Let $u,v,m,n \in \mathbb{N}$ satisfy:
(1) $vmn \neq 0$ (namely, perhaps $u=0$, but each of $\{v,m,n\}$ is non-zero).
(2) $u+m$ is odd.
(3) $v+n$ is odd.
Is it possible to find $1 < L \in \mathbb{N}$, such that the greatest common divisor of $u+Lm$ and $v+Ln$ is $1$?
It seems to me that the answer should be positive, but I do not know how to prove this. Perhaps it is better to first concentrate on the case $u=0$, and then consider arbitrary $u \in \mathbb{N}$.
Thank you very much!
The conditions as given allow the case where $u,v,m,n$ are all divisible by some prime $p>2$, for example $(u,m,v,n) = (5,10,15,20)$. This would mean that for any choice of $L$, $g:=\gcd(u+Lm,v+Ln)$ is divisible by $p$ (and so $g\ne 1$).
You could also have the case where $(u,m,v,n) = (ka,kb,\ell a, \ell b)$ leading to $g$ being divisible by $a+Lb$ - for example, $(u,m,v,n) = (15,20,21,28)$.
If the conditions were changed to:
(1) $vmn\ne 0$
(2) $\gcd(u,m)=1$
(3) $\gcd(v,n)=1$
then I think there will be a choice of $L$ that gives $g=1$.