All Random Solved Random Open
6 solved out of 17 shown (show only solved or open)
OPEN - $5000
If $A\subseteq \mathbb{N}$ has $\sum_{n\in A}\frac{1}{n}=\infty$ then must $A$ contain arbitrarily long arithmetic progressions?
This is essentially asking for good bounds on $r_k(N)$, the size of the largest subset of $\{1,\ldots,N\}$ without a non-trivial $k$-term arithmetic progression. For example, a bound like \[r_k(N) \ll_k \frac{N}{(\log N)(\log\log N)^2}\] would be sufficient.

Even the case $k=3$ is non-trivial, but was proved by Bloom and Sisask [BlSi20]. Much better bounds for $r_3(N)$ were subsequently proved by Kelley and Meka [KeMe23]. Green and Tao [GrTa17] proved $r_4(N)\ll N/(\log N)^{c}$ for some small constant $c>0$. Gowers [Go01] proved \[r_k(N) \ll \frac{N}{(\log\log N)^{c_k}},\] where $c_k>0$ is a small constant depending on $k$. The current best bounds for general $k$ are due to Leng, Sah, and Sawhney [LSS24], who show that \[r_k(N) \ll \frac{N}{\exp((\log\log N)^{c_k})}\] for some constant $c_k>0$ depending on $k$.

Curiously, Erdős [Er83c] thought this conjecture was the 'only way to approach' the conjecture that there are arbitrarily long arithmetic progressions of prime numbers, now a theorem due to Green and Tao [GrTa08].

In [Er81] Erdős makes the stronger conjecture that \[r_k(N) \ll_C\frac{N}{(\log N)^C}\] for every $C>0$ (now known for $k=3$ due to Kelley and Meka [KeMe23]) - see [140].

See also [139] and [142].

OPEN - $500
Does every set of $n$ distinct points in $\mathbb{R}^2$ determine $\gg n/\sqrt{\log n}$ many distinct distances?
A $\sqrt{n}\times\sqrt{n}$ integer grid shows that this would be the best possible. Nearly solved by Guth and Katz [GuKa15] who proved that there are always $\gg n/\log n$ many distinct distances.

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].

OPEN - $500
Does every set of $n$ distinct points in $\mathbb{R}^2$ contain at most $n^{1+O(1/\log\log n)}$ many pairs which are distance 1 apart?
The unit distance problem. In [Er94b] Erdős dates this conjecture to 1946.

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.

See also [92], [96], and [605].

Let $h(n)$ be such that any $n$ points in $\mathbb{R}^2$, with no three on a line and no four on a circle, determine at least $h(n)$ distinct distances. Does $h(n)/n\to \infty$?
Erdős could not even prove $h(n)\geq n$. Pach has shown $h(n)<n^{\log_23}$. Erdős, Füredi, and Pach have improved this to \[h(n) < n\exp(c\sqrt{\log n})\] for some constant $c>0$.
OPEN - $500
Let $f(n)$ be minimal such that any $f(n)$ points in $\mathbb{R}^2$, no three on a line, contain $n$ points which form the vertices of a convex $n$-gon. Prove that $f(n)=2^{n-2}+1$.
The Erdős-Klein-Szekeres 'Happy Ending' problem. The problem originated in 1931 when Klein observed that $f(4)=5$. Turán and Makai showed $f(5)=9$. Erdős and Szekeres proved the bounds \[2^{n-2}+1\leq f(n)\leq \binom{2n-4}{n-2}+1.\] ([ErSz60] and [ErSz35] respectively). There were several improvements of the upper bound, but all of the form $4^{(1+o(1))n}$, until Suk [Su17] proved \[f(n) \leq 2^{(1+o(1))n}.\] The current best bound is due to Holmsen, Mojarrad, Pach, and Tardos [HMPT20], who prove \[f(n) \leq 2^{n+O(\sqrt{n\log n})}.\]

In [Er97e] Erdős clarifies that the \$500 is for a proof, and only offers \$100 for a disproof.

This problem is #1 in Ramsey Theory in the graphs problem collection.

See also [216] and [651].

Additional thanks to: Casey Tompkins
SOLVED - $1000
Let $r_k(N)$ be the size of the largest subset of $\{1,\ldots,N\}$ which does not contain a non-trivial $k$-term arithmetic progression. Prove that $r_k(N)=o(N)$.
Proved by Szemerédi [Sz74]. The best known bounds are due to Kelley and Meka [KeMe23] for $k=3$ (with further slight improvements in [BlSi23]), Green and Tao [GrTa17] for $k=4$, and Leng, Sah, and Sawhney [LSS24] for $k\geq 5$.

See also [3].

Additional thanks to: Zachary Chase
In any $2$-colouring of $\mathbb{R}^2$, for all but at most one triangle $T$, there is a monochromatic congruent copy of $T$.
For some colourings a single equilateral triangle has to be excluded, considering the colouring by alternating strips. Shader [Sh76] has proved this is true if we just consider a single right-angled triangle.
A finite set $A\subset \mathbb{R}^n$ is called Ramsey if, for any $k\geq 1$, there exists some $d=d(A,k)$ such that in any $k$-colouring of $\mathbb{R}^d$ there exists a monochromatic copy of $A$. Characterise the Ramsey sets in $\mathbb{R}^n$.
Erdős, Graham, Montgomery, Rothschild, Spencer, and Straus [EGMRSS73] proved that every Ramsey set is 'spherical': it lies on the surface of some sphere. Graham has conjectured that every spherical set is Ramsey. Leader, Russell, and Walters [LRW12] have alternatively conjectured that a set is Ramsey if and only if it is 'subtransitive': it can be embedded in some higher-dimensional set on which rotations act transitively.

