Some arithmetic...
just asking things:
I'm guessing it's known that [TEX](2^{p1}\eq 1 \text { mod p^2}) == (2^{2n1}1\eq 2n^2+2n \text { mod 4n^2+4n+1})[/TEX] where 2n+1 =p; if so has anything useful come out of it ? 
Better than that, the Wieferich condition is equivalent to [TEX]2^{(p1)/2} \equiv \pm 1\pmod{p^2}[/TEX]. 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. 
[QUOTE=ZetaFlux;378104]Better than that, the Wieferich condition is equivalent to [TEX]2^{(p1)/2} \equiv \pm 1\pmod{p^2}[/TEX]. 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.[/QUOTE] Okay, Thanks for that, I just thought it might be useful for trying to pin down what n are possible to create such p instead of trying any p ( which as far as I know, and yes I don't know much if any, is how it works). 
[QUOTE=ZetaFlux;378104] [TEX]2^{(p1)/2} \equiv \pm 1\pmod{p^2}[/TEX].[/QUOTE]
Sorry for quoting this twice. One thing that just came to me is that this is equivalent of saying [TEX]2^n \eq \pm 1\pmod{4n^2+4n+1}[/TEX] and this equals [TEX]2^n1 \eq 0 \pmod{4n^2+4n+1}[/TEX] or [TEX]2^n+1 \eq 0 \pmod{4n^2+4n+1}[/TEX] which when you consider that if 2m+1 divides 2k+1, [TEX] k \eq m \pmod{2m+1}[/TEX] we can bring this down to [TEX]2^{n1}1 \eq 2n^2+2n \pmod{4n^2+4n+1} [/TEX] or [TEX]2^{n1} \eq 2n^2+2n\pmod{4n^2+4n+1}[/TEX] I'm I getting better or just making it worse ? 
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.

[QUOTE=science_man_88;378101]just asking things:
I'm guessing it's known that [TEX](2^{p1}\eq 1 \text { mod p^2}) == (2^{2n1}1\eq 2n^2+2n \text { mod 4n^2+4n+1})[/TEX] where 2n+1 =p; if so has anything useful come out of it ?[/QUOTE] Would a moderator please move this (and succeeding) gibberish to the crank math subforum? 
[QUOTE=R.D. Silverman;378140]Would a moderator please move this (and succeeding) gibberish to the crank
math subforum?[/QUOTE] can you explain to me why you think it's gibberish ? 
[QUOTE=ZetaFlux;378114]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.[/QUOTE]
I found the title but not the paper itself the real problem after that is to translate it to English if I can't find an English copy. Do you know where I can find a free copy of Zum letzten Fermat'schen Theorem ? 
[QUOTE=science_man_88;378172]I found the title but not the paper itself the real problem after that is to translate it to English if I can't find an English copy. Do you know where I can find a free copy of Zum letzten Fermat'schen Theorem ?[/QUOTE]
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 [TEX]m\leq 113[/TEX] 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. 
[QUOTE=R.D. Silverman;378140]Would a moderator please move this (and succeeding) gibberish to the crank
math subforum?[/QUOTE] I decided I better show you how I got that: [LIST][*]the fact that if 2n+12k+1 ( meaning divides), [TEX]k \eq n \pmod{2n+1}[/TEX] can be shown by showing 2n+1 divides when k=n and that 2n+1 divides one in every 2n+1 odd numbers. [*]based on 2n+1=p we get: [TEX]2^{p1}\eq 1 \pmod {p^2} == 2^{2n+11}\eq 1 \pmod {{2n+1}^2} == 2^{2n+11}\eq 1 \pmod{4n^2+4n+1} == 2^{2n}\eq 1 \pmod{4n^2+4n+1} == 2^{2n}1\eq 0 \pmod{4n^2+4n+1[/TEX][*]based on these other two we can show that if [TEX]2^{2n}1\eq 0 \pmod{4n^2+4n+1}[/TEX] then [TEX]2^{2n1}1 \eq 2n^2+2n \pmod{4n^2+4n+1}[/TEX] [/LIST] 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 [TEX]2^{p2}1 \eq {\frac{p1}{2}(p+1)} \pmod {p^2}[/TEX] 
potential proof ( partial checked for errors along the way)
1 Attachment(s)
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. 
All times are UTC. The time now is 08:56. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2020, Jelsoft Enterprises Ltd.