All Random Solved Random Open
10 solved out of 28 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].

SOLVED - $500
If $f:\mathbb{N}\to \{-1,+1\}$ then is it true that for every $C>0$ there exist $d,m\geq 1$ such that \[\left\lvert \sum_{1\leq k\leq m}f(kd)\right\rvert > C?\]
The Erdős discrepancy problem. This is true, and was proved by Tao [Ta16], who also proved the more general case when $f$ takes values on the unit sphere.

In [Er81] it is further conjectured that \[\max_{md\leq x}\left\lvert \sum_{1\leq k\leq m}f(kd)\right\rvert \gg \log x.\]

Let the van der Waerden number $W(k)$ be such that whenever $N\geq W(k)$ and $\{1,\ldots,N\}$ is $2$-coloured there must exist a monochromatic $k$-term arithmetic progression. Improve the bounds for $W(k)$ - for example, prove that $W(k)^{1/k}\to \infty$.
When $p$ is prime Berlekamp [Be68] has proved $W(p+1)\geq p2^p$. Gowers [Go01] has proved \[W(k) \leq 2^{2^{2^{2^{2^{k+9}}}}}.\]

In [Er81] Erdős further asks whether $W(k+1)/W(k)\to \infty$, or $W(k+1)-W(k)\to \infty$.

Let $F(N)$ be the size of the largest subset of $\{1,\ldots,N\}$ which does not contain any set of the form $\{n,2n,3n\}$. What is \[ \lim_{N\to \infty}\frac{F(N)}{N}?\] Is this limit irrational?
This limit was proved to exist by Graham, Spencer, and Witsenhausen [GrSpWi77]. Similar questions can be asked for the density or upper density of infinite sets without such configurations.
Additional thanks to: Jonathan Chapman
Let $k\geq 3$ and $f(k)$ be the supremum of $\sum_{n\in A}\frac{1}{n}$ as $A$ ranges over all sets of positive integers which do not contain a $k$-term arithmetic progression. Estimate $f(k)$.

Is \[\lim_{k\to \infty}\frac{f(k)}{\log W(k)}=\infty\] where $W(k)$ is the van der Waerden number?

Gerver [Ge77] has proved \[f(k) \geq (1+o(1))k\log k.\] It is trivial that \[\frac{f(k)}{\log W(k)}\geq \frac{1}{2},\] but improving the right-hand side to any constant $>1/2$ is open.
Is it true that for every $\epsilon>0$ and integer $t\geq 1$, if $N$ is sufficiently large and $A$ is a subset of $[t]^N$ of size at least $\epsilon t^N$ then $A$ must contain a combinatorial line $P$ (a set $P=\{p_1,\ldots,p_t\}$ where for each coordinate $1\leq j\leq t$ the $j$th coordinate of $p_i$ is either $i$ or constant).
The 'density Hales-Jewett' problem. This was proved by Furstenberg and Katznelson [FuKa91]. A new elementary proof, which gives quantitative bounds, was proved by the Polymath project [Po09].
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 $N(k,\ell)$ be the minimal $N$ such that for any $f:\{1,\ldots,N\}\to\{-1,1\}$ there must exist a $k$-term arithmetic progression $P$ such that \[ \left\lvert \sum_{n\in P}f(n)\right\rvert> \ell.\] Find good upper bounds for $N(k,\ell)$. Is it true that for any $c>0$ there exists some $C>1$ such that \[N(k,ck)\leq C^k?\] What about \[N(k,1)\leq C^k\] or \[N(k,\sqrt{k})\leq C^k?\]
No decent bound is known even for $N(k,1)$. Probabilistic methods imply that, for every fixed constant $c>0$, we have $N(k,ck)>C_c^k$ for some $C_c>1$.
Find the smallest $h(d)$ such that the following holds. There exists a function $f:\mathbb{N}\to\{-1,1\}$ such that, for every $d\geq 1$, \[\max_{P_d}\left\lvert \sum_{n\in P_d}f(n)\right\rvert\leq h(d),\] where $P_d$ ranges over all finite arithmetic progressions with common difference $d$.
Cantor, Erdős, Schreiber, and Straus [Er66] proved that $h(d)\ll d!$ is possible. Van der Waerden's theorem implies that $h(d)\to \infty$. Beck [Be17] has shown that $h(d) \leq d^{8+\epsilon}$ is possible for every $\epsilon>0$. Roth's famous discrepancy lower bound [Ro64] implies that $h(d)\gg d^{1/2}$.
Let $A_1,A_2,\ldots$ be an infinite collection of infinite sets of integers, say $A_i=\{a_{i1}<a_{i2}<\cdots\}$. Does there exist some $f:\mathbb{N}\to\{-1,1\}$ such that \[\max_{m, 1\leq i\leq d} \left\lvert \sum_{1\leq j\leq m} f(a_{ij})\right\rvert \ll_d 1\] for all $d\geq 1$?
Erdős remarks 'it seems certain that the answer is affirmative'. This was solved by Beck [Be81]. Recently Beck [Be17] proved that one can replace $\ll_d 1$ with $\ll d^{4+\epsilon}$ for any $\epsilon>0$.
Let $1\leq k<\ell$ be integers and define $F_k(N,\ell)$ to be minimal such that every set $A\subset \mathbb{N}$ of size $N$ which contains at least $F_k(N,\ell)$ many $k$-term arithmetic progressions must contain an $\ell$-term arithmetic progression. Find good upper bounds for $F_k(N,\ell)$. Is it true that \[F_3(N,4)=o(N^2)?\] Is it true that for every $\ell>3$ \[\lim_{N\to \infty}\frac{\log F_3(N,\ell)}{\log N}=2?\]
Erdős remarks the upper bound $o(N^2)$ is certainly false for $\ell >\epsilon \log N$. The answer is yes: Fox and Pohoata [FoPo20] have shown that, for all fixed $1\leq k<\ell$, \[F_k(N,\ell)=N^{2-o(1)}\] and in fact \[F_{k}(N,\ell) \leq \frac{N^2}{(\log\log N)^{C_\ell}}\] where $C_\ell>0$ is some constant.
Let $F(N)$ be the maximal size of $A\subseteq \{1,\ldots,N\}$ which is 'non-averaging', so that no $n\in A$ is the arithmetic mean of at least two elements in $A$. What is the order of growth of $F(N)$?
Originally due to Straus. It is known that \[N^{1/4}\ll F(N) \ll N^{\sqrt{2}-1+o(1)}.\] The lower bound is due to Bosznay [Bo89] and the upper bound to Conlon, Fox, and Pham [CFP23] (improving on earlier bound due to Erdős and Sárközy [ErSa90] of $\ll (N\log N)^{1/2}$).

