parmenides51 wrote: Let $n \ge 3$ be an integer and $A$ be a subset of the real numbers of size n. Denote by $B$ the set of real numbers that are of the form $ x \cdot y$, where $x, y \in A$ and $x\ne y$. At most how many distinct positive primes could $B$ contain (depending on $n$)? I used graph in the contest, but fake solved to get answer $n-1$ apparently the answer is $n$, after reading the official solution

parmenides51 wrote: Let $n \ge 3$ be an integer and $A$ be a subset of the real numbers of size n. Denote by $B$ the set of real numbers that are of the form $ x \cdot y$, where $x, y \in A$ and $x\ne y$. At most how many distinct positive primes could $B$ contain (depending on $n$)? [redacted, thanks @below]

GreenTea2593 wrote: Clearly we cannot have any cycles in graph $G$, since multiplication of some distinct primes cannot be a perfect square. Bro i think there's something wrong here. $A$ is a subset of real numbers, then you cannot call $x_1^2x_2^2\cdots x_n^2$ a perfect square. In fact, let $x_1=\sqrt{\frac{pq}{r}}, x_2=\sqrt{\frac{qr}{p}}, x_3=\sqrt{\frac{rp}{q}}$, where $p,q,r$ are distinct primes, then $x_1x_2=q, x_2x_3=r, x_3x_1=p$, then it forms a cycle in the graph.

ABCDTNT__ wrote: GreenTea2593 wrote: Clearly we cannot have any cycles in graph $G$, since multiplication of some distinct primes cannot be a perfect square. Bro i think there's something wrong here. $A$ is a subset of real numbers, then you cannot call $x_1^2x_2^2\cdots x_n^2$ a perfect square. In fact, let $x_1=\sqrt{\frac{pq}{r}}, x_2=\sqrt{\frac{qr}{p}}, x_3=\sqrt{\frac{rp}{q}}$, where $p,q,r$ are distinct primes, then $x_1x_2=q, x_2x_3=r, x_3x_1=p$, then it forms a cycle in the graph. thanks for pointing that out, so it can only imply there is no cycle of even length