An alternative, simpler, proof was given by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22], who improved the bound on the smallest modulus to $616000$.
Hough and Nielsen [HoNi19] proved that at least one modulus must be divisible by either $2$ or $3$. A simpler proof of this fact was provided by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22].
Selfridge has shown (as reported in [Sc67]) that such a covering system exists if a covering system exists with moduli $n_1,\ldots,n_k$ such that no $n_i$ divides any other $n_j$ (but the latter has been shown not to exist, see [586]).
The sequence of such numbers is A006286 in the OEIS.
Granville and Soundararajan [GrSo98] have conjectured that at most $3$ powers of 2 suffice for all odd integers, and hence at most $4$ powers of $2$ suffice for all even integers. (The restriction to odd integers is important here - for example, Bogdan Grechuk has observed that $1117175146$ is not the sum of a prime and at most $3$ powers of $2$, and pointed out that parity considerations, coupled with the fact that there are many integers not the sum of a prime and $2$ powers of $2$ (see [9]) suggest that there exist infinitely many even integers which are not the sum of a prime and at most $3$ powers of $2$).
Granville and Soundararajan [GrSo98] have proved that this is very related to the problem of finding primes $p$ for which $2^p\equiv 2\pmod{p^2}$ (for example this conjecture implies there are infinitely many such $p$).
This is equivalent to asking whether every $n$ not divisible by $4$ is the sum of a squarefree number and a power of two. Erdős thought that proving this with two powers of 2 is perhaps easy, and could prove that it is true (with a single power of two) for almost all $n$.
Is it true that \[\sum_{n\in A}\frac{1}{n}<\infty?\]
An example of an $A$ with this property where \[\liminf \frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N^{1/2}}\log N>0\] is given by the set of $p^2$, where $p\equiv 3\pmod{4}$ is prime.
For the finite version see [13].
For the infinite version see [12].
Erdös and Freud investigated the finite analogue in 'a recent Hungarian paper', proving that there exists $A\subseteq \{1,\ldots,N\}$ such that the number of integers not representable in exactly one way as the sum of two elements from $A$ is $<2^{3/2}N^{1/2}$, and suggest the constant $2^{3/2}$ is perhaps best possible.
Tao [Ta23] has proved that this series does converge assuming a strong form of the Hardy-Littlewood prime tuples conjecture.
There is likely nothing special about the integers in this question, and indeed Erdős and Szemerédi also ask a similar question about finite sets of real or complex numbers. The current best bound for sets of reals is the same bound of Rudnev and Stevens above. The best bound for complex numbers is \[\max( \lvert A+A\rvert,\lvert AA\rvert)\gg\lvert A\rvert^{\frac{5}{4}},\] due to Solymosi [So05].
One can in general ask this question in any setting where addition and multiplication are defined (once one avoids any trivial obstructions such as zero divisors or finite subfields). For example, it makes sense for subsets of finite fields. The current record is that if $A\subseteq \mathbb{F}_p$ with $\lvert A\rvert <p^{5/8}$ then \[\max( \lvert A+A\rvert,\lvert AA\rvert)\gg\lvert A\rvert^{\frac{11}{9}+o(1)},\] due to Rudnev, Shakan, and Shkredov [RSS20].
There is also a natural generalisation to higher-fold sum and product sets. For example, in [ErSz83] (and in [Er91]) Erdős and Szemerédi also conjecture that for any $m\geq 2$ and finite set of integers $A$ \[\max( \lvert mA\rvert,\lvert A^m\rvert)\gg \lvert A\rvert^{m-o(1)}.\] See [53] for more on this generalisation and [808] for a stronger form of the original conjecture. See also [818] for a special case.
Erdős and Szemerédi proved that there exist arbitrarily large sets $A$ such that the integers which are the sum or product of distinct elements of $A$ is at most \[\exp\left(c (\log \lvert A\rvert)^2\log\log\lvert A\rvert\right)\] for some constant $c>0$.
See also [52].
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.
Szemerédi conjectured (see [Er97e]) that this stronger variant remains true if we only assume that no three points are on a line, and proved this with the weaker bound of $n/3$.
See also [660].
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].
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.
Erdős (and independently Hall [Ha96] and Montgomery) also asked about $F(N)$, the size of the largest $A\subseteq\{1,\ldots,N\}$ such that the product of no odd number of $a\in A$ is a square. Ruzsa [Ru77] observed that $1/2<\lim F(N)/N <1$. Granville and Soundararajan [GrSo01] proved an asymptotic \[F(N)=(1-c+o(1))N\] where $c=0.1715\ldots$ is an explicit constant.
This problem was answered in the negative by Tao [Ta24], who proved that for any $k\geq 4$ there is some constant $c_k>0$ such that $F_k(N) \leq (1-c_k+o(1))N$.
In [Er92b] Erdős wrote 'last year I made the following silly conjecture': every integer $n$ can be written as the sum of distinct integers of the form $2^k3^l$, none of which divide any other. 'I mistakenly thought that this was a nice and difficult conjecture but Jansen and several others found a simple proof by induction.' This simple proof is as follows: one proves the stronger fact that such a representation always exists, and moreover if $n$ is even then all the summands can be taken to be even: if $n=2m$ we are done applying the inductive hypothesis to $m$. Otherwise if $n$ is odd then let $3^k$ be the largest power of $3$ which is $\leq n$ and apply the inductive hypothesis to $n-3^k$ (which is even).
In [Er92b] Erdős makes the stronger conjecture (for $a=2$, $b=3$, and $c=5$) that, for any $\epsilon>0$, all large integers $n$ can be written as the sum of distinct integers $b_1<\cdots <b_t$ of the form $2^k3^l5^m$ where $b_t<(1+\epsilon)b_1$.
See also [845].
See also [125].
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'.
This problem is #2 in Ramsey Theory in the graphs problem collection.
Surányi was the first to conjecture that the only non-trivial solution to $a!b!=n!$ is $6!7!=10!$.
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 \[c=\frac{48-14e}{55-16e}=0.864137\cdots.\]
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].
This qualitative statement follows from the density Hales-Jewett theorem proved by Furstenberg and Katznelson [FuKa91]. A quantitative proof (yet with very poor bounds) was given by Solymosi [So04].
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.
The bound $q(n,k)<(1+o(1))k\log n$ is easy. It may be true this improved bound holds even up to $k=o(\log n)$.
See also [457].