Mathematical Reflections 2017, Issue 3 - Problem O414

Problem:
Characterize all positive integers $n$ with the following property: for any two coprime divisors $a<b$ of $n$, $b-a+1$ is also a divisor of $n$.

Proposed by Vlad Matei, University of Wisconsin, Madison, USA

Solution:
It's easy to see that if $n=p^k$, where $p$ is a prime and $k \in \mathbb{Z}^+$, then $n$ satisfies the condition. Assume that $n$ has at least two prime divisors. Let $n=mp^k$, where $p$ is the smallest prime dividing $n$ and $(m,p)=1$. Clearly $p<m$ and both $p$ and $m$ are divisors of $n$. If $n$ satisfies the condition, then $m-p+1$ is also a divisor of $n$. Let $q$ be a prime such that $q \mid m$. Since $m-q<m-p+1<m$, then $q$ doesn't divide $m-p+1$. It follows that the only prime factor of $m-p+1$ is $p$. Hence, $m-p+1=p^a$ for some positive integer $a \leq k$, i.e. $$m=p^a+p-1.$$ Assume that $a \geq 2$. Since $p^{a-1} \mid n$ and $p^{a-1}<m$, then $m-p^{a-1}+1=p(p^{a-1}-p^{a-2}+1)$ is also a divisor of $n$. As $p^{a-1}-p^{a-2}+1$ is not divisible by $p$, then it must divide $m$. As
$$m=p^a+p-1=(p+1)(p^{a-1}-p^{a-2}+1)+p^{a-2}-2,$$ then $(p^{a-1}-p^{a-2}+1) \mid m$ if and only if $(p^{a-1}-p^{a-2}+1) \mid (p^{a-2}-2)$. We have $$p^{a-1}-p^{a-2}+1>p^{a-2}-2 \iff p^{a-2}(p-2)+3>0,$$ so there are no solutions if $a \geq 2$. If $a=1$, we get $m=2p-1$.

(i) If $k \geq 2$, then $p^2 \mid n$ and $p^2>2p-1$, so $p^2-(2p-1)+1=p^2-2p+2$ is also a divisor of $n$. If $p>2$, then $p^2-2p+2$ is not divisible by $p$, which gives $(p^2-2p+2) \mid m$, i.e. $(p^2-2p+2) \mid (2p-1)$ and this is true only if $p=3$. So, $m=5$ and $n=5\cdot 3^k$. Since $3^k>5$, then also $3^k-5+1=3^k-4$ is a divisor of $n$, which forces $3^k-4=5$, i.e. $k=2$ and $n=45$. If $p=2$, then $m=3$ and $n=3\cdot 2^k$. Since $2^k>3$, then also $2^k-3+1=2(2^{k-1}-1)$ is a divisor of $n$ and this implies $k=2$ or $k=3$. An easy check shows that indeed $n \in \{12,24\}$.

(ii) If $k=1$, then $n=(2p-1)p$. Let $q$ be the smallest prime divisor of $2p-1$. Then $q \mid n$, $p<q$ and so $q-p+1$ is a divisor of $n$. So, $(q-p+1) \mid (2p-1)$ or $(q-p+1) \mid p$. In the first case, by the minimality of $q$, it must be $q<q-p+1$, contradiction. In the second case, $q-p+1=1$ or $q-p+1=p$, which gives $q=2p-1$.

In conclusion, we get $n=p^k$, where $k \in \mathbb{Z}^+$, or $n=(2p-1)p$, where $p$ is prime and $2p-1$ is prime, or $n \in \{12,24,45\}$.