Reused writeup on Math-813 (gpt pro 5.2)

Erdős Problem 428 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/428

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/428.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/428.tex.

FORMAL RESTATEMENT Does there exist a set $A\subseteq\mathbb N$ such that: 1) There are infinitely many integers $n\ge 1$ with the property that for every $a\in A$ satisfying $0<a<n$, the integer $n-a$ is prime. 2) The set $A$ has positive lower density relative to the primes, i.e. \[ \liminf_{x\to\infty}\frac{|A\cap[1,x]|}{\pi(x)}>0. \] Here $\pi(x)$ is the number of primes $\le x$.

Equivalently, for infinitely many $n$ we have \[ A\cap[1,n-1]\subseteq \{n-p: p\text{ prime},\ 1\le p\le n-1\}. \]

Edge cases: $a=0$ is excluded by hypothesis; $n-a$ being prime includes the possibility $n-a=2$.

QUICK LITERATURE/CONTEXT CHECK The problem statement notes that Erdős and Graham can show the analogous statement with $\limsup$ (in place of $\liminf$) assuming the prime $k$-tuple conjecture. No other external results are used here.

ATTACK PLAN Proof track ideas. 1) Identify unavoidable local obstructions (parity, mod $3$, etc.) forcing $A$ into certain residue classes. 2) Try to build $A$ inside a single residue class (e.g. even integers) and search for infinitely many $n$ where the reflected set $n-A$ hits only primes.

Disproof track ideas. 1) Use parity and small-modulus constraints to show that if $A$ is large enough (comparable to primes) then the constraints for $n-a$ to all be prime become mutually incompatible for large $n$. 2) Show that for any “dense relative to primes” set $A$, the set of shifts $n$ with $n-A$ all prime must be finite.

WORK Fast reality check (finite toy models). For a finite set $A$, the condition “$n-a$ is prime for all $a\in A$ with $a<n$” is exactly a prime-constellation condition. For example: - If $A=\{2\}$, the condition is that $n-2$ is prime, which holds for infinitely many $n$ (assuming infinitely many primes). - If $A=\{2,4\}$, the condition is that $n-4$ and $n-2$ are both prime, which is equivalent to the existence of twin primes. A brute-force check up to $n\le 200$ shows that for $A=\{2,4\}$ there are $14$ such $n$ (corresponding to twin prime pairs $\le 198$), while for $A=\{2,4,6\}$ there is exactly one such $n$, namely $n=9$ (corresponding to primes $3,5,7$).

Lemma 428.1 (parity obstruction for infinitely many $n$). If $A$ contains both an even integer and an odd integer, then there are only finitely many $n$ satisfying condition $1$. In particular, no such $A$ can satisfy the “infinitely many $n$” requirement.

Proof. Assume $a_e\in A$ is even and $a_o\in A$ is odd. Let $n$ be an integer with $n>\max\{a_e,a_o\}+2$. Then both $n-a_e$ and $n-a_o$ are positive integers $>2$. Also, $n-a_e$ has the same parity as $n$ (since $a_e$ is even), whereas $n-a_o$ has the opposite parity to $n$ (since $a_o$ is odd). Thus one of $n-a_e$ and $n-a_o$ is even. But any even prime is equal to $2$, and we have just seen both numbers are $>2$. Therefore, at least one of $n-a_e$ and $n-a_o$ is an even integer greater than $2$, hence composite. So for every $n>\max\{a_e,a_o\}+2$, condition $1$ fails. Thus there are at most finitely many $n$ satisfying $1$, namely $n\le \max\{a_e,a_o\}+2$. $\square$

Lemma 428.2 (eventual parity of the witnessing $n$ for infinite $A$). Suppose $A$ is infinite (in particular, condition $2$ holds) and $A$ satisfies condition $1$ for infinitely many $n$. Then either: - $A$ is contained in the even integers and all sufficiently large witnessing $n$ must be odd, or - $A$ is contained in the odd integers and all sufficiently large witnessing $n$ must be even.

Proof. By Lemma 428.1, $A$ cannot contain both parities, so $A\subseteq 2\mathbb Z$ or $A\subseteq 2\mathbb Z+1$.

Assume first that $A\subseteq 2\mathbb Z$. Since $A$ is infinite, choose two distinct elements $a_1<a_2$ of $A$. Consider any even integer $n>a_2$. Then $a_1,a_2<n$, so condition $1$ would require both $n-a_1$ and $n-a_2$ to be prime. But $n-a_1$ and $n-a_2$ are even (even minus even), and they are $>2$ because $n>a_2\ge a_1+2$. An even integer $>2$ is composite, so condition $1$ fails for every even $n>a_2$. Therefore, among the infinitely many witnessing $n$, only finitely many can be even; equivalently, all sufficiently large witnessing $n$ are odd.

The case $A\subseteq 2\mathbb Z+1$ is analogous. Choose two distinct odd elements $b_1<b_2$ of $A$. For any odd $n>b_2$, the differences $n-b_1$ and $n-b_2$ are even integers $>2$, hence composite, so such $n$ cannot witness condition $1$. Thus all sufficiently large witnessing $n$ must be even. $\square$

VERIFICATION - Lemma 428.1: the bound $n>\max\{a_e,a_o\}+2$ ensures both differences exceed $2$, so “even prime” $=2$ is excluded. - Lemma 428.2: the argument uses only the fact that an even prime is $2$ and that $A$ is fixed while witnessing $n$ vary. - Toy computations: the finite-set checks align with known necessary constraints (e.g. $\{2,4,6\}$ forces a prime triplet and only $3,5,7$ works because of mod $3$).

FINAL UNRESOLVED (i) Strongest proved partial result: any $A$ satisfying the “infinitely many $n$” requirement must lie entirely in one parity class (Lemma 428.1), and then the witnessing $n$ must eventually have the opposite parity (Lemma 428.2). (ii) First gap (crisp): construct (or rule out) an infinite set $A$ with $|A\cap[1,x]|\gg \pi(x)$ and infinitely many $n$ such that $\{n-a: a\in A,\ a<n\}$ are all prime. (iii) Top 3 next moves: 1. Local obstruction analysis beyond parity: prove that $A$ must avoid additional residue patterns mod small primes (in the spirit of admissible prime $k$-tuples) uniformly across all initial segments $A\cap[1,n)$. 2. Constructive search: restrict to $A\subseteq 2\mathbb Z$ (or $2\mathbb Z+1$) and use computation to find large initial segments $A\cap[1,x]$ that admit some $n$ with all $n-a$ prime, then study whether these initial segments can be extended indefinitely. 3. Sieve-theoretic barrier: attempt to show that if $|A\cap[1,x]|\ge c\pi(x)$ for fixed $c>0$, then for large $n$ the congruence conditions imposed by $\{n-a\}$ being prime force a contradiction (e.g. too many residues hit modulo a small prime). (iv) Minimal counterexample structure: if the statement is false, then for every $A\subseteq\mathbb N$ with $|A\cap[1,x]|\gg \pi(x)$ there would exist $N(A)$ such that for all $n\ge N(A)$ some $a\in A$ with $a<n$ makes $n-a$ composite; a minimal obstruction would likely come from a fixed small modulus $q$ for which $A$ hits too many residue classes so that $n-A$ cannot avoid multiples of $q$ for all large $n$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.