PROVED
This has been solved in the affirmative.
Let $G$ be a graph on $n$ vertices, $\alpha_1(G)$ be the maximum number of edges that contain at most one edge from every triangle, and $\tau_1(G)$ be the minimum number of edges that contain at least one edge from every triangle.
Is it true that\[\alpha_1(G)+\tau_1(G) \leq \frac{n^2}{4}?\]
A problem of Erdős, Gallai, and Tuza
[EGT96], who observe that this is probably quite difficult since there are different examples where equality hold: the complete graph, the complete bipartite graph, and the graph obtained from $K_{m,m}$ by adding one vertex joined to every other.
This is true, and was proved by Norin and Sun
[NoSu16], who in fact proved that\[\alpha_1(G)+\tau_B(G) \leq \frac{n^2}{4},\]where $\tau_B(G)$ is the minimum number of edges that need to be removed to make the graph bipartite. (Note that clearly $\tau_1(G)\leq \tau_B(G)$.) Problem
[23] can be phrased as the conjecture $\tau_B(n)\leq n^2/25$.
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This page was last edited 14 October 2025.
Additional thanks to: msellke
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T. F. Bloom, Erdős Problem #621, https://www.erdosproblems.com/621, accessed 2026-01-17