20170808, 17:05  #1 
"Sam"
Nov 2016
311 Posts 
Mersenne Primes p which are in a set of twin primes is finite?
I want to bring up the of the Mersenne Prime 2^n1 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^p1 and 2^p3 are prime).
First off 2^n1 and 2^n+1 cannot both be prime for n > 2, therefore we only focus on 2^n1 and 2^n3 both being primes. Second, if 2^n1 and 2^n3 are both prime, n must be prime because if n is composite = ab, then 2^n1 = (2^a1)*(1 + 2^a + 2^(2*a) + 2^(3*a) .... + 2^(b*aa) Third, if 2^n1 and 2^n3 are both prime, n = 1 (mod 4), because if n = 3 (mod 4), 2^n3 = 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^n1 and 2^n3 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. 
20170808, 20:31  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20170810, 13:36  #3  
Feb 2017
Nowhere
DA1_{16} Posts 
Quote:
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) Nondivisors 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 qth 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. 

20170810, 13:47  #4  
Sep 2002
Database er0rr
3,413 Posts 
Quote:
I was interested in 2PRP being enough for these PRPs. In general 2^n2^k1. Most general aPRP for: "For integers a>1, s>=0, all r>0, all t>0, odd and irreducible {a^s\times\prod{(a^r1)^t}}1 is aPRP, except for the cases a^2a1 and a2 and a1 and 1." This includes a^22, for which I have done a cursory check for a < 10^13. Last fiddled with by paulunderwood on 20170810 at 13:53 

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