MY RESPONSE OF AUGUST 12, 2017

I spent some time thinking about Prop. P between meetings
and events in LA last week. There really should be no confusion about when P is true and
when it is false. I will attempt to clarify this while answering your latest
comments, because they do reflect what I see as confusion about what FLT65 says
and does. I shall also attempt to show you how distinguishing

*when*Prop. P is true and*when*it is false, and when the identity sign is appropriate, leads to a better understanding of FLT65.
From
your email earlier today:

Q: Do you believe that Prop P is ever false?

Let’s consider this. Here is Prop P as you stated it:

P: Equating a
polynomial to a constant, for the purpose of finding a

specific solution, automatically turns
the polynomial itself into a

constant.

This should not be a matter of “belief”, because it is
easy to show that P can be either true or false, depending upon the nature of
the polynomial, and the circumstances. And, it is an added benefit that the
circumstances also allow us to clarify the proper use of the identity symbol ≡.

P is true,

*if and only if*, all variables are specified as constants. For example, we know that, if Y in the FLT equation is an integer, then for any Y there is some integer factor A such that z – a = A. If we should also know that Z and Y are the*specific*integers Z_{1}and Y_{1}, then a is determined, and we have:**Z**, an_{1}– a_{1 }≡_{ }A*identity*.
But in FLT65, z is an unknown. Clearly, integers can
be assumed for X and Y, and if so, then whether or not z can be an integer is as
yet, in the proof, unknown. To assume that it is an integer at the beginning, is
premature and leads to

*circular reasoning*. For the polynomial z – a in FLT65, z is a real number, but its specific value is unknown, so when we set the polynomial z –a = A, a known integer factor of Y,**P is false**, because z and a can take on an infinite number of values. The only requirement imposed by setting z – a = A is that their difference is always A. The value of z can vary, and thus this equation is not an identity.
I once sent you the
judgement of a mathematician who had spent his entire working life as a
mathematics professor in a university. You dismissed his judgement that FLT65
was invalid as “something you had seen before.”

To
be clear, I had seen this line of reasoning several times before, and have given
it all the serious thought it deserves, so I saw it as the same knee-jerk
reaction I’ve seen numerous times, and refuted every time. I felt that you should
have recognized the circularity of the argument. But I apologize for my abrupt
manner. I should not have been so abrupt, and I certainly should not have been condescending.
I’ll copy the math professor’s comment here and try to respond more
appropriately. He said:

“Here
is a possible way of putting it that might convince Mr. Close. In his
argument he sets a = Z - A and considers the divisor
polynomial g(Z) = Z - a, which he says is a polynomial of degree 1
in Z. But g(Z) = Z - a = Z - (Z - A) = A, which is not
of degree 1 but of degree 0. When the divisor A is a
(nonzero) constant, the polynomial division algorithm over the reals just says
there exists a (unique) polynomial q(Z) such that f(Z) =
Aq(Z) + 0, where, of course, q(Z) = f(Z)/A. so there is no f(a).”

This argument completely misses the point of FLT65 by
assuming the definition a = Z – A which is not the case in FLT65. His statement
that

*I set*a = Z – A is false. You will not find this anywhere in FLT65. Z is the unknown, the dependent variable. Z – a is defined as a polynomial of the 1^{st}degree, meaning that the value of Z depends on the value of A, an integer factor of Y and the value of a, which is an integer if there is an integer solution of the FLT equation, which is as yet, in the FLT65 chain of logic, unknown. By assuming that Z is an integer by definition, the conclusion that Z –a is a constant, and therefore of 0 degree, is of course, circular reasoning, as mentioned above, and stated in my more abrupt response. By the way, I meant no disrespect to your professor friend by dismissing his comments. A university professor known as a number theory expert made the same mistake, but he quickly acknowledged that it was circular reasoning when I pointed it out to him.
Thank you for your statement of what you see as a disproof of FLT65. It enables me to better understand why you kept coming up with propositions that had nothing to do with FLT65. I think other critics may have had this same misconception about how FLT65 goes about proving FLT.

Your
statement has the logic of FLT65

*completely*backward:
“The

**, f(Z**__equation of constants___{1}) = A^{p}= (Z_{1}- a)^{p}, does**imply that the**__not__**(Z - a) is a**__variable__**of the**__variable factor__**f(Z).**__variable__
I agree! However, what FLT65 says is the converse: the fact that f(z) cannot contain z –
a as polynomial factor for real number values of the variables, implies that,
if there were an integer solution for the Fermat equation, then there would be
an integer version of f(z)/(z – a) = q(z) + f(a) where f(a) would equal zero,
and that would violate the ‘if and only if’ condition of the division algorithm
for polynomials.

