Re: Minimizing (a^n + b^n - c^n)^2: Was: Re: Existence of reals and observation of them

On Dec 19, 1:06 pm, quasi <qu...@xxxxxxxx> wrote:
On Wed, 19 Dec 2007 08:05:49 -0800 (PST), galathaea

<galath...@xxxxxxxxx> wrote:
On Dec 19, 2:29 am, quasi <qu...@xxxxxxxx> wrote:
On Wed, 19 Dec 2007 02:01:47 -0500, quasi <qu...@xxxxxxxx> wrote:
On Tue, 18 Dec 2007 22:38:29 -0800 (PST), Proginoskes
<CCHeck...@xxxxxxxxx> wrote:

On Dec 17, 1:52 am, Proginoskes <CCHeck...@xxxxxxxxx> wrote:
On Dec 13, 6:14 am, glenn <glenn...@xxxxxxxxx> wrote:

[...]
Perhaps you want to read this:

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/real.pdf

I re-read this recently. (It sounds a lot like AP's ideas, but ends up
with different conclusions and doesn't sound as crazy as AP does.)

In it, Zeilberger, mentions that FLT would follow if there is a
polynomial with nonnegative coefficients with a certain additional
property related to the function

W = W(n; a, b, c) := (a^n + b^n - c^n)^2.
[...]
Yes -- Zeilberg's polynomial P is a figment of his imagination.
[...]
The basic concept of Zeilberg's idea does makes sense, just not his
actual proposal.
[...]
As far as I can see, he's taking a natural idea and mangling it badly.

He mentioned this approach to proving FLT in one of his "April Fool's
Day" posts/journal entries, suggesting that he's not taking it
entirely serious. However, the natural idea is (at first glance)
workable; it's just that the details are too messy, and he's too
"respectable", to actually try to work on the problem.

(On April 1, 2006, I posted to sci.physics the "results" of an
experiment intended to show that the universe has more than 3 physical
dimensions, and that the 3-dimensional part we have obssed about is
actually curved, not "affine". It's the same idea; the concept is
reasonable enough but the implementation is ludicrous.)

The idea that some polynomial identity might show self-evidently that
a solution to FLT(n) to implies either a "smaller" solution to FLT(n)
or a solution to FLT(n') where n' is less than n (but greater than 2),
makes sense, and is a nice idea, even if not terribly ingenious. Of
course, descent itself _is_ ingenious, but that's an old idea -- it
was Fermat's idea (or perhaps it goes all the way back to Euclid or
Diophantus)

So I have no problem with the underlying idea. My problem is with his
formulation.

For one thing, the positivity of a,b,c cannot be forced by the
polynomial identity, [...]

And it isn't. It's only when you get beyond the algebraic manipulation
that you use positivity, the fact that n is at least 3, etc. But I
think it's been mentioned elsewhere.

Here are a few of the principles that must be adhered to:

(1) A proof of FLT better not disprove the existence of real
solutions, or of complex solutions. Thus, any such proof must actually
_use_ the hypothesis that a,b,c are integers.

So one of the coefficients could be b*(b-1), for instance; certainly
this expression is nonnegative for all *integers* b, but it isn't
*always* nonnegative, and this opens up the possibility of solutions
to Fermat's equation with non-integer numbers.

(2) A proof of FLT better not imply impossibility for all integers
a,b,c since trivial solutions can't be avoided.

and this can be avoided by allowing coefficients to be (b-2), for
instance, which is nonnegative when b >= 2, i.e., when there is a non-
trivial solution.

(3) A proof of FLT better not imply impossibility for all n, since
there are evident nontrivial solutions for n = 1 and n = 2.

and again this can be taken care of by coefficients like (n-3), which
is nonnegative when n >= 3.

I'm not saying all of the above are violated by Zeilberg's imprecise
formulation, but as far as I can see, he does not appear to provide
any defense against (2) or (3), thus apparently falling into to some
of the same traps that catch the most oblivious of the FLT cranks. The
more sophisticated FLT cranks at least know enough to make sure their
claimed proofs avoid these simple traps.

All of these objections can be taken care of if we knew that, for
instance,

W(n; a, b, c)
= a * (a - 1.5) * W(n-1; a-1, b, c)
+ b * (b - 1) * W(n-1; a, b-1, c)
+ c * (c - 2.71828) * W(n-2; a, b, c-1)
+ (n - 3)* (n - 3.5)^2 * W(n-1; a, b, c).

The algebra almost surely doesn't work out; but let's assume that it
does (regardless of whether a, b, c, n are integers). Note that this
equality does NOT rule out the possibility of solutions when n is 2,
it does NOT rule out the possibility of trivial solutions (a=1, b=0),
etc. It is simply an algebraic formula.

Then a proof of FLT would proceed as follows:

Theorem. [If the equation above is correct, then] W(n; a, b, c) > 0
for all n>= 3, and all positive integers a, b, c. I.e., FLT is true.

Proof. Induction on n. The case n = 3 has been taken care of, so
suppose n >= 4.

This means W(n'; a', b', c') > 0 whenever n' < n and whenever a', b',
c' are integers.

Now suppose a, b, c are positive integers; we want to show that
W(n'; a, b, c) > 0.

The case where a=1 or b=1 is easy to take care of, so we may
assume a, b >= 2 (and hence c >= 3).

Now, a * (a - 1.5) > 0
b * (b - 1) > 0
c * (c - 2.71828) > 0 and
(n - 3) * (n - 3.5)^2 > 0

are all trivially true. Since we are using induction,

W(n-1; a-1, b, c) > 0
W(n-1; a, b-1, c) > 0
W(n-2; a, b, c-1) > 0 and
W(n-1; a, b, c) > 0,

the right-hand side of the equality is positive. (Note that we need
the facts that a,b are at least 2 here!) But this means W(n; a, b,
c) is positive, QED, end of "Proof".

(Of course, if X is an integer, then X > 0 means X >= 1, a fact which
makes some induction hypotheses stronger than they first appear. This
fact might *also* be incorporated in a proof of FLT.)

--- Christopher Heckman
.

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