SOLVED - $100

Let $d_n=p_{n+1}-p_n$. Are there infinitely many $n$ such that $d_n<d_{n+1}<d_{n+2}$?

Conjectured by Erdős and Turán [ErTu48]. Shockingly Erdős offered \$25000 for a disproof of this, but as he comments, it 'is certainly true'.

Indeed, the answer is yes, as proved by Banks, Freiberg, and Turnage-Butterbaugh [BFT15] with an application of the Maynard-Tao machinery concerning bounded gaps between primes [Ma15]. They in fact prove that, for any $m\geq 1$, there are infinitely many $n$ such that \[d_n<d_{n+1}<\cdots <d_{n+m}\] and infinitely many $n$ such that \[d_n> d_{n+1}>\cdots >d_{n+m}.\]

SOLVED - $100

Let $A\subseteq \{1,\ldots,N\}$ be such that there are no $a,b,c\in A$ such that $a\mid(b+c)$ and $a<\min(b,c)$. Is it true that $\lvert A\rvert\leq N/3+O(1)$?

Asked by Erdős and Sárközy, who observed that $(2N/3,N]\cap \mathbb{N}$ is such a set. The answer is yes, as proved by Bedert [Be23].

SOLVED - $100

An $\epsilon$-almost covering system is a set of congruences $a_i\pmod{n_i}$ for distinct moduli $n_1<\ldots<n_k$ such that the density of those integers which satisfy none of them is $\leq \epsilon$. Is there a constant $C>1$ such that for every $\epsilon>0$ and $N\geq 1$ there is an $\epsilon$-almost covering system with $N\leq n_1<\cdots <n_k\leq CN$?

By a simple averaging argument the set of moduli $[m_1,m_2]\cap \mathbb{N}$ has a choice of residue classes which form an $\epsilon(m_1,m_2)$-almost covering system with
\[\epsilon(m_1,m_2)=\prod_{m_1\leq m\leq m_2}(1-1/m).\]
A $0$-covering system is just a covering system, and so by Hough [Ho15] these only exist for $n_1<10^{18}$.

The answer is no, as proved by Filaseta, Ford, Konyagin, Pomerance, and Yu [FFKPY07], who (among other results) prove that if \[1< C \leq N^{\frac{\log\log\log N}{4\log\log N}}\] then, for any $N\leq n_1<\cdots< n_k\leq CN$, the density of integers not covered for any fixed choice of residue classes is at least \[\prod_{i}(1-1/n_i)\] (and this density is achieved for some choice of residue classes as above).

SOLVED - $100

Is there an explicit construction of a set $A\subseteq \mathbb{N}$ such that $A+A=\mathbb{N}$ but $1_A\ast 1_A(n)=o(n^\epsilon)$ for every $\epsilon>0$?

The existence of such a set was asked by Sidon to Erdős in 1932. Erdős (eventually) proved the existence of such a set using probabilistic methods. This problem asks for a constructive solution.

An explicit construction was given by Jain, Pham, Sawhney, and Zakharov [JPSZ24].

OPEN - $100

If $A,B\subset \{1,\ldots,N\}$ are two Sidon sets such that $(A-A)\cap(B-B)=\{0\}$ then is it true that
\[ \binom{\lvert A\rvert}{2}+\binom{\lvert B\rvert}{2}\leq\binom{f(N)}{2}+O(1),\]
where $f(N)$ is the maximum possible size of a Sidon set in $\{1,\ldots,N\}$? If $\lvert A\rvert=\lvert B\rvert$ then can this bound be improved to
\[\binom{\lvert A\rvert}{2}+\binom{\lvert B\rvert}{2}\leq (1-c)\binom{f(N)}{2}\]
for some constant $c>0$?

SOLVED - $100

If $\delta>0$ and $N$ is sufficiently large in terms of $\delta$, and $A\subseteq\{1,\ldots,N\}$ is such that $\sum_{a\in A}\frac{1}{a}>\delta \log N$ then must there exist $S\subseteq A$ such that $\sum_{n\in S}\frac{1}{n}=1$?

