---- ---- ----- validity of an equation
in [Math]
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From: david on 1 Aug 2010 20:46 On Aug 1, 4:23 pm, Rob Johnson <r...(a)trash.whim.org> wrote: > In article <bf6419c3-0449-4e33-8ec2-0984a7d69...(a)s24g2000pri.googlegroups..com>, > > > > > > david <dlee...(a)yahoo.com> wrote: > >On Aug 1, 12:48 am, Rob Johnson <r...(a)trash.whim.org> wrote: > >> In article <92e181c5-a6b2-4280-8892-90d3ef0f9...(a)q22g2000yqm.googlegroups.com>, > > >> david <dlee...(a)yahoo.com> wrote: > >> >Consider the following equation for the given condition. > > >> > 2^[(k-1)(k-2)] = 1 mod k^2 (1) > > >> >Condition: prime k > 3. > > >> >Question: For what value of k (1) is valid? > > >> >A helpful reply will be appreciated. > > >> If k is prime, then phi(k^2) = k(k-1). If k is odd, gcd(2,k) = 1. > >> These two conditions imply > > >> k(k-1) 2 > >> 2 = 1 mod k [1] > > >> Using this, we get that > > >> (k-1)(k-2) -2(k-1) 2 > >> 2 = 2 mod k [2] > > >> So, your question reduces to > > >> k-1 2 > >> 4 = 1 mod k [3] > > >> Since 2^{k-1} = 1 mod k^2 for Wieferich primes, [3] must be true > >> also, but there could be other primes that satisfy [3]. I have run > >> a check on the first 5,000,000 primes, and the only k less than or > >> equal to 86,028,121 which satisfy [3] are the Wieferich primes: > >> 1093 and 3511. > > >> Rob Johnson <r...(a)trash.whim.org> > >> take out the trash before replying > >> to view any ASCII art, display article in a monospaced font > > >---- ----- ----- > >So you agree that (A) implies (B) pl see below > > > 2^[(k-1)(k-2)] = 1 mod k^2 (A) > > > k-1 2 > > 4 = 1 mod k (B) > > > Where k is Wieferich prime. Otherwise, A does not imply B. > > > Right or wrong ? > > >--- ---- ---- ---- -- > > Wrong. Assuming k is an odd prime, (A) and (B) are equivalent. > That's because for any odd prime k, > > k(k-1) 2 > 2 = 1 mod k (C) > > Rob Johnson <r...(a)trash.whim.org> > take out the trash before replying > to view any ASCII art, display article in a monospaced font- Hide quoted text - > > - Show quoted text - ---- ---- ---- ---- Please refer to (A), (B), (C) above 2(k-1) 2 2 = 1 mod k (BB) (B) is written as (BB) (A), (BB), (C) are valid for any prime >3, regardless of the fact that k is a Wieferich prime OR not Agree or not? If not why not? Thanks for your additional comments ---- ---- ----
From: Rob Johnson on 2 Aug 2010 11:05
In article <0efa7452-911e-46f4-8d0c-3d22b24ae38f(a)i4g2000prf.googlegroups.com>, david <dlee753(a)yahoo.com> wrote: >On Aug 1, 4:23 pm, Rob Johnson <r...(a)trash.whim.org> wrote: >> In article <bf6419c3-0449-4e33-8ec2-0984a7d69...(a)s24g2000pri.googlegroups.com>, >> >> >> >> >> >> david <dlee...(a)yahoo.com> wrote: >> >On Aug 1, 12:48 am, Rob Johnson <r...(a)trash.whim.org> wrote: >> >> In article <92e181c5-a6b2-4280-8892-90d3ef0f9...(a)q22g2000yqm.googlegroups.com>, >> >> >> david <dlee...(a)yahoo.com> wrote: >> >> >Consider the following equation for the given condition. >> >> >> > 2^[(k-1)(k-2)] = 1 mod k^2 (1) >> >> >> >Condition: prime k > 3. >> >> >> >Question: For what value of k (1) is valid? >> >> >> >A helpful reply will be appreciated. >> >> >> If k is prime, then phi(k^2) = k(k-1). If k is odd, gcd(2,k) = 1. >> >> These two conditions imply >> >> >> k(k-1) 2 >> >> 2 = 1 mod k [1] >> >> >> Using this, we get that >> >> >> (k-1)(k-2) -2(k-1) 2 >> >> 2 = 2 mod k [2] >> >> >> So, your question reduces to >> >> >> k-1 2 >> >> 4 = 1 mod k [3] >> >> >> Since 2^{k-1} = 1 mod k^2 for Wieferich primes, [3] must be true >> >> also, but there could be other primes that satisfy [3]. I have run >> >> a check on the first 5,000,000 primes, and the only k less than or >> >> equal to 86,028,121 which satisfy [3] are the Wieferich primes: >> >> 1093 and 3511. >> >> >> Rob Johnson <r...(a)trash.whim.org> >> >> take out the trash before replying >> >> to view any ASCII art, display article in a monospaced font >> >> >---- ----- ----- >> >So you agree that (A) implies (B) pl see below >> >> > 2^[(k-1)(k-2)] = 1 mod k^2 (A) >> >> > k-1 2 >> > 4 = 1 mod k (B) >> >> > Where k is Wieferich prime. Otherwise, A does not imply B. >> >> > Right or wrong ? >> >> >--- ---- ---- ---- -- >> >> Wrong. Assuming k is an odd prime, (A) and (B) are equivalent. >> That's because for any odd prime k, >> >> k(k-1) 2 >> 2 = 1 mod k (C) >> >> Rob Johnson <r...(a)trash.whim.org> >> take out the trash before replying >> to view any ASCII art, display article in a monospaced font- Hide quoted text - >> >> - Show quoted text - > >---- ---- ---- ---- >Please refer to (A), (B), (C) above > > 2(k-1) 2 > 2 = 1 mod k (BB) > >(B) is written as (BB) > >(A), (BB), (C) are valid for any prime >3, regardless of the fact that >k is a > Wieferich prime OR not > >Agree or not? If not why not? > >Thanks for your additional comments > >---- ---- ---- (C) is true for any prime k. If k is a prime, (A) and (B) are equivalent but true if and only if k is a Wieferich prime. What is false is the statement that "if k is not a Wieferich prime, A does not imply B". For example, k = 7 is not a Wieferich prime, but A and B are both false, so A implies B. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |