 Solution

There is an error in the printed request: the cubic is not irreducible at $n=2$, and three elements cannot be a field basis in that case. The integral-basis assertion is valid under the necessary hypothesis $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$.

Under that hypothesis, $\alpha$ is an algebraic integer and $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_k$ has finite index $m$. Change of basis in the trace pairing gives the discriminant-index identityThe field discriminant is an integer. If the left side is square-free, $m^2$ cannot divide it unless $m=1$. Thus $1,\alpha ,{\alpha }^2$ is an integral basis when the cubic is irreducible and its discriminant is square-free.

For any root, ${\alpha }^3=n\alpha +1$ givesSince $\alpha +\beta +\gamma =0$, $\alpha \beta +\beta \gamma +\gamma \alpha =-n$ and $\alpha \beta \gamma =1$,On the other hand, the product of $r{f}^{\prime }(r)$ is $-\Delta$: the product of the roots is one, and each of the three unordered root pairs contributes a negative squared difference. Therefore, without needing irreducibility for this polynomial calculation,The three-root product is the field norm only when the cubic is the minimal polynomial.

At $n=1$, the polynomial has no rational root (the only candidates are $\pm 1$), so it is irreducible, and its discriminant is $-23$. Hence $1,\alpha ,{\alpha }^2$ is an integral basis.

At $n=2$,If $\alpha =-1$, then $k=\mathbb{Q}$ with integral basis $\{1\}$. If $\alpha =(1\pm \sqrt{5})/2$, then $k=\mathbb{Q}(\sqrt{5})$ and $\{1,\alpha \}$ is an integral basis, since this quadratic order has square-free discriminant five; also ${\alpha }^2=\alpha +1$. In neither case is the three-element list a basis. Its discriminant five is instead that of the degree-three algebra $\mathbb{Q}[x]/(x^3-2x-1)$; a basis there does not repair the printed assertion about the field $k$.