Reused writeup on Math-814 (gpt pro 5.2)

Erdős Problem 431 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/431

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_5.2/431.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/431.tex.

Are there two infinite sets $A$ and $B$ such that $A+B$ agrees with the set of prime numbers up to finitely many exceptions? A problem of Ostmann, sometimes known as the 'inverse Goldbach problem'. The answer is surely no. The best result in this direction is due to Elsholtz and Harper [ElHa15], who showed that if $A,B$ are such sets then for all large $x$ we must have\[\frac{x^{1/2}}{\log x\log\log x} \ll \lvert A \cap [1,x]\rvert \ll x^{1/2}\log\log x\]and similarly for $B$. Elsholtz [El01] has proved there are no sets $A,B,C$ (all of size at least $2$) such that $A+B+C$ agrees with the set of prime numbers up to finitely many exceptions. Granville [Gr90] proved, conditional on the prime $k$-tuples conjecture, that there are infinite sets $B$ and $C$ such that\[\{ \tfrac{b+c}{2}: b\in B, c\in C\}\]is a subset of the primes. Tao and Ziegler [TaZi23] gave an unconditional proof that there are infinite sets $B=\{b_1<\cdots\}$ and $C=\{c_1<\cdots\}$ such that\[\{ b_i+c_j : b_i\in B, c_j\in C, i<j\}\]is a subset of the primes. See also [429] and [432] . References [El01] Elsholtz, Christian, The inverse Goldbach problem . Mathematika $2001$, 151-158. [ElHa15] Elsholtz, Christian and Harper, Adam J., Additive decompositions of sets with restricted prime factors . Trans. Amer. Math. Soc. $2015$, 7403-7427. [Gr90] Granville, Andrew, A note on sums of primes . Canad. Math. Bull. $1990$, 452--454. [TaZi23] Tao, Terence and Ziegler, Tamar, Infinite partial sumsets in the primes . J. Anal. Math. $2023$, 375--389.

FORMAL RESTATEMENT

Let $\mathbb N:=\{1,2,3,\dots\}$. For sets $A,B\subseteq\mathbb N$ define the (binary) sumset \[ A+B := \{a+b : a\in A,\ b\in B\}\subseteq \mathbb N. \] Let $\mathcal P\subseteq\mathbb N$ be the set of prime numbers. The phrase "$A+B$ agrees with the primes up to finitely many exceptions" means \[ \lvert (A+B)\,\triangle\,\mathcal P\rvert <\infty \quad\text{equivalently,}\quad \exists N_0\ \forall n\ge N_0:\ \left(n\in A+B \iff n\in\mathcal P\right). \] Question: do there exist infinite $A,B\subseteq\mathbb N$ with this property?

Stress points: (i) parity (all primes except $2$ are odd); (ii) congruence obstructions mod small primes $q$ (primes contain at most one multiple of $q$); (iii) "infinite" means we cannot use singleton tricks.

QUICK LITERATURE/CONTEXT CHECK

I am constrained to not use any external literature beyond what is explicitly stated in the problem text. The problem text reports: - Elsholtz--Harper give size constraints $\lvert A\cap[1,x]\rvert,\lvert B\cap[1,x]\rvert\asymp x^{1/2}$ up to polylog factors if such $A,B$ exist. - Elsholtz proves there are no $A,B,C$ of size at least $2$ with $A+B+C$ agreeing with primes up to finitely many exceptions. - Granville and Tao--Ziegler give constructions of infinite partial sumsets contained in the primes (different from full $A+B$).

ATTACK PLAN

Proof track (show nonexistence): 1) Push modular obstructions: for each small prime $q$, large sums in $A+B$ cannot hit $0\pmod q$; force $A$ and $B$ into sparse residue patterns. 2) Try to combine many moduli to force $A,B$ into near-arithmetic progressions, then show $A+B$ cannot model primes.

Disproof/construction track (try to build $A,B$): 1) Attempt to build $A,B$ with opposite parity and with residue supports avoiding $0$ modulo many small primes, while still having enough sums to hit almost all primes. (Heuristically this seems extremely hard; I did not find such a construction.)

I pursue the modular-obstruction direction to obtain necessary conditions (partial progress).

WORK

FAST REALITY CHECK (finite truncations). I brute-forced a small finite analogue: search for finite $A\subseteq\{2,4,\dots,30\}$ and $B\subseteq\{1,3,\dots,29\}$ with $2\le |A|,|B|\le 5$ such that \[(A+B)\cap[1,30] = \mathcal P\cap[1,30].\] Result: no such pairs were found.

N=30, M=30, sizes 2..5, parity constrained: num_solutions 0
elapsed_seconds 24.225643157958984

This does not prove anything about infinite sets, but it sanity-checks that the constraint is already tight in small ranges.

Lemma 431.1 (eventual parity separation). Assume $A,B\subseteq\mathbb N$ are infinite and there exists $N_0$ such that for all $n\ge N_0$, \[n\in A+B \iff n\in\mathcal P.\] Then there exist $\varepsilon_A,\varepsilon_B\in\{0,1\}$ with $\varepsilon_A+\varepsilon_B\equiv 1\pmod 2$ and finite subsets $F_A\subseteq A$, $F_B\subseteq B$ such that every $a\in A\setminus F_A$ satisfies $a\equiv \varepsilon_A\pmod 2$ and every $b\in B\setminus F_B$ satisfies $b\equiv \varepsilon_B\pmod 2$.

Proof. Suppose first that $A$ contains infinitely many even integers and $B$ contains infinitely many even integers. Choose an even $b_0\in B$ and an infinite increasing sequence of even $a_1<a_2<\cdots$ in $A$. Then all sums $a_i+b_0$ are even and strictly increasing, hence infinitely many of them exceed $N_0$. For such large $i$ we have $a_i+b_0\in A+B$, hence (by the agreement hypothesis) $a_i+b_0$ must be prime. But every even integer greater than $2$ is composite, a contradiction. Therefore, at least one of the sets $A,B$ contains only finitely many even integers.