Solved by Bloom [Bl21], who showed that the quantitative threshold
\[\sum_{n\in A}\frac{1}{n}\gg \frac{\log\log\log N}{\log\log N}\log N\]
is sufficient. This was improved by Liu and Sawhney [LiSa24] to
\[\sum_{n\in A}\frac{1}{n}\gg (\log N)^{4/5+o(1)}.\]
Erdős speculated that perhaps even $\gg (\log\log N)^2$ might be sufficient. (A construction of Pomerance, as discussed in the appendix of [Bl21], shows that this would be best possible.)

SOLVED - $100

A set of integers $A$ is Ramsey $2$-complete if, whenever $A$ is $2$-coloured, all sufficiently large integers can be written as a monochromatic sum of elements of $A$.

Burr and Erdős [BuEr85] showed that there exists a constant $c>0$ such that it cannot be true that \[\lvert A\cap \{1,\ldots,N\}\rvert \leq c(\log N)^2\] for all large $N$ and that there exists a Ramsey $2$-complete $A$ such that for all large $N$ \[\lvert A\cap \{1,\ldots,N\}\rvert < (2\log_2N)^3.\] Improve either of these bounds.

SOLVED - $100

Is there a set $A\subset \mathbb{N}$ of density $0$ and a constant $c>0$ such that every graph on sufficiently many vertices with average degree $\geq c$ contains a cycle whose length is in $A$?

Bollobás proved that such a $c$ does exist if $A$ is an infinite arithmetic progression containing even numbers. Erdős was 'almost certain' that if $A$ is the set of powers of $2$ then no such $c$ exists (although conjectures that $n$ vertices and average degree $\gg (\log n)^{C}$ suffices for some $C=O(1)$). If $A$ is the set of squares (or the set of $p\pm 1$ for $p$ prime) then he had no guess.

Solved by Verstraëte [Ve05], who gave a non-constructive proof that such a set $A$ exists.

Liu and Montgomery [LiMo20] proved that in fact this is true when $A$ is the set of powers of $2$ (more generally any set of even numbers which doesn't grow too quickly) - in particular this contradicts the previous belief of Erdős.

OPEN - $100

Give a constructive proof that $R(k)>C^k$ for some constant $C>1$.

Erdős gave a simple probabilistic proof that $R(k) \gg k2^{k/2}$.
Equivalently, this question asks for an explicit construction of a graph on $n$ vertices which does not contain any clique or independent set of size $\geq c\log n$ for some constant $c>0$. Cohen [Co15] (see the introduction for further history) constructed a graph on $n$ vertices which does not contain any clique or independent set of size
\[\geq 2^{(\log\log n)^{C}}\]
for some constant $C>0$. Li [Li23b] has recently improved this to
\[\geq (\log n)^{C}\]
for some constant $C>0$.

OPEN - $100

Let $Q_n$ be the $n$-dimensional hypercube graph (so that $Q_n$ has $2^n$ vertices and $n2^{n-1}$ edges). Is it true that every subgraph of $Q_n$ with
\[\geq \left(\frac{1}{2}+o(1)\right)n2^{n-1}\]
many edges contains a $C_4$?

The best known result is due to Balogh, Hu, Lidicky, and Liu [BHLL14], who proved that $0.6068 n2^{n-1}$ edges suffice.

A similar question can be asked for other even cycles.

SOLVED - $100

For any $\epsilon>0$ there exists $\delta=\delta(\epsilon)>0$ such that if $G$ is a graph on $n$ vertices with no independent set or clique of size $\geq \epsilon\log n$ then $G$ contains an induced subgraph with $m$ edges for all $m\leq \delta n^2$.

