74th Romanian Mathematical Olympiad

$$4d_{i+1}^2-4d_{i+2}\cdot d_i\ge 0\iff \frac{d_{i+1}}{d_i}\ge \frac{d_{i+2}}{d_{i+1}}(\ast )$$ for any $i\in \{1,2,\dots ,k-2\}$. As $d_2$ is the smallest proper divisor of $n$, it means that the number $\frac{n}{d_2}$ is the greatest proper divisor of $n$, so $\frac{n}{d_2}=d_{k-1}$. We have: $$\frac{n}{d_2}=d_{k-1}=\prod _{i=1}^{k-2}\frac{d_{i+1}}{d_i}\ge \prod _{i=1}^{k-2}\frac{d_{i+2}}{d_{i+1}}=\frac{n}{d_2}.$$ It follows that all inequalities (*) turn into equalities. Thus, $d_{i+1}^2=d_{i+2}\cdot d_i$ for any $i\in \{1,2,\dots ,k-2\}$. It follows that the numbers $1=d_1,d_2,d_3,\dots ,d_k=n$ (in this order) are consecutive terms of a geometric progression of ratio $d_2$, so $n=d_2^{k-1}$. The lowest proper divisor of the composite number $n$ is a prime number $p$, so $d_2=p$, and $n=p^{k-1}$.