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

 Messages: [ Previous | Next ]
 jstevh@my-deja.com Posts: 348 Registered: 12/13/04
FLT Discussion: Simplifying
Posted: Jan 15, 2001 6:23 PM

Like I've said before, I like simplicity, and I'm committed to the
truth.

What I propose to do in this post is give what I hope is a simpler
perspective for understanding my proof of Fermat's Last Theorem. If I
don't succeed in convincing you at least I hope to outline the points
of disagreement clearly.

Since there's been a lot of discussion about whether or not I've
included enough steps in my proof, I'll start by quickly going over
another controversial proof.

Given x^2 + y^2 = 0, x and y nonzero integers, show that no solution
exists.

(x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2 = 0, so

x = sqrt(-1)y *or* x = -sqrt(-1)y.

There doesn't exist an integers that multiplies times itself to give a
negative number, and an integer can't be the product of an integer and
a non integer, so there's a contradiction.

First off, it's worth noting that I'm treating the sqrt(-1) as an
*operation*. I think some of you live under the false notion that
something like sqrt(2) is the actual number. Nope. It's an operation
that tells you something about that number, like that it multiplies
times itself to give 2. We also use it handily as a shorthand
representation for the actual number, which is also sometimes
represented by 1.414...

No human being has ever seen or ever will see the actual number that we
can conveniently represent by sqrt(2).

We're just limited that way.

So, I use sqrt(-1) in a proof that's dealing with integers, and I don't
care that there's such a thing as i = sqrt(-1). All I care about is
that I've found this thing that isn't an integer.

Ok, jumping to my proof of FLT for p=5. I prove (and no one disputes
this) that

[(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] [(v^5 + 1) z^2 -(5v^3 -
sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h), when x^5 + y^5 = z^5, and
for the simplest case h = (x+y)^{1/5}.

(Yup, a good deal more complicated than the simple example I started
with, but remember, now we're talking about Fermat's Last Theorem, so
that shouldn't be a surprise.)

The v in the above is just some integer that can't equal -1, other than
that, it's not really all that limited.

The point I want to bring back to you is that sqrt(5v^6 - 20v) is an
operation, like before.

Now, since I have x,y,z and v as integers, it makes sense that I get to
this point where I notice an operation that can't be fulfilled by
integers.

It might not suprise you that now I go ahead and say that

[(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h)

*or*

[(v^5 + 1) z^2 +(5v^3 + sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h.

If the patterns that you see compel you to accept that then you
basically understand the proof.

Now, some of you have said that the above step is unjustified.

The equivalent step with easy polynomials would be

[x^2 + y^2][x^2 - y^2] = 0(mod (x+y)), so

x^2 + y^2 = 0 (mod (x+y))

*or*

x^2 - y^2 = 0(mod (x+y))

(which is actually a bit misleading but I won't go into why at this
point to try and lessen the confusion level).

What's at issue then?

It has to do with that modulus (x+y+vz)/h.

It turns out that all you have to do to be rigorous is note that your
exponents for x,y and z are less than the highest exponent in, say,

[(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2], which is 2, and also
less than or equal to your lowest exponent which is 1.

Now, there probably wouldn't have been all the arguing back and forth
if I didn't have that h in there complicating things up a bit.

However, all you have to do is ask yourself a simple question:

Can you imagine some factor that can multiply times (x+y+vz)/h to give
something like

[(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] ?

Well, it turns out that isn't very easy to do. But it can be done.

It turns out that such a thing exists, but we human beings can't look
at it in its entirety. Sort of like with sqrt(2).

Now, people have apparently become used to that fact with something
like the sqrt(2), but they've given me all kinds of trouble because I
can't just write that other factor out.

What I've found annoying is that some of them then try to jump to the
bold claim that what I've presented is wrong, even though there's no
mathematical basis for that claim. It's just that they can't quite see
it.

Back to that modulus.

The modulus to be meaningful, has to be a smaller element than what
it's being compared to. For instance, 4 = 0(mod 2) is meaningful, 2=0
(mod 5) is not (but notice I can still write it) because 2 is less than
5.

For polynomials and expressions like them, you have to look at the
power of the exponents, so x+y is smaller than x^2 + 2xy + y^2.

(x+y+vz)/h is smaller than

[(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2].

And it does depend on the exponents.

In case you're getting way distracted, let me remind you about what we

[(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] [(v^5 + 1) z^2 -(5v^3 -
sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h)

is true (with our given conditions that I won't repeat).

Some have questioned whether or not the given modulus, (x+y+vz)/h, is
small enough to be a factor of other two terms (yes, I'm simplifying
again).

As I've mentioned, this question wouldn't come up with a polynomial and
all polynomial factors, and probably wouldn't come up here if it
weren't FLT we were talking about.

Worse, even if someone found my chatty manner unconvincing, simple
logic would lead them to the same conclusion with a modicum of effort.

After all, it make far *less* sense for the alternative.

James Harris

Sent via Deja.com
http://www.deja.com/

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