GPT 6 Astra Ultra attempt on Math-5721

Erdős problem 537. Must every positive-density subset of $[1,N]$, for sufficiently large $N$, contain $a_1,a_2,a_3$ with $a_1p_1=a_2p_2=a_3p_3$ for three distinct primes?

Status and attribution. No. We reconstruct Ruzsa's counterexample recorded at https://www.erdosproblems.com/537, checked 24 September 2026, including the positive-density step. The only number-theoretic estimate used below is the classical Chebyshev bound $\pi(x)=O(x/\log x)$.

A positive-density set of separated prime factors. Let $B$ consist of squarefree positive integers whose consecutive prime factors have ratio greater than $2$. Equivalently, an integer belongs to $B$ if it is divisible by no $p^2$ and by no $pq$ with primes $p<q\leq2p$. The sum of the reciprocal forbidden divisors is finite: \[ \sum_p\frac1{p^2}+\sum_{p<q\leq2p}\frac1{pq}<\infty. \tag{1} \] Indeed, for $2^j\leq p<2^{j+1}$, Chebyshev bounds the number of possible pairs $(p,q)$ by $O(4^j/j^2)$, while $pq\geq4^j$. Summing $O(j^{-2})$ proves the second convergence.

For a finite list of primes, the valuations truncated to $\{0,1,\geq2\}$ of a uniformly chosen residue modulo the product of their squares are independent, by the Chinese remainder theorem. Each of the conditions $p^2\nmid n$ and $pq\nmid n$ is decreasing in these coordinates. Decreasing events in independent ordered coordinates are positively correlated. Here is a short justification: in one coordinate, for decreasing functions $f,g$ and independent identically distributed $X,X'$, \[ 2\operatorname{Cov}(f(X),g(X)) =\mathbb E[(f(X)-f(X'))(g(X)-g(X'))]\geq0. \] Conditioning on the last coordinate and inducting proves the product-space assertion; iterating it applies to any finite intersection of decreasing events.

It follows that the density of integers obeying any finite collection of the forbidden-divisibility conditions is at least the product of their individual probabilities. By $1$, the infinite product \[ c=\prod_p(1-p^{-2}) \prod_{p<q\leq2p}(1-(pq)^{-1}) \] converges to a positive number. Let $B_y$ impose just the conditions involving primes at most $y$. Its natural density $d_y$ exists by periodicity and satisfies $d_y\geq c$. The upper density of $B_y\setminus B$ is at most \[ \sum_{p>y}p^{-2} +\sum_{\substack{p<q\leq2p\\q>y}}(pq)^{-1}, \] which tends to zero. This estimate is uniform in the counting endpoint, since the proportion of multiples of an integer $d$ up to $X$ is at most $1/d$. Thus $B$ has a natural density $d=\lim_y d_y\geq c>0$.

The obstruction to three prime multiples. Take $A_N=B\cap(N/2,N]$. Then $|A_N|=(d/2+o(1))N$. Suppose $a_1p_1=a_2p_2=a_3p_3=M$ with distinct primes. Since $a_1=M/p_1$ and $p_2p_3\mid M$, both $p_2$ and $p_3$ divide $a_1$. Relabel them so $p_2<p_3$. The defining property of $B$ forces $p_3/p_2>2$, even if these primes are not consecutive factors. On the other hand \[ \frac{p_3}{p_2}=\frac{a_2}{a_3}<2 \] because both $a_2,a_3$ lie in $(N/2,N]$. This is a contradiction. For example, fixing $\epsilon=d/4$ gives a counterexample with $|A_N|\geq\epsilon N$ for every sufficiently large $N$.

Completion estimate. Both the positive natural density and the absence of the required configuration are proved above. COMPLETION: 100%