gpt 5.2 attempt on Math-5417

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. - For an $n$-vertex graph $G$, let $\delta(G)$ be its minimum degree. - Let $C_4$ denote the 4-cycle. - Define \[f(n):=\min\{d\in\mathbb N:\ \text{every graph }G\text{ on }n\text{ vertices with }\delta(G)\ge d \text{ contains a }C_4\}.\] Equivalently, \[f(n)=1+\max\{\delta(G):\ |V(G)|=n,\ G\text{ is }C_4\text{-free}\}.\]

Question. Is $f(n)$ eventually nondecreasing, i.e. does there exist $n_0$ such that $f(n+1)\ge f(n)$ for all $n\ge n_0$?

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. From the problem text only: - Asymptotically $f(n)=(1+o(1))\sqrt n$. I do not use external results from [552] here.

3) ATTACK PLAN Proof track ideas: 1. Derive general upper bounds on $f(n)$ from double-counting in $C_4$-free graphs (gives $f(n)\le \sqrt n + O(1)$). 2. Try to compare extremal $C_4$-free graphs on $n$ and $n+1$ vertices to get monotonicity up to bounded error.

Disproof track ideas: 1. Search for sporadic $n$ where the maximum possible minimum degree of a $C_4$-free graph jumps down, causing $f(n+1)<f(n)$.

Chosen path in this attempt: prove two basic $C_4$-free graph lemmas giving the elementary bound $f(n)\le \sqrt n+O(1)$, and do a small brute-force check for $n\le 7$.

4) WORK Lemma 1 (Common-neighbor bound in $C_4$-free graphs). If $G$ is $C_4$-free, then any two distinct vertices $u\ne v$ have at most one common neighbor: \[|N(u)\cap N(v)|\le 1.\]

Proof. If $u$ and $v$ had two distinct common neighbors $x\ne y$, then the four vertices $u,x,v,y$ form the cycle \[u-x-v-y-u,\] which is a $C_4$. This contradicts $C_4$-freeness. $\square$

Lemma 2 (A global constraint on 2-paths). If $G$ is $C_4$-free on $n$ vertices with degrees $d(w)$, then \[\sum_{w\in V(G)} \binom{d(w)}{2} \le \binom{n}{2}.\]

Proof. Count the set of pairs $(w,\{u,v\})$ where $u$ and $v$ are distinct neighbors of $w$ (equivalently, length-2 paths $u-w-v$ with midpoint $w$). For fixed $w$, there are $\binom{d(w)}{2}$ such unordered pairs $\{u,v\}$, so the total is $\sum_w \binom{d(w)}{2}$. On the other hand, for a fixed unordered pair $\{u,v\}$ of vertices, the number of vertices $w$ adjacent to both $u$ and $v$ is exactly $|N(u)\cap N(v)|$. By Lemma 1 this is at most $1$, so each pair $\{u,v\}$ contributes at most one triple $(w,\{u,v\})$. There are $\binom{n}{2}$ choices for $\{u,v\}$, hence the total number of such triples is at most $\binom{n}{2}$. This gives the inequality. $\square$

Corollary 3 (Elementary upper bound $f(n)\le \sqrt n+O(1)$). If $G$ is $C_4$-free on $n$ vertices, then \[\delta(G)\,(\delta(G)-1)\le n-1.\] In particular, any graph on $n$ vertices with minimum degree at least $\lfloor \sqrt{n-1}\rfloor+2$ contains a $C_4$, so \[f(n)\le \lfloor \sqrt{n-1}\rfloor+2 < \sqrt n + 2.\]

Proof. Apply Lemma 2 and use $d(w)\ge \delta(G)$ for all $w$: \[n\binom{\delta(G)}{2} \le \sum_w \binom{d(w)}{2}\le \binom{n}{2}.\] Multiply by $2/n$ to obtain $\delta(G)(\delta(G)-1)\le n-1$. If $\delta(G)\ge \lfloor \sqrt{n-1}\rfloor+2$, then $\delta(G)(\delta(G)-1)> n-1$, contradiction. Thus such a graph cannot be $C_4$-free, proving the bound on $f(n)$. $\square$

FAST REALITY CHECK (small $n$ by exhaustive search). Using brute force over all graphs for $n\le 7$, I computed the maximum possible minimum degree among $C_4$-free graphs: \[ \begin{array}{c|c|c} n & \max\{\delta(G): |V(G)|=n,\ C_4\nsubseteq G\} & f(n)\\\hline 4 & 1 & 2\\ 5 & 2 & 3\\ 6 & 2 & 3\\ 7 & 2 & 3 \end{array} \] So $f(4)=2$ and $f(5)=f(6)=f(7)=3$, consistent with monotonicity at small $n$.

5) VERIFICATION - Lemma 1 is a standard "two common neighbors create a $C_4$" argument. - Lemma 2 is a clean double-counting of 2-paths using Lemma 1. - Corollary 3 is algebraic and matches the stated asymptotic $f(n)=O(\sqrt n)$. - Computation is exhaustive only for $n\le 7$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: I proved an elementary upper bound $f(n)<\sqrt n+2$ via double counting (Corollary 3).

(ii) First gap: Prove eventual monotonicity $f(n+1)\ge f(n)$ (or even the bounded-drop version $f(m)\ge f(n)-c$ for $m>n$). The counting lemmas control the scale of $f(n)$ but do not compare adjacent values.

(iii) Top 3 next moves (concrete): 1. Relate $f(n)$ to the maximum minimum degree in $C_4$-free graphs and study how that extremal quantity behaves under adding a vertex with carefully chosen adjacency. 2. Use known near-extremal constructions for $\mathrm{ex}(n,C_4)$ (polarity graphs) to estimate $f(n)$ more sharply and see whether the step behavior can be controlled. 3. Compute $f(n)$ for moderate $n$ using integer programming / SAT (maximizing minimum degree under $C_4$-free constraints) to look for nonmonotone behavior.

(iv) What a minimal counterexample would likely look like (structure/parameters): An integer $n$ such that the maximum minimum degree among $C_4$-free graphs on $n+1$ vertices is strictly smaller than the corresponding maximum on $n$ vertices, forcing $f(n+1)<f(n)$.