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from every problem. erdosproblems.com no longer shows statuses (Bloom's reasons); this is the database's record, not our verdict.
908 of 1221 statements are formalized in Lean (Formal Conjectures) · 107 carry an Erdős prize · 3 are independent of ZFC and 9 more can't be proved, or can't be disproved, from ZFC alone · “open” counts the computable and Lean-checked ones too.
If \(n\) distinct points in \(\mathbb{R}^2\) form a convex polygon then they determine at least \(\lfloor \frac{n}{2}\rfloor\) distinct distances. Solved by Altman [Al63].
Suppose \(n\) points in \(\mathbb{R}^2\) determine a convex polygon and the set of distances between them is \(\{u_1,\ldots,u_t\}\). Suppose \(u_i\) appears as the distance between \(f(u_i)\) many pairs of points. Then \[\sum_i f(u_i)^2…
Let \(x_1,\ldots,x_n\in\mathbb{R}^2\) determine the set of distances \(\{u_1,\ldots,u_t\}\). Suppose \(u_i\) appears as the distance between \(f(u_i)\) many pairs of points. Then for all \(\epsilon>0\) \[\sum_i f(u_i)^2 \ll_\epsilon…
Let \(f(n)\) be minimal such that any \(f(n)\) points in \(ℝ^2\), no three on a line, contain \(n\) points which form the vertices of a convex \(n\)-gon. Prove that \(f(n) = 2^{n-2} + 1\).
Let \(x_1, \ldots, x_n \in \mathbb{R}^3\) be the vertices of a convex polyhedron. Are there at least \[(1 - o(1)) \frac{n}{2}\] many distinct distances between the \(x_i\)? The \((1 - o(1)) \frac{n}{2}\) lower bound is formalised as: for…
If \(n\) distinct points in \(\mathbb{R}^2\) form a convex polygon then some vertex has at least \(\lfloor\frac{n}{2}\rfloor\) different distances to other vertices.
Source: teorth/erdosproblems (its own interactive table), Apache-2.0, pulled 2026-10-21. A status here is what that database records, not an independent verification by us, except where a problem page says so.