mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2017-08-08, 17:05   #1
carpetpool
 
carpetpool's Avatar
 
"Sam"
Nov 2016

28010 Posts
Post Mersenne Primes p which are in a set of twin primes is finite?

I want to bring up the of the Mersenne Prime 2^n-1 being the second prime p+2 in a twin prime pair {p, p+2} are there finitely many Mersenne Primes which hold this condition (this is the same as primes p such that 2^p-1 and 2^p-3 are prime).

First off 2^n-1 and 2^n+1 cannot both be prime for n > 2, therefore we only focus on 2^n-1 and 2^n-3 both being primes.

Second, if 2^n-1 and 2^n-3 are both prime, n must be prime because if n is composite = ab, then 2^n-1 = (2^a-1)*(1 + 2^a + 2^(2*a) + 2^(3*a) .... + 2^(b*a-a)

Third, if 2^n-1 and 2^n-3 are both prime, n = 1 (mod 4), because if n = 3 (mod 4), 2^n-3 = 0 (mod 5) cannot be a prime. This follows from 2^(4*n+3) = 3 (mod 5) - 3 = 0 (mod 5).

The only known exponents for which 2^n-1 and 2^n-3 are 3 and 5 (up to the Same Limit the Mersenne Numbers were tested). This is conjectured to be finite unless anyone brings up an arguments as to maybe why not.

Are there any more restrictions to this? Thanks for help.
carpetpool is offline   Reply With Quote
Old 2017-08-08, 20:31   #2
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

Quote:
Originally Posted by carpetpool View Post
I want to bring up the of the Mersenne Prime 2^n-1 being the second prime p+2 in a twin prime pair {p, p+2} are there finitely many Mersenne Primes which hold this condition (this is the same as primes p such that 2^p-1 and 2^p-3 are prime).

First off 2^n-1 and 2^n+1 cannot both be prime for n > 2, therefore we only focus on 2^n-1 and 2^n-3 both being primes.

Second, if 2^n-1 and 2^n-3 are both prime, n must be prime because if n is composite = ab, then 2^n-1 = (2^a-1)*(1 + 2^a + 2^(2*a) + 2^(3*a) .... + 2^(b*a-a)

Third, if 2^n-1 and 2^n-3 are both prime, n = 1 (mod 4), because if n = 3 (mod 4), 2^n-3 = 0 (mod 5) cannot be a prime. This follows from 2^(4*n+3) = 3 (mod 5) - 3 = 0 (mod 5).

The only known exponents for which 2^n-1 and 2^n-3 are 3 and 5 (up to the Same Limit the Mersenne Numbers were tested). This is conjectured to be finite unless anyone brings up an arguments as to maybe why not.

Are there any more restrictions to this? Thanks for help.
the restriction for exponents of form 1 mod 4 also comes from mersenne primes greater than 7, are 7 and 31 mod 120, the restriction that 2^n-3 also be prime restricts to exponents that give rise to 31 mod 120.
science_man_88 is offline   Reply With Quote
Old 2017-08-10, 13:36   #3
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

3,251 Posts
Default

Quote:
Originally Posted by carpetpool View Post
I want to bring up the of the Mersenne Prime 2^n-1 being the second prime p+2 in a twin prime pair {p, p+2} are there finitely many Mersenne Primes which hold this condition (this is the same as primes p such that 2^p-1 and 2^p-3 are prime).
The question of 2^p - 3 and 2^p - 1 both being prime, doesn't seem very interesting. After all, 2^p - 1 is very seldom prime.

The question of when 2^n - 3 alone might be prime may be of some interest in its own right. It wouldn't surprise me at all if someone had compiled a factor table for n into the hundreds, and a list of pseudoprimes for larger n's.

By way of keeping in practice with this sort of thing, I note the following:

Values of n < 1000 for which 2^n - 3 tests as a pseudoprime:

n = 3, 4, 5, 6, 9, 10, 12, 14, 20, 22, 24, 29, 94, 116, 122, 150, 174, 213, 221, 233, 266, 336, 452, 545, 689, 694, 850

Prime values of 2^n - 3 seem to occur for composite n much more often than for prime n.

2^p - 3 is (pseudo)prime, but 2^p - 1 is not, for the primes p = 29 and 233.

Recurrent prime divisors p < 100 of 2^n - 3. Primes followed by an asterisk only divide 2^n - 3 for composite exponents n.

5 (n = 4*k + 3)
11* (n = 10*k + 8)
13* (n = 12*k + 4)
19 (n = 18*k + 13)
23 (n = 11*k + 8)
29 (n = 28*k + 5)
37* (n = 36*k + 26)
47 (n = 23*k + 19)
53 (n = 52*k + 17)
59* (n = 58*k + 50)
61* (n = 60*k + 6)
67* (n = 66*k + 39)
71 (n = 35*k + 16)
83* (n = 82*k + 72)
97 (n = 48*k + 19)

Non-divisors p of 2^n - 3: There are the obvious cases p = 2 and 3. For p > 3, we have the following: If 2 is a q-th power residue of the prime p but 3 is not, then no power of 2 can be congruent to 3 (mod p). In the case q = 2 we can say this is the case for p congruent to 7 or 17 (mod 24), e.g. p = 7, 17, 31, 41, 79, 89. An example with q = 3 is p = 43.
Dr Sardonicus is offline   Reply With Quote
Old 2017-08-10, 13:47   #4
paulunderwood
 
paulunderwood's Avatar
 
Sep 2002
Database er0rr

3·29·37 Posts
Default

Quote:
Originally Posted by Dr Sardonicus View Post
The question of when 2^n - 3 alone might be prime may be of some interest in its own right. It wouldn't surprise me at all if someone had compiled a factor table for n into the hundreds, and a list of pseudoprimes for larger n's.
http://www.primenumbers.net/prptop/s...&action=Search

I was interested in 2-PRP being enough for these PRPs. In general 2^n-2^k-1. Most general a-PRP for:

"For integers a>1, s>=0, all r>0, all t>0, odd and irreducible {a^s\times\prod{(a^r-1)^t}}-1 is a-PRP, except for the cases a^2-a-1 and a-2 and a-1 and -1."

This includes a^2-2, for which I have done a cursory check for a < 10^13.

Last fiddled with by paulunderwood on 2017-08-10 at 13:53
paulunderwood is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Distribution of Mersenne primes before and after couples of primes found emily Math 34 2017-07-16 18:44
What to do with 16 digit twin, non-Mersenne primes? RienS Miscellaneous Math 15 2014-11-18 01:48
Twin Primes Hugh Math 64 2011-01-26 08:45
twin primes! sghodeif Miscellaneous Math 9 2006-07-19 03:22
Twin Mersenne Primes jinydu Math 23 2004-06-11 00:35

All times are UTC. The time now is 07:32.

Fri Jun 5 07:32:40 UTC 2020 up 72 days, 5:05, 0 users, load averages: 1.01, 1.02, 1.03

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.