gpt pro 5.2 attempt on Math-5738

1) FORMAL RESTATEMENT

Let $n\in\mathbb{N}$ and $\alpha\in[0,1/2]$. Consider the complete simple graph $K_n$ on vertex set $[n]={1,2,\dots,n}$. A 2‑colouring is a map \[ \chi:\binom{[n]}{2}\to{\text{red},\text{blue}}. \] For any $X\subseteq[n]$ with $|X|=s$, let

* $e_R(X)$ be the number of red edges with both endpoints in (X), * $e_B(X)$ be the number of blue edges with both endpoints in (X).

Then \[ e_R(X)+e_B(X)=\binom{s}{2}. \]

Define $F(n,\alpha)$ to be the smallest integer (m) such that there exists a 2‑colouring $\chi$ of $E(K_n)$ for which the following holds:

\[ \forall X\subseteq[n]\ \text{with}\ |X|\ge m:\quad e_R(X)>\alpha\binom{|X|}{2}\ \text{and}\ e_B(X)>\alpha\binom{|X|}{2}. \tag{(*)} \]

The claim to prove (as written) is:

For every $\alpha\in[0,1/2]$, there exists a constant $c_\alpha>0$ such that as $n\to\infty$, \[ F(n,\alpha)\sim c_\alpha\log n. \tag{Claim} \]

Ambiguity/misstated points (must resolve)

1. Is (m) allowed to exceed (n)? The definition “smallest (m) such that … every $X\subseteq[n]$ with $|X|\ge m$ …” allows $m>n$, in which case the condition $(*)$ is vacuous (there are no such (X)). This is how the ErdősProblems page states it, without restricting $m\le n$. ([Erdős Problems][1])

2. Endpoint $\alpha=1/2$ with strict “more than”: For any $s\ge2$, it is impossible to have both $e_R(X)>\tfrac12\binom{s}{2}$ and $e_B(X)>\tfrac12\binom{s}{2}$ because their sum is $\binom{s}{2}$. This will force $F(n,1/2)=n+1$ (if (m) may exceed (n)), which contradicts $\sim c\log n$.

Minimal corrected statement (standard convention)

The minimal correction consistent with the intended “$\Theta(\log n)$” behaviour is to restrict to \[ 0\le \alpha < \tfrac12. \tag{Corrected domain} \] (Alternatively one could replace “$>$” by “$\ge$” and then treat parity issues, but that is not the given statement.)

---

2) QUICK LITERATURE/CONTEXT CHECK (web browsing used)

The ErdősProblems statement for #563 matches your “smallest (m)” definition and also states the asymptotic conjecture $F(n,\alpha)\sim c_\alpha\log n$ for $0\le\alpha\le1/2$. ([Erdős Problems][1]) It also notes that for $\alpha=0$ this recovers the classical diagonal Ramsey setting (avoid a monochromatic clique of size (m)). ([Erdős Problems][1])

Separately, a standard survey on Ramsey numbers remarks that “the limit of $R(k,k)^{1/k}$, if it exists …” indicating that even the existence of a clean exponential growth constant for diagonal Ramsey numbers is not settled in general. ([combinatorics.org][2]) This matters because $\alpha=0$ is embedded in the present statement.

---

3) ATTACK PLAN

Disproof / counterexample strategies

1. Endpoint attack: test $\alpha=\tfrac12$. With strict “$>$”, the requirement demands both colours exceed half the edges inside each large (X), which is arithmetically impossible for any $|X|\ge2$. This should force $F(n,1/2)$ to be $n+1$ (or undefined if one forbids $m>n$), contradicting $\sim c\log n$. 2. Tiny‑(n) brute force sanity check: compute $F(n,\alpha)$ for small (n) and $\alpha$ to detect definitional edge cases.

Proof strategies (for the corrected $\alpha<1/2$ version)

1. Upper bound: random 2‑colouring + Chernoff/Hoeffding + union bound to get $F(n,\alpha)\le C(\alpha)\log n$. 2. Lower bound: since $(*)$ implies no monochromatic $K_m$, derive $F(n,\alpha)\ge F(n,0)$ and hence $F(n,\alpha)\ge c\log n$ using standard Ramsey bounds.

I will pursue Disproof Strategy 1 because it yields a complete, gap-free disproof of the statement as written.

---

4) WORK

Phase 1 — FAST REALITY CHECK (hand + computation)

Hand check at $\alpha=1/2$: Take any (X) with $|X|=s\ge2$. Then $e_R(X)+e_B(X)=\binom{s}{2}$. If both were $>\tfrac12\binom{s}{2}$, their sum would be $>\binom{s}{2}$, impossible.

Brute force (exact) for small (n): I exhaustively enumerated all $2^{\binom{n}{2}}$ 2‑colourings for $n\le 6$ and, for each colouring, tested whether there exists an $m\in{1,\dots,n}$ satisfying $(*)$. For $\alpha=0.5$, the result was “no such $m\le n$” for every $n=2,3,4,5,6$. (This matches the hand argument.)

(For comparison: for $\alpha=0,0.1,0.25$, the computed values were $F(3,\alpha)=3$, $F(4,\alpha)=3$, $F(5,\alpha)=3$, $F(6,\alpha)=4$, consistent with the classical $R(3,3)=6$ phenomenon.)

