GPT 6 Astra Ultra attempt on Math-5642

Erdős Problem #429: arbitrarily sparse admissible sets without prime translates

Complete diagonal construction, including the squarefree variant, 24 September 2026.

1) Statement and the meaning of arbitrary sparsity

Does sufficient sparsity of an infinite admissible $A\subseteq\mathbb N$ guarantee some integer translate $A+n$ consisting only of primes? Admissible means that $A$ omits a residue class modulo each prime. We prove a stronger negative statement: for every nondecreasing unbounded function $f:\mathbb N\to\mathbb Z_{\ge0}$ there is an infinite $A$ with $A(N)\le f(N)$ for all $N$, omitting a residue class modulo every prime and every prime square, such that every integer translate contains a positive nonsquarefree composite number.

2) Construction by compatible congruences

Enumerate the integers as $z_1,z_2,\ldots$ without omission, for example $0,1,-1,2,-2,\ldots$, and enumerate the primes as $p_1<p_2<\cdots$. Put \[P_j=\prod_{i=1}^j p_i^2,\qquad q_j=p_{j+1}. \] The Chinese remainder theorem supplies an infinite arithmetic progression of integers $a$ satisfying \[a\equiv0\pmod{P_j},\qquad a\equiv-z_j\pmod{q_j^2},\tag{1}\] because $P_j$ and $q_j^2$ are coprime. Choose $a_j$ from that progression sufficiently large that \[a_j>a_{j-1},\qquad a_j+z_j>q_j^2,\qquad f(a_j)\ge j, \tag{2}\] with $a_0=0$. Each condition is eventually satisfied along the progression: for the last one, use monotonicity and unboundedness of $f$. Define $A=\{a_j:j\ge1\}$.

3) Verify sparsity and admissibility for every prime

If $N<a_1$, then $A(N)=0\le f(N)$. Otherwise let $j=A(N)$, so $a_j\le N<a_{j+1}$. By $2$ and monotonicity, $A(N)=j\le f(a_j)\le f(N)$. Thus the desired bound holds at every integer cutoff, not just along the selected sequence.

Fix a prime $p_i$. Every $a_j$ with $j\ge i$ is divisible by $p_i^2$, by $1$. The entire sequence therefore occupies at most $i$ residue classes modulo $p_i^2$: at most $i-1$ classes from the earlier elements, together with zero. The same bound holds modulo $p_i$. Since $i<p_i<p_i^2$, at least one residue class is omitted for both moduli. This counts the whole infinite set and does not merely establish finite-prefix admissibility.

4) Every translate is excluded

Given any $z\in\mathbb Z$, choose $j$ with $z=z_j$. Equations $1$ and $2$ say that $a_j+z$ is a positive multiple of $q_j^2$, strictly greater than $q_j^2$. It is therefore composite and not squarefree. Consequently $A+z$ is neither a subset of the primes nor a subset of the squarefree integers. Negative shifts cause no exception: the selected witness was explicitly chosen positive.

5) Outcome and attribution

SOLVED in the negative, for both formulations. The construction works under every prescribed monotone unbounded counting envelope and defeats all integer shifts simultaneously while preserving both residue-omission conditions. It is a Chinese-remainder diagonal treatment of the known counterexamples, with every compatibility and quantifier checked.

Sources: https://www.erdosproblems.com/429, checked 24 September 2026; D. Weisenberg, Sparse Admissible Sets and a Problem of Erdős and Graham, https://arxiv.org/abs/2405.12310. The negative resolution is due to Weisenberg. No novelty or independent Lean verification is claimed.

The estimate applies to the full sparsity and squarefree assertions addressed here.

6) COMPLETION ESTIMATE

COMPLETION: 100%