Let $n \geqslant 2$ be an integer and $k \geqslant 1$ be an integer. Arthur and Thomas play the following game. Arthur chooses two sets of natural numbers, $A$ and $B$, of cardinality $n$, and $k$ integers $x_{1}, \ldots, x_{k}$ (not necessarily distinct) that can be written as the product of an element of $A$ and an element of $B$. Thomas wins if he can choose three indices $i_{1}, i_{2}, i_{3}$ pairwise distinct such that $x_{i_{1}}$ divides $x_{i_{2}} x_{i_{3}}$. For which values of $k$, in terms of $n$, does Arthur have a winning strategy?
Let $a_{1}, \ldots, a_{n}$ be the elements of $A$ and $b_{1}, \ldots, b_{n}$ the elements of $B$. Note that if $a_{j} b_{l}$ is chosen by Arthur, and there exists another element of the form $a_{j} b_{l^{\prime}}$ chosen by Arthur and one of the form $a_{j^{\prime}} b_{l}$, then Thomas can find a triplet $(x, y, z)$ such that $x$ divides $y z$. If Arthur chooses the same integer twice, Bob wins. Suppose Arthur has a winning strategy for $k$.

Consider the graph where the vertices are the $(a_{i})$ and the $(b_{j})$ such that $a_{i}$ and $b_{j}$ are connected if $a_{i} b_{j}$ is chosen by Arthur (if $a_{i} b_{j}$ can be written as multiple products of elements of $A$ and $B$, we only consider one product) for all $(i, j) \in\{1, \ldots, n\}^{2}$. The previous remark tells us that if we consider any edge, one of the two vertices cannot have a degree of at least 2, so it must have a degree of 1. Let $G_{1}$ be the set of vertices in $G$ from which exactly one edge departs. According to the previous remark, $k$ being the number of edges in $G$, it is smaller than the cardinality of $\left|G_{1}\right|$, so it is less than or equal to $2 n$. One option to have many edges is to connect $a_{1}$ to all $b_{i}$ except $b_{1}$, and connect $b_{1}$ to all $a_{i}$ except $a_{1}$, which would give $2 n-2$ edges. We can therefore try to show that there are at most $2 k-2$ edges.

Suppose $G$ contains strictly more than $2 n-2$ edges. In particular, $\left|G_{1}\right|$ is of size $2 n$ or $2 n-1$. If $G_{1}$ is of cardinality at least $2 n-1$, all vertices except possibly 1 are in $G_{1}$. Thus $G_{1}$ contains all the $(a_{i})$ or all the $(b_{i})$, say all the $(a_{i})$ by symmetry. Since each edge departs from an $(a_{i})$, there are at most $n < 2 n-2$ edges, which concludes. Thus $k \leqslant 2 n-2$. In particular, if $k \geqslant 2 n-1$, Bob can win.

Let's try to create an example similar to our graph with a maximum number of edges. We denote $p_{1}, \ldots, p_{2 n}$ the $2 k$ smallest prime numbers. We set $A=\left\{p_{1}, \ldots, p_{n}\right\}$ and $B=\left\{p_{n+1}, \ldots, p_{2 n}\right\}$, if Arthur chooses elements of the form $p_{1} p_{j}$ with $n+2 \leqslant j \leqslant 2 n$, and $p_{n+1} p_{l}$ for $2 \leqslant l \leqslant n$, it has $2 n-2$ distinct elements. Since every chosen element has a prime factor that it alone possesses, Thomas cannot find a triplet $(x, y, z)$ satisfying the statement, so Arthur wins. By removing elements if necessary, if $k \leqslant 2 n-2$, Arthur can win.