Show that $x^3+3y^3+9z^3-9xyz=1$ has infinitely many integer solutions. [duplicate]

Question

Show that $x^3+3y^3+9z^3-9xyz=1$ has infinitely many integer solutions.

I have found that (1,0,0) and (1,-18,12) are two solutions and tried (1,-18+n,12-n).

There is a hint saying that I should try to decompose the left hand side into complex polynomials. Any idea how I can do this?

Answer 1

Since you found a nontrivial solution $(x,y,z)=(1,-18,12)$, I define for each $n\in\mathbb{Z}$ the triple $(x_n,y_n,z_n)\in\mathbb{Z}$ to be such that $$x_n+\alpha\,y_n+\alpha^2\,z_n=(1-18\alpha+12\alpha^2)^n\,,$$ with $\alpha:=\sqrt[3]{3}$. (Since $1-18\alpha+12\alpha^2$ is not a primitive root of unity, no two triples of this sequence are equal.) Then, it follows that the norm of $x_n+\alpha\, y_n+\alpha^2\,z_n$ is then $$x_n^3+3\,y_n^3+9\,z_n^3-9\,x_n\,y_n\,z_n=\big((1)^3+3\,(-18)^3+9\,(12)^3-9\,(1)\,(-18)\,(12)\big)^n=1\,.$$