# For any $x\in \mathbb{N}$ does there exist $m\in \mathbb{N}$ such that $2x+1+2m, 2x+1+4m$ are both prime?

Could someone please give me a proof (or counter example) for this (I believe it is true):

For any $x$ (Whole Number) there exists some $m$ (Also Whole) such that $2x+1+2m$ and $2x+1+4m$ are both prime.

An equivalent is for any odd $n$ there exists some even $y$ such that $n+y$ and $n+2y$ are both prime. I am pretty sure a proof would be based around Dirichlet's Theorem.

• @ritwiksinha Depends on your definition of $\Bbb N$. It is in no way established convention that $0\notin \Bbb N$. – Arthur Jul 9 '16 at 17:04
• An equivalent question is: for any odd $n$, does there exist a solution to the linear equation $2y-z=n$ where both $y$ and $z$ are prime. This is probably as hard as the Goldach conjecture (which asserts that for any even $n$, there exists a solution to the linear equation $y+z=n$ where both $y$ and $z$ are prime). – Greg Martin Jul 9 '16 at 17:08
• Green Tao theorem implies that there are infinitely many prime numbers $n$ such that there is even $y$ with $n$, $n+y$, and $n+2y$ are all primes. – Sungjin Kim Jul 9 '16 at 17:08
• Dickson's conjecture seems to imply the claim. – Peter Jul 9 '16 at 17:11
• Not quite yet. Green Tao theorem implies infinitely many 3-term arithmetic progression consisted of all primes. That's why I stated infinitely many prime numbers $n$. But, OP's claim is for all odd $n$. – Sungjin Kim Jul 9 '16 at 17:11