Problem 8-5. 8 8-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number.
Answer: $12(2,3,4,6)$.

Solution. One of two cases applies.

A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to the choice of $q$, we get $q=3$ and $n=12$.

B) Suppose the smallest divisors are of the form $p$ and $p^{2}$, where $p$ is a prime number. This case is analyzed similarly.

Remark. It is possible that a number has only three proper divisors. Then the difference mentioned in the condition is the difference between the largest and the smallest of the proper divisors. But any number with three proper divisors is a power of a prime $p^{4}$, and the difference $p^{3}-p$ cannot be a prime number.

| Evaluation | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 16 | Minor errors |
| $+/ 2$ | 10 | Significant gaps in the solution |
| .- | 6 | The idea that $p n q=(p+q)(n-p q)$ is present |
| - | 0 | The solution is completely incorrect/only the answer |