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.

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

This seems to be equivalent to asking whether every even positive integer is the difference of two primes, rather similar in spirit to Goldbach's conjecture. — Geoff Robinson, Nov 05 '21 at 20:45
Related: Name of a conjecture on difference of prime numbers? — Glorfindel, Nov 05 '21 at 21:13
@GeoffRobinson: An important difference in spirit is that the "expected number" of pairs of primes summing to an even $n$ is finite, while the expected number of pairs of primes with difference $n$ is infinite. — Anthony Quas, Nov 05 '21 at 21:13
It has long been conjectured that every even integer is a difference of two primes, indeed, is infinitely often a difference of two primes, indeed, is infinitely often a difference of two consecutive primes. [exception to the last for the even integer zero] — Gerry Myerson, Nov 05 '21 at 22:16

score 4 · Answer 1 · edited Nov 06 '21 at 02:31

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 $C(k) = 2\prod_{p>2} \frac{p(p-2)}{(p-1)^2} \prod_{\substack{p\mid k\\ p>2}} \frac{p-1}{p-2}.$

score 0 · Answer 2 · answered Nov 05 '21 at 23:43

0

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 !

answered Nov 05 '21 at 23:43

Lagrida Yassine

521

4

The k-tuple conjecture is a much stronger statement than the one you mention. – Wojowu Nov 06 '21 at 00:08

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

2 Answers2

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