20200817, 18:37  #23  
Aug 2020
3^{2} Posts 
Quote:
But before I doare you familiar with the traditional proof that there are infinitely many primes? My proof is ultimately just a small revision of the traditional proof that there are infinitely many primes to show that there must also be infinitely many nonMersenne primes. So I want to be sure you are familiar with that traditional proof to make sure we are starting from the same place. Are you? 

20200817, 21:32  #24 
"Jeppe"
Jan 2016
Denmark
2^{5}·5 Posts 
There are infinitely many primes of the form 4*x + 1. (This is more elementary than the more general Dirichlet theorem on arithmetic progressions, and certainly more elementary than Bertrand's postulate and the prime number theorem.)
Since no Mersenne prime is of form 4*x + 1, it follows that infinitely many primes are not Mersenne primes (A138837). (Not to be confused with the unsolved problem: Are there infinitely many primes p such that 2^p  1 is a composite Mersenne number? (A054723)) Edit: I also like literka's second proof above, about the convergence of the reciprocals of the Mersenne primes. /JeppeSN Last fiddled with by JeppeSN on 20200817 at 21:46 
20200817, 21:37  #25  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×4,357 Posts 
Quote:


20200817, 22:26  #26 
"Jane Sullivan"
Jan 2011
Beckenham, UK
11101010_{2} Posts 

20200817, 22:39  #27  
"Rashid Naimi"
Oct 2015
Remote to Here/There
3603_{8} Posts 
Quote:
https://primes.utm.edu/notes/proofs/...e/euclids.html And please feel free to substitute 2^n1 with Mn to make things easier to digest and easier to type. Thanks for your patience. 

20200818, 13:22  #28 
"Rashid Naimi"
Oct 2015
Remote to Here/There
783_{16} Posts 
So I had a few days to think about this.
If we assume that there are a finite n nonMersenne Primes, then there can be a finite n' Mersenne primes where n'<=n And There can be a finite n'' double Mersenne primes where n''<=n' There can be a finite n''' triple Mersenne primes where n'''<=n'' There can be a finite n'''' quadrupole Mersenne primes where n'''<=n'' .... So unless we can somehow eliminate the "=" from (more than n of) the above inequalities, there could be argumentaly finite nonMersenne primes. ETA Now if multiply any subset of these primes together and add or subtract 1 from it we will end up with a number that is coprime to all of them but not necessarily with prime factors that are smaller than the largest one of them. I still fail to see the contradiction here. Last fiddled with by a1call on 20200818 at 13:41 
20200818, 15:40  #29  
Aug 2020
1001_{2} Posts 
Quote:
Let's find all primes strictly less than MM13 (that is a double Mersenne prime), multiply them together, add 1, and see what the divisor's areexactly the same approach that Euclid used. Well, what are those primes? They include the nondouble Mersenne primes 2,3,5,31. We were assuming those are the only nondouble Mersenne primes in existence. Of course that assumption is false but this is just an example to illustrate the concept. Then there are the potential double Mersenne primes MM2,MM3,...,MM12. Those numbers2,3,5,31,MM2,MM3,...,MM12 are the only possible primes less than MM13. In fact, some of them aren't prime, but if we multiply all the primes together and add 1 (call that number L), we will get the following: L<=1+2*3*5*31*MM2*MM3*...*MM12 A little bit of arithmetic will then allow us to say: L<=2^2^2*2^2^3*2^2^4*...*2^2^12 Remember that when you multiply two powers with the same base, you get the base to the sum of the exponents. So this becomes: L<=2^(2^2+2^3+2^4+...+2^12) Adding up the elements of a geometric series this becomes: L<=2^(2^134) So by the definition of MM13, L<MM13 Remember that L was defined to be the product of all primes strictly less than MM13 plus one. So L must be either prime itself, or have a prime divisorin either case the prime involved must be at least MM13 based on the way L was defined. L cannot have any prime divisor less than MM13. But we just showed L<MM13, so we have a contradiction. 

20200818, 15:46  #30  
Aug 2020
3^{2} Posts 
Quote:
If there are only finitely many nonMersenne primes, then there must only finitely many Mersenne primes that aren't double. If there are only finitely many nonMersenne primes, then there must only finitely many Mersenne primes that are double but not triple. If there are only finitely many nonMersenne primes, then there must only finitely many Mersenne primes that are triple but not quadruple. And so on. However it isn't necessary for my proof to go beyond talking about double Mersenne primes. We don't need to invoke triple, etc, Mersenne primes for my proof to work. The key argument that I made is that double Mersenne primes grow fast enough for Euclid's original argument, with some modification, to work. Moving up to triple Mersenne primes and beyond doesn't really add anything of value to my proof. 

20200818, 16:00  #31  
"Rashid Naimi"
Oct 2015
Remote to Here/There
3×641 Posts 
Quote:
If we assume that there are a finite n nonMersenne Primes, then there can be a finite n' SingleMersennePrimes (where the exponent is not a Mersenne number/Prime) where n'<=n (We could get rid of the "=" here because we know not it in this case) And There can be a finite n'' double Mersenne primes where n''<=n' There can be a finite n''' triple Mersenne primes where n'''<=n'' There can be a finite n'''' quadrupole Mersenne primes where n'''<=n'' .... So unless we can somehow eliminate the "=" from more than n1 of the above infinite inequalities, there could be argumentatively finite nonMersenne primes. ETA Now if multiply any subset of these primes together and add or subtract 1 from it we will end up with a number that is coprime to all of them but not necessarily with prime factors that are smaller than the largest one of them. I still fail to see the contradiction here. Last fiddled with by a1call on 20200818 at 16:01 

20200818, 16:11  #32  
"Rashid Naimi"
Oct 2015
Remote to Here/There
3×641 Posts 
Quote:
Last fiddled with by a1call on 20200818 at 16:17 

20200818, 16:27  #33  
Aug 2020
9_{10} Posts 
Quote:
And that observation does in fact lead to a proofmy proof in fact. I've now invested some time in making several follow up posts trying to explain my proof to you in more detail. If you insist on investing your own time not in trying to understand my proof but instead in exploring your own direction (which isn't fruitful) that is on you. 

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