Conjectured by Erdős and McKay, who proved it with $\delta n^2$ replaced by $\delta (\log n)^2$. Solved by Kwan, Sah, Sauermann, and Sawhney [KSSS22]. Erdős' original formulation also had the condition that $G$ has $\gg n^2$ edges, but an old result of Erdős and Szemerédi says that this follows from the other condition anyway.

OPEN - $100

Does every convex polygon have a vertex with no other $4$ vertices equidistant from it?

Erdős originally conjectured this with no $3$ vertices equidistant, but Danzer found a convex polygon on 9 points such that every vertex has three vertices equidistant from it (but this distance depends on the vertex), and Fishburn and Reeds [FiRe92] have found a convex polygon on 20 points such that every vertex has three vertices equidistant from it (and this distance is the same for all vertices).

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.

OPEN - $100

Let $A\subseteq\mathbb{R}^2$ be a set of $n$ points with minimum distance equal to 1, chosen to minimise the diameter of $A$. If $n$ is sufficiently large then must there be three points in $A$ which form an equilateral triangle of size 1?

The minimal such diameter is achieved (asymptotically) by the points in a triangular lattice intersected with a circle. In general Erdős believed such a set must have very large intersection with the triangular lattice (perhaps as many as $(1-o(1))n$).

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].

OPEN - $100

Given $n$ points in $\mathbb{R}^2$, no five of which are on a line, the number of lines containing four points is $o(n^2)$.

There are examples of sets of $n$ points with $\sim n^2/6$ many collinear triples and no four points on a line. Such constructions are given by Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84].

Grünbaum [Gr76] constructed an example with $\gg n^{3/2}$ such lines. Erdős speculated this may be the correct order of magnitude. This is false: Solymosi and Stojaković [SoSt13] have constructed a set with no five on a line and at least \[n^{2-O(1/\sqrt{\log n})}\] many lines containing exactly four points.

See also [102]. A generalisation of this problem is asked in [588].

SOLVED - $100

Let $z_i$ be an infinite sequence of complex numbers such that $\lvert z_i\rvert=1$ for all $i\geq 1$, and for $n\geq 1$ let
\[p_n(z)=\prod_{i\leq n} (z-z_i).\]
Let $M_n=\max_{\lvert z\rvert=1}\lvert p_n(z)\rvert$. Is it true that $\limsup M_n=\infty$? Is it true that there exists $c>0$ such that for infinitely many $n$ we have $M_n > n^c$, or even that for all $n$
\[\sum_{k\leq n}M_k > n^{1+c}?\]

The weaker conjecture that $\limsup M_n=\infty$ was proved by Wagner, who show that there is some $c>0$ with $M_n>(\log n)^c$ infinitely often.

This was solved by Beck [Be91], who proved that there exists some $c>0$ such that \[\max_{n\leq N} M_n > N^c.\]

OPEN - $100

Let $A\subseteq\mathbb{R}$ be an infinite set. Must there be a set $E\subset \mathbb{R}$ of positive measure which does not contain any set of the shape $aA+b$ for some $a,b\in\mathbb{R}$ and $a\neq 0$?

The Erdős similarity problem.

This is true if $A$ is unbounded or dense in some interval. It therefore suffices to prove this when $A=\{a_1>a_2>\cdots\}$ is a countable strictly monotone sequence which converges to $0$.

Steinhaus [St20] has proved this is false whenever $A$ is a finite set.

This conjecture is known in many special cases (but, for example, it is is open when $A=\{1,1/2,1/4,\ldots\}$. For an overview of progress we recommend a nice survey by Svetic [Sv00] on this problem.

OPEN - $100

Let $A\subset \mathbb{R}^2$ be a set of $n$ points. Must there be two distances which occur at least once but between at most $n$ pairs of points? Must the number of such distances $\to \infty$ as $n\to \infty$?

Asked by Erdős and Pach. Pannowitz and Hopf proved that the largest distance between points of $A$ can occur at most $n$ times, but it is unknown whether a second such distance must occur.

