Erdős Problem #97
Still open — but if it is false, a finite counterexample exists, so a computer search could in principle disprove it. A search can't prove it.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
A set of points \(A\) has n equidistant points at \(p\) if there exist at least \(n\) other points in \(A\) that are equidistant from \(p\).
A set of points \(A\) has n equidistant points on a set of points \(B\) if for every point in \(B\), there exist at least \(n\) other points in \(A\) that are equidistant from it.
A set of points \(A\) has n equidistant property if for every point in \(A\), there exist at least \(n\) other points in \(A\) that are equidistant from it.
Attempts
submit one →The one that beat every model at every budget in our cross-model bench. Best outcome: an honest decline.
WALL