Positive intenger $n\geq3$. $a_1,a_2,\cdots,a_n$ are $n$ positive intengers that are pairwise coprime, satisfying that there exists $k_1,k_2,\cdots,k_n\in\{-1,1\}, \sum_{i=1}^{n}k_ia_i=0$. Are there positive intengers $b_1,b_2,\cdots,b_n$, for any $k\in\mathbb{Z}_+$, $b_1+ka_1,b_2+ka_2,\cdots,b_n+ka_n$ are pairwise coprime?
Problem
Source: 2017 China Northern MO, Grade 11, Problem 4
Tags: number theory
XbenX
24.02.2020 18:44
Very nice problem. The answer is yes for $n=3$ and no for $n\geq 4$. Claim. If $n\geq 4$ then there does not exist such $b_i$'s. Proof. If $n\geq 4$, note that at most $1$ of $a_i$'s will be even, and at most one of $b_i$'s will be even since it must also hold $\gcd(b_i,b_j)=1$ for $1\leq i,j \leq n$. So, there exists $i,j$ such that $b_i,b_j$ and $a_i,a_j$ are both odd, and taking $k$ odd, then both $b_i+ka_i$ and $b_j+ka_j$ will be even, which is impossible. Claim $n=3$ works. Proof. Let $p,q,r$ be pairwise relatively prime positive integers, such that $r=p+q$, then we'll show there exist $a,b,c$ such that $a+kp,b+kq,c+kr$ are pairwise relatively prime for all positive integers $k$. First, we will show that there exist $a,b$ such that $\gcd(a+kp,b+kq)=1$ for all $k$ and then take $c=a+b$ and by Euclidean Algorithm the result will hold. Indeed since $\gcd(p,q)=1$ there exist positive integers $X,Y$ such that $pX-qY=1$, now, we'll show that taking $a=Y, b=X$ we have the desired result. Indeed, $-q(Y+kp)+p(X+kq)=-qY+pX=1$, so by the converse of the Bezout theorem we have that $\gcd(a+kp,b+kq)=1$ for all $k$. $\blacksquare$