Task 2. The sets $A$ and $B$ are subsets of the positive integers. The sum of any two different elements from $A$ is an element of $B$. The quotient of any two different elements of $B$ (where we divide the larger by the smaller) is an element of $A$. Determine the maximum number of elements in $A \cup B$.
Solution. Suppose $A$ contains at least three elements, say $a<b<c$, so $b-a$ is positive, thus it must hold that $a+c \leq b-a$. This gives $c \leq b-2a$. Therefore, $A$ contains at most two elements.

Suppose $B$ contains at least four elements, say $a<b<c<d$. Then $A$ contains the three distinct elements $\frac{d}{a}, \frac{d}{b}$, and $\frac{d}{c}$. But $A$ cannot contain three distinct elements, contradiction. Therefore, $B$ contains at most three elements.

In total, $A \cup B$ contains at most 5 elements. This is possible, for example with $A=\{2,4\}$ and $B=\{3,6,12\}$. Now $2+4=6 \in B$ and $\frac{12}{6}=\frac{6}{3}=2 \in A$ and $\frac{12}{3}=4 \in A$, so this pair of sets satisfies the conditions. We conclude that $A \cup B$ contains at most 5 elements. Finding a pair of sets that satisfies the conditions can be done as follows. Suppose $B$ contains the elements $a<b<c$. Then $A$ contains the elements $\frac{c}{b}$, $\frac{b}{a}$, and $\frac{c}{a}$, with $\frac{c}{a}$ being the largest. Since $A$ contains only two elements, it must hold that $\frac{c}{b}=\frac{b}{a}$. Furthermore, the sum of the two elements in $A$ must be in $B$, so $\frac{b}{a}+\frac{c}{a} \in B$. Write $b=t a$, then $c=t b=t^{2} a$ and $\frac{b}{a}+\frac{c}{a}=t+t^{2}$. Now you see that $b$ can equal $t+t^{2}$ by choosing $a=t+1$. With $t=1$ you get $b=a$, so that does not work; with $t=2$ you get the solution mentioned above.