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Topic: --- --- --- Find the error
Replies: 13   Last Post: Jun 17, 2011 3:21 PM

 Messages: [ Previous | Next ]
 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^(k-1) is not valid.
>
> > >A clear and helpful response will be appreciated.
>
> > >Odd = Even is not mathematically justified.
>
>
> > 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^(k-1)
>
> > 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)
>
>
> > 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 -
>
> ***  ***  *** 5-17-11
> 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 2k|y
>
> Again, many many thanks.
>
> ***  ***  ****

"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.

Date Subject Author
6/16/11 david
6/16/11 quasi
6/16/11 Pubkeybreaker
6/16/11 david
6/16/11 Virgil
6/16/11 Pubkeybreaker
6/16/11 Tim Little
6/16/11 tommyrjensen@gmail.com
6/16/11 Tim Little
6/17/11 quasi
6/17/11 david
6/17/11 achille
6/17/11 quasi
6/17/11 Frederick Williams