achille
Posts:
575
Registered:
2/10/09


Re:    Find the error
Posted:
Jun 17, 2011 10:57 AM


On Jun 17, 8:32 pm, david <dlee...@yahoo.com> wrote: > On Jun 17, 3:57 am, quasi <qu...@null.set> wrote: > > > > > On Thu, 16 Jun 2011 15:59:17 0700 (PDT), david <dlee...@yahoo.com> > > wrote: > > > >On Jun 16, 4:03 pm, Pubkeybreaker <pubkeybrea...@aol.com> wrote: > > > >> On Jun 16, 3:08 pm, quasi <qu...@null.set> wrote: > > > >> > On Thu, 16 Jun 2011 09:14:37 0700 (PDT), david > > >> > <dlee...@yahoo.com> wrote: > > >> >> ... > > >> > ... > > > >> > Remarks: > > > >> > You've been shown similar counterexamples many times in > > >> > the past, and yet you continue to post the same flawed > > >> > "assertions" again and again, showing that you learn > > >> > nothing from your past errors. > > > >> According to a standard definition, he is clearly insane: > > >> Doing the same thing twice and expecting a different result > > >> the second time. > > > >> He is a classic crank: obsessed with unsound ideas, > > >> unwilling to listen to others, and unwilling to learn. > > > >With due respect to all who are trying to educate and help > > >me kindly justify why the following assertion is not valid: > > > >Assertion: z^k  x^k + 1 = 2^(k1) is not valid. > > > >A clear and helpful response will be appreciated. > > > >Odd = Even is not mathematically justified. > > > >Please comment > > > Assume x,y,k are integers with k > 1 and x,y both odd. > > > Let "!=" denote "not equal". > > > Then of course > > > z^k  x^k + 1 != 2^(k1) > > > since the LHS is odd and the RHS is even. > > > However, the fact that LHS != RHS does allow you > > to claim > > > LHS != RHS (mod k^2) > > > That's your error. > > > You've made the same error in the past and the error has > > previously been explained to you. > > > I'll try again ... > > > Suppose a,b,n are integers with n > 1 and suppose you > > know that a = b. From the fact that a = b can you > > conclude that a = b (mod n)? Yes, of course. > > > Next suppose a,b,n are integers with n > 1 and suppose > > you know that a != b. From the fact that a != b can you > > conclude that a != b (mod n). No, definitely not. > > > As an example, 33 != 8 but 33 = 8 (mod 25). > > > Thus, although equality trivially implies congruence, > > negated equality does _not_ imply negated congruence. > > > quasi Hide quoted text  > > >  Show quoted text  > > *** *** *** 51711 > Thank you very much quasi > > If your time and interest permit kindly offer some hints to prove the > following conjecture. > > Conjecture: (1) cannot be satisfied if k is a Wieferich Prime: > > z^k  x^k = y^2 (1); x, y, z are coprime integers each > 3, prime > k > 3 and 2ky > > Again, many many thanks. > > *** *** ****
Please read Darmon and Merel's famous paper:
"Winding quotients and some variants of Fermat's Last Theorem" (URL: http://people.math.jussieu.fr/~merel/winding.pdf )
yourself and stop bothering us, thanks for your attention.

