Task 10.3. Determine all natural numbers $n$ that have exactly $\sqrt{n}$ natural divisors (including 1 and the number $n$ itself).
Answer: 1 and 9.

Solution. First, note that $n=1$ works, and we will assume that $n \geqslant 2$. Second, from the condition it follows that $n=k^{2}$ for some natural number $k$. The squares of natural numbers have an odd number of divisors, since all divisors, except the base of the square, are paired as $d$ and $n / d$. Thus, $k$ is an odd number greater than 1. Let $k=2 m+1$ for some natural number $m$. The number $n$ has $m$ divisors less than $k$, and $m$ divisors greater than $k$, and all divisors are odd. This means that $n$ is divisible by all odd natural numbers less than $k$, since there are exactly $m$ of them.

In particular, $k^{2}$ is divisible by $k-2$. Since $k^{2}-4=(k-2)(k+2)$ is also divisible by $k-2$, the difference $4=k^{2}-\left(k^{2}-4\right)$ is divisible by the number $k-2$. It remains to note that $k-2$ is an odd divisor of the number 4, that is, $k-2=1$, from which $k=3$ and $n=9$.

## Criteria

Any correct solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed up:

1 6. It is noted that all divisors of the number $n$ are odd.

1 b. The oddness of the number $n$ is established.

## 1 6. It is noted that the number is a perfect square.

A correct answer without a correct solution is not scored.