2. For a natural number $n$, we will say it is superprime if the difference between any two of its consecutive positive divisors is a prime number. Determine all superprime numbers.
Solution. Clearly, the numbers $n=1$ and $n=3$ are super-prime, which we verify by direct inspection.

We will prove that there are no other super-prime numbers. Let $n \geq 2$ be a super-prime number. Let $d$ be the smallest natural divisor of $n$, different from 1. If $d$ is not a prime number, then $d=ab$, where $1<a<d$, so $a$ is a divisor of $n$ which is different from 1, and strictly less than $d$, which contradicts the minimality of $d$. We conclude that $d=p$ for some prime number $p$.

Then 1 and $p$ are consecutive divisors of $n$, so $p-1$ is some prime number $q$. If $q \geq 3$, then $q$ is an odd number, so $p$ is an even number greater than 2, which means it is not a prime number, a contradiction. Therefore, $q=2$ and $p=3$. Since $p$ is the smallest divisor of $n$ different from 1, the number $\frac{n}{3}$ is the second largest divisor of $n$, which means that $n-\frac{n}{3}=\frac{2n}{3}$ must be a prime number, let's denote it by $r$. Therefore, $2n=3r$. But $r$ is a prime number, so it must be $r=2$ and $n=3$.