After reading the answer to this question, I wish to ask the following question:
Let $u,v,m,n \in \mathbb{N}$ satisfy:
(1) $m \neq n$.
(2) $vmn \neq 0$ (namely, perhaps $u=0$, but each of $\{v,m,n\}$ is non-zero).
(3) $\gcd(u,m)=1$.
(4) $\gcd(v,n)=1$.
Is it possible to find $1 < L \in \mathbb{N}$, such that the $\gcd(u+Lm,v+Ln)=1$?
Thank you very much!
Edit: After reading the answer, if I am not wrong, the following claims are true:
(i) If $m < n$, then we can dismiss of condition $(3)$ (condition $(4)$ yields $v+Ln$, for large enough $L$, is prime).
(ii) If $m < n$, and we know that condition $(3)$ is satisfied but do not know if condition $(4)$ is satisfied or not, then for large enough $L$, $\gcd(u+Lm,v+Ln) \in \{1,p\}$, for some prime $p$.
Suppose $0<m<n$.
There is $L_0\in\mathbb{N}$ such that $$0<u+Lm < v+Ln$$for any $L\geq L_0$.
By Dirichlet's Theorem on Primes in Arithmetic Progression, there is $L\geq L_0$ such that $v+Ln=p$ for some prime number $p$.
Then this prime $p$ cannot divide $u+Lm$.