Subfield: Geometry


Question: A social club has $n$ members. They have the membership numbers $1,2, \ldots, n$, respectively. From time to time members send presents to other members, including items they have already received as presents from other members. In order to avoid the embarrassing situation that a member might receive a present that he or she has sent to other members, the club adds the following rule to its statutes at one of its annual general meetings:

"A member with membership number $a$ is permitted to send a present to a member with membership number $b$ if and only if $a(b-1)$ is a multiple of $n . "$

Prove that, if each member follows this rule, none will receive a present from another member that he or she has already sent to other members.

Alternative formulation: Let $G$ be a directed graph with $n$ vertices $v_{1}, v_{2}, \ldots, v_{n}$, such that there is an edge going from $v_{a}$ to $v_{b}$ if and only if $a$ and $b$ are distinct and $a(b-1)$ is a multiple of $n$. Prove that this graph does not contain a directed cycle.

Solution:
Suppose there is an edge from $v_{i}$ to $v_{j}$. Then $i(j-1)=i j-i=k n$ for some integer $k$, which implies $i=i j-k n$. If $\operatorname{gcd}(i, n)=d$ and $\operatorname{gcd}(j, n)=e$, then $e$ divides $i j-k n=i$ and thus $e$ also divides $d$. Hence, if there is an edge from $v_{i}$ to $v_{j}$, then $\operatorname{gcd}(j, n) \mid \operatorname{gcd}(i, n)$.

If there is a cycle in $G$, say $v_{i_{1}} \rightarrow v_{i_{2}} \rightarrow \cdots \rightarrow v_{i_{r}} \rightarrow v_{i_{1}}$, then we have

$$
\operatorname{gcd}\left(i_{1}, n\right)\left|\operatorname{gcd}\left(i_{r}, n\right)\right| \operatorname{gcd}\left(i_{r-1}, n\right)|\ldots| \operatorname{gcd}\left(i_{2}, n\right) \mid \operatorname{gcd}\left(i_{1}, n\right)
$$

which implies that all these greatest common divisors must be equal, say be equal to $t$.

Now we pick any of the $i_{k}$, without loss of generality let it be $i_{1}$. Then $i_{r}\left(i_{1}-1\right)$ is a multiple of $n$ and hence also (by dividing by $t$ ), $i_{1}-1$ is a multiple of $\frac{n}{t}$. Since $i_{1}$ and $i_{1}-1$ are relatively prime, also $t$ and $\frac{n}{t}$ are relatively prime. So, by the Chinese remainder theorem, the value of $i_{1}$ is uniquely determined modulo $n=t \cdot \frac{n}{t}$ by the value of $t$. But, as $i_{1}$ was chosen arbitrarily among the $i_{k}$, this implies that all the $i_{k}$ have to be equal, a contradiction.
If $a, b, c$ are integers such that $a b-a$ and $b c-b$ are multiples of $n$, then also $a c-a=a(b c-b)+(a b-a)-(a b-a) c$ is a multiple of $n$. This implies that if there is an edge from $v_{a}$ to $v_{b}$ and an edge from $v_{b}$ to $v_{c}$, then there also must be an edge from $v_{a}$ to $v_{c}$. Therefore, if there are any cycles at all, the smallest cycle must have length 2 . But suppose the vertices $v_{a}$ and $v_{b}$ form such a cycle, i. e., $a b-a$ and $a b-b$ are both multiples of $n$. Then $a-b$ is also a multiple of $n$, which can only happen if $a=b$, which is impossible.
Suppose there was a cycle $v_{i_{1}} \rightarrow v_{i_{2}} \rightarrow \cdots \rightarrow v_{i_{r}} \rightarrow v_{i_{1}}$. Then $i_{1}\left(i_{2}-1\right)$ is a multiple of $n$, i. e., $i_{1} \equiv i_{1} i_{2} \bmod n$. Continuing in this manner, we get $i_{1} \equiv i_{1} i_{2} \equiv$ $i_{1} i_{2} i_{3} \equiv i_{1} i_{2} i_{3} \ldots i_{r} \bmod n$. But the same holds for all $i_{k}$, i. e., $i_{k} \equiv i_{1} i_{2} i_{3} \ldots i_{r} \bmod n$. Hence $i_{1} \equiv i_{2} \equiv \cdots \equiv i_{r} \bmod n$, which means $i_{1}=i_{2}=\cdots=i_{r}$, a contradiction.
Let $n=k$ be the smallest value of $n$ for which the corresponding graph has a cycle. We show that $k$ is a prime power.

If $k$ is not a prime power, it can be written as a product $k=d e$ of relatively prime integers greater than 1. Reducing all the numbers modulo $d$ yields a single vertex or a cycle in the corresponding graph on $d$ vertices, because if $a(b-1) \equiv 0 \bmod k$ then this equation also holds modulo $d$. But since the graph on $d$ vertices has no cycles, by the minimality of $k$, we must have that all the indices of the cycle are congruent modulo $d$. The same holds modulo $e$ and hence also modulo $k=d e$. But then all the indices are equal, which is a contradiction.

Thus $k$ must be a prime power $k=p^{m}$. There are no edges ending at $v_{k}$, so $v_{k}$ is not contained in any cycle. All edges not starting at $v_{k}$ end at a vertex belonging to a non-multiple of $p$, and all edges starting at a non-multiple of $p$ must end at $v_{1}$. But there is no edge starting at $v_{1}$. Hence there is no cycle.
Suppose there was a cycle $v_{i_{1}} \rightarrow v_{i_{2}} \rightarrow \cdots \rightarrow v_{i_{r}} \rightarrow v_{i_{1}}$. Let $q=p^{m}$ be a prime power dividing $n$. We claim that either $i_{1} \equiv i_{2} \equiv \cdots \equiv i_{r} \equiv 0 \bmod q$ or $i_{1} \equiv i_{2} \equiv \cdots \equiv i_{r} \equiv$ $1 \bmod q$.

Suppose that there is an $i_{s}$ not divisible by $q$. Then, as $i_{s}\left(i_{s+1}-1\right)$ is a multiple of $q, i_{s+1} \equiv$ $1 \bmod p$. Similarly, we conclude $i_{s+2} \equiv 1 \bmod p$ and so on. So none of the labels is divisible by $p$, but since $i_{s}\left(i_{s+1}-1\right)$ is a multiple of $q=p^{m}$ for all $s$, all $i_{s+1}$ are congruent to 1 modulo $q$. This proves the claim.

Now, as all the labels are congruent modulo all the prime powers dividing $n$, they must all be equal by the Chinese remainder theorem. This is a contradiction.