we-solve (beta)

Math-842-A1 · revision r1 · on Math-842

Reused writeup on Math-842 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem 531 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. FORMAL RESTATEMENT
  5. QUICK LITERATURE/CONTEXT CHECK
  6. ATTACK PLAN
  7. WORK
  8. FAST REALITY CHECK (exact computation for small parameters).
  9. VERIFICATION
  10. FINAL
  11. COMPLETION ESTIMATE

Erdős Problem 531 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/531

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_5.2/531.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/531.tex.

Let $F(k)$ be the minimal $N$ such that if we two-colour $\{1,\ldots,N\}$ there is a set $A$ of size $k$ such that all subset sums $\sum_{a\in S}a$ (for $\emptyset\neq S\subseteq A$) are monochromatic. Estimate $F(k)$. The existence of $F(k)$ was established by Sanders and Folkman, and it also follows from Rado's theorem. It is commonly known as Folkman's theorem . Erdős and Spencer [ErSp89] proved that\[F(k) \geq 2^{ck^2/\log k}\]for some constant $c>0$. Balogh, Eberhrad, Narayanan, Treglown, and Wagner [BENTW17] have improved this to\[F(k) \geq 2^{2^{k-1}/k}.\] References [BENTW17] Balogh, József and Eberhard, Sean and Narayanan, Bhargav and Treglown, Andrew and Wagner, Adam Zsolt, An improved lower bound for Folkman's theorem . Bull. Lond. Math. Soc. $2017$, 745-747. [ErSp89] Erdős, Paul and Spencer, Joel, Monochromatic sumsets . J. Combin. Theory Ser. A $1989$, 162-163.

FORMAL RESTATEMENT

Fix $k\ge 1$. For $N\ge 1$, a 2-colouring of $[N]:=\{1,2,\dots,N\}$ is a map $\chi:[N]\to\{0,1\}$. For a finite set $A\subseteq [N]$, define its nonempty subset-sum set \[ \mathrm{FS}(A):=\Big\{\sum_{a\in S}a:\ \emptyset\ne S\subseteq A\Big\}\subseteq \mathbb N. \] We say $A$ is monochromatic-sum for $\chi$ if
  1. $\mathrm{FS}(A)\subseteq [N]$ (so every sum is assigned a colour), and
  2. $\chi$ is constant on $\mathrm{FS}(A)$.
Define $F(k)$ to be the minimal $N$ (if it exists) such that every 2-colouring of $[N]$ contains some $A\subseteq [N]$ with $|A|=k$ that is monochromatic-sum.

Ambiguity in the literal statement. The original phrasing does not explicitly require $\mathrm{FS}(A)\subseteq [N]$, but since colours are only defined on $[N]$, the natural convention is to require it; otherwise the property is ill-posed. I adopt that convention.

QUICK LITERATURE/CONTEXT CHECK

The statement records large lower bounds for $F(k)$ from [ErSp89,BENTW17]. Per the integrity rule I do not add or use any additional external results.

ATTACK PLAN

Proof-track ideas. (i) Compute exact values for small $k$ by exhaustive search over all colourings (reality check). (ii) Prove structural reductions (monotonicity; necessary constraints on $A$ such as $\sum A\le N$).

Disproof-track ideas. For any proposed explicit upper bound on $F(k)$, attempt to construct a colouring of $[N]$ avoiding monochromatic-sum sets of size $k$ using backtracking/SAT.

WORK

Lemma (Necessary sum constraint): If $A\subseteq [N]$ satisfies $\mathrm{FS}(A)\subseteq [N]$, then $\sum_{a\in A} a \le N$.
Proof: The full sum $\sum_{a\in A}a$ is an element of $\mathrm{FS}(A)$ (take $S=A$). If $\mathrm{FS}(A)\subseteq [N]$, then in particular $\sum_{a\in A}a\le N$. ∎
Proposition (Exact value for $k=2$): Under the convention above (distinct elements in $A$), one has $F(2)=9$.
Proof: For $k=2$, a set $A=\{x,y\}$ with $x<y$ is monochromatic-sum iff $x,y,x+y\in [N]$ and $\chi(x)=\chi(y)=\chi(x+y)$. By Lemma Lemma (Necessary sum constraint), necessarily $x+y\le N$.

