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SOLVED
Let $F_{k}(N)$ be the size of the largest $A\subseteq \{1,\ldots,N\}$ such that the product of no $k$ many distinct elements of $A$ is a square. Is $F_5(N)=(1-o(1))N$? More generally, is $F_{2k+1}(N)=(1-o(1))N$?
Conjectured by Erdős, Sós, and Sárkzözy [ESS95], who proved \[F_2(N)=\left(\frac{6}{\pi^2}+o(1)\right)N,\] \[F_3(N) = (1-o(1))N,\] and also established asymptotics for $F_k(N)$ for all even $k\geq 4$ (for which $F_k(N)\asymp N/\log N$ for all even $k\geq 4$. Erdős [Er38] earlier proved that $F_4(N)=o(N)$ (indeed, if $\lvert A\rvert \gg N$ and $A\subseteq \{1,\ldots,N\}$ then there is a non-trivial solution to $ab=cd$ with $a,b,c,d\in A$.)

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 $F_k(N) \leq (1-c_k+o(1))N$.

OPEN
Let $n_1<n_2<\cdots$ be the sequence of integers which are the sum of two squares. Explore the behaviour of (i.e. find good upper and lower bounds for) the consecutive differences $n_{k+1}-n_k$.
Erdős [Er51] proved that, for infinitely many $k$, \[ n_{k+1}-n_k \gg \frac{\log n_k}{\sqrt{\log\log n_k}}.\] Richards [Ri82] improved this to \[\limsup_{k\to \infty} \frac{n_{k+1}-n_k}{\log n_k} \geq 1/4.\] The constant $1/4$ here has been improved, most lately to $0.868\cdots$ by Dietmann, Elsholtz, Kalmynin, Konyagin, and Maynard [DEKKM22]. The best known upper bound is due to Bambah and Chowla [BaCh47], who proved that \[n_{k+1}-n_k \ll n_k^{1/4}.\]