GPT 6 Astra Ultra attempt on Math-5412

Erdős Problem 76: packing monochromatic triangles

1) Statement and scope

Must every red--blue coloring of $K_n$ have a family of edge-disjoint monochromatic triangles of size $(1+o(1))n^2/12$? Gruslys and Letzter proved the affirmative answer. This attempt gives a self-contained quadratic lower bound with a smaller leading constant and proves that the requested constant $1/12$ is optimal. It does not reproduce their sharp packing theorem.

2) A maximal packing leaves a $K_6$-free graph

Take a maximal family of edge-disjoint monochromatic triangles, and let its size be $t$. Remove their $3t$ edges, retaining the original colors on the remaining graph $H$. There is no monochromatic triangle in $H$, by maximality.

Every red--blue $K_6$ contains a monochromatic triangle. Indeed, some vertex has three incident edges of one color, say red, to $x,y,z$. If any edge within $\{x,y,z\}$ is red, it completes a red triangle with that vertex; otherwise those three edges are all blue. Consequently the uncolored graph $H$ contains no $K_6$.

Here is an elementary density estimate sufficient for our purpose. In a $K_6$-free graph on $v$ vertices, choose a maximum clique of size $s\leq5$. Every vertex outside it has at most $s-1$ neighbors in it. Hence the sum of degrees of clique vertices is at most $(s-1)v$, and one of them has degree at most $(s-1)v/s\leq4v/5$. This includes $s=1$, when the graph has no edges. Repeated vertex deletion, applied to the remaining induced graphs, now gives \[ e(H)\leq\frac45\sum_{v=1}^n v=\frac25 n(n+1). \] Since $e(H)=\binom n2-3t$, we conclude \[ t\geq\frac{n^2}{30}-\frac{3n}{10}. \] For small $n$ a negative right side is simply a vacuous lower bound. This argument uses maximality, not an unproved claim that a greedy packing is maximum.

3) A sharp obstruction to improving the target constant

For even $n=2m$, split the vertices into two sets of size $m$. Color edges within each part blue and edges between the parts red. There are no red triangles, and every blue triangle lies inside a part. Thus any permitted packing has at most \[ \frac{2\binom m2}{3} =\frac{n^2}{12}-\frac n6 \] triangles. In particular no universal leading constant greater than $1/12$ is possible.

For the infinite sequence $m=3^r$, this upper bound is achieved exactly. Identify each part with the vector space $\mathbb F_3^r$. An affine line has three points. Each pair of distinct points lies on exactly one such line, since its direction is their nonzero difference and the only nonzero scalars in $\mathbb F_3$ are $1,-1$. The triangles formed by all affine lines therefore partition the edges of each complete graph on a part. Using them in both parts gives exactly the displayed number of triangles.

4) Gap and sources

Outcome of this attempt: UNRESOLVED. The lower constant proved here is $1/30$, whereas the problem requires $1/12$. The upper example establishes optimality but does not improve the universal lower bound. The needed additional packing structure is the substantive missing part of this reconstruction. The batch check enumerates all colorings on six vertices for the Ramsey step and verifies the affine-line decompositions for small dimensions.

• V. Gruslys and S. Letzter, Monochromatic triangle packings in red-blue graphs, https://arxiv.org/abs/2008.05311; cited for the known resolution, not used in the partial proof above.
• T. F. Bloom, Problem 76, https://www.erdosproblems.com/76, checked 23 September 2026.

No novelty is claimed. The estimate concerns the supplied proof, not confidence in the published sharp theorem.

5) COMPLETION ESTIMATE

COMPLETION: 20%