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

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  • $\begingroup$ Set $(u, v, m, n) = (3, 4, 1, 1)$; there will be no such $L$. $\endgroup$ – user44191 Sep 25 '18 at 23:37
  • $\begingroup$ 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). $\endgroup$ – user237522 Sep 25 '18 at 23:40
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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?".

It would seem reasonable to think that the latter type of question is easier than the former, but in general that doesn't seem to be the case. Hence this is an open problem.

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  • $\begingroup$ Thank you very much for your answer! (So my question is equivalent to a generalization of the twin primes conjecture?). Actually, I have a weaker version of my question, which I will soon add; hopefully, it will have a known answer. $\endgroup$ – user237522 Sep 25 '18 at 17:39
  • $\begingroup$ Actually, this more like a conjecture of Schinzel about simultaneous solution in primes of a system of polynomial forms. Gerhard "Is Thinking Of Conjecture H?" Paseman, 2018.09.25. $\endgroup$ – Gerhard Paseman Sep 25 '18 at 18:33
  • $\begingroup$ @GerhardPaseman, thank you for the comment! (I did not understand your second sentence). $\endgroup$ – user237522 Sep 25 '18 at 19:12
  • $\begingroup$ Indeed. Wikipedia has articles on a hierarchy of conjectures, including twin prime, Bateman-Horn, Schinzel Hypothesis H, and others. I think you will be interested in all of them. Gerhard "Almost Had The Name Right" Paseman, 2018.09.25. $\endgroup$ – Gerhard Paseman Sep 25 '18 at 19:21
  • $\begingroup$ Thank you for the clarification. I will take a look at the Wikipedia articles that you mentioned. $\endgroup$ – user237522 Sep 25 '18 at 19:32

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