See also [789].

Additional thanks to: Zachary Chase
Find the best function $f(d)$ such that, in any 2-colouring of the integers, at least one colour class contains an arithmetic progression with common difference $d$ of length $f(d)$ for infinitely many $d$.
Originally asked by Cohen. Erdős observed that colouring according to whether $\{ \sqrt{2}n\}<1/2$ or not implies $f(d) \ll d$ (using the fact that $\|\sqrt{2}q\| \gg 1/q$ for all $q$, where $\|x\|$ is the distance to the nearest integer). Beck [Be80] has improved this using the probabilistic method, constructing a colouring that shows $f(d)\leq (1+o(1))\log_2 d$. Van der Waerden's theorem implies $f(d)\to \infty$ is necessary.
What is the smallest $k$ such that $\mathbb{R}^2$ can be red/blue coloured with no pair of red points unit distance apart, and no $k$-term arithmetic progression of blue points?
Juhász [Ju79] has shown that $k\geq 5$. It is also known that $k\leq 10000000$ (as Erdős writes, 'more or less').
If $\mathbb{R}^2$ is finitely coloured then must there exist some colour class which contains the vertices of a rectangle of every area?
Graham [Gr80] has shown that this is true if we replace rectangle by right-angled triangle. The same question can be asked for parallelograms. It is not true for rhombuses.

This is false; Kovač [Ko23] provides an explicit (and elegantly simple) colouring using 25 colours such that no colour class contains the vertices of a rectangle of area $1$. The question for parallelograms remains open.

Additional thanks to: Ryan Alweiss, Vjekoslav Kovac
Let $H(k)$ be the smallest $N$ such that in any finite colouring of $\{1,\ldots,N\}$ (into any number of colours) there is always either a monochromatic $k$-term arithmetic progression or a rainbow arithmetic progression (i.e. all elements are different colours). Estimate $H(k)$. Is it true that \[H(k)^{1/k}/k \to \infty\] as $k\to\infty$?
This type of problem belongs to 'canonical' Ramsey theory. The existence of $H(k)$ follows from Szemerédi's theorem, and it is easy to show that $H(k)^{1/k}\to\infty$.
Let $C>0$ be arbitrary. Is it true that, if $n$ is sufficiently large depending on $C$, then in any $2$-colouring of $\binom{\{2,\ldots,n\}}{2}$ there exists some $X\subset \{2,\ldots,n\}$ such that $\binom{X}{2}$ is monochromatic and \[\sum_{x\in X}\frac{1}{\log x}\geq C?\]
The answer is yes, which was proved by Rödl [Ro03].

In the same article Rödl also proved a lower bound for this problem, constructing, for all $n$, a $2$-colouring of $\binom{\{2,\ldots,n\}}{2}$ such that if $X\subseteq \{2,\ldots,n\}$ is such that $\binom{X}{2}$ is monochromatic then \[\sum_{x\in X}\frac{1}{\log x}\ll \log\log\log n.\]

This bound is best possible, as proved by Conlon, Fox, and Sudakov [CFS13], who proved that, if $n$ is sufficiently large, then in any $2$-colouring of $\binom{\{2,\ldots,n\}}{2}$ there exists some $X\subset \{2,\ldots,n\}$ such that $\binom{X}{2}$ is monochromatic and \[\sum_{x\in X}\frac{1}{\log x}\geq 2^{-8}\log\log\log n.\]

Additional thanks to: Mehtaab Sawhney
Let $A=\{a_1,a_2,\ldots\}\subset \mathbb{R}^d$ be an infinite sequence such that $a_{i+1}-a_i$ is a positive unit vector (i.e. is of the form $(0,0,\ldots,1,0,\ldots,0)$). For which $d$ must $A$ contain a three-term arithmetic progression?
This is true for $d\leq 3$ and false for $d\geq 4$.