---

Core disproof: $\alpha=\tfrac12$ breaks the statement

Lemma 1 (Half+half is impossible on any nontrivial induced subgraph)

Let $s\ge2$. For any 2‑colouring of $E(K_s)$, there is no way to have simultaneously \[ e_R([s])>\tfrac12\binom{s}{2} \quad\text{and}\quad e_B([s])>\tfrac12\binom{s}{2}. \]

Proof. In any 2‑colouring of $K_s$, \[ e_R([s]) + e_B([s]) = \binom{s}{2}. \] If both summands were $>\tfrac12\binom{s}{2}$, then adding gives \[ e_R([s]) + e_B([s]) > \binom{s}{2}, \] contradiction. $\square$

Lemma 2 (Exact value of $F(n,1/2)$ under the literal definition)

Assume $n\ge1$ and interpret $F(n,\alpha)$ as “the smallest integer (m)” with no restriction $m\le n$ (this matches the literal quantifier structure as written on the ErdősProblems page). ([Erdős Problems][1]) Then \[ F(n,1/2)=n+1. \]

Proof.

* Step 1: No $m\le n$ can work.

* If $m=1$, then the condition $(*)$ must hold for (X) of size $1$. But $\binom{1}{2}=0$ and $e_R(X)=e_B(X)=0$, so the strict inequalities \[ 0> \tfrac12\cdot 0 \quad\text{and}\quad 0> \tfrac12\cdot 0 \] both fail.

* If $2\le m\le n$, then take any subset $X\subseteq[n]$ with $|X|=m$. The condition $(*)$ requires \[ e_R(X)>\tfrac12\binom{m}{2} \quad\text{and}\quad e_B(X)>\tfrac12\binom{m}{2}. \] This is impossible by Lemma 1 (applied to $s=m$).

Hence no $m\le n$ satisfies the defining property.

* Step 2: $m=n+1$ always works (vacuously).

If $m=n+1$, then there is no subset $X\subseteq[n]$ with $|X|\ge m$. Therefore the universal condition $(*)$ is vacuously true for every 2‑colouring of $K_n$. So such an (m) exists.

* Step 3: Minimality. From Step 1, no smaller (m) works; from Step 2, $m=n+1$ works. Therefore $F(n,1/2)=n+1$. $\square$

Consequence: the claimed asymptotic fails

From Lemma 2, \[ \frac{F(n,1/2)}{\log n}=\frac{n+1}{\log n}\longrightarrow \infty \quad(n\to\infty), \] so there is no finite constant $c_{1/2}$ with $F(n,1/2)\sim c_{1/2}\log n$.

Thus the statement “for every $0\le\alpha\le1/2$” is false.

---

(Optional but relevant) Minimal corrected statement and what can be proved quickly

Corrected statement: Restrict to $0\le\alpha<1/2$. ([Erdős Problems][1]) Then the ErdősProblems page notes the easy (probabilistic) bounds $F(n,\alpha)\asymp_\alpha \log n$. ([Erdős Problems][1]) A standard proof is: random 2‑colouring + Chernoff gives an upper bound $O_\alpha(\log n)$; and since the property implies “no monochromatic $K_m$”, one gets a lower bound $\Omega(\log n)$ from classical Ramsey upper bounds. (This does not resolve the stronger $\sim c_\alpha\log n$ limit statement.)

I am not labeling this as a full proof of the corrected “$\sim$” claim, because the endpoint disproof already fully resolves the literal statement, and proving the corrected “$\sim$” would in particular settle $\alpha=0$, which connects to deep questions about the exponential growth constant of diagonal Ramsey numbers. ([combinatorics.org][2])

---

5) VERIFICATION (adversarial checks)

1. Quantifier check: The disproof uses only the fact that the claim asserts the asymptotic for every $\alpha\in[0,1/2]$. Exhibiting one $\alpha$ for which the asymptotic is false is sufficient.

2. Edge cases $n=1$: For $n=1$, any $m\le 1$ forces checking $X={1}$ and fails because strict “$>$” demands $0>0$. Thus the smallest (m) is $2=n+1$, consistent with Lemma 2.

3. Potential “escape” via redefining (F) only for $m\le n$: If one imposed $m\le n$, then for $\alpha=1/2$ there is no such (m), so $F(n,1/2)$ would be undefined or $+\infty$. Either way, the asymptotic $F(n,1/2)\sim c\log n$ still fails.

4. Cross-check with computation: Exact brute force for $n\le 6$ finds no satisfying $m\le n$ for $\alpha=1/2$, matching Lemma 1’s impossibility argument.

---

6) FINAL

FULL SOLUTION COUNTEREXAMPLE/DISPROOF

Take $\alpha=\tfrac12$. For any subset (X) with $|X|\ge2$, the total number of edges is $\binom{|X|}{2}$, so it is impossible for both colours to have strictly more than half of these edges (Lemma 1). Therefore no $m\le n$ can satisfy the defining property, while $m=n+1$ makes the condition vacuous, giving $F(n,1/2)=n+1$ (Lemma 2). Hence \[ F(n,1/2)\not\sim c_{1/2}\log n, \] and the statement “for every $0\le\alpha\le1/2$” is false as written.

[1]: https://www.erdosproblems.com/563 "

Erdős Problem #563

" [2]: https://www.combinatorics.org/files/Surveys/ds1/ds1v16-2021.pdf "https://www.combinatorics.org/files/Surveys/ds1/ds1v16-2021.pdf"