Let $K/\Bbb Q$ be a finite field extension and $R$ be the ring of integers of $K$. In order to show that a given $\Bbb Q$-Basis $(\alpha_1, ...,\alpha_n)$ of $K$ whose elements are already in $R$ is an integral basis, there are a couple of useful tricks, saving us from exhausting calculation of traces: Let $d_K$ be the discriminant of $K/\Bbb Q$ Then:
Lemma 1: $d(\alpha_1,...,\alpha_n)$ = $c^2 d_K$ for some $c\in
\Bbb Z$.
Lemma 2: if $d(\alpha_1,...,\alpha_n)=d_K$, then $(\alpha_1,...,\alpha_n)$ is an integral basis.
Lemma 3: $d_K \equiv 0,1 \mod 4$
Lemma 4: If $\sigma_1,...\sigma_n$ are the homomorphisms $K\rightarrow \Bbb C$, then $d(\alpha_1,...,\alpha_n) = \det((\sigma_i \alpha_j)_{i,j})^2$
So let's consider the basis you gave: $(1,\alpha,\alpha^2/2)$ is a $\Bbb Q$-Basis that is already in $R$. Fine. We know that $d(1,\alpha,\alpha^2)$ coincides with the discriminant $\Delta f=2^4\cdot 5\cdot 7$ of $f$. Also, by Lemma 4 the discriminant has some linearity in its arguments, more precisely:
$$d(1,\alpha,\frac{\alpha^2}{2})=\frac{1}{4}d(1,\alpha,\alpha^2)=\frac{1}{4}\Delta f=2^2\cdot 5 \cdot 7.$$
By Lemma 1 we have $d_K=2^2 \cdot 5\cdot 7$ or $d_K = 5\cdot 7$. But the last is a contradiction to Lemma 3, so $$d_K=2^2 \cdot 5\cdot 7 = d(1,\alpha,\frac{\alpha^2}{2}).$$
By Lemma 2, we conclude that $(1,\alpha,\alpha^2/2)$ is an integral basis, which completes the proof.
The discriminant is $\Delta(f)$.
By definition, the discriminant of an order is the determinant of the trace form. In this case, this is the determinant of the $d\times d$ matrix $M$ whose entries are given by the formula $M_{ij} = \operatorname{tr} ( \beta_{i-1} \beta_{j-1})$, taking $\beta_{d-1} =1$.
The entries $M_{ij}$ are polynomials in $a_0,\dots, a_d$. For example $$M_{dd} = \operatorname{tr}(1) =d,$$ $$M_{d,d-1} = \operatorname{tr} (a_d \alpha + a_{d-1} ) = - a_{d-1} + d a_{d-1} =(d-1) a_{d-1},$$ $$M_{d,d-2} = \operatorname{tr} ( a_d \alpha^2 + a_{d-1} \alpha + a_{d-2}) = \frac{a_{d-1}^2}{a_d} -2 a_{d-2} - \frac{ a_{d-1}^2}{a_d} + d a_{d-2} = (d-2) a_{d-2} $$
$$M_{d-1,d-1} =\operatorname{tr} (a_d^2 \alpha^2 + 2 a_d a_{d-1} \alpha + a_{d-1}^2) = a_{d-1}^2 - 2 a_{d-2} a_d - 2 a_{d-1}^2 + d a_{d-1}^2 = (d-1)a_{d-1}^2 - 2 a_d a_{d-2} $$
So the determinant is a polynomial in $a_0,\dots, a_d$.
Now your same argument shows that this polynomial divides $a_0 ^{ (d-1) (d-2)} \Delta(f)$ and $a_d ^{ (d-1) (d-2)} \Delta(f)$ in the ring of polynomials in $a_0,\dots, a_d$ over $\mathbb Z$. But in that ring, the only common factor of these polynomials is $\Delta(f)$ (because the ring has unique factorization into irreducibles), so this polynomial divides $\Delta(f)$. Then because $\Delta(f)$ is an irreducible polynomial, this polynomial is $\pm \Delta(f)$.
Best Answer
As in the link of Gerry, one is supposed to check if there are any algebraic integers among the seven :$\Sigma_{i=0}^2 a_i \alpha^i/2$, where $a_i$ are either $0$ or $1$, and not all of them are $0$.
Now, after some computations(some minutes maybe), one finds that the only one among them which is an algebraic integer is: $(\alpha+\alpha²)/2$, satisfying the irreducible polynomial: $x^3-x²-3x-2$.
Replace $\alpha²$ by $(\alpha+\alpha²)/2$ in the basis, one then finds that the discriminant becomes $-107$ by the transition formula. Hence this is an integral basis, as required.
P.S. Since this is subject to a certain amount of calculations, of which I think quite tediously, it is quite expectable that some errors penetrated in the above arguments. Thus, if there are some mistakes, please tell me. Thanks in advance.