7.4. Determine all natural numbers $n$, which have exactly four positive divisors, knowing that the sum of all positive divisors of the number $n$ is twice as large as $n$.


## Solution:

Since the natural number $n$ has exactly four positive divisors, it follows that the number has the form $n=p \cdot q$, $p \neq q$ or $n=p^{3}$, where $p$ and $q$ are prime numbers.

Assume that $n=p \cdot q$. Then the positive divisors of the number $n$ are $1, p, q$, and $p \cdot q$. From the problem data, we obtain the equation $1+p+q+p q=2 p q$, which is equivalent to

$p q-p-q=1 \quad \Leftrightarrow p(q-1)-q+1=2 \Leftrightarrow(q-1)(p-1)=2$.

The last relation can only hold in two cases: 1) $q-1=1$ and $p-1=2$ or 2) $q-1=2$ and $p-1=1$. Thus, $q=2, p=3$ or $q=3, p=2$. In both cases, we get $n=6$.

Assume that $n=p^{3}$. Then the positive divisors of the number $n$ are $1, p, p^{2}$, and $p^{3}$. From the problem data, we obtain the equation $1+p+p^{2}+p^{3}=2 p^{3}$. Since $2 p^{3} \vdots p$ and $\left(p+p^{2}+p^{3}\right) \vdots p$, it follows that $1 \vdots p$. Contradiction since $p \geq 2$.

Answer: $n=6$.