I think that this recent exchange with a friend who has been looking at FLT65 for some time may be of interest to those who are following the subject of Fermat's Last theorem, which as they may know, is importent to the explanation of quantum phenomena in TDVP. This conversation may be helpful for anyone interested in understanding the logic of the 1965 proof. For this reason, I am posting my latest response here without identification of my friend or his math professor friend.

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**_{1} – a_{1 }≡_{ }A, an *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 __equation of constants__,
f(Z_{1})
= A^{p} =
(Z_{1} -
a)^{p},
does __not__ imply that the __variable__ (Z
- a) is a __variable factor__ of the __variable__ f(Z).

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**^{p-1} + z^{p-2}x + z^{p-3}x^{2} +•••+
x^{p-1})/(z-a) = q(z) + f(a)/(z-a). Multiplying through by **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**^{p-1} +
a^{p-2}x + a^{p-3}x^{2} +•••+ x^{p-1}, which *cannot* equal zero because **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)**^{p}, where **Z-a = A**, a single integer factor of **Y**^{p}. So now we have:

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**^{p-1} + Z^{p-2}X
+ Z^{p-3}X^{2} +•••+ Z^{p-1}) = q(Z)(Z-a) + M(Z-a),
from which we have:** 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**_{1}(Z).
Thus for hypothetical integer solutions of the FLT equation, we have the
variable polynomial** 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