gpt pro 5.2 attempt on Math-942

1) FORMAL RESTATEMENT

Fix an integer $k\ge 3$. For each integer $n\ge 1$, define $F_k(n)$ to be the minimum integer $r\ge 1$ such that there exist

* a simple undirected graph $G=(V,E)$ with $|V|=n$ and $|E|=\lfloor n^2/4\rfloor+1$, and * an edge-colouring $\varphi:E\to [r]={1,2,\dots,r}$,

with the property that every subgraph $H\subseteq G$ isomorphic to the odd cycle $C_{2k+1}$ is rainbow under $\varphi$; i.e. $\varphi$ is injective on (E(H)) (no colour repeats among the $2k+1$ edges of that cycle).

Question (Burr–Erdős–Graham–Sós): \[ F_k(n)\sim \frac{n^2}{8}\quad (n\to\infty), \] meaning $\displaystyle \lim_{n\to\infty}\frac{F_k(n)}{n^2}=\frac18$.

Conventions/edge cases:

* If $n<2k+1$, then (G) contains no $C_{2k+1}$, so the rainbow constraint is vacuous. * Logs are irrelevant here. * Graphs are finite, simple, undirected.

Stress points:

* The edge count $\lfloor n^2/4\rfloor+1$ is just one above Mantel’s bound for triangle-free / bipartite extremality, so (G) is “barely non-bipartite”. * A single $C_{2k+1}$ already forces $r\ge 2k+1$.

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2) QUICK LITERATURE/CONTEXT CHECK (web browsing)

* The ErdosProblems entry for #809 states: Burr–Erdős–Graham–Sós proved $F_k(n)\gg n^2$. * For context: for (n) large compared to (k), the extremal number for odd cycles satisfies \[ \mathrm{ex}(n,C_{2k+1})=\left\lfloor\frac{n^2}{4}\right\rfloor \quad\text{(for } n\ge 4k-2\text{)}, \] attributed to Füredi–Gunderson. Hence for $n\ge 4k-2$, any graph with $\lfloor n^2/4\rfloor+1$ edges must contain at least one $C_{2k+1}$, so the rainbow condition is asymptotically non-vacuous.

I did not find any indication (up to the site’s 2025 update cadence) that the asymptotic $\sim n^2/8$ has been resolved.

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3) ATTACK PLAN

Proof strategies (to try to prove $F_k(n)\sim n^2/8$)

1. Upper bound construction: build (G) with $\lfloor n^2/4\rfloor+1$ edges and an edge-colouring using $(1/8+o(1))n^2$ colours. Natural route: make the “cycle-conflict graph” on edges sparse enough that many edges can share colours safely (no two same-coloured edges lie on a common $C_{2k+1}$). 2. Lower bound via conflicts: show that in every such (G), there is a large set of edges pairwise forced to have distinct colours because any two lie together in some $C_{2k+1}$. Turning this into a sharp constant $1/8$ is the hard part. 3. Stability + supersaturation: use that (G) is near-bipartite (by stability around Mantel) and argue that the forced “odd-cycle mechanism” creates a dense conflict relation on at least $\sim n^2/4$ edges, yielding a quadratic colour lower bound with a constant.

Disproof/construction strategies (to falsify $\sim n^2/8$)

1. Produce a family with $F_k(n)\le (1/8-\varepsilon)n^2$ (a better colouring than conjectured). 2. Prove a universal lower bound $F_k(n)\ge (1/8+\varepsilon)n^2$ (show conjectured constant too small).

Best path chosen in the work below: I focused on (i) small-case sanity and (ii) analyzing the most natural “one edge above bipartite extremal” graph, to see what colour lower bounds it forces. This yields a rigorous (but negative-for-construction) structural lemma.

