It looks like an elliptic curve as its genus is one. So one can transform it to Weierstrass form and hopefully that leads to a solution.

puuhikki wrote: It looks like an elliptic curve as its genus is one. So one can transform it to Weierstrass form and hopefully that leads to a solution. Maybe you can try it? Does it work? Can you post the solution?

I guess we search only integer solutions. Therefore the possible solutions can be only different numbers - $ x = y$ gives us irrational solution. Notice that if $ (x,y)$ is a solution, then $ (y, - x - y)$ is a solution, too. $ (y^3 - 3y(y + x)^2 + (y + x)^3 =$ $ y^3 - 3(y^3 + 2xy^2 + x^2y) + (y^3 + 3y^2x + 3yx^2 + x^3) =$ $ x^3 - 3xy^2 - y^3)$. $ \Rightarrow ( - x - y,x)$ is also solution of the equation. 1. If $ y = 0$ $ \Rightarrow (1,0)$ is a solution, $ \Rightarrow (0, - 1)$ and $ ( - 1,1)$ are solutions, too. 2. Let $ y\not = 0$. Then we can rewrite the equation as $ y^3(t^3 - 3t - 1) = 1$ when $ t = \frac {x}{y}$. Let us assume we have a solution $ (x_0,y_0)$ for which $ |x_0| < |y_0|$. Then $ ( - x_0 - y_0,x_0)$ and $ (y_0, - x_0 - y_0)$ are also solutions, so without losing of generality we are able to take $ |X| > |Y|$ (If $ |x_0 + y_0| > |x_0|$, it's obvious. If not, then $ |y_0| > |x_0| > |x_0 + y_0|$ ). Therefore, if there exists some solution, we can suppose $ |t|\ge 2$ $ \Rightarrow t^3 - 3t - 1\ge 1$ when $ t\ge 2$ or $ t^3 - 3t - 1\le - 3$ when $ t\le - 2$. Therefore the only possible solution is $ (2,1)$, which generates $ (1, - 3)$ and $ ( - 3,2)$.

Peter wrote: puuhikki wrote: It looks like an elliptic curve as its genus is one. So one can transform it to Weierstrass form and hopefully that leads to a solution. Maybe you can try it? Does it work? Can you post the solution? I don't know enough the theory of elliptic curves. I need to study how to transform curves to the Weierstrass form. It works if someone is able to compute the rank of that curve, but its is not a trivial thing to do. But I promise that I will help people in this forum as much as possible. It takes just time for me to learn new things.

I found that its Weierstrass equation is $ x^3 - \frac {81}{4} + y^2 = 0$, its rank is one, and its torsion subgroup is isomorphic to trivial group.

Actually we don't need the torsion group nor the rank. The discriminant of the Weierstrass form is $ - 3^{11}$ so $ y = 0$ or $ y = 3^k$ where $ 0\leq k\leq 11$. Substitute these values into Weierstrass equation and solve $ x$. Then transform the equation back to original and see what points maps to integer points. The result should be $ (x,y)\in \{(1,0),(0, - 1),( - 1,1),(1, - 3),( - 3,2),(2,1)\}$.

Sorry. That approach won't work. Nagell-Lutz requires the coefficients to be integers.

We can see this as a cubic in x and use the method of the cubic formula to solve it. $x^3+x(-3y^2) = 1+y^3 \rightarrow 3MN = -3y^2, M^3-N^3 = 1+y^3 \rightarrow M^3 = \frac{1+y^3\pm\sqrt{1+2y^3-3y^6}}{2} \rightarrow y = -1, 0, 1$ from which the solutions follow.