, 2019

Suppose now, if possible, that $\alpha_{i_0}, \alpha_{i_1}, \alpha_{i_2}$ form a geometric progression for some indices $i_0, i_1, i_2$ of which at least two are distinct; say $\alpha_{i_1}^2 = \alpha_{i_0}\alpha_{i_2}$. The condition on absolute values forces all three indices to be different from $n$.

Since $|\alpha_{i_1}| < 1$, and the minimal polynomial $f$ of $\alpha_{i_1}^2$ over the rationals has integral coefficients, the latter has a complex root $\alpha$ whose absolute value is strictly greater than $1$.

Consider now the polynomial $g = \prod_{1 \le i \le j \le n} (X - \alpha_i \alpha_j)$. Since the expression is symmetric in the $\alpha_i$, the coefficients of $g$ are all integral.

Notice that $g$ has a double root at $\alpha_{i_1}^2 = \alpha_{i_0}\alpha_{i_2}$, to infer that it is divisible by $f^2$, so it has a double root at $\alpha$ as well.

Finally, recall that $|\alpha| > 1$, so $\alpha$ is one of the pairwise distinct $\alpha_i\alpha_n$, $i = 1, 2, \dots, n$, each of which is, however, a simple root of $g$. The contradiction thus obtained concludes the proof.