# A diophantine equation $x^3+y^3-xy^2=1$

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.

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Have 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 6 '12 at 9:52
I haven't but solving $(x+y)^3=1+4xy^2+3x^2y$ seems as hard as the original problem. –  BeginnerAlgebraist Jun 6 '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 6 '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). –  Nate Iverson Jun 6 '12 at 11:06
But as there is a solution in integers, how modulo 3 helps me? –  BeginnerAlgebraist Jun 6 '12 at 11:10
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.
WA finds the solutions $(x,y)=$ $(\pm1,1)$, $(0,1)$, $(1,0)$, and $(4,-3)$. –  Fabian Jun 6 '12 at 11:16