Lower bound $F(2)>8$. Consider the colouring of $[8]$ given (in order $1,2,\dots,8$) by \[ BBRBRRRB, \] where $B$ means colour $0$ and $R$ means colour $1$. I checked all pairs $1\le x<y\le 8$ with $x+y\le 8$ (there are finitely many) and verified that none satisfy $\chi(x)=\chi(y)=\chi(x+y)$ under this colouring. Hence there is a 2-colouring of $[8]$ with no monochromatic-sum $2$-set, so $F(2)\ge 9$.

Upper bound $F(2)\le 9$. I exhaustively checked all $2^9=512$ colourings of $[9]$; in every case there exists a pair $1\le x<y\le 9$ with $x+y\le 9$ and $\chi(x)=\chi(y)=\chi(x+y)$. Therefore every colouring of $[9]$ contains a monochromatic-sum set of size $2$, so $F(2)\le 9$.

Combining yields $F(2)=9$. ∎

FAST REALITY CHECK (exact computation for small parameters).
The bounds in Proposition prop:531-F2 were verified by exhaustive search over all colourings of $[N]$ for $N\le 9$ and all candidate pairs $\{x,y\}$ with $x+y\le N$.

For $k=3$, exhaustive search over all colourings becomes infeasible for large $N$ because there are $2^N$ colourings, but I did check the existence of one obstruction colouring for each $N\le 25$; in particular, \[ F(3)>25. \] One explicit witness colouring of $[25]$ avoiding monochromatic-sum $3$-sets is (in order $1,2,\dots,25$) \[ RRBBRBRRRRBBBBBBBBBBBBBBB. \] This was verified by enumerating all triples $A=\{a<b<c\}\subseteq [25]$ with $a+b+c\le 25$ (Lemma Lemma (Necessary sum constraint)) and checking that the seven subset sums in $\mathrm{FS}(A)$ are not all the same colour.

VERIFICATION

  • Definition check: $\mathrm{FS}(A)$ uses distinct elements, so $k=2$ forbids using $x=x$; this is why $F(2)$ differs from the classical "monochromatic $x+y=z$" condition that allows $x=y$.
  • Proposition prop:531-F2: both the lower-bound witness on $[8]$ and the universal property on $[9]$ were established by exhaustive enumeration over finite search spaces.
  • For $F(3)>25$: the verification checks all eligible triples for the displayed colouring, so it is exact for that $N$.

FINAL

UNRESOLVED

(i) Strongest proved partial result. Exact small-$k$ information: $F(2)=9$ (Proposition prop:531-F2) and an explicit computed lower bound $F(3)>25$ with a witness colouring. Additionally, any candidate $A$ must satisfy $\sum A\le N$ (Lemma Lemma (Necessary sum constraint)).

(ii) First gap (crisp statement). Determine $F(3)$ (even within reasonable magnitude), or prove any explicit general upper bound on $F(k)$ as a function of $k$.

(iii) Top 3 next moves. 1. Implement SAT/ILP with symmetry-breaking for $k=3$ to push explicit lower-bound colourings to much larger $N$, narrowing the range for $F(3)$. 2. Seek a finitary proof of existence of $F(k)$ with explicit quantitative bounds by encoding the problem as a finite system of partition-regular equations and tracking constants. 3. Attempt an iterative "finite sums" construction that produces a monochromatic $\mathrm{FS}(A)$ set while explicitly controlling $\max \mathrm{FS}(A)\le N$, yielding constructive upper bounds.

(iv) Minimal counterexample structure. A colouring witnessing large $F(k)$ must avoid monochromaticity on every $\mathrm{FS}(A)$ with $|A|=k$, forcing a dense pattern of colour alternation across many additive configurations; for $k=3$ this means avoiding monochromaticity on all $7$-term sets $\mathrm{FS}(\{a,b,c\})$ with $a+b+c\le N$.

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.