Similarly, suppose $A$ contains infinitely many odd integers and $B$ contains infinitely many odd integers. Choose an odd $b_1\in B$ and an infinite increasing sequence of odd $a'_1<a'_2<\cdots$ in $A$. Then $a'_i+b_1$ are even, strictly increasing, and eventually exceed $N_0$, giving the same contradiction. Thus, at least one of $A,B$ contains only finitely many odd integers.

Because $A$ is infinite, it cannot have only finitely many evens and only finitely many odds. Hence exactly one of these parity classes occurs infinitely often in $A$; call that parity $\varepsilon_A$. Likewise $B$ has an eventual parity $\varepsilon_B$. Finally, if $\varepsilon_A=\varepsilon_B$ then for all sufficiently large $a\in A$ and $b\in B$ we have $a+b$ even, so $A+B$ contains infinitely many even integers exceeding $N_0$, contradicting agreement with primes. Therefore $\varepsilon_A\ne\varepsilon_B$, i.e. $\varepsilon_A+\varepsilon_B\equiv 1\pmod 2$. Taking $F_A$ and $F_B$ to be the finite sets of exceptional parity elements completes the proof. $\Box$

Lemma 431.2 (residue-support obstruction modulo a prime). Let $q$ be a fixed prime. Let $A,B\subseteq\mathbb N$ be infinite. Define \[ R_A(q):=\{r\in\mathbb Z/q\mathbb Z : \text{infinitely many }a\in A\text{ satisfy }a\equiv r\pmod q\}, \] and define $R_B(q)$ analogously. If $A+B$ contains only finitely many multiples of $q$, then \[ R_A(q)\cap\left(-R_B(q)\right)=\varnothing, \quad\text{equivalently}\quad 0\notin R_A(q)+R_B(q) \quad\text{and hence}\quad |R_A(q)|+|R_B(q)|\le q. \]

Proof. Assume for contradiction that there exist residues $r\in R_A(q)$ and $s\in R_B(q)$ with $r+s\equiv 0\pmod q$. By definition of $R_A(q)$ and $R_B(q)$, there exist infinitely many $a\in A$ with $a\equiv r\pmod q$ and infinitely many $b\in B$ with $b\equiv s\pmod q$. Fix one such $b=b_*$.

Consider the infinite set $\{a+b_*: a\in A,\ a\equiv r\pmod q\}\subseteq A+B$. All these sums are distinct (because $a$ are distinct) and each satisfies \[(a+b_*)\equiv r+s\equiv 0\pmod q, \] so they are multiples of $q$. Hence $A+B$ contains infinitely many multiples of $q$, contradicting the hypothesis that $A+B$ contains only finitely many multiples of $q$. Therefore $R_A(q)\cap(-R_B(q))=\varnothing$.

The inequality $|R_A(q)|+|R_B(q)|\le q$ follows because $-R_B(q)$ has the same size as $R_B(q)$ and both subsets lie in a set of size $q$ with empty intersection. $\Box$

Application to Problem 431. If $A+B$ agrees with primes beyond some $N_0$, then for any fixed prime $q$ we have: all sufficiently large multiples of $q$ are composite and hence cannot lie in $A+B$; therefore $A+B$ contains only finitely many multiples of $q$, and Lemma 431.2 applies.

VERIFICATION

- Lemma 431.1: The only prime of even parity is $2$, but it is a single exception. The argument only uses that there are infinitely many even sums if both parities persist, producing infinitely many composite even numbers $>2$; this is correct. - Lemma 431.2: The proof uses a fixed $b_*$ and infinitely many distinct $a$ in a residue class, ensuring infinitely many distinct sums. No step assumes $A$ or $B$ has positive density. - Edge case: if $q\ge N_0$, the set $A+B$ may contain the prime $q$ itself, but that is still only one multiple of $q$; "finitely many" is unaffected.

FINAL

UNRESOLVED

(i) Strongest proved partial result: If $A+B$ agrees with primes beyond $N_0$, then (a) $A$ and $B$ are eventually of opposite parity (Lemma 431.1), and (b) for every prime $q$, the sets of residue classes occurring infinitely often in $A$ and $B$ satisfy $0\notin R_A(q)+R_B(q)$ (Lemma 431.2).

(ii) First gap (crisp): Prove that there do not exist infinite $A,B\subseteq\mathbb N$ with $\lvert (A+B)\triangle\mathcal P\rvert<\infty$.

(iii) Top 3 next moves: 1. Strengthen Lemma 431.2 to show that for many small primes $q$, one of $R_A(q),R_B(q)$ must be very small (ideally a singleton), and quantify how these constraints interact across many moduli. 2. Combine the modular restrictions with additive-combinatorial structure theorems (e.g. small-doubling type statements for $A,B$ implied by $A+B$ being "thin") to force rigid structure, then contradict the irregular distribution of primes. 3. Computation: search for finite sets $A,B$ with $|A|,|B|\ge 2$ such that $(A+B)\cap[1,N]$ matches primes for larger $N$ and examine forced residue patterns; attempt to extrapolate an obstruction.

(iv) What a minimal counterexample would likely look like: $A$ and $B$ would have opposite parity, and for each small prime $q$ the infinite residue supports $R_A(q)$ and $R_B(q)$ would avoid additive inverses. Moreover, according to the problem text, such a counterexample would have to satisfy $|A\cap[1,x]|,|B\cap[1,x]|$ on the order of $x^{1/2}$ up to polylog factors.

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.