# Almost simultaneous Dirichlet's theorem on arithmetic progressions

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 $$k=2$$, $$b_1=b_2=1$$.

• Set $(u, v, m, n) = (3, 4, 1, 1)$; there will be no such $L$. – user44191 Sep 25 '18 at 23:37
• Thank you for your comment. (Indeed, this counterexample is also mentioned by Wojowu in one of the comments of the question that I have added in the edit). – user237522 Sep 25 '18 at 23:40

This is simply a reformulation of (a weaker version of) the prime pairs conjecture, a generalization of the twin primes conjecture. For example, when $$m=n=1$$ and $$u=1$$, $$v=3$$, this is exactly the twin primes conjecture, except that such conjectures are usually formulated as "are there infinitely many?" rather than "does there exist one?".