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 science_man_88 2014-07-14 23:06

Some arithmetic...

I'm guessing it's known that [TEX](2^{p-1}\eq 1 \text { mod p^2}) == (2^{2n-1}-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 ?

 Zeta-Flux 2014-07-14 23:30

Better than that, the Wieferich condition is equivalent to [TEX]2^{(p-1)/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.

 science_man_88 2014-07-14 23:35

[QUOTE=Zeta-Flux;378104]Better than that, the Wieferich condition is equivalent to [TEX]2^{(p-1)/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).

 science_man_88 2014-07-15 01:23

[QUOTE=Zeta-Flux;378104] [TEX]2^{(p-1)/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^n-1 \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^{n-1}-1 \eq 2n^2+2n \pmod{4n^2+4n+1} [/TEX] or [TEX]2^{n-1} \eq 2n^2+2n\pmod{4n^2+4n+1}[/TEX] I'm I getting better or just making it worse ?

 Zeta-Flux 2014-07-15 03:55

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.

 R.D. Silverman 2014-07-15 16:26

I'm guessing it's known that [TEX](2^{p-1}\eq 1 \text { mod p^2}) == (2^{2n-1}-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 sub-forum?

 science_man_88 2014-07-15 19:03

[QUOTE=R.D. Silverman;378140]Would a moderator please move this (and succeeding) gibberish to the crank
math sub-forum?[/QUOTE]

can you explain to me why you think it's gibberish ?

 science_man_88 2014-07-15 20:41

[QUOTE=Zeta-Flux;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 ?

 science_man_88 2014-07-15 23:52

[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.

 science_man_88 2014-07-28 22:04

[QUOTE=R.D. Silverman;378140]Would a moderator please move this (and succeeding) gibberish to the crank
math sub-forum?[/QUOTE]

I decided I better show you how I got that:
[LIST][*]the fact that if 2n+1|2k+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^{p-1}\eq 1 \pmod {p^2} == 2^{2n+1-1}\eq 1 \pmod {{2n+1}^2} == 2^{2n+1-1}\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^{2n-1}-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^{p-2}-1 \eq {\frac{p-1}{2}(p+1)} \pmod {p^2}[/TEX]

 science_man_88 2014-07-30 19:38

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.

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