Fix a positive integer $n\geq 3$. Does there exist infinitely many sets $S$ of positive integers $\lbrace a_1,a_2,\ldots, a_n$, $b_1,b_2,\ldots,b_n\rbrace$, such that $\gcd (a_1,a_2,\ldots, a_n$, $b_1,b_2,\ldots,b_n)=1$, $\lbrace a_i\rbrace _{i=1}^n$, $\lbrace b_i\rbrace _{i=1}^n$ are arithmetic progressions, and $\prod_{i=1}^n a_i = \prod_{i=1}^n b_i$?
Problem
Source: 2019 China TST Day 1 Q2
Tags: arithmetic sequence, number theory
Hamel
05.03.2019 15:34
Well how about $a_i=a_{i-1}+t$ , $b_i=b_{i-1}+t$ and $b_1=a_1=s$ where $(s,t)=1$?
XbenX
05.03.2019 15:48
@above , there cannot be 2 equal elements in the same set.
SchuesterCHNones
05.03.2019 16:40
@61plus What about P1? Thank you for sharing.
stroller
06.03.2019 23:24
In the case $n=3$ the answer is positive. $2k(2k-1),4k^2+1,2(2k^2+k+1); 2,k(4k^2+1),2(2k^2+k+1)(2k-1)$ works.
liekkas
09.03.2019 07:24
The answer is positive for all $n$, while the construction is rather simple: Let $m>1$, $m \equiv 1 \pmod n$. Thus $m^n \equiv 1 \pmod n$. Let $a_i=1+i \cdot \frac{m^n-1}{n}$, and $b_i=m(1+(i-1)\cdot \frac{m^n-1}{n})$ , $1 \le i \le n$ ,Easy to check is satisfy the condition.
PepsiCola
10.04.2019 02:23
Am I making a mistake? Set $a_{i} = ik + (i-1)$ and $b_i = (n+1- i)k + (n - i)$, which shows there are infinite by letting k towards infinity.
Alapan1729
08.06.2019 05:33
There are infinitely many sets (a_1, a_2, ....a_n) for any n s.t gcd(a_1, a_2, ....a_n)=1 and {a_i} are in arithmetic progression. Taking {b_i} the opposite of (a_1, a_2, ....a_n) we will fill the asked criterion. Thus, there are infinitely many such sets.
m_problemsolver
18.09.2020 08:39
PepsiCola wrote: Am I making a mistake? Set $a_{i} = ik + (i-1)$ and $b_i = (n+1- i)k + (n - i)$, which shows there are infinite by letting k towards infinity. yes, $a_{i}$ and $b_{j}$ can't equal to each other since they are in the same set S.