A stronger form (see [604]) may be true: is there a single point which determines $\gg n/\sqrt{\log n}$ distinct distances, or even $\gg n$ many such points, or even that this is true averaged over all points.
See also [661].
This would be the best possible, as is shown by a set of lattice points. It is easy to show that there are $O(n^{3/2})$ many such pairs. The best known upper bound is $O(n^{4/3})$, due to Spencer, Szemerédi, and Trotter [SST84]. In [Er83c] and [Er85] Erdős offers \$250 for an upper bound of the form $n^{1+o(1)}$.
Part of the difficulty of this problem is explained by a result of Valtr (see [Sz16]), who constructed a metric on $\mathbb{R}^2$ and a set of $n$ points with $\gg n^{4/3}$ unit distance pairs (with respect to this metric). The methods of the upper bound proof of Spencer, Szemerédi, and Trotter [SST84] generalise to include this metric. Therefore to prove an upper bound better than $n^{4/3}$ some special feature of the Euclidean metric must be exploited.
See a survey by Szemerédi [Sz16] for further background and related results.
Is it true that $f(n)\leq n^{o(1)}$? Or even $f(n) < n^{c/\log\log n}$ for some constant $c>0$?
The set of lattice points imply $f(n) > n^{c/\log\log n}$ for some constant $c>0$. Erdős offered \$500 for a proof that $f(n) \leq n^{o(1)}$ but only \$100 for a counterexample.
It is trivial that $f(n) \ll n^{1/2}$. A result of Pach and Sharir implies $f(n) \ll n^{2/5}$.
Fishburn (personal communication to Erdős) proved that $6$ is the smallest $n$ such that $f(n)=3$ and $8$ is the smallest $n$ such that $f(n)=4$, and suggested that the lattice points may not be best example.
See also [754].
Edelsbrunner and Hajnal [EdHa91] have constructed $n$ such points with $2n-7$ pairs distance $1$ apart. (This disproved an early stronger conjecture of Erdős and Moser, that the true answer was $\frac{5}{3}n+O(1)$.)
If this fails for $4$, perhaps there is some constant for which it holds?
Erdős suggested this as an approach to solve [96]. Indeed, if this problem holds for $k+1$ vertices then, by induction, this implies an upper bound of $kn$ for [96].
The answer is no if we omit the requirement that the polygon is convex (I thank Boris Alexeev and Dustin Mixon for pointing this out), since for any $d$ there are graphs with minimum degree $d$ which can be embedded in the plane such that each edge has length one (for example one can take the $d$-dimensional hypercube graph on $2^d$ vertices). One can then connect the vertices in a cyclic order so that there are no self-intersections and no three consecutive vertices on a line, thus forming a (non-convex) polygon.
Erdős [Er94b] wrote 'I could not prove it but felt that it should not be hard. To my great surprise both B. H. Sendov and M. Simonovits doubted the truth of this conjecture.' In [Er94b] he offers \$100 for a counterexample but only \$50 for a proof.
The stated problem is false for $n=4$, for example taking the points to be vertices of a square. The behaviour of such sets for small $n$ is explored by Bezdek and Fodor [BeFo99].
See also [103].
See also [99].
It may be true that there are at least $n^{1-o(1)}$ many such distances. In [Er97e] Erdős offers \$100 for 'any nontrivial result'.
Answered in the negative by Tao [Ta24c], who proved that for any large $n$ there exists a set of $n$ points in $\mathbb{R}^2$ such that any four points determine at least give distinct distances, yet there are $\ll n^2/\sqrt{\log n}$ distinct distances in total. Tao discusses his solution in a blog post.
Indeed, Shaffaf and Tao actually proved that such a rational distance set must be contained in a finite union of real algebraic curves. Solymosi and de Zeeuw [SdZ10] then proved (unconditionally) that a rational distance set contained in a real algebraic curve must be finite, unless the curve contains a line or a circle.
Ascher, Braune, and Turchet [ABT20] observed that, combined, these facts imply that a rational distance set in general position must be finite (conditional on the Bombieri-Lang conjecture).
Ascher, Braune, and Turchet [ABT20] have shown that there is a uniform upper bound on the size of such a set, conditional on the Bombieri-Lang conjecture. Greenfeld, Iliopoulou, and Peluse [GIP24] have shown (unconditionally) that any such set must be very sparse, in that if $S\subseteq [-N,N]^2$ has no three on a line and no four on a circle, and all pairwise distances integers, then \[\lvert S\rvert \ll (\log N)^{O(1)}.\]
See also [130].
Estimate \[m_1=\sup \overline{\delta}(A),\] where $A$ ranges over all measurable subsets of $\mathbb{R}^2$ without two points distance $1$ apart. In particular, is $m_1\leq 1/4$?
