20150531, 00:33  #1 
Jul 2014
2^{2}·3·37 Posts 
Good mathematical tinkerers are in the Zen room
Hi,
we all probably know that 2^p 1 are mostly composite but I'd like to know a reason why some are prime. I guess a proof by contradiction might prove this. Can someone suggest anything? I'd be very grateful for any suggestions, thoughts. 
20150531, 01:01  #2 
∂^{2}ω=0
Sep 2002
República de California
2×13×443 Posts 
Speaking roughly at the gradeschool math level:
 They're all odd and no 'these cannot possibly be prime' condition (as e.g. for oddcompositeexponent 2^n1) applies to them, thus (naively) have at least the same odds of being prime as a randomly chosen odd number.  Less naively, the special form of the possible factors q of 2^p1 (q = 2*k*p+1) greatly restricts the candidate odd numbers which can possibly divide 2^p1, so in fact a randomly chosen 2^p1 has far greater odds of being prime than a randomly chosen odd numberofnospecialalgebraicform of similar size. Note that there are good heuristic reasons to expect both an oo number of Mprimes to exist and further as to their largescale statistical distribution (the avg Mprime has p roughly 1.4x larger than its immediate predecessor), but no rigorous proofs along these lines. Last fiddled with by ewmayer on 20150531 at 01:02 
20150531, 01:51  #3 
Jul 2014
2^{2}·3·37 Posts 
Can you point me to a proof of the necessary form of a prime factor of a composite, primeexponent Mersenne number??
Thanks for the reply. 
20150531, 02:25  #5 
May 2013
East. Always East.
11·157 Posts 
That is actually a very interesting question.
Like you say, there are examples by contradiction (48 of them, to be precise ) but all that really tells us is that there is no possible proof that there can't be any Mersenne primes. If we took a different number form, say [a * (b + c + 1) + (d * e)  1] for any positive integers a, ..., e > 1, as something I have literally made up on the spot, and ask "can there possibly be any prime numbers of this form?", I don't think there's much we can do about it. It's more like the opposite: Prove that there can't be any primes and assume until then that we can have primes. 
20150531, 13:13  #6  
Nov 2003
2^{2}·5·373 Posts 
What does it mean for a number to be "mostly composite"?
I think you mean that most 2^p1 are composite. If you want to do mathematics you have to learn the correct use of quantifiers. Quote:
theory. No disrespect intended, but you don't know enough math. (1) 2^p1 does not have any algebraic factors that force it to be composite. Consider instead, x^p1, for p prime. Since p = 2k+1 for some k, we have the algebraic factor (x1) of (x^(2k+1)  1). Now, when x equals 2, this algebraic factor has the value 1 whereas for all other x > 1, the algebraic factor is nontrivial. (2) sum from p=2 to infinity of 1/log(2^p1) diverges. The latter implies that there should be infinitely many Mersenne Primes via the Prime Number Theorem. Quote:


20150531, 13:17  #7 
Nov 2003
1110100100100_{2} Posts 

20150531, 15:30  #8  
Jul 2014
2^{2}×3×37 Posts 
Quote:
Quote:
I have problems with plurality. 

20150531, 15:46  #9  
Nov 2003
1110100100100_{2} Posts 
Quote:
Quote:
not apply. And your followup statement is still wrong. There is a difference between "numbers .... are mostly composite" and "most numbers.... are composite". I will offer a hint: In English, your wording would consist of a "misplaced modifier". 

20150531, 15:52  #10 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20150531, 15:54  #11 
Nov 2003
2^{2}×5×373 Posts 

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