Does there exist an even positive integer $n$ such that, for each prime number $p>2$, $p+n$ is not a prime number?

Question

I don't know if this is a known problem, but I didn't find any similar question.

Let's do some example to explain what I'm searching.

1. Take $n=10$. We have $p=3$ odd prime number and also $p+n = 3+10=13$ prime, so $n=10$ is not valid
2. Take $n=30$. We have $p=7$ odd prime number and also $p+n=7+30=37$ prime, so $n=30$ is not valid
3. Take $n=138$. We have $p=11$ odd prime number and also $p+n=11+138=149$ prime, so $n=138$ is not valid

I wonder if there exist an even positive integer $n$ such that, for each odd prime number $p$, $p+n$ is not itself a prime.

The challenge is to prove that such an integer must exist, or prove that it cannot exist at all.

Any answers or comments will be appreciated.

Answer 1

The answer is no, if you believe the Hardy-Littlewood $k$-Tuple Conjecture. Let $\pi_{k}(x)$ denote the number of primes $p\leq x$ such that $p+2k$ is also prime. Then the conjecture predicts $$ \pi_{k}(x) \sim C(k) \int_{2}^{x} \frac{dt}{(\ln t)^{2}}, ​ $$ where $$ C(k) = 2\prod_{p>2} \frac{p(p-2)}{(p-1)^2} \prod_{\substack{p\mid k\\ p>2}} \frac{p-1}{p-2}. $$

Answer 2

The k-tupe conjecture states that for a fixed even number $k$, they are infinitly primes $p$ satisfying $p$ and $p+k$ are both primes.

In your case you asked if exists an even number $n$ with $n+p$ it not prime for all $p\in\mathbb{P}$, well the answer depending on the k-tuple conjecture is no !