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4) WORK

PHASE 1 — FAST REALITY CHECK

Vacuous regime. If $n<2k+1$, then no $C_{2k+1}$ exists in any (n)-vertex graph, so the rainbow condition is vacuous. Any graph (G) with $\lfloor n^2/4\rfloor+1$ edges (exists for all $n\ge 3$) can be coloured with one colour. Hence: \[ \boxed{n<2k+1\ \Rightarrow\ F_k(n)=1.} \]

Non-vacuous regime for large (n). If $n\ge 4k-2$, then $\mathrm{ex}(n,C_{2k+1})=\lfloor n^2/4\rfloor$. So any (G) with $\lfloor n^2/4\rfloor+1$ edges contains at least one $C_{2k+1}$, and a rainbow $C_{2k+1}$ needs at least $2k+1$ distinct colours. Therefore: \[ \boxed{n\ge 4k-2\ \Rightarrow\ F_k(n)\ge 2k+1.} \]

Trivial upper bound: \[ F_k(n)\le \lfloor n^2/4\rfloor+1 \] by colouring each edge distinctly.

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PHASE 2 — A rigorous “stress test” on the obvious near-extremal graph

A natural candidate (G) with $\lfloor n^2/4\rfloor+1$ edges is: \[ G_0 := K_{a,b}\ \cup\ {uv}, \] where $a=\lfloor n/2\rfloor$, $b=\lceil n/2\rceil$, $A\sqcup B$ is the bipartition of $K_{a,b}$, and $u,v\in A$ are distinct.

This is “one extra edge on top of Mantel”.

I now prove that in $G_0$, the rainbow requirement actually forces almost all bipartite edges to have distinct colours, i.e. (r) must be $\sim n^2/4$. This shows $G_0$ cannot be close to achieving an $n^2/8$ upper bound.

Lemma (two bipartite edges are co-cyclic with (uv))

Fix $k\ge 3$. Suppose $a\ge k+1$ and $b\ge k$. Let $G_0=K_{a,b}\cup{uv}$ as above with $u,v\in A$.

Let $e_1,e_2$ be two distinct edges of the complete bipartite part with endpoints in $(A\setminus{u,v})\times B$. Then there exists a cycle $C\subseteq G_0$ with \[ C\cong C_{2k+1}\quad\text{and}\quad {uv,e_1,e_2}\subseteq E(C). \]

Proof. It suffices to construct a simple (u)-to-(v) path (P) of length $2k$ inside $K_{a,b}$ that contains both $e_1$ and $e_2$. Then adding the edge (uv) closes (P) into a $C_{2k+1}$.

Because the bipartite part is complete, any alternating sequence of distinct vertices \[ u\in A,\ B,\ A,\ B,\ \dots,\ B,\ v\in A \] gives a valid path. So we only must ensure:

1. alternation, 2. no vertex repetition, 3. the two required edges appear somewhere among the consecutive pairs.

Since $k\ge 3$, the path has at least (k) (B)-vertices and $k-1$ intermediate (A)-vertices. Our assumptions $b\ge k$, $a\ge k+1$ guarantee there are enough distinct vertices available to realize any prescribed constant-size “pattern” near the start (containing $e_1,e_2$) and then complete the remainder with fresh vertices.

Concretely: build an initial alternating prefix starting at (u) that traverses $e_1$ and $e_2$ (if they share an endpoint, place them consecutively through that shared endpoint; if they are disjoint, place them as two separate consecutive $B!-!A$ steps). This uses at most 2 intermediate (A)-vertices from $A\setminus{u,v}$ and at most 2 vertices from (B). Then extend the alternating walk to reach (v) using fresh vertices; completeness ensures all required edges exist. The result is a simple (u)-to-(v) path of length exactly $2k$ containing $e_1,e_2$. Adding (uv) yields the desired $C_{2k+1}$. ∎

Corollary (large forced number of colours in $G_0$)