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'.

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]).

OPEN - $100

Let $f(N)$ be the maximum size of $A\subseteq \{1,\ldots,N\}$ such that the sums $a+b+c$ with $a,b,c\in A$ are all distinct (aside from the trivial incidences). Is it true that
\[ f(N)\sim N^{1/3}?\]

Originally asked to Erdős by Bose. Bose and Chowla [BoCh62] provided a construction proving one half of this, namely
\[(1+o(1))N^{1/3}\leq f(N).\]
The best upper bound known to date is due to Green [Gr01],
\[f(N) \leq ((7/2)^{1/3}+o(1))N^{1/3}\]
(note that $(7/2)^{1/3}\approx 1.519\cdots$).

OPEN - $100

We say $H$ is a unique subgraph of $G$ if there is exactly one way to find $H$ as a subgraph (not necessarily induced) of $G$. Is there a graph on $n$ vertices with
\[\gg \frac{2^{\binom{n}{2}}}{n!}\]
many distinct unique subgraphs?

A problem of Erdős and Entringer [EnEr72], who constructed a graph with
\[\gg 2^{\binom{n}{2}-O(n^{3/2+o(1)})}\]
many unique subgraphs. This was improved by Harary and Schwenk [HaSc73] and then by Brouwer [Br75], who constructed a graph with
\[\gg \frac{2^{\binom{n}{2}-O(n)}}{n!}\]
many unique subgraphs.

Note that there are $\sim 2^{\binom{n}{2}}/n!$ many non-isomorphic graphs on $n$ vertices (folklore, often attributed to Pólya), and hence the bound in the problem statement is trivially best possible.

Erdős believed Brouwer's construction was essentially best possible, but Spencer suggested that $\gg \frac{2^{\binom{n}{2}}}{n!}$ may be possible. Erdős offered \$100 for a construction and \$25 for a proof that no such construction is possible.

SOLVED - $100

Does there exist a graph $G$ with at most $10^{10}$ many vertices which contains no $K_4$, and yet any $2$-colouring of the edges produces a monochromatic $K_3$?

Erdős and Hajnal [ErHa67] first asked for the existence of any such graph. Existence was proved by Folkman [Fo70], but with very poor quantitative bounds. (As a result these quantities are often called Folkman numbers.) Let this particular Folkman number be denoted by $N$.

Frankl and Rödl [FrRo86] proved $N\leq 7\times 10^{11}$, which Spencer [Sp88] improved to $\leq 3\times 10^{9}$. This resolved the initial challenge of Erdős [Er75d] to beat $10^{10}$.

Lu [Lu07] proved $N\leq 9697$ vertices. The current record is due to Dudek and Rödl [DuRo08] who proved $N\leq 941$ vertices. For further information we refer to a paper of Radziszowski and Xu [RaXu07], who prove that $N\geq 19$ and speculate that $N\leq 127$.

OPEN - $100

Let $f_k(n)$ be minimal such that if $n$ points in $\mathbb{R}^2$ have no $k+1$ points on a line then there must be at most $f_k(n)$ many lines containing at least $k$ points. Is it true that
\[f_k(n)=o(n^2)\]
for $k\geq 4$?

A generalisation of [101] (which asks about $k=4$).

The restriction to $k\geq 4$ is necessary since Sylvester has shown that $f_3(n)= n^2/6+O(n)$. (See also Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84] for constructions which show that $f_3(n)\geq(1/6+o(1))n^2$.)

For $k\geq 4$, Kárteszi [Ka] proved \[f_k(n)\gg_k n\log n.\] Grünbaum [Gr76] proved \[f_k(n) \gg_k n^{1+\frac{1}{k-2}}.\] Erdős speculated this may be the correct order of magnitude, but Solymosi and Stojaković [SoSt13] give a construction which shows \[f_k(n)\gg_k n^{2-O_k(1/\sqrt{\log n})}\]