
Problem I-4. Find all positive integers $n$ which satisfy the following two conditions:

(i) $n$ has at least four different positive divisors;

(ii) for any divisors $a$ and $b$ of $n$ satisfying $1<a<b<n$, the number $b-a$ divides $n$.


Solution. Clearly primes, squares of primes and the number 1 have the given property. We will exclude these numbers from further considerations.

First assume that $n$ is even; thus $n=2 x$ for some integer $x$. Then $x-2$ divides $n$. Any divisor of $n$ smaller then $x=n / 2$ is at most $n / 3$. Therefore, $x-2 \leq n / 3$, yielding $x \leq 6$. Checking all possibilities for $x$, we arrive to three new satisfactory values of $n$, namely, 6,8 , and 12 .

Next assume that $n$ is odd. Let $n=p x$, where $p$ is the smallest divisor of $n$ greater than one. Obviously $p$ is an odd prime. Note that $p+1 \nmid n$ because $p+1$ is even; thus $x \neq p+1$. Since $1<p<x<n$, we have

$$
x-p \mid p x
$$

If $p \nmid x$ then $x-p$ and $x$ are coprime; hence $x-p \mid p$. Then $x-p \leq p$. On the other hand, $x \neq p+1$; therefore, $x-p \geq p$ since $p$ is the smallest nontrivial divisor. Hence $x=2 p$, contradicting the assumption that $n$ is odd.

If $p \mid x$, then $x=p y$ for some integer $y$ greater than one. The choice of $p$ implies that $y \geq p$. Moreover, $p y-p \mid p^{2} y$; thus $y-1 \mid p y$. Since $y-1$ and $y$ are coprime, $y \leq p+1$. If $y=p+1$, then $y$ is an even divisor of $n$; this contradicts the assumption that $n$ is odd. Otherwise $y=p$ since $y \geq p$. This contradicts the condition $y-1 \mid p y$.

All the solutions are primes, squares of primes and the numbers 1, 6, 8 and 12 .