It appears to me that the confusion comes from considering Z to be a specific integer before it is known whether z can be an integer or
not. Maybe this will be clearer if we go step-by-step:

Dividing

**f(z)**, a polynomial of degree p-1 by**z - a,**a 1^{st}degree polynomial, we have unique polynomials**q(z)**and**f(a)**of degree less than p such that:**(z**. Multiplying through by

^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+•••+ x^{p-1})/(z-a) = q(z) + f(a)/(z-a)**z-a**, we have:

**f(z) = (z**.

^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+•••+ x^{p-1}) = q(z)(z-a) + f(a)
From this we see that the polynomial

**f(z)**is factorable into two polynomial factors,**q(z)**and**z-a**,*if and only if***f(a) = 0**. But**f(a) =****a**, which^{p-1}+ a^{p-2}x + a^{p-3}x^{2}+•••+ x^{p-1}**equal zero because***cannot***a**and**x**are positive for any integer solution of the FLT equation.**.**

*Therefore, the FLT equation factor polynomial f(z) cannot be factored into two polynomials of degree less than p, one of which is z - a*
If there is

**an integer solution, then with the term-by-term substitution of the integer variables into the variable polynomial****f(z)**,**we obtain****the variable polynomial****f(Z)**for any integer solution of the Fermat equation. But, for any integer solution,**f(Z) = (Z-a)**, where^{p}**Z-a = A**, a single integer factor of**Y**. So now we have:^{p}
The hypothetical integer polynomial

**f(Z) = (Z**.^{p-1}+ Z^{p-2}X + Z^{p-3}X^{2}+•••+ Z^{p-1}) = q(Z)(Z-a) + f(a) = (Z-a)^{p}
By inspection of this integer polynomial
equation we see that

**f(a)**contains**Z - a**as a factor, and although we don’t know what the specific values of**a**and**X**are for an integer solution, we know that they are positive constants. So**f(a) = M(Z-a)**where**M**is an integer constant, and we have**f(Z) ) = (Z-a)**,^{p}**= (Z**), from which we have:^{p-1}+ Z^{p-2}X + Z^{p-3}X^{2}+•••+ Z^{p-1}) = q(Z)(Z-a) + M(Z-a**f(Z) = [q(Z) + M](Z-a).**
And

**q(Z) + M**is a variable polynomial in**Z**of degree less than p, call it**q**Thus for hypothetical integer solutions of the FLT equation, we have the variable polynomial_{1}(Z).**f(Z) = q**._{1}(Z)(Z-a)**But this is a violation of corollary III of the division algorithm****,**which tells us that the variable polynomial**f(Z)**cannot be divided into the factors**Z - a**and another polynomial of degree less than p. The only way we can avoid this contradiction is for z to be an irrational real, not an integer.
I have
demonstrated in at least three different ways, including Fermat’s favorite
method of proof, infinite descent, that if we ignore this contradiction, and assume
that two of the three variables x, y and z are integers, and solve the FLT
equation for the third variable, then that third variable cannot be an integer.

I don’t consider adding such demonstrations to
FLT65 to be necessary because the contradiction f(a) ≠ 0

*versus*f(a) = R = 0 is sufficient to prove FLT by itself. This contradiction is valid because it is obtained by applying the division algorithm and corollaries to variable polynomials, not constants, and a single contradiction is sufficient to prove there can be no integer solutions for**z**.^{p}– x^{p}= y^{p}
I
believe the whole confusion for most critics arises from assuming that the
division algorithm and corollaries are inappropriately applied to constants.
They think this because they jump to the conclusion that all three variables
must be treated as integer constants from the beginning of the proof.

**Thanks to you, this confusion has been made clear with the analysis of Prop P!**
Once you
see that P can be true

__or__false, depending upon the nature of the polynomial, and that in FLT65, f(Z) and Z – a are still polynomials of the variable Z, even though for a hypothetical integer solution, they are equal to the constant integer factors of Y^{p}, the logic of FLT65 becomes clear, and the search for counter examples and counter propositions becomes unnecessary and irrelevant.
Edward R. Close August 12, 2017

hmm i see a methodological problem here, let's assume i (if i would look more into detail into the EC 'proof) that i would find a mistake, and thus refute the 'proof', then there are these scenario's

ReplyDelete1) you EC would not accept my refutation; thus you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect.

2) you (EC) see the problem, try to fix it, come up with some modification; refuting that again would take time, and in the meantime, you you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect.

3) but if after a while i would again point out the mistake in your modification, then there are a few scenario's

a) you EC would not accept my refutation; thus you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect. BACK TO 1) above

b) 2) you (EC) see the problem, try to fix it, come up with some modification; refuting that again would take time, and in the meantime, you you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect.

c) but if after a while i would again point out the mistake in your modification, then there are a few scenario's: BACK TO 3) above (yes circular and endless spirals of thought)

CONCLUSION: this could lead to endless discussion. Thus it's safer for a spectator like me to look at what the math community at large is saying, namely, accepting the proof by Andrew Wiles, and dismissing all other 'proofs' as crank stuff. Meanwhile ofcourse it might be interesting for some odd math researchers, or math/sociology researchers to do a broad study on all (hundreds, or thousands) of crackpot 'proofs' but unless this leads to a clear result, another proof, accepted by the math community at large, i'm not going to waste my time on some other unconventional 'proofs', whether they may be true (very small chance) or not.

PS Now you EC can ofcourse delete this comment again, but i've saved it this time and if you delete it might popup some time elsewhere; lets say eg . on a certain ning forum, or something like that (youknowwhattamean)

I don't think I have ever deleted one of your comments, unless, of course it contained crude language, which you seem sometimes to like to use. But your analysis is interesting. I agree that if you start by looking for an error that you believe might lie behind methodological problems, rather than trying to understand the logic of the proof, it would be a waste of time.

DeleteDr Ed and Jaqui Close, I have to thank you for sharing your remedies for various toxic molds. My house is clean and my indoor garden harvest had only one spot of mold within the fattest flower top that was leaning into a wall! That is success on any day where I am from . Off subject perhaps, maybe not, but I feel good about the non-caustic, non-threatening solution you re-discovered and shared! It works for me and my indoor garden of flowers. sorry if post is inappropriate, you know what I mean. gwz

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