LV OM - II - Task 4

Determine all positive integers $ n $ that have exactly $ \sqrt{n} $ positive divisors.
The number $ n = 1 $ satisfies the conditions of the problem. Let's assume, for the rest of the solution, that $ n > 1 $.
From the conditions of the problem, it follows that $ n = k^2 $ for some positive integer $ k $. The number of positive divisors of $ n $ that are less than $ k $ is equal to the number of its divisors that are greater than $ k $. Indeed: to each divisor $ d < k $ of $ n $, there corresponds a divisor $ n/d $ of $ n $, which is greater than $ k $. Additionally, the number $ k $ itself is a divisor of $ n $, which implies that $ n $ has an odd number of divisors. Therefore, $ k $, and thus $ n $, is an odd number.
The problem thus reduces to finding such odd numbers $ n = k^2 $ $ (n > 1) $, for which the number of divisors less than $ k $ is equal to $ \frac{1}{2} (k-1) $.
Since $ k^2 $ is an odd number, no even number is a divisor of $ k^2 $. Therefore, for the number of divisors less than $ k $ of $ k^2 $ to be equal to $ \frac{1}{2} (k-1) $, every odd number less than $ k $ must be a divisor of $ k^2 $. In particular, $ k^2 $ must be divisible by $ k-2 $. Since $ k^2 = (k-2)(k+2)+4 $, the number $ k-2 $ is an odd divisor of 4. This means that $ k - 2 = 1 $, or $ k = 3 $.
We directly verify that the number $ n = 3^2 = 9 $ satisfies the conditions of the problem. Therefore, the sought numbers are $ n = 1 $ and $ n = 9 $.