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Claim. Call a finite set isosceles when every three of its points form an isosceles triangle. The solution of Monthly problem E735, posed by Erdős and solved by Kelly, proves that no isosceles set in the plane has seven points and that the vertices of a regular pentagon together with its center form an isosceles set of six points. A set of $m$ points contains three points with three distinct distances exactly when it is not isosceles, so in the notation of Problem 1088 this gives $f_2(3)=7$. The same solution exhibits an isosceles set of eight points in $\mathbb{R}^3$, the lower bound for $f_3(3)$. The theorem and its uniqueness statement are recorded, as the answer for $d=2$ of the isosceles-set question, on [[problems/distance_problems/E0503/claims/1947_04_01_kelly|Kelly's claim page for Problem 503]].

Covers. The value $f_2(3)=7$, the instance $d=2$, $n=3$; and the lower bound $f_3(3)\ge9$, whose matching upper bound is on [[problems/discrete_geometry/E1088/claims/1962_01_01_croft|Croft's claim page]].

Depends on. No page of this wiki.

Acceptance. Refereed: P. Erdős and L. M. Kelly, E735, Amer. Math. Monthly 54 (1947), no. 4, 227–229, in the journal's problems section, Erdős as proposer and Kelly as solver. The site's remarks say that Erdős could prove $f_2(3)=7$, but the site labels the problem OPEN, so the remark is not reviewed evidence. The page is dated to the issue, April 1947.