2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I am using Bertrand's Postulate to show that $\{p_1, p_2, \dots, p_n\}$ is not an empty set. $\endgroup$ – Larry Freeman Jan 7 '15 at 5:47
  • $\begingroup$ There are no counterexamples for $x/2$ under one million $\endgroup$ – Jorge Fernández Hidalgo Jan 7 '15 at 5:58
  • 4
    $\begingroup$ Believable, but I doubt there is a known proof. After all, it is not even known whether there are infinitely many Sophie Germain primes. $\endgroup$ – André Nicolas Jan 7 '15 at 5:59
  • 1
    $\begingroup$ 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. $\endgroup$ – Jorge Fernández Hidalgo Jan 7 '15 at 6:02
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.