2019 ROMANIAN MATHEMATICAL OLYMPIAD

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.