This problem is equivalent to one on 'abelian squares' (see [231]). In particular $A$ can be interpreted as an infinite string over an alphabet with $d$ letters (each letter describining which of the $d$ possible steps is taken at each point). An abelian square in a string $s$ is a pair of consecutive blocks $x$ and $y$ appearing in $s$ such that $y$ is a permutation of $x$. The connection comes from the observation that $p,q,r\in A\subset \mathbb{R}^d$ form a three-term arithmetic progression if and only if the string corresponding to the steps from $p$ to $q$ is a permutation of the string corresponding to the steps from $q$ to $r$.

This problem is therefore equivalent to asking for which $d$ there exists an infinite string over $\{1,\ldots,d\}$ with no abelian squares. It is easy to check that in fact any finite string of length $7$ over $\{1,2,3\}$ contains an abelian square.

An infinite string without abelian squares was constructed when $d=4$ by Keränen [Ke92]. We refer to a recent survey by Fici and Puzynina [FiPu23] for more background and related results, and a blog post by Renan for an entertaining and educational discussion.

Additional thanks to: Boris Alexeev and Dustin Mixon
Let $S\subseteq \mathbb{Z}^3$ be a finite set and let $A=\{a_1,a_2,\ldots,\}\subset \mathbb{Z}^3$ be an infinite $S$-walk, so that $a_{i+1}-a_i\in S$ for all $i$. Must $A$ contain three collinear points?
Originally conjectured by Gerver and Ramsey [GeRa79], who showed that the answer is yes for $\mathbb{Z}^2$, and for $\mathbb{Z}^3$ that the largest number of collinear points can be bounded.
Additional thanks to: Terence Tao
Let $k\geq 3$. Must any ordering of $\mathbb{R}$ contain a monotone $k$-term arithmetic progression, that is, some $x_1<\cdots<x_k$ which forms an increasing or decreasing $k$-term arithmetic progression?
The answer is no, even for $k=3$, as shown by Ardal, Brown, and Jungić [ABJ11].

See also [195] and [196].

What is the smallest $k$ such that in any permutation of $\mathbb{Z}$ there must exist a monotone $k$-term arithmetic progression $x_1<\cdots<x_k$?
Geneson [Ge19] proved that $k\leq 5$. Adenwalla [Ad22] proved that $k\leq 4$.

See also [194] and [196].

Additional thanks to: Boris Alexeev and Dustin Mixon
Must every permutation of $\mathbb{N}$ contain a monotone 4-term arithmetic progression $x_1<x_2<x_3<x_4$?
Davis, Entringer, Graham, and Simmons [DEGS77] have shown that there must exist a monotone 3-term arithmetic progression and need not contain a 5-term arithmetic progression.

See also [194] and [195].

Additional thanks to: Boris Alexeev and Dustin Mixon
Can $\mathbb{N}$ be partitioned into two sets, each of which can be permuted to avoid monotone 3-term arithmetic progressions?
If three sets are allowed then this is possible.
Additional thanks to: Boris Alexeev and Dustin Mixon
If $A\subset \mathbb{N}$ is a Sidon set then must the complement of $A$ contain an infinite arithmetic progression?
The answer is yes, as shown by Baumgartner [Ba75].
If $A\subset \mathbb{R}$ does not contain a 3-term arithmetic progression then must $\mathbb{R}\backslash A$ contain an infinite arithmetic progression?
The answer is no, as shown by Baumgartner [Ba75] (whose construction uses the axiom of choice).
Does the longest arithmetic progression of primes in $\{1,\ldots,N\}$ have length $o(\log N)$?
It follows from the prime number theorem that such a progression has length $\leq(1+o(1))\log N$.
Let $G_k(N)$ be such that any set of $N$ integers contains a subset of size at least $G_k(N)$ which does not contain a $k$-term arithmetic progression. Determine the size of $G_k(N)$. How does it relate to $R_k(N)$, the size of the largest subset of $\{1,\ldots,N\}$ without a $k$-term arithmetic progression? Is it true that \[\lim_{N\to \infty}\frac{R_3(N)}{G_3(N)}=1?\]
First asked and investigated by Riddell [Ri69]. It is trivial that $G_k(N)\leq R_k(N)$, and it is possible that $G_k(N) <R_k(N)$ (for example with $k=3$ and $N=14$). Komlós, Sulyok, and Szemerédi [KSS75] have shown that $R_k(N) \ll_k G_k(N)$.
Additional thanks to: Zachary Chase