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Math-5375-A1 · revision r1 · on Math-5375

GPT 6 Astra Ultra attempt on Math-5375

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

Manuscript

Contents
  1. Erdős Problem #22: a barrier for complete-join constructions
  2. 1) Target and known solution
  3. 2) Complete joins are too rigid
  4. 3) Why merely choosing a dense multipartite graph fails
  5. 4) Exact threshold bookkeeping
  6. 5) Outcome and source check
  7. 6) COMPLETION ESTIMATE

Erdős Problem #22: a barrier for complete-join constructions

Independent attempt, 23 September 2026.

1) Target and known solution

For every $\epsilon>0$ and all sufficiently large $N$, construct a $K_4$-free graph with at least $N^2/8$ edges and independence number at most $\epsilon N$. The problem page records the affirmative theorem of Fox, Loh, and Zhao. Their paper addresses the exact critical density; a construction with only $(1/8-o(1))N^2$ edges does not automatically reach the requested threshold. We examine a natural elementary construction and prove that it cannot solve the question when $\epsilon<1/12$.

2) Complete joins are too rigid

Suppose a graph $G$ has a partition into two nonempty sets $X,Y$, with every edge between $X$ and $Y$ present. If both induced graphs contain an edge, those two edges and the four cross edges form a $K_4$. Thus, in a $K_4$-free join, at least one side, say $X$, is independent. Also $G[Y]$ is triangle-free: any triangle in $Y$, together with any vertex in $X$, would give a $K_4$.

Write $a=|X|$, $b=|Y|$, $N=a+b$, and $\alpha=\alpha(G)$. We have $a\le\alpha$. Every neighbourhood in the triangle-free graph $G[Y]$ is independent in $G$, so its maximum degree is at most $\alpha$. The degree sum now gives \[ e(G)=ab+e(G[Y])\le ab+\frac{\alpha b}{2} \le\frac32\alpha b\le\frac32\alpha N.\tag{1} \] In particular, $e(G)\ge N^2/8$ forces \[ \boxed{\alpha(G)\ge N/12.} \] This is uniform in the choice of both parts and their internal graphs. It rules out the entire complete-join strategy for sufficiently small $\epsilon$, even when the second part is an excellent triangle-free Ramsey graph.

The obstruction can be phrased constructively. Start with a triangle-free graph $H$ on $b$ vertices and join it to an independent set of size $a$. Then the resulting graph is $K_4$-free, has $ab+e(H)$ edges, and independence number exactly $\max(a,\alpha(H))$: an independent set cannot use vertices on both sides. Thus a complete join does preserve $K_4$-freeness, but $1$ shows the density--independence trade-off it imposes.

3) Why merely choosing a dense multipartite graph fails

The complete balanced three-partite graph is $K_4$-free and has roughly $N^2/3$ edges, comfortably above the requested density. Its largest part is an independent set of size at least $N/3$, so its edge count does not solve the problem. More generally, in a blow-up of a loopless graph on $r$ vertices, the $r$ vertex classes are independent. The largest class has at least $N/r$ vertices. A fixed template therefore cannot give arbitrarily small independence ratio.

This observation does not rule out templates whose size depends on $\epsilon$. Such a family would itself have to solve the finite density--independence trade-off. In particular it would be wrong to argue that all blow-up constructions are impossible merely because every fixed one has a positive independence ratio.

4) Exact threshold bookkeeping

Suppose one only knows graphs $H_N$ with $e(H_N)\ge(1/8-\delta_N)N^2$ and $\delta_N\to0$. Adding isolated vertices can only reduce the edge density and cannot turn this lower bound into $N^2/8$. Blowing up every vertex by the same integer factor preserves the ratio $e(H_N)/N^2$ exactly, so it also cannot repair a shortfall. Adding edges can repair a count, but then $K_4$-freeness requires a separate proof. These elementary facts explain why the exact threshold in the original question contains additional content beyond the classical asymptotic construction.

5) Outcome and source check

ATTEMPT UNRESOLVED. The complete-join obstruction and the threshold audit are proved here. They do not reproduce the affirmative construction, and do not improve the known quantitative bounds. The original problem is already resolved; this label refers to the limited reconstruction in this file.

Sources: https://www.erdosproblems.com/22, checked 23 September 2026; J. Fox, P.-S. Loh, and Y. Zhao, The critical window for the classical Ramsey--Turán problem, https://arxiv.org/abs/1208.3276. The known resolution is attributed to that paper; none of its deep construction results is silently assumed in the proof of $1$. No novelty is claimed.

This subjective estimate credits a rigorous obstruction to an attempted construction, while the central positive construction remains absent.

6) COMPLETION ESTIMATE

COMPLETION: 5%