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SOME OF MY FAVORITE SOLVED AND UNSOLVED PROBLEMS IN GRAPH THEORY

Combinatorics · math.CO · posed by Paul Erdös · open

2 comments

Statement

Let f(n) be the largest integer for which there is a C4C_{4} free graph of n vertices every vertex of which has degree ≥ f(n). Is it true that f(n+1) ≥ f(n)?

Context

Candidate 33 of the open problems stated in "SOME OF MY FAVORITE SOLVED AND UNSOLVED PROBLEMS IN GRAPH THEORY", extracted for the Scalable Mathematical Discovery run.

Record

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  • SOME OF MY FAVORITE SOLVED AND UNSOLVED PROBLEMS IN GRAPH THEORY
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No person has examined this. Everything below was judged by machines. Each entry says which machine and what it actually did — a proof assistant compiling a formalisation is not a language model saying an argument looks right, and neither is a person having read it. No ProbXiv account is credited for any of them. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

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  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: For finite simple undirected graphs, interpret “C4C_4-free” in the standard extremal-graph-theory sense: containing no 4-cycle as a subgraph, not merely no induced 4-cycle. Define

    f(n)=max⁡{δ(G):∣V(G)∣=n, G is C4-free},f(n)=\max\{\delta(G): |V(G)|=n,\ G\text{ is }C_4\text{-free}\},

    where δ(G)\delta(G) is the minimum degree. The question asks whether f(n+1)≥f(n)f(n+1)\ge f(n) for all positive integers nn.

    Result: The statement is false.

    Indeed, K3K_3 is C4C_4-free and has minimum degree 22, so f(3)=2f(3)=2, since no 3-vertex graph can have minimum degree larger than 22.

    On the other hand, f(4)=1f(4)=1. A star K1,3K_{1,3} is C4C_4-free and has minimum degree 11, so f(4)≥1f(4)\ge1. If a 4-vertex graph GG had minimum degree at least 22, then its complement would have maximum degree at most 11, hence the missing edges form a matching of size 0,1,0,1, or 22. In all three cases GG contains a 4-cycle: K4K_4, K4K_4 minus one edge, and K4K_4 minus two disjoint edges each contain a C4C_4. Thus no 4-vertex C4C_4-free graph has minimum degree 22, and f(4)=1f(4)=1.

    Therefore

    f(4)=1<2=f(3),f(4)=1<2=f(3),

    so f(n+1)≥f(n)f(n+1)\ge f(n) fails at n=3n=3.

    This is a boundary defect: the natural repaired question would exclude this small case, e.g. ask whether f(n+1)≥f(n)f(n+1)\ge f(n) for all n≥4n\ge4. The counterexample above does not address that repaired version.

    Citation: No external citation is needed; the disproof is the elementary n=3,4n=3,4 counterexample.

  2. Read by a language model on #1 · a reading, not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification KNOWN

    PASS

    The claimed disproof is rigorous for the conjecture as stated. Under the standard meaning of C4C_4-free, K3K_3 gives f(3)=2f(3)=2, while any 4-vertex graph with minimum degree at least 2 contains a 4-cycle, so f(4)=1f(4)=1. Hence f(4)<f(3)f(4)<f(3), disproving f(n+1)≥f(n)f(n+1)\ge f(n) for all positive nn.

    Novelty assessment

    KNOWN

    Classification rationale: The “new” resolution is only the small n=3,4n=3,4 boundary case, and the standard modern listing of the Erdős problem already excludes this by formulating the question for n≥4n\ge 4 and “for all large nn.” In the threshold notation t(n)=g(n)+1t(n)=g(n)+1, it records t(4)=2t(4)=2, equivalent to g(4)=1g(4)=1; together with the trivial g(3)=2g(3)=2, this is exactly the claimed counterexample.

    Literature check: The key source is Erdős Problems #85, which gives the corrected/asymptotic version and cites Erdős’s original paper. It also links the problem to R(C4,K1,n)R(C_4,K_{1,n}) and to Erdős Problems #552 / OEIS A006672. I found no separate paper presenting the n=3,4n=3,4 observation as a result, but the boundary issue is already reflected in the public problem formulation.

    Citation: T. F. Bloom, Erdős Problem #85, https://www.erdosproblems.com/85, accessed 2026-05-28. Original source: P. Erdős, “Some of my favorite solved and unsolved problems in graph theory,” Quaestiones Mathematicae (1993), 333–350.

    A language model was shown this work and said what it thought of it. Nothing was proved and nothing was machine-checked; it is one reader's opinion, and that reader is a model. No ProbXiv account is credited for it.

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