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Old 2014-07-14, 23:06   #1
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Default Some arithmetic...

just asking things:

I'm guessing it's known that (2^{p-1}\eq 1 \text { mod p^2}) == (2^{2n-1}-1\eq 2n^2+2n \text { mod 4n^2+4n+1}) where 2n+1 =p; if so has anything useful come out of it ?

Last fiddled with by science_man_88 on 2014-07-14 at 23:13 Reason: took a +1 away after realizing the error.
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Old 2014-07-14, 23:30   #2
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Better than that, the Wieferich condition is equivalent to 2^{(p-1)/2} \equiv \pm 1\pmod{p^2}. 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.
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Old 2014-07-14, 23:35   #3
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Quote:
Originally Posted by Zeta-Flux View Post
Better than that, the Wieferich condition is equivalent to 2^{(p-1)/2} \equiv \pm 1\pmod{p^2}. 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.
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).

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Old 2014-07-15, 01:23   #4
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Quote:
Originally Posted by Zeta-Flux View Post
2^{(p-1)/2} \equiv \pm 1\pmod{p^2}.
Sorry for quoting this twice. One thing that just came to me is that this is equivalent of saying 2^n \eq \pm 1\pmod{4n^2+4n+1} and this equals 2^n-1 \eq 0 \pmod{4n^2+4n+1} or 2^n+1 \eq 0 \pmod{4n^2+4n+1} which when you consider that if 2m+1 divides 2k+1,  k \eq m \pmod{2m+1} we can bring this down to 2^{n-1}-1 \eq 2n^2+2n \pmod{4n^2+4n+1} or 2^{n-1} \eq  2n^2+2n\pmod{4n^2+4n+1} I'm I getting better or just making it worse ?

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Old 2014-07-15, 03:55   #5
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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.
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Old 2014-07-15, 16:26   #6
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Quote:
Originally Posted by science_man_88 View Post
just asking things:

I'm guessing it's known that (2^{p-1}\eq 1 \text { mod p^2}) == (2^{2n-1}-1\eq 2n^2+2n \text { mod 4n^2+4n+1}) where 2n+1 =p; if so has anything useful come out of it ?
Would a moderator please move this (and succeeding) gibberish to the crank
math sub-forum?
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Old 2014-07-15, 19:03   #7
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Quote:
Originally Posted by R.D. Silverman View Post
Would a moderator please move this (and succeeding) gibberish to the crank
math sub-forum?
can you explain to me why you think it's gibberish ?
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Old 2014-07-15, 20:41   #8
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Quote:
Originally Posted by Zeta-Flux View Post
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.
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 ?
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Old 2014-07-15, 23:52   #9
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Quote:
Originally Posted by science_man_88 View Post
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 ?
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 m\leq 113 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.

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Old 2014-07-28, 22:04   #10
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Quote:
Originally Posted by R.D. Silverman View Post
Would a moderator please move this (and succeeding) gibberish to the crank
math sub-forum?
I decided I better show you how I got that:
  • the fact that if 2n+1|2k+1 (| meaning divides), k \eq n \pmod{2n+1} 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: 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
  • based on these other two we can show that if 2^{2n}-1\eq 0 \pmod{4n^2+4n+1} then 2^{2n-1}-1 \eq 2n^2+2n \pmod{4n^2+4n+1}

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 2^{p-2}-1 \eq {\frac{p-1}{2}(p+1)} \pmod {p^2}

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Old 2014-07-30, 19:38   #11
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Default 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.
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