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
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Quote:
Last fiddled with by science_man_88 on 20140714 at 23:39 

20140715, 01:23  #4 
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Jul 2009
Dumbassville
20300_{8} Posts 
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
3013_{8} 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
2^{6}×131 Posts 
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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