GPT 6 Astra Ultra attempt on Math-6063
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Contents
Erdős Problem 1009
Statement and result.
For every fixed $c>0$, there is a constant $f(c)$ such that every graph on $n$ vertices with at least $\lfloor n^2/4\rfloor+k$ edges, where $0\leq k<cn$ is an integer, has at least $k-f(c)$ edge-disjoint triangles. We give the precise deduction from Győri's stronger packing estimate, together with an elementary general bound, exact examples, and the obstruction to taking $f(c)=0$ for every $c$.The external theorem and its quantifiers.
The packing theorem of Győri, in the triangle case, states that if \[ e(G)=\lfloor n^2/4\rfloor+k,\qquad k=o(n^2), \] then \[ \nu_3(G)\geq k-O(k^2/n^2), \tag{1} \] where $\nu_3$ is the maximum number of edge-disjoint triangles. We use the uniform graph-theoretic meaning of this estimate: for a fixed bound $k\leq cn$, there are constants $C_c,N_c$ such that \[ \nu_3(G)\geq k-C_c k^2/n^2 \quad(n\geq N_c,\ 0<k<cn). \tag{2} \] For clarity, this uniformity also follows from the sequential formulation of $1$. If $2$ failed for a fixed $c$, one could choose $n_j\to\infty$, $k_j<cn_j$, and graphs with $(k_j-\nu_3(G_j))n_j^2/k_j^2\to\infty$. Then $k_j=o(n_j^2)$, contradicting $1$.This substantial packing theorem is the external dependency; its proof is not replaced by the elementary arguments below. It is stated as Theorem 1.6 of Balogh--Wigal, Packing edge disjoint cliques in graphs, https://arxiv.org/abs/2502.16683, which cites Győri's original work. In particular the error constant is not asserted to be arbitrarily small.
Deduction for every order, including small orders.
If the host has more than the specified edge count, discard edges until equality holds. For $n\geq N_c$, $2$ gives $\nu_3(G)\geq k-C_c c^2$. For $n<N_c$, use the trivial inequality $\nu_3(G)\geq0$ and $k<cN_c$. Thus a concrete permissible choice in terms of the theorem's constants is \[ f(c)=\left\lceil\max(C_c c^2,cN_c)\right\rceil. \] The case $k=0$ is immediate. This proves exactly the requested existence of a constant independent of $n$.An elementary comparison and a sharp family.
For reference, a maximal family of $r$ edge-disjoint triangles leaves a triangle-free graph after its $3r$ edges are removed. Mantel's bound therefore gives \[ \lfloor n^2/4\rfloor+k-3r\leq\lfloor n^2/4\rfloor, \qquad r\geq k/3. \] Mantel's bound itself follows by observing that in a triangle-free graph $d(u)+d(v)\leq n$ on every edge. Summing and using Cauchy--Schwarz gives $4e^2/n\leq\sum_vd(v)^2\leq ne$, hence $e\leq n^2/4$. The factor $1/3$ argument alone does not prove $1$.Start with $K_{a,b}$, where $a=\lceil n/2\rceil$, $b=\lfloor n/2\rfloor$, and add $k\leq b$ edges within the $a$-part, when that many edges are available. Assign a different vertex of the $b$-part to each added edge. The corresponding $k$ triangles are edge-disjoint: different apices ensure distinct cross edges. Every triangle uses at least one added edge, so no edge-disjoint packing has more than $k$ triangles. Thus $\nu_3=k$ in this entire family.