On the equation $ n_1 ... n_{k+1} =n_{k+2}... n_{2(k+1)} $

Sanying Shi, Michel Weber

Published 2016-07-21Version 1

Let $k\ge 1$ and let $N_{k}(B)$ denote the number of solutions of the equation $n_1\ldots n_{k+1} =n_{k+2}\ldots n_{2(k+1)}$ with unknowns verifying $1\le n_i\le B$,$1\le i\le 2(k+1)$. When $k=1$, very precise estimates are known. When $k=2$, we show that $B^3(\log^3 B)(\log\log B)^{-1}\ll N_2(B )\ll B^3 \log^4 B$. For $k>2$ we also show that $B^{k+1}(\log B)^{[\log_2k]}\ll_k N_{k}(B) \ll_k B^{k+1}\big(\log B\big)^{k^2+2k-2}$. We further study the number of solutions $N(B,F)$ of the equation $n_1n_2=n_3n_4$, where $1\le n_1\le B$, $1\le n_3\le B$, $n_2, n_4\in F$ and $F\subset [1,B]$ is a factor closed set. Let $F= \big\{m=p_1^{\epsilon_1}\ldots p_k^{\epsilon_k}, \ \epsilon_j\in \{0,1\}, 1\le j\le k\big\}$, where $p_1<\ldots<p_k$ are prime numbers, and let $B\ge p_1p_2\cdots p_k$. We prove that $ 2^kB\big(1+2\sum_{n\in F\atop n\neq 1} \frac{1}{2^{ \omega(n)} n } \big)\ \le \ N(B,F)\le B\big(\frac{5}{2}\big)^{k }\big( 1+ C \sum_{n\in F\atop n\neq 1} \frac{1}{5^{ \omega(n)} n } \big)+ 4^k$, where $\omega(n)$ is the prime divisor function and $C$ is a numerical constant.