$\underline{\text { Berlov S.L. }}$.

A natural number $n$ is called good if each of its natural divisors, increased by 1, is a divisor of the number $n+1$.

Find all good natural numbers.
Clearly, $n=1$ satisfies the condition. Also, all odd prime numbers satisfy it: the divisors of such a number $p$, increased by 1, are 2 and

$p+1$; both of them divide $p+1$.

On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisible by $1+$ 1, which means $n$ is odd.

Suppose some composite $n=ab$, where $a \geq b \geq 2$, satisfies the condition. Then the number $n+1$ is divisible by $a+1$ and the number $n+b=(a+1)b$ is also divisible by $a+1$. Therefore, the number $b-1=(n+b)-(n+1)$ is also divisible by $a+1$. Since $b-1>0$, we get that $b-1 \geq a+1$. But this contradicts the inequality $b \leq a$.

## Answer

One and all odd prime numbers.