The trivial upper bound is $m_1\leq 1/2$, since for any unit vector $u$ the sets $A$ and $A+u$ must be disjoint. Erdős' question was solved by Ambrus, Csiszárik, Matolcsi, Varga, and Zsámboki [ACMVZ23] who proved that $m_1\leq 0.247$.
Bannai, Bannai, and Stanton [BBS83] have proved that \[\lvert A\rvert \leq \binom{n+2}{2}.\] A simple proof of this upper bound was given by Petrov and Pohoata [PePo21].
Shengtong Zhang has observed that a simple lower bound of $\binom{n}{2}$ is given by considering all points with exactly two coordinates equal to $1$ and all others equal to $0$.
It may be true that there are $\gg n$ many such points, or that this is true on average. In [Er97e] Erdős offers \$500 for a solution to this problem, but it is unclear whether he intended this for proving the existence of a single such point or for $\gg n$ many such points.
In [Er97e] Erdős wrote that he initially 'overconjectured' and thought that the answer to this problem is the same as for the number of distinct distances between all pairs (see [89]), but this was disproved by Harborth. It could be true that the answers are the same up to an additive factor of $n^{o(1)}$.
The best known bound is \[\gg n^{c-o(1)},\] due to Katz and Tardos [KaTa04], where \[c=\frac{48-14e}{55-16e}=0.864137\cdots.\]
Erdős, Hickerson, and Pach [EHP89] proved that $u_{\sqrt{2}}(n)\asymp n^{4/3}$ and $u_D(n)\gg n\log^*n$ for all $D>1$ and $n\geq 2$ (where $\log^*$ is the iterated logarithm function).
This lower bound was improved by Swanepoel and Valtr [SwVa04] to $u_D(n) \gg n\sqrt{\log n}$. The best upper bound for general $D$ is $u_D(n)\ll n^{4/3}$.
Zach Hunter has observed that taking $n$ points equally spaced on a circle disproves this conjecture. In the spirit of related conjectures of Erdős and others, presumably some kind of assumption that the points are in general position (e.g. no three on a line and no four on a circle) was intended.
More generally, one can ask how many distances $A$ must determine if every set of $p$ points determines at least $q$ points.
See also [657].
In [Er73] Erdős says it is not even known in $\mathbb{R}$ whether $f(n)\to \infty$. Straus has observed that if $2^k\geq n$ then there exist $n$ points in $\mathbb{R}^k$ which contain no isosceles triangle and determine at most $n-1$ distances.
See also [656].
More generally, if $F(2n)$ is the minimal number of such distances, and $f(2n)$ is minimal number of distinct distances between any $2n$ points in $\mathbb{R}^2$, then is $f \ll F$?
See also [89].
Let $x_1,\ldots,x_n\in \mathbb{R}^2$ be such that $d(x_i,x_j)\geq 1$ for all $i\neq j$. Is it true that, provided $n$ is sufficiently large depending on $t$, the number of distances $d(x_i,x_j)\leq t$ is less than or equal to $f(t)$ with equality perhaps only for the triangular lattice?
In particular, is it true that the number of distances $\leq \sqrt{3}-\epsilon$ is less than $1$?
This is essentially verbatim the problem description in [Er97e], but this does not make sense as written; there must be at least one typo. Suggestions about what this problem intends are welcome.
Erdős also goes on to write 'Perhaps the following stronger conjecture holds: Let $t_1<t_2<\cdots$ be the set of distances occurring in the triangular lattice. $t_1=1$ $t_2=\sqrt{3}$ $t_3=3$ $t_4=5$ etc. Is it true that there is an $\epsilon_n$ so that for every set $y_1,\ldots,$ with $d(y_i,y_j)\geq 1$ the number of distances $d(y_i,y_j)<t_n$ is less than $f(t_n)$?'
Again, this is nonsense interpreted literally; I am not sure what Erdős intended.
Is it true that $f(n)\leq \frac{n}{2}+O(1)$?
See also [753].
The answer is yes: Bhowmick [Bh24] constructs a set of $n$ points in $\mathbb{R}^2$ such that $\lfloor\frac{n}{4}\rfloor$ distances occur at least $n+1$ times. More generally, they construct, for any $m$ and large $n$, a set of $n$ points such that $\lfloor \frac{n}{2(m+1)}\rfloor$ distances occur at least $n+m$ times.
Erdős and Sós proved that $c\geq 1/2$. Gyárfás and Lehel [GyLe95] proved \[\frac{1}{2}<c<\frac{3}{5}.\] (The example proving the upper bound is the set of the first $n$ Fibonacci numbers.)