Let $\epsilon>0$ and $N$ be sufficiently large. If $A\subseteq \{1,\ldots,N\}$ has $\lvert A\rvert \geq \epsilon N$ then must there exist $a_1,a_2,a_3\in A$ and distinct primes $p_1,p_2,p_3$ such that $$a_1p_1=a_2p_2=a_3p_3?$$

Let $\epsilon>0$ and $N$ be sufficiently large. If $A\subseteq \{1,\ldots,N\}$ has $\lvert A\rvert \geq \epsilon N$ then must there exist $a_1,a_2,a_3\in A$ and distinct primes $p_1,p_2,p_3$ such that $$a_1p_1=a_2p_2=a_3p_3?$$

A positive answer would imply [536].

Erdős describes a construction of Ruzsa which disproves this: consider the set of all squarefree numbers of the shape $p_1\cdots p_r$ where $p_{i+1}>2p_i$ for $1\leq i<r$. This set has positive density, and hence if $A$ is its intersection with $(N/2,N)$ then $\lvert A\rvert \gg N$ for all large $N$. Suppose now that $p_1a_1=p_2a_2=p_3a_3$ where $a_i\in A$ and $p_1,p_2,p_3$ are distinct primes. Without loss of generality we may assume that $a_2>a_3$ and hence $p_2<p_3$, and so since $p_2p_3\mid a_1\in A$ we must have $2<p_3/p_2$. On the other hand $p_3/p_2=a_2/a_3\in (1,2)$, a contradiction.