# Question about $2p-1$ and $2p+1$, where $p$ is a prime.

Let $x+1$ be any prime greater than $3$.

By Bertrand's Postulate, there is at least one prime between $\frac{x}{2}$ and $x$.

Let $\{p_1,p_2,\dots, p_n\}$ be the primes between $\frac{x}{2}$ and $x$

In all cases that I've checked, there exists $p_i$ in this set where either $2p_i-1$ or $2p_i+1$ is a prime.

Is this always true?

• I am using Bertrand's Postulate to show that $\{p_1, p_2, \dots, p_n\}$ is not an empty set. – Larry Freeman Jan 7 '15 at 5:47
• There are no counterexamples for $x/2$ under one million – Jorge Fernández Hidalgo Jan 7 '15 at 5:58
• Believable, but I doubt there is a known proof. After all, it is not even known whether there are infinitely many Sophie Germain primes. – André Nicolas Jan 7 '15 at 5:59
• Although sophie germain primes are a little more restrictive than these primes, I called these supa-primes. Although I see your point, if there was a finite amount of primes so that $2p+1$ is prime there would probably be a finite amount of primes so that $2p-1$ is prime. Where here probably means it would make sense to my brain. – Jorge Fernández Hidalgo Jan 7 '15 at 6:02

Pretty sure it is open. We know Sophie Germain is open, and this mentions a theorem of Chen regarding the $2 \cdot p - 1$ case which is strictly weaker (suggesting that infinite $2 \cdot p - 1$ primes is open). Unless there is some way we can prove the disjunction without proving either case individually then this is also open.