Romanian Mathematical Olympiad

From $b^2 = ac$ it follows that $\frac{b}{a} = \frac{c}{b} = \frac{x}{y}$, where $x, y \in \mathbb{N}^*, (x, y) = 1$ and $x > y$. Therefore, $b = \frac{a x}{y}$, $c = a \left(\frac{x}{y}\right)^2$.
Since $c = a \left(\frac{x}{y}\right)^2 \in \mathbb{N}^*$ and $(x, y) = 1$, it follows that $\frac{a}{y^2} \in \mathbb{N}^*$, so $a = p y^2$, where $p \in \mathbb{N}^*$. The numbers are $a = p y^2$, $b = p x y$, $c = p x^2$, where $x, y \in \mathbb{N}^*, x > y$, $p \in \mathbb{N}^*$.
From $a \in A$ follows that $y \sqrt{p} \ge n$ and from $c \in A$ follows that $\sqrt{p} x \le n + 1$, hence $\sqrt{p}(x-y) \le 1$ and, since $x-y \ge 1$, it follows that $1 \ge \sqrt{p}(x-y) \ge \sqrt{p}$ and, since $p \in \mathbb{N}^*$, we deduce $p = 1$.
We obtain $n \le y < x \le n + 1$, so $y = n, x = n + 1$ and the solution $a = n^2$, $b = n(n + 1)$, $c = (n + 1)^2$.