Find all natural numbers $n$ such that for any two coprime divisors $a$ and $b$ of $n$, the number $a + b - 1$ is also a divisor of $n$.
It is easy to check that numbers of the form $p^{k}$, where $p$ is a prime, and the number 12 satisfy the condition of the problem.

We will show that no other numbers satisfy the condition.

The case of odd $n$ is considered in 109752. Let $n$ be even and not a power of two; represent it in the form $n=2^{m} k$, where $m \geq 1$, and $k>1$ is odd.

Notice that $k+2-1=k+1$ is a divisor of $n$. Since $(k+1, k)=1$, we have $k+1=2^{\alpha}, \alpha>1$.

Therefore, $2^{2}+k-1=k+3$ is also a divisor of $n$. Notice that $k+3=(k+1)+2=2^{\alpha}+2$ is not divisible by 4. Additionally, $(k+3, k)=(3, k) \leq 3$.

From this, it follows that $k+3 \leq 2 \cdot 3=6$, and $k \leq 3$.

Thus, $n=2^{m} \cdot 3$. But $m=1$ does not work; $m \geq 3$ also does not work, as in this case we would get that $2^{3}+3-1=10$ is also a divisor of $n$.

## Answer

$n$ is a power of a prime number or $n=12$.