1- Every prime divisor of $2^p-1$, where $p$ is a prime, is greater than $p$.

2- Every prime divisor of $\frac{1}{3}(2^p+1)$, where $p$ is a prime $> 3$, is greater than $p$.

I'm curious to know if there is other expression with primes that is divisible by a greater prime. I think this is fantastic because it's another way to prove that there are infinitely many prime numbers.

  • $\begingroup$ If $p$ is an odd prime, then $2^p\equiv 2\pmod{3}$, so $\frac13(2^p-1)$ is not an integer. $\endgroup$ – G Tony Jacobs May 17 '17 at 17:57
  • $\begingroup$ You're right. It's $2^p+1$. I corrected. $\endgroup$ – Daniel Cintra May 17 '17 at 21:43
  • $\begingroup$ @DanielCintra do you have a reference for the proof of #2? I need it for another problem. $\endgroup$ – Χpẘ May 22 '17 at 18:16
  • $\begingroup$ @Xpw here is the reference: link $\endgroup$ – Daniel Cintra May 23 '17 at 21:18

Every prime divisor of $p!+1$ is greater than $p$. That's one well-known example, and it would be possible to create many variations on it.

  • $\begingroup$ That's true! Thanks! The only problem is that $p!$ grows really fast, but it is a nice expression $\endgroup$ – Daniel Cintra May 17 '17 at 18:01
  • $\begingroup$ If $p!$ is too big, you could also just use the product of all the primes up to and including $p$. $\endgroup$ – G Tony Jacobs May 17 '17 at 18:02

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