A number of improvements of the constant have been given (see [St23] for a history), with the current record $\sqrt{2/\pi}$ first proved in unpublished work of Elkies and Gleason. Two proofs achieving this constant are provided by Dubroff, Fox, and Xu [DFX21], who in fact prove the exact bound $N\geq \binom{n}{\lfloor n/2\rfloor}$.
In [Er73] and [ErGr80] the generalisation where $A\subseteq (0,N]$ is a set of real numbers such that the subset sums all differ by at least $1$ is proposed, with the same conjectured bound. (The second proof of [DFX21] applies also to this generalisation.)
This problem appears in Erdős' book with Spencer [ErSp74] in the final chapter titled 'The kitchen sink'. As Ruzsa writes in [Ru99] "it is a rich kitchen where such things go to the sink".
The sequence of minimal $N$ for a given $n$ is A276661 in the OEIS.
See also [350].
An alternative, simpler, proof was given by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22], who improved the upper bound on the smallest modulus to $616000$.
The best known lower bound is a covering system whose minimum modulus is $42$, due to Owens [Ow14].
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] (see [219]).
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 [687].
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 $2$).
Kahn [Ka92] proved that $\chi(G)\leq (1+o(1))n$ (for which Erdős gave him a 'consolation prize' of \$100). Hindman has proved the conjecture for $n<10$. Kang, Kelly, Kühn, Methuku, and Osthus [KKKMO21] have proved the answer is yes for all sufficiently large $n$.
In [Er97d] Erdős asks how large $\chi(G)$ can be if instead of asking for the copies of $K_n$ to be edge disjoint we only ask for their intersections to be triangle free, or to contain at most one edge.
In [Er81] offered \$1000 for a proof or disproof even just in the special case when $k=3$, which he expected 'contains the whole difficulty'. He also wrote 'I really do not see why this question is so difficult'.
The usual focus is on the regime where $k=O(1)$ is fixed (say $k=3$) and $n$ is large, although for the opposite regime Kostochka, Rödl, and Talysheva [KRT99] have shown \[f(n,k)=(1+O_n(k^{-1/2^n}))k^n.\]
Another stronger conjecture would be that the hypothesis $\lvert A\cap [1,N]\rvert \gg N^{1/2}$ for all large $N$ suffices.
Erdős and Sárközy conjectured the stronger version that if $A=\{a_1<a_2<\cdots\}$ and $B=\{b_1<b_2<\cdots\}$ with $a_n/b_n\to 1$ are such that $A+B=\mathbb{N}$ then $\limsup 1_A\ast 1_B(n)=\infty$.
See also [40].
An explicit construction was given by Jain, Pham, Sawhney, and Zakharov [JPSZ24].
Erdős and Rényi have constructed, for any $\epsilon>0$, a set $A$ such that \[\lvert A\cap \{1\ldots,N\}\rvert \gg_\epsilon N^{1/2-\epsilon}\] for all large $N$ and $1_A\ast 1_A(n)\ll_\epsilon 1$ for all $n$.
This was solved in the affirmative if the minimum degree is larger than some absolute constant by Liu and Montgomery [LiMo20] (therefore disproving the above stronger conjecture of Erdős and Gyárfás). Liu and Montgomery prove a much stronger result: if the average degree of $G$ is sufficiently large then there is some large integer $\ell$ such that for every even integer $m\in [(\log \ell)^8,\ell]$, $G$ contains a cycle of length $m$.
An infinite tree with minimum degree $3$ shows that the answer is trivially false for infinite graphs.
In [Er81] it is further conjectured that \[\max_{md\leq x}\left\lvert \sum_{1\leq k\leq m}f(kd)\right\rvert \gg \log x.\]
In [Er85c] Erdős also asks about the special case when $f$ is multiplicative.
Erdős was 'almost certain' that if $A$ is the set of powers of $2$ then no such $c$ exists (although he conjectured 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.
Rödl [Ro82] has proved this for hypergraphs, and also proved there is such a graph (with chromatic number $\aleph_0$) if $f(n)=\epsilon n$ for any fixed constant $\epsilon>0$.
It is open even for $f(n)=\sqrt{n}$. Erdős offered \$500 for a proof but only \$250 for a counterexample. This fails (even with $f(n)\gg n$) if the graph has chromatic number $\aleph_1$ (see [111]).
A shorter and simpler proof of an upper bound of the strength $4-c$ for some constant $c>0$ (and a generalisation to the case of more than two colours) was given by Balister, Bollobás, Campos, Griffiths, Hurley, Morris, Sahasrabudhe, and Tiba [BBCGHMST24].
This problem is #3 in Ramsey Theory in the graphs problem collection.
In [Er69b] Erdős asks for even a construction whose largest clique or independent set has size $o(n^{1/2})$, which is now known.
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 $C>0$.
This problem is #4 in Ramsey Theory in the graphs problem collection.
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].
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].
It is easy to see that $h_c(n) \ll_c n^{1/2}$, and Erdős originally suggested that perhaps a similar lower bound $h_c(n)\gg_c n^{1/2}$ holds. Zach Hunter has pointed out that this is false, even replacing $>3$ points on each line with $>k$ points: consider the set of points in $\{1,\ldots,m\}^d$ where $n\approx m^d$. These intersect any line in $\ll_d n^{1/d}$ points, and have $\gg_d n^2$ many pairs of points each of which determine a line with at least $k$ points. This is a construction in $\mathbb{R}^d$, but a random projection into $\mathbb{R}^2$ preserves the relevant properties.
This construction shows that $h_c(n) \ll n^{1/\log(1/c)}$.
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.
Wagner [Wa88] proves, for $n\geq 3$, the existence of such polynomials with \[\mu(A) \ll_\epsilon (\log\log n)^{-1/2+\epsilon}\] for all $\epsilon>0$.
In [Er82c] he further conjectures that, if $m,k$ are fixed and $n$ is sufficiently large, then there must be at least $k$ distinct primes $p$ such that \[p\mid m(m+1)\cdots (m+n)\] and yet $p^2$ does not divide the right-hand side.
See also [364].
See also [3].
The existence of such progressions for small $k$ has been verified for $k\leq 10$, see the Wikipedia page. It is open, even for $k=3$, whether there are infinitely many such progressions.
See also [219].
The best known upper bounds for $r_k(N)$ are due to Kelley and Meka [KeMe23] for $k=3$, Green and Tao [GrTa17] for $k=4$, and Leng, Sah, and Sawhney [LSS24] for $k\geq 5$. An asymptotic formula is still far out of reach, even for $k=3$.
In [Er73] mentions an unpublished proof of Haight that \[\lim \frac{\lvert A\cap [1,x]\rvert}{x}=0\] holds if the elements of $A$ are independent over $\mathbb{Q}$.
See also [858].
See also [544].
This problem is #5 in Ramsey Theory in the graphs problem collection.
Is there some constant $c>0$ such that \[R_3(n) \geq 2^{2^{cn}}?\]
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 \[c=\frac{48-14e}{55-16e}=0.864137\cdots.\]
The theory of Pell equations implies that there are infinitely many pairs $n,d$ with $(n,d)=1$ such that $n(n+d)(n+2d)$ is a square.
Considering the question of whether the product of an arithmetic progression of length $k$ can be equal to an $\ell$th power:
Jakob Führer has observed this is possible for integers in general, for example $(-6)\cdot(-1)\cdot 4\cdot 9=6^3$.