GPT 6 Astra Ultra attempt on Math-6072

Erdős Problem 1018

Statement and result.

For every fixed $\epsilon>0$, all sufficiently large graphs with $n$ vertices and at least $n^{1+\epsilon}$ edges contain a nonplanar subgraph on at most $C_\epsilon$ vertices. We derive this from Janzer's subdivision theorem, obtaining a possible bound \[ C_\epsilon\leq 20/\epsilon+15\qquad(0<\epsilon<1). \] We also give a probabilistic construction proving that no absolute bound on $C_\epsilon$ can work as $\epsilon$ decreases to zero. The main subdivision theorem is an explicit external dependency.

The external input and the parameter choice.

For a fixed graph $F$ and a fixed even integer $k\geq2$, let $F^{k-1}$ mean the graph formed by replacing every edge of $F$ by a path of exactly $k$ edges, with mutually disjoint internal vertices. Janzer's Theorem 1.6 states \[ \operatorname{ex}(n,F^{k-1})=O_{F,k}(n^{1+1/k}). \tag{1} \] The source permits multigraphs $F$; the simple graph $F=K_5$ suffices. Equation $1$, proved by the source's path-counting argument, is used as a theorem, not reproved here.

Given $0<\epsilon<1$, take $k=2\lceil1/\epsilon\rceil$. Then $1/k\leq\epsilon/2<\epsilon$, and $1$ gives a constant $D_k$ such that \[ \operatorname{ex}(n,K_5^{k-1})\leq D_k n^{1+1/k} \] for all sufficiently large $n$. Once $n^{\epsilon-1/k}>D_k$, the edge count $n^{1+\epsilon}$ exceeds this bound. Thus the host contains $K_5^{k-1}$, whose vertex count is \[ 5+10(k-1)=10k-5\leq20/\epsilon+15. \] All constants are fixed before letting $n$ grow; the unknown size of $D_k$ affects the threshold on $n$, not the displayed bound on the subgraph's order.

Why this subgraph is nonplanar.

The graph $K_5$ is nonplanar because a simple planar graph on five vertices has at most $3\cdot5-6=9$ edges, whereas $K_5$ has ten. Subdividing its edges cannot make it planar: from a hypothetical plane drawing of the subdivision, concatenate the arcs along each replacement path and suppress its internal degree-two vertices. Internally disjoint paths then give a plane drawing of $K_5$, a contradiction. This verifies precisely the nonplanarity required by the question.

When $\epsilon\geq1$, a simple graph cannot have $n^{1+\epsilon}$ edges for large $n$, so the implication is vacuous.

The bound must depend on $\epsilon$.

Fix any integer $L\geq3$, and take $0<\epsilon<1/(2L)$. In the random graph $G(n,p)$, set $p=n^{-1+2\epsilon}$. The expected number of edges is $(1/2+o(1))n^{1+2\epsilon}$, and its variance is at most its expectation. Chebyshev's inequality therefore implies that, with probability tending to one, \[ e(G)\geq\tfrac14n^{1+2\epsilon}. \] The expected number $Z$ of cycles of lengths $3,\ldots,L$ is at most \[ \sum_{j=3}^{L}\frac{(np)^j}{2j} =O_L(n^{2\epsilon L})=o(n^{1+2\epsilon}). \] Markov's inequality gives $Z\leq\tfrac18n^{1+2\epsilon}$ with probability tending to one. Choose an outcome satisfying both bounds and remove one edge from each short cycle. At least $\tfrac18n^{1+2\epsilon}>n^{1+\epsilon}$ edges remain for large $n$, while the girth exceeds $L$. Every subgraph on at most $L$ vertices is then a forest and is planar. Thus, for every fixed $L$, all sufficiently small positive $\epsilon$ require $C_\epsilon>L$.

Attribution and assessment.

Kostochka and Pyber originally answered the problem. The deduction above uses O. Janzer, The extremal number of longer subdivisions, Theorem 1.6, https://arxiv.org/abs/1905.08001; the original result is also described in its introduction. See https://www.erdosproblems.com/1018, checked 24 September 2026. This is a complete deduction relative to the explicitly stated subdivision theorem, supplemented by the supplied dependence obstruction. No novelty or local formal verification is claimed.

Completion estimate.

COMPLETION: 100%