20100226, 07:13  #232 
Jan 2010
379 Posts 
It is the same, Just wanted to let people see it as a Merssene and 2^n structure which can help.
The "Prooooooof": If there are infintely many primes of the form 2p+1 where p a prime, Then 2(2p+1)+1 primes are infinitely too. By induction: 2^np+2^n1. And now, If you have a prime q=2Q+1 where Q is not nessecarily a prime number, you have a infinitely many primes 4Q+3. By induction 2^nQ+2^n1. So my "conjecture" (which is the same as the old one) is: If there infinitely many primes g=2^nQ+2^n1, there are infinitely many primes m=2^{n+1}Q+2^{n+1}1 too. Which includes 2^{0}p+2{0}1=p, 2^{1}p+2^{1}1=2p+1... infinitely many primes p so infinitely many primes 2p+1. That's all...same thing. Why is it weaker? Last fiddled with by blob100 on 20100226 at 07:58 
20100226, 11:05  #233 
Nov 2003
2^{2}×5×373 Posts 
The original conjecture was that p and 2p+1 are both prime i.o.
The new conjecture is that p = 2^n R 1 and 2p+1 are both prime i.o. n is a free variable and adds an additional degree of freedom. The conjecture did not say that 2^n R  1 and 2^(n+1)R  1 are prime i.o FOR ALL n. (which is why I said to be careful about quantifiers). It could be the case that for SOME n, 2^n R  1 and 2^(n+1) R  1 are both prime only finitely often, yet the conjecture could still be correct. 
20100226, 12:39  #234  
Jan 2010
379 Posts 
Quote:
I'm not talking about your conjecture (with the 2^nR1 which I can't understand). The original conjecture says: there are infinitely many primes q=2p+1. So by that there are infinitely many primes 2q+1 which is 4p+3. By induction: 2^np+2^n1 right? So I say that there are infinitely many primes of the form 2^np+2^n1 where 2^{n1}p+2^{n1}1 is also a prime (by the original conjecture, these two forms contain infinitey many primes). Last fiddled with by blob100 on 20100226 at 12:39 

20100226, 17:12  #235  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Quote:
conjecture, or the "conjecture" that you presented. Quote:
Induction??? What Induction??? You have done nothing even remoting resembling an induction proof. 4p+3 = 2^n p + 2^n1, iff n=2. Your conjecture did not specify a value for n. 

20100226, 21:31  #236 
Jan 2010
379 Posts 
I suggest to leave this conjecture as it is, becuase it won't come with results here.
What I tried to do isn't to prove as you said. Look: 2p+1 is infinitely by the conjecture. So there are infinitely many prmes 4p+3 (because there are infinitely many primes 2p+1). Then there are infinitely many primes 8p+7. ........ 16p+15. ................ ............. BY INDUCTION: 2^{n}p+2^{n}1 primes are infinitely where 2^{n1}p+2^{n1}1 is also a prime. That's all what I said. I only tried to let you see my point of view to this conjecture, with a question if it is true... Ha, and I'll say again, there is no "my" conjecture, I am talking about the original one. And "my conjecture" (as you say) is just another way to show the original. It doesn't specify anything about n, becuase it is the same thing as the original. Anyways, I just wanted to move it to a Merssene and 2^n form.. It seems to me legitem... As I use induction here isn't as a way of proving. I use it as a way to show 2p+1 with p a prime one of the "collection" 2^{n}p+2^{n}1. Last fiddled with by blob100 on 20100226 at 21:40 
20100226, 21:31  #237 
Dec 2008
833_{10} Posts 
@ Tomer (or anyone else who would like to respond):
Let's say you come up with a conjecture that is nontrivial and important. Where would one even publish this conjecture? To my knowledge, there are NO mathematics journals that solely publish conjectures (other than Exp. Math. but the paper would have to be quite polished). If anyone finds a journal that does, then let us know. Last fiddled with by flouran on 20100226 at 21:33 
20100226, 21:37  #238  
Nov 2003
1110100100100_{2} Posts 
Quote:
This isn't even wrong. It is nonsense. And your so called induction is not how an an induction argument works. 

20100226, 21:41  #239 
Jan 2010
379_{10} Posts 
Why not? So which tool did I use?

20100226, 21:52  #240 
Nov 2003
2^{2}·5·373 Posts 
You didn't use ANY tool. You say nothing that made any sense.
An induction argument works as follows (and is equivalent, by the way to the fact that the nonnegative integers are wellordered). Given a function f: Z+ > Z+. We want to prove that a certain statement about f(n) is true. [actually it can be applied to ANY function on [A, oo) to (oo, oo) for any integer A. It need not be strictly on Z+] It can also be applied on a domain (oo, B] but now instead of step 2 below, you must show that if it is true for n, then it is true for n1.] (1) Show that it is true for n = 1. (2) Show that if it is true for an arbitrary integer n, then it is true for n+1. QED. Note that the domain is integers. ALL of the positive integers. Not just a subset of the integers. Your argument did not even have Z+ as its domain. Your function p > 2p+1 was defined on PRIMES as the domain. Not integers. 
20100226, 21:53  #241  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{7}×47 Posts 
Quote:
Perhaps the original conjecture fails when p is of the form p=2q+1? 

20100226, 22:02  #242 
Jan 2010
379 Posts 
No, it does not fail.
I just say that if 2p+1 primes are infinitely and p is a prime, the by the same idea there are infinitely many primes 4p+3. By that: 8p+7......2^{n}p+2^{n}1 primes are infinitely where 2{n1}p+2^{n1}1 a prime. BTW: The discussion/argument will stop here and will continue tommorow becuase I need to go to sleep, good night.. Last fiddled with by blob100 on 20100226 at 22:07 
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