Problem 1421

1. Given Information and Initial Setup:
- Let $n$ be a composite positive integer.
- Let $1 = d_1 < d_2 < \cdots < d_k = n$ be the positive divisors of $n$.
- The quadratic equations $d_{i+2}x^2 - 2d_{i+1}x + d_i = 0$ for $i = 1, \ldots, k-2$ all have real solutions.

2. Condition for Real Solutions:
- For the quadratic equation $d_{i+2}x^2 - 2d_{i+1}x + d_i = 0$ to have real solutions, the discriminant must be non-negative:
$$\Delta = (2d_{i+1})^2 - 4 \cdot d_{i+2} \cdot d_i \geq 0$$
Simplifying the discriminant condition:
$$4d_{i+1}^2 - 4d_{i+2}d_i \geq 0 \implies d_{i+1}^2 \geq d_i d_{i+2}$$

3. Equality Condition:
- We have $d_{i+1}^2 \geq d_i d_{i+2}$ for $1 \leq i \leq k-2$.
- On the other hand, for the product of divisors, we know:
$$d_1 d_k = d_2 d_{k-1} = \cdots = n$$
- This implies:
$$\frac{n^2}{d_{k-i}^2} \geq \frac{n}{d_{k+1-i}} \cdot \frac{n}{d_{k-1-i}} \implies d_{k+1-i} d_{k-i-1} \geq d_{k-i}^2$$

4. Reindexing and Simplifying:
- Setting $k-i = j+1$, we get:
$$d_{j+2} d_j \geq d_{j+1}^2$$
- For $j \in \{1, \ldots, k-2\}$, all the inequalities must actually be equalities:
$$d_{i+1}^2 = d_i d_{i+2}$$

5. Conclusion:
- The equality $d_{i+1}^2 = d_i d_{i+2}$ implies that the sequence of divisors forms a geometric progression.
- For $n$ to have divisors in geometric progression, $n$ must be a power of a prime number.
- Therefore, $n = p^{k-1}$ for some prime number $p$.

$\blacksquare$