inequalities and congruences without success. I also found on Wikipedia that this might be a Thue equation but I have no idea what is a bivariate form.
Asked
Active
Viewed 402 times
4
-
1Have you tried completing the cube? As in adding $(x+y)^3 - x^3 -3x^2y-3xy^2-y^3$ on the left side. Don't know if it helps in this case, but you might get something out of it. – Arthur Jun 06 '12 at 09:52
-
I haven't but solving $(x+y)^3=1+4xy^2+3x^2y$ seems as hard as the original problem. – BeginnerAlgebraist Jun 06 '12 at 10:00
-
Well, it's a technique that works on some polynomial diophantine equations. There might be other cubes out there as well. I don't know if any of them will do the trick. – Arthur Jun 06 '12 at 10:06
-
The completed cube looks more complicated but may be nicer modulo 3. Perhaps you can lift a solution from there (just a guess). – N8tron Jun 06 '12 at 11:06
-
But as there is a solution in integers, how modulo 3 helps me? – BeginnerAlgebraist Jun 06 '12 at 11:10
-
If $x \in {0,1,2}$ is a solution mod 3, then a lifted solution in the integers would look like $x+3k$ for some integer $k$. You'd need to substitute that into the equation and see if you can solve for $k$-value(s) that work. – N8tron Jun 06 '12 at 12:59
-
I suppose since you're bivariate you'd have to do the same to $y$, maybe I'm wrong here... looks kinda complicated. – N8tron Jun 06 '12 at 13:03
2 Answers
2
You are correct, this is a Thue equation. "bivariate" just means "having two variables", and "form" refers to the polynomial being homogeneous, which it is since every term has degree 3. Consequently, there are only finitely many solutions, but finding them all can be quite involved. The general procedure requires working over $\mathbf Q(\alpha)$ where $\alpha$ in this case is a root of $x^3-x+1$. You might want to try the Thue solver in Pari/GP, rather than carry this out directly.
The fact that (1,0),(0,1),(1,1) are solutions suggests you wouldn't be able to rule out too much by congruences.
Erick Wong
- 25,198
- 3
- 37
- 91
-
Thanks. I tried to compute on the Internet using Sage but when I typed gp_console() the program output nothing instead of GP/PARI CALCULATOR ... – BeginnerAlgebraist Jun 06 '12 at 11:02
-
1WA finds the solutions $(x,y)=$ $(\pm1,1)$, $(0,1)$, $(1,0)$, and $(4,-3)$. – Fabian Jun 06 '12 at 11:16
-
1
I came across here accidentally when I searched for a Thue solver today.
I think you can find solution in the literature. I believe Bennett's paper "On the representation of unity by binary forms" is a good starting point for finding original articles. It is in Trans. AMS vol 353 (year 2000) and is accessible for every one. I reached this paper by searching for the key word "Thue" and author "Bennett". If you get the paper, you can read page 1509 for finding references.