In any edge-colouring $\varphi$ of $G_0$ in which every $C_{2k+1}$ is rainbow, all edges between $A\setminus{u,v}$ and (B) must have pairwise distinct colours. Hence \[ r \ \ge\ |E(A\setminus{u,v},B)|=(a-2)b. \] For balanced $a=\lfloor n/2\rfloor$, $b=\lceil n/2\rceil$, \[ (a-2)b=\left\lfloor\frac{n^2}{4}\right\rfloor-\Theta(n), \] so this particular $G_0$ requires \[ \boxed{r\ge \left\lfloor\frac{n^2}{4}\right\rfloor-O(n).} \]

Proof. If two distinct edges $e_1,e_2$ in $E(A\setminus{u,v},B)$ shared a colour, then by the lemma there is a $C_{2k+1}$ containing both, so that cycle would repeat a colour—contradiction. Therefore all those edges have distinct colours, so (r) is at least their count. ∎

Interpretation: the “obvious” near-extremal graph (K_{a,b}+uv) is extremely hostile to colour reuse; it forces $\sim n^2/4$ colours, not $\sim n^2/8$. So any graph witnessing $F_k(n)\sim n^2/8$ (if true) must have a much sparser “two edges lie on a common $C_{2k+1}$” relation on a positive fraction of its edges.

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5) VERIFICATION (adversarial self-check)

* Vacuous case $n<2k+1$: correct; no $C_{2k+1}$ subgraph exists. * Non-vacuous trigger $n\ge 4k-2$: depends on $\mathrm{ex}(n,C_{2k+1})=\lfloor n^2/4\rfloor$ for that range, cited. * Key lemma on $G_0$: the only delicate point is ensuring the two specified edges can be embedded in a simple alternating (u)-to-(v) path of length $2k$ without vertex repetition. The availability conditions $a\ge k+1$, $b\ge k$ ensure we can choose all intermediate vertices distinct while reserving the endpoints needed for the forced edges. Completeness of $K_{a,b}$ guarantees the necessary adjacencies.

No step uses hidden assumptions beyond simplicity and completeness.

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6) FINAL: UNRESOLVED

I did not produce either:

* a complete proof that $F_k(n)\sim n^2/8$, or * an explicit counterexample disproving that asymptotic.

(i) Strongest fully proved partial results here

1. $F_k(n)=1$ for $n<2k+1$. 2. For $n\ge 4k-2$, $F_k(n)\ge 2k+1$. 3. For the canonical graph (G_0=$K_{\lfloor n/2\rfloor,\lceil n/2\rceil}$+uv), any valid colouring needs \[ r\ge (a-2)b = \left\lfloor\frac{n^2}{4}\right\rfloor - O(n), \] so $G_0$ cannot witness a $\sim n^2/8$ upper bound.

(ii) First gap

Construct (or rule out) an (n)-vertex graph with $\lfloor n^2/4\rfloor+1$ edges whose edge-set can be partitioned into $\sim n^2/8$ colour classes such that no two edges in a class lie on a common $C_{2k+1}$.

(iii) Top 3 next moves

1. Identify minimizers of $F_k(n)$ structurally: what graphs with $\lfloor n^2/4\rfloor+1$ edges minimize the “cycle conflict graph” on edges? 2. Attempt an explicit upper-bound construction where every $C_{2k+1}$ must pass through a sparse “bottleneck” set of edges (so colours can be reused heavily outside that set). 3. Compute small-(n) optimizers (fixed (k)) to guess the extremal pattern for (G) and for the colouring (even a heuristic can suggest the right structure).

(iv) What a minimal counterexample would likely look like

If the conjecture $F_k(n)\sim n^2/8$ is false, then for some $\varepsilon>0$ infinitely many (n) would satisfy either:

* $F_k(n)\le (1/8-\varepsilon)n^2$ via a graph where most edges fall into “non-co-cyclic” pairs (allowing more reuse), or * $F_k(n)\ge (1/8+\varepsilon)n^2$ via a universal argument that every such (G) has a conflict relation too dense to permit $\sim m/2$ colouring.

My proved lemma shows the most obvious near-Mantel graph (K_{a,b}+uv) lies in the opposite extreme, forcing $\sim n^2/4$ colours.