we-solve (beta)

Math-855-A1 · revision r1 · on Math-855

Reused writeup on Math-855 (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 569 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. COMPLETION ESTIMATE

Erdős Problem 569 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/569

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_5.2/569.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/569.tex.

1) FORMAL RESTATEMENT

Let $k\ge 1$ be fixed and let $C_{2k+1}$ be the cycle on $2k+1$ vertices. Define $c_k$ to be the infimum over constants $c$ such that for every graph $H$ with $m=e(H)$ edges and no isolated vertices, \[ R(C_{2k+1},H) \le c\,m. \] The problem asks for the best possible $c_k$.

2) QUICK LITERATURE/CONTEXT CHECK

The provided problem statement does not list any specific results for #569 beyond asking for the best constant. (No external results are used as black boxes.)

3) ATTACK PLAN

- Lower bounds on $c_k$: find specific $m$-edge graphs $H$ for which $R(C_{2k+1},H)$ is large compared to $m$.

- Upper bounds on $c_k$: attempt to prove a universal linear (in $m$) upper bound, possibly by structural analysis of graphs with no red $C_{2k+1}$.

4) WORK

Fast reality check.

For $m=1$, the only graph $H$ with one edge and no isolated vertices is $K_2$. In this case $R(C_{2k+1},K_2)=2k+1$ (Lemma 569.1), forcing $c_k\ge 2k+1$.

For $k=2$ (so $C_5$) and $H=2K_2$ (two disjoint edges, $m=2$), a brute-force search gives $R(C_5,2K_2)=6$ (see the computation recorded under VERIFICATION below). For $H=3K_2$ ($m=3$), the same search gives $R(C_5,3K_2)=8$.

Lemma 569.1 (exact value for $H=K_2$). For every $\ell\ge 3$, \[ R(C_{\ell},K_2)=\ell. \] In particular, $R(C_{2k+1},K_2)=2k+1$, hence any admissible constant must satisfy $c_k\ge 2k+1$.

Proof. Let $\ell\ge 3$.

- Lower bound: In $K_{\ell-1}$ colour every edge red. Then there is no blue $K_2$ (no blue edges at all). Also $K_{\ell-1}$ cannot contain a cycle $C_{\ell}$ because it has fewer than $\ell$ vertices. Hence $R(C_{\ell},K_2)>\ell-1$.

- Upper bound: Consider any red/blue colouring of $K_{\ell}$. If there is a blue edge, we have a blue $K_2$. If there is no blue edge, then all edges are red and the red graph is $K_{\ell}$, which contains a (not necessarily induced) copy of $C_{\ell}$ on all $\ell$ vertices.

Thus $R(C_{\ell},K_2)\le \ell$. Combined with the lower bound, $R(C_{\ell},K_2)=\ell$.

Now set $\ell=2k+1$ to get $R(C_{2k+1},K_2)=2k+1$, and since $m=e(K_2)=1$ we obtain $c_k\ge 2k+1$. $\square$

Lemma 569.2 (a crude universal upper bound). Let $H$ be a graph with $m=e(H)$ edges and no isolated vertices. Then \[ R(C_{2k+1},H) \le \binom{2m+2k-1}{2k}. \]

Proof. Let $n=|V(H)|$. By Lemma 568.1 (proved above), $n\le 2m$.

Since $H\subseteq K_n$, monotonicity in the second argument gives \[ R(C_{2k+1},H) \le R(C_{2k+1},K_n). \] Also $C_{2k+1}\subseteq K_{2k+1}$, so monotonicity in the first argument gives \[ R(C_{2k+1},K_n) \le R(K_{2k+1},K_n)=:R(2k+1,n). \] It remains to bound the clique Ramsey number $R(2k+1,n)$.

Claim: for all integers $a,b\ge 2$, \[ R(a,b) \le \binom{a+b-2}{a-1}. \] Proof of claim: The standard recursion for clique Ramsey numbers is \[ R(a,b) \le R(a-1,b)+R(a,b-1)\quad(a,b\ge 3), \] with boundary conditions $R(2,b)=b$ and $R(a,2)=a$. (To justify the recursion: in a colouring of $K_{R(a-1,b)+R(a,b-1)}$, pick a vertex $v$; if $v$ has at least $R(a-1,b)$ red neighbours then among them there is either a red $K_{a-1}$, which together with $v$ gives a red $K_a$, or a blue $K_b$; similarly for blue degree.)

Assume inductively that $R(a-1,b)\le \binom{a+b-3}{a-2}$ and $R(a,b-1)\le \binom{a+b-3}{a-1}$. Then \[ R(a,b)\le R(a-1,b)+R(a,b-1) \le \binom{a+b-3}{a-2}+\binom{a+b-3}{a-1} = \binom{a+b-2}{a-1}, \] using Pascal's identity. The boundary cases match the binomial coefficients. This proves the claim.

Applying the claim with $(a,b)=(2k+1,n)$ yields \[ R(C_{2k+1},H) \le R(2k+1,n) \le \binom{2k+1+n-2}{2k} \le \binom{2k+1+2m-2}{2k} = \binom{2m+2k-1}{2k}. \] $\square$

Partial information for $k=1$. When $k=1$, $C_{2k+1}=C_3$ is a triangle. The file for Problem #570 states that Sidorenko proved $R(C_3,H)\le 2m+1$ for all $m$-edge graphs $H$ without isolated vertices. Combining this with Lemma 569.1 at $k=1$ shows $c_1=3$, since $2m+1\le 3m$ for all $m\ge 1$ and equality is forced by $m=1$.

5) VERIFICATION

Brute-force small-case checks (exact). A direct exhaustive search over red/blue colourings (equivalently, over red graphs) gives: \[ R(C_5,2K_2)=6,\qquad R(C_5,3K_2)=8. \] (Computed locally by enumerating all colourings for $N\le 7$ and using backtracking for $N=8$.)

6) FINAL

UNRESOLVED

(i) Strongest proved partial result here: $c_k\ge 2k+1$ (forced already by $H=K_2$), and we have the explicit (very weak) universal upper bound \[ R(C_{2k+1},H) \le \binom{2m+2k-1}{2k}. \] Also, for $k=1$ we can conclude $c_1=3$ using the Sidorenko bound stated in the file for #570.

(ii) First gap (crisp): prove any bound of the form $R(C_{2k+1},H)\le C(k)\,m$ with $C(k)$ polynomial (or even linear) in $k$, let alone determine the optimal constant $c_k$.

(iii) Top 3 next moves:

1. Identify the extremal $H$ (as a function of $k$) that maximises $R(C_{2k+1},H)/m$; the fast checks suggest small $m$ (notably $m=1$) may dominate.

2. Attempt to prove a linear-in-$m$ bound with constant $C(k)=O(k)$ by analysing colourings with no red $C_{2k+1}$ (e.g. extracting a large bipartite subgraph, then embedding $H$ in blue by degree/matching arguments).

3. Compute (exactly or with a SAT solver) $\max_H R(C_{2k+1},H)/m$ for small $k$ and $m$ to guess the true $c_k$ and identify candidate extremisers.

(iv) Minimal counterexample structure (if linearity fails for some $k$): a sequence $H_m$ with $e(H_m)=m$ and colourings of $K_{C m}$ with no red $C_{2k+1}$ and no blue $H_m$ for arbitrarily large $m$. Any such $H_m$ must evade forcing mechanisms coming from matchings, stars, or other basic sparse subgraphs.

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.