74th Romanian Mathematical Olympiad

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