Sets known to be Ramsey include vertices of $k$-dimensional rectangles [EGMRSS73], non-degenerate simplices [FrRo90], trapezoids [Kr92], and regular polygons/polyhedra [Kr91].

Let $f(n)$ be minimal such that the following holds. For any $n$ points in $\mathbb{R}^2$, not all on a line, there must be at least $f(n)$ many lines which contain exactly 2 points (called 'ordinary lines'). Does $f(n)\to \infty$? How fast?
Conjectured by Erdős and de Bruijn. The Sylvester-Gallai theorem states that $f(n)\geq 1$. The fact that $f(n)\geq 1$ was conjectured by Sylvester in 1893. Erdős rediscovered this conjecture in 1933 and told it to Gallai who proved it.

That $f(n)\to \infty$ was proved by Motzkin [Mo51]. Kelly and Moser [KeMo58] proved that $f(n)\geq\tfrac{3}{7}n$ for all $n$. This is best possible for $n=7$. Motzkin conjectured that for $n\geq 13$ there are at least $n/2$ such lines. Csima and Sawyer [CsSa93] proved a lower bound of $f(n)\geq \tfrac{6}{13}n$ when $n\geq 8$. Green and Tao [GrTa13] proved that $f(n)\geq n/2$ for sufficiently large $n$. (A proof that $f(n)\geq n/2$ for large $n$ was earlier claimed by Hansen but this proof was flawed.)

The bound of $n/2$ is best possible for even $n$, since one could take $n/2$ points on a circle and $n/2$ points at infinity. Surprisingly, Green and Tao [GrTa13] show that if $n$ is odd then $f(n)\geq 3\lfloor n/4\rfloor$.

SOLVED - $100
Let $1\leq k<n$. Given $n$ points in $\mathbb{R}^2$, at most $n-k$ on any line, there are $\gg kn$ many lines which contain at least two points.
In particular, given any $2n$ points with at most $n$ on a line there are $\gg n^2$ many lines formed by the points. Solved by Beck [Be83] and Szemerédi and Trotter [SzTr83].

In [Er84] Erdős speculates that perhaps there are $\geq (1+o(1))kn/6$ many such lines, but says 'perhaps [this] is too optimistic and one should first look for a counterexample'. The constant $1/6$ would be best possible here, since there are arrangements of $n$ points with no four points on a line and $\sim n^2/6$ many lines containing three points (see Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84]).

Is there a dense subset of $\mathbb{R}^2$ such that all pairwise distances are rational?
Conjectured by Ulam. Erdős believed there cannot be such a set. This problem is discussed in a blogpost by Terence Tao, in which he shows that there cannot be such a set, assuming the Bombieri-Lang conjecture. The same conclusion was independently obtained by Shaffaf [Sh18].

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).

Let $n\geq 4$. Are there $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, such that all pairwise distances are integers?
Anning and Erdős [AnEr45] proved there cannot exist an infinite such set. Harborth constructed such a set when $n=5$. The best construction to date, due to Kreisel and Kurz [KK08], has $n=7$.

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].

Let $S\subset \mathbb{R}^2$ be such that no two points in $S$ are distance $1$ apart. Must the complement of $S$ contain four points which form a unit square?
The answer is yes, proved by Juhász [Ju79], who proved more generally that the complement of $S$ must contain a congruent copy of any set of four points. This is not true for arbitrarily large sets of points, but perhaps is still true for any set of five points.
Additional thanks to: Bhavik Mehta
Does there exist $S\subseteq \mathbb{R}^2$ such that every set congruent to $S$ (that is, $S$ after some translation and rotation) contains exactly one point from $\mathbb{Z}^2$?
An old question of Steinhaus. Erdős was 'almost certain that such a set does not exist'.

In fact, such a set does exist, as proved by Jackson and Mauldin [JaMa02]. Their construction depends on the axiom of choice.

Additional thanks to: Vjekoslav Kovac
Let $g(k)$ be the smallest integer (if any such exists) such that any $g(k)$ points in $\mathbb{R}^2$ contains an empty convex $k$-gon (i.e. with no point in the interior). Does $g(k)$ exist? If so, estimate $g(k)$.
A variant of the 'happy ending' problem [107], which asks for the same without the 'no point in the interior' restriction. Erdős observed $g(4)=5$ (as with the happy ending problem) but Harborth [Ha78] showed $g(5)=10$. Nicolás [Ni07] and Gerken [Ge08] independently showed that $g(6)$ exists. Horton [Ho83] showed that $g(n)$ does not exist for $n\geq 7$.

This problem is #2 in Ramsey Theory in the graphs problem collection.

Additional thanks to: Zachary Hunter
For which $n$ are there $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, which determine $n-1$ distinct distances and so that (in some ordering of the distances) the $i$th distance occurs $i$ times?
An example with $n=4$ is an isosceles triangle with the point in the centre. Erdős originally believed this was impossible for $n\geq 5$, but Pomerance constructed a set with $n=5$ (see [Er83c] for a description), and Palásti has proved such sets exist for all $n\leq 8$. Erdős believes this is impossible for all sufficiently large $n$.
OPEN - $500
Given $n$ distinct points $A\subset\mathbb{R}^2$ must there be a point $x\in A$ such that \[\#\{ d(x,y) : y \in A\} \gg n^{1-o(1)}?\] Or even $\gg n/\sqrt{\log n}$?
The pinned distance problem, a stronger form of [89]. The example of an integer grid show that $n/\sqrt{\log n}$ would be best possible.

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 \[c=\frac{48-14e}{55-16e}=0.864137\cdots.\]