20140714, 23:06  #1 
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Jul 2009
Dumbassville
2^{6}×131 Posts 
Some arithmetic...
just asking things:
I'm guessing it's known that where 2n+1 =p; if so has anything useful come out of it ? Last fiddled with by science_man_88 on 20140714 at 23:13 Reason: took a +1 away after realizing the error. 
20140714, 23:30  #2 
May 2003
7×13×17 Posts 
Better than that, the Wieferich condition is equivalent to . This makes it slightly easier to test for the condition.
The only place I know of where the Wieferich condition is really useful is in the first case of Fermat's last theorem and in the solution to Catalan's conjecture. 
20140714, 23:35  #3  
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Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
Last fiddled with by science_man_88 on 20140714 at 23:39 

20140715, 01:23  #4 
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Jul 2009
Dumbassville
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Sorry for quoting this twice. One thing that just came to me is that this is equivalent of saying and this equals or which when you consider that if 2m+1 divides 2k+1, we can bring this down to or I'm I getting better or just making it worse ?
Last fiddled with by science_man_88 on 20140715 at 01:29 
20140715, 03:55  #5 
May 2003
60B_{16} Posts 
I certainly don't want to discourage you from pursuing these ideas, as not much progress has been made in understanding Wieferich primes! I would recommend reading the original paper of Wieferich, or possibly those that followed (I seem to remember one by Mirimanoff (sp?)), which introduced how these primes play a role in Fermat's last theorem. You may find a connection between what you are seeing and that problem.

20140715, 16:26  #6 
Nov 2003
2^{2}·5·373 Posts 

20140715, 19:03  #7 
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Jul 2009
Dumbassville
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20140715, 20:41  #8  
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Jul 2009
Dumbassville
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Quote:


20140715, 23:52  #9 
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Jul 2009
Dumbassville
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nevermind I found a free preview of some of it and read a bit on wikipedia. I did fail so far to find specific n candidates that work, I'll give RDS that. edit: the part about sounds interesting for me since I see a way to generalize my first post to use those m. but I don't think it will help.
Last fiddled with by science_man_88 on 20140716 at 00:27 
20140728, 22:04  #10  
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Jul 2009
Dumbassville
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Quote:
therefore the original congruence plus the basics of division can show that what I said is true. edit: this last result can be rewritten as Last fiddled with by science_man_88 on 20140728 at 22:13 

20140730, 19:38  #11 
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Jul 2009
Dumbassville
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potential proof ( partial checked for errors along the way)
I'd like a double check on this but based on my last congruence I believe I've found a way to prove p can not be the second prime in a twin prime pair. (Q,p)
I'd love your feedback since I feel like I'm talking to myself. Last fiddled with by science_man_88 on 20140730 at 19:39 Reason: forgot to attach the file the first time. 
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