11.5. 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.

(S. Berlov)
Answer. One and all odd prime numbers.

Solution. It is clear that $n=1$ satisfies the condition. Also, all odd primes satisfy it: if $n=p$, then its divisors 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.

Now suppose that some composite $n$ satisfies the condition. We have $n=a b$, where $a \geqslant b \geqslant 2$. Then the number $n+1$ is divisible by $a+1$; moreover, 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 $b-1 \geqslant a+1$. But this contradicts the inequality $b \leqslant a$.