 mersenneforum.org > Math how to know if my ideas didnt tought before?
 Register FAQ Search Today's Posts Mark Forums Read  2010-02-26, 07:13 #232 blob100   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^n-1. 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^n-1. So my "conjecture" (which is the same as the old one) is: If there infinitely many primes g=2^nQ+2^n-1, 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 2010-02-26 at 07:58   2010-02-26, 11:05   #233
R.D. Silverman

Nov 2003

22×5×373 Posts Quote:
 Originally Posted by CRGreathouse Is it? It seems to be exactly the same to me.
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.   2010-02-26, 12:39   #234
blob100

Jan 2010

379 Posts Quote:
 Originally Posted by blob100 I found this conjecture: there are ifinitely many primes q of the form 2p+1 where p a prime. And after some playing with this conjecture, I found a stronger one: conjecture: there are infinitely many primes such that is also a prime, where Q is not neccesarily a prime number.
As you see, my conjecture is the same as the original conjecture.
I'm not talking about your conjecture (with the 2^nR-1 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^n-1 right?
So I say that there are infinitely many primes of the form 2^np+2^n-1 where 2^{n-1}p+2^{n-1}-1 is also a prime (by the original conjecture, these two forms contain infinitey many primes).

Last fiddled with by blob100 on 2010-02-26 at 12:39   2010-02-26, 17:12   #235
R.D. Silverman

Nov 2003

22·5·373 Posts Quote:
 Originally Posted by blob100 As you see, my conjecture is the same as the original conjecture. I'm not talking about your conjecture (with the 2^nR-1 which I can't understand).
What don't you understand? Tell me and I will try to help.

Quote:
 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.
This last observation is trivial and adds nothing to either the original
conjecture, or the "conjecture" that you presented.

Quote:
 By induction: 2^np+2^n-1 right?

Induction??? What Induction??? You have done nothing even
remoting resembling an induction proof.

4p+3 = 2^n p + 2^n-1, iff n=2.
Your conjecture did not specify a value for n.   2010-02-26, 21:31 #236 blob100   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^{n-1}p+2^{n-1}-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 2010-02-26 at 21:40   2010-02-26, 21:31 #237 flouran   Dec 2008 83310 Posts @ Tomer (or anyone else who would like to respond): Let's say you come up with a conjecture that is non-trivial 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 2010-02-26 at 21:33   2010-02-26, 21:37   #238
R.D. Silverman

Nov 2003

11101001001002 Posts Quote:
 Originally Posted by blob100 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^{n-1}p+2^{n-1}-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...

This isn't even wrong. It is nonsense. And your so called induction
is not how an an induction argument works.   2010-02-26, 21:41 #239 blob100   Jan 2010 37910 Posts Why not? So which tool did I use?   2010-02-26, 21:52   #240
R.D. Silverman

Nov 2003

22·5·373 Posts Quote:
 Originally Posted by blob100 Why not? So which tool did I use?
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 non-negative integers are well-ordered).

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 n-1.]

(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.   2010-02-26, 21:53   #241
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

27×47 Posts Quote:
 Originally Posted by blob100 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).
I don't see how you can just extend it like that.

Perhaps the original conjecture fails when p is of the form p=2q+1?   2010-02-26, 22:02 #242 blob100   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{n-1}p+2^{n-1}-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 2010-02-26 at 22:07   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post paul0 Factoring 3 2015-03-14 19:55 Dubslow NFS@Home 13 2015-02-02 22:25 Mini-Geek No Prime Left Behind 16 2008-03-01 23:32 TTn 15k Search 15 2003-09-23 16:28 Xyzzy Lounge 17 2003-03-24 16:20

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