Denote the set of prime numbers by $P$. Let $u,v,m,n \in \mathbb{N}-\{0\}$ satisfy: $m \leq n$, $\gcd(u,m)=1$ and $\gcd(v,n)=1$.

Is it possible to find $L \in \mathbb{N}$, such that $u+Lm \in P$ and $v+Ln \in P$? (different primes, probably).

Of course, by Dirichlet's theorem on arithmetic progressions, we can find $L$ such that one of $\{u+Lm,v+Ln\}$ is prime. But are those $L$'s 'dense enough' to guarantee that both $\{u+Lm,v+Ln\}$ are primes?

A similar, hopefully a less difficult question, is as follows:

Is it possible to find $L \in \mathbb{N}$, such that:

(i) $u+Lm < v+Ln$;

(ii) $u+Lm = p \in P$;

(iii) $p$ does not divide $v+Ln$?

From the answer to this question, it is clear that there exists $L \in \mathbb{N}$ such that: (i) $u+Lm < v+Ln$ and (ii) $u+Lm = p \in P$. The problem is how to guarantee that $p$ will not divide $v+Ln$?

Thank you very much!

**A remark about the answer:** Let us concentrate on the special case mentioned in the answer: $m=n=1$, $u=1$, $v=3$, so $A:=1+L$ and $B:=3+L$. There is a major difference between asking: 'Are there infinitely many $L$'s such that $A,B \in P$' (very very difficult question) and 'Does there exist $L$ such that $A,B \in P$',
which is a very easy question that has a positive answer, for example $L=2$ yields $(A,B)=(3,5) \in P^2$. Therefore, I expect that my (first) question has a positive answer, or maybe I am missing something, and even finding only one such $L$ is a difficult task? (I do not require that $A$ and $B$ will be greater than a given number).

**Edit:** I have now noticed that my (first) question has already been asked before on MO, here. Now I see that my (first) question is just Dickson's conjecture with $k=2$, $\gcd(a_1,b_1)=1$ and $\gcd(a_2,b_2)=1$ (in Wikipedia notations). In order to avoid 'trivial' counterexamples such as the one mentioned in the comments (namely, $3+L$ and $4+L$, with always one of the two necessarily not a prime number, since it is even), additional conditions
("congruence condition") must be imposed.

This is a relevant paper that I have now found (it is from 2015); interestingly, it mentions a result of Maynard-Tao that almost answers Dickson's Conjecture in case $k=2$, $b_1=b_2=1$.