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Topic: FLT Discussion: Simplifying
Replies: 65   Last Post: Mar 17, 2001 11:59 PM

 Messages: [ Previous | Next ]
 Dik T. Winter Posts: 7,899 Registered: 12/6/04
Re: FLT Discussion: Simplifying
Posted: Jan 19, 2001 10:11 AM

In article <94810q\$6o1\$1@nnrp1.deja.com> jstevh@my-deja.com writes:
...
> Some have tried to find fault with my method using the n=2 case, which
> *does* use x+y+vz. I think those are interesting, so I'll go over them
> quickly.

Interesting that you find them interesting now, but not in a reply to the
posting where I wrote about that example.

> x+y+vz = 0(mod (x+y+vz)), jumping ahead
> x^2 + 2xy + y^2 = v^2z^2, and jumping to the end,
> (v^2 - 1)z^2 - 2xy = 0(mod (x+y+vz)), with x^2 + y^2 = z^2.
> Some have argued, why not factor what's on the left to get
> (sqrt(v^2 - 1)z - sqrt(2xy))(sqrt(v^2 - 1)z + sqrt(2xy)) = 0,
> so (sqrt(v^2 - 1)z - sqrt(2xy)) = 0(mod (x+y+vz)?

But this is a bit of a distortion, as I came with:
[(1-v^2)z - sqrt((1-v^2).2xy][(1-v^2)z + sqrt((1-v^2).2xy)] = 0.
>
> Is there a really simple answer that's rigorous? I guess so, but the
> answer I give is you look at the exponents.

Yup, I do and see that there are no integral exponents for x, y and v.
So what? In your factorisation v does not have an integral exponent.
(Yes, I know that you say that sqrt(x^2 + y^2) can be written as
x + ... + y, but you do not bother to fill in the ellipses, and the
only thing I can fill in there sensibly is [sqrt(x^2+y^2)-x-y]. Like,
sqrt(x^2 + y^2) is a polynomial, which is *false*.)
>
> To me it's a simple matter of asking what expression can multiply times
> (x+y+vz) to get (sqrt(v^2 - 1)z - sqrt(2xy))?

Yup, so what is that expression?

> Again, the statements are true if x^p + y^p = z^p. Note, as written
> that is saying that that statement is true. That is, that we have some
> x, y and z that fit.

No. Not all statements are true in your proof. In particular one statement
is *not* true, but false. As I parafrased it in my "proof" (replace 'n' by
'1' to get your proof, and note that this is a "valid" statement for *all*
values of 'n'):
[(v^5+n)z^2 - (5v^3+sqrt(5v^6-20nv)*xy/2)] *
[(v^5+n)z^2 - (5v^3-sqrt(5v^6-20nv)*xy/2)] = 0 (mod (x+y+vz)/h), when
x^5 + y^5 = nz^5.
Fill in n=68101, x=15, y=17, z=2, h=2 (I once asked whether that was the
only coprime solution for n=68101, and indeed, it has been proven) to get:
[4v^5+272404 - (5v^3+sqrt(5v^6-1362020v).255/2] *
[4v^5+272404 - (5v^3-sqrt(5v^6-1362020v).255/2] = 0 (mod v+16).
and note that even this is "valid". But none of the factors is
divisible by (v+16) by any definition that I know. (Note that when
you fill in v=-16 on the left the product is indeed 0, proving that
is is "valid".)

> http://www.mindspring.com/~jstev/FLTp5.htm.

I still await your response about which statement in the "proof" on
http://www.cwi.nl/~dik/mathematics/jsh.html is exactly wrong, and why.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Date Subject Author
1/15/01 jstevh@my-deja.com
1/15/01 Dik T. Winter
1/16/01 Charles H. Giffen
1/16/01 jstevh@my-deja.com
1/16/01 Randy Poe
1/18/01 jstevh@my-deja.com
1/18/01 Michael Hochster
1/18/01 Peter Johnston
1/18/01 Randy Poe
1/18/01 Doug Norris
1/16/01 Doug Norris
1/16/01 Randy Poe
1/16/01 Dik T. Winter
1/18/01 jstevh@my-deja.com
1/19/01 Dik T. Winter
1/19/01 Randy Poe
1/20/01 jstevh@my-deja.com
1/20/01 oooF
1/21/01 hale@mailhost.tcs.tulane.edu
1/21/01 Peter Percival
1/21/01 Randy Poe
1/26/01 Franz Fritsche
1/19/01 gus gassmann
1/20/01 jstevh@my-deja.com
1/20/01 Doug Norris
1/26/01 Franz Fritsche
1/16/01 hale@mailhost.tcs.tulane.edu
1/16/01 Randy Poe
1/17/01 hale@mailhost.tcs.tulane.edu
1/18/01 jstevh@my-deja.com
1/19/01 hale@mailhost.tcs.tulane.edu
1/20/01 jstevh@my-deja.com
1/21/01 hale@mailhost.tcs.tulane.edu
1/18/01 Peter Percival
1/19/01 hale@mailhost.tcs.tulane.edu
3/17/01 Ross A. Finlayson
1/16/01 hale@mailhost.tcs.tulane.edu
1/18/01 jstevh@my-deja.com
1/19/01 hale@mailhost.tcs.tulane.edu
1/29/01 jstevh@my-deja.com
1/19/01 Dik T. Winter
1/21/01 Dennis Eriksson
1/15/01 Michael Hochster
1/16/01 jstevh@my-deja.com
1/16/01 Michael Hochster
1/18/01 jstevh@my-deja.com
1/18/01 Peter Percival
1/18/01 Randy Poe
1/19/01 oooF
1/21/01 Dik T. Winter
1/21/01 oooF
1/18/01 Edward Carter
1/19/01 W. Dale Hall
1/19/01 Michael Hochster
1/16/01 Randy Poe
1/16/01 Randy Poe
1/17/01 W. Dale Hall
1/17/01 W. Dale Hall
1/19/01 oooF
1/16/01 Charles H. Giffen
1/16/01 David Bernier
1/16/01 jstevh@my-deja.com
1/18/01 Arthur
1/30/01 plofap@my-deja.com
1/30/01 plofap@my-deja.com
1/30/01 plofap@my-deja.com