10.4. A proper divisor of a natural number is any of its divisors, different from 1 and the number itself. Find all natural numbers for which the maximum proper divisor is 2 more than the square of the minimum proper divisor.
Answer: $n=12$ and $n=33$.

Solution. Let the required number be $n$, and its smallest proper divisor, which is a prime number, be $p$. The largest proper divisor of $n$ is $\frac{n}{p}$, which, according to the condition, equals $p^{2}+2$, hence $n$ has the form $n=p\left(p^{2}+2\right)$, where $p$ is its smallest proper divisor. This is equivalent to the number $p^{2}+2$ not having proper divisors smaller than $p$. Clearly, this is true for $p=2, n=12$ and $p=3, n=33$. Any prime number greater than 3, when divided by 3, gives a remainder of 1 or 2, so its square, when divided by 3, gives a remainder of 1, and after adding 2, it is divisible by 3. Therefore, the smallest proper divisor of the number $n=p\left(p^{2}+2\right)$ for $p>3$ will be 3, not $p$, and it does not satisfy the condition of the problem.

Grading criteria. Noting that the smallest proper divisor of $n$ is a prime number $p: 1$ point.

Noting that the largest proper divisor of $n$ is the number $\frac{n}{p}: 1$ point.

Finding the formula $n=p\left(p^{2}+2\right): 1$ point.

Indicating the answer in the form $n=p\left(p^{2}+2\right)$, where $p$ is any prime number: 3 points.

If it is noticed that not all primes $p$ work and attempts are made to eliminate them, but not brought to the correct answer: plus another 1 point.

For each correct answer found by trial and error with verification: 1 point each. Only for guessed answers, 2 points in total.