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

On Dec 20, 12:39 am, quasi <qu...@xxxxxxxx> wrote:
On Wed, 19 Dec 2007 23:02:52 -0800 (PST), Proginoskes



<CCHeck...@xxxxxxxxx> wrote:
On Dec 19, 12:01 am, 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.

This polynomial would allow for a proof by induction that
W > 0, for all positive integers a, b, c, and n >= 3, but it would
also put a lower bound on how fast the minimum value of W increases.
So the question here is:

What is the minimum value M(n) of W(n; a, b, c), where
n >= 3 is fixed, and a, b, c are positive integers?

For instance,

0 < M(3) <= (6^3 + 8^3 - 9^3)^2 = 1,

so M(3) = 1.

Does anyone know how fast M(n) grows?

Okay, since no one wants to embarass me, I'll embarass myself. The
answer is: It doesn't, since
(1^n + 1^n - 1^n)^2 = 1 for all n.

That should say something about Zeilberg's polynomial P, though.

Yes -- Zeilberg's polynomial P is a figment of his imagination.

Not really, if you allow the coefficients to involve a, b, c, and n;
then (a-1) would be zero if a=1 and not grow at all.

I have no idea what you're talking about, but in any case, it's silly
to think a solution (a,b,c) would induce a solution (a-1,b,c) or any
such combination. It suffices to note that there are _real_ solutions
(a,b,c) which do not induce _any_ "priors" of the form (a-1,b,c).

you are still misunderstanding the recurrence

this has to do with the polynomial
f_n(a,b,c) = (a^n + b^n - c^n)^2

this has nothing to do with "solutions"
f_n = 0

look at the simpler
f_n(a) = a + n

then there are the recurrences
f_n(a) = f_n(a-1) + 1
and
f_n(a) = f_{n-1}(a) + 1
(among many others)

when used in inequality proving
we can immediately deduce

if f_n(a) > 0 then f_n(a+1) > 0 and f_{n+1}(a) > 0
and so on

this has nothing to do with relating one solution to another

(Not that I'm going to drop everything and search for this
polynomial.)

Good choice.

For one thing, you have no definite target, but more importantly,
there is no valid target matching Zeilberger's form. In my opinion,
Zeilberger's formulation is essentially a pipe dream.

reconsider the possibility with the above information

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galathaea: prankster, fablist, magician, liar
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