20200825, 16:04  #34  
Sep 2009
3574_{8} Posts 
Quote:
There are infinitely many primes of form 3n+2, none of which are Mersenne primes. Lemma 1: No Mersenne numbers are of form 3n+2. 2^n is never divisible by 3 so 2^n mod 3 is 1 or 2 (1 if n is even, 2 if n is odd). Hence 2^n1 mod 3 is 0 or 1. Lemma 2: There are infinitely many primes of form 3n+2. Proof by contradiction. Assume there are a finite number, p1, p2, ... pn. Contruct M = 3*p1*p2*...*pn1. M is not divisible by any of p1, p2, ... pn because of the 1 and M mod 3 will be 2. The product of 2 numbers of the form 3n+1 will also be of form 3n+1 so any number of form 3n+2 must have at least 1 prime factor of form 3n+2 (possibly itself). So M must have a prime factor of form 3n+2 that is not in p1, p2, ... pn. Hence there are an infinite number of such primes, none being Mersenne primes. Chris 

20200825, 17:04  #35 
Aug 2006
17·349 Posts 
Another different proof (a 'sledgehammer proof'):
There are x/log x + O(x/log^2 x) primes up to x by the Prime Number Theorem. There are log(x)/log(2) + O(1) Mersenne numbers up to x, and hence O(log x) Mersenne primes up to x. Hence there are x/log x + O(x/log^2 x) nonMersenne primes up to x. 
20200826, 08:08  #36 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
16BF_{16} Posts 
Another proof.
Yitang Zhang showed that there are an infinite number of primes separated by less than 10,000,000. The current best known proven bound has been improved to 246. So If we choose any pair of two large primes separated by 246, if one of them is a Mersenne number, then the other one won't be a Mersenne number. Last fiddled with by retina on 20200826 at 08:23 
20200907, 00:53  #37 
Feb 2017
Nowhere
3×5×239 Posts 
Adding to xilman's note that the Fermat numbers are pairwise relatively prime, I note that Lucas proved that for k > 1, every prime factor of the Fermat number F_{k} = 2^(2^{k}) + 1 is congruent to 1 modulo 2^{k+2}.
Taking k = M_{p}  2 for p > 2, we have that the smallest prime factor of F_{M_p  2} is greater than M_{p} (and, being congruent to 1 (mod 4), is not a Mersenne number). We can thus (in theory, if not in practice) easily specify, for each M_{p}, p > 2, a nonMersenne prime greater than M_{p}, and these primes will be different for different p. For example, the smallest prime factor of F_{M_82589933  2} is a nonMersenne prime which is greater than M_{82589933}. 
20200907, 02:44  #38  
Aug 2006
17·349 Posts 
Quote:


20200907, 12:45  #39  
Feb 2017
Nowhere
3·5·239 Posts 
Quote:
OK, PariGP, give me nextprime(2^82589933  1). Huh, I've never seen an error message like that before: Being prone to mentally transposing things inadvertently, I realized that you can also ask about the "largest Mersenne nonprime," i.e. the largest M_{p} with prime exponent for which a proper factor is known. The largest "Mersenne nonprime" I know of for sure is with p = 2618163402417 ยท 2^1290000  1, the largest known Sophie Germain prime. Since p is congruent to 3 (mod 4), q = 2*p + 1 is a prime congruent to 7 (mod 8), whence 2^{(q1)/2} == 1 (mod q), i.e. q divides M_{p}. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
(M48) NEW MERSENNE PRIME! LARGEST PRIME NUMBER DISCOVERED!  dabaichi  News  571  20201026 11:02 
probable largest prime.  sudaprime  Miscellaneous Math  11  20180205 08:10 
Largest known prime  Unregistered  Information & Answers  24  20081213 08:13 
Largest 64 bit prime?  amcfarlane  Math  6  20041226 23:15 
need Pentium 4s for 5th largest prime search (largest proth)  wfgarnett3  Lounge  7  20021125 06:34 