A recent email to an ISPE friend:

A good friend does not give up trying to convince
someone he considers to be his friend of the truth. I believe that you are my
friend, because if that were not the case, you would have given up trying to
convince me that my FLT65 proof is flawed long ago. So, in return, I must not tire
of trying to convince you that FLT65 is valid, as long as I see it that way.

After responding to your recent email containing the
short circular argument put forth by your retired math professor friend, I had
an inspiration. I believe I see a more direct way to explain FLT65. Here it is:

If one number is divisible by another, then dividing the smaller one
into the larger one produces a zero remainder, while if they are not divisible,
the remainder is non-zero. These simple facts are expressed by the division
algorithm and its corollaries and they are true for all polynomials over the
field of real numbers, whether reducible to integers or to any other real
number.

Recalling FLT65, we see that when p ≥ 3, p a prime number, the FLT
equation can be factored and expressed in the form (z-x)( z

^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+ ••• + x^{p-1}) = y^{p}. If there are integer solutions for z^{p}– x^{p}= y^{p}, then, with specific integer values of x, y and z, g(z) = z–x and f(z)= z^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+•••+ x^{p-1}must be equal to relatively prime integers raised to the pth power. That is to say that, if there is an integer solution, then y^{p }will be equal to B^{p}A^{p}, where B^{p}= z –x, and A^{p}= z^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+•••+ x^{p-1}, A and B relatively prime integers. Furthermore, for an integer solution, the fact that integers are closed with respect to addition guarantees that there is always an integer a, such that z –a = A.
Of course A

^{p}is divisible by A, so A^{p}= z^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+•••+ x^{p-1}must be divisible by A = z – a, and the division algorithm Cor. III says that f(z) is divisible by z – a, IF AND ONLY IF, the remainder, f(a), is equal to zero. Therefore, to find the value of x for any given values of z and a, we must set f(a) = a^{p-1}+ a^{p-2}x + a^{p-3}x^{2}+•••+ x^{p-1}= 0 and solve for x. There are exactly p-1 solutions to this equation and for all of them, x is non-integer. This proves that for z and y equal to integers, x cannot be an integer, and FLT is proved.
I
believe that the argument above is a more direct way to see FLT65, and it is
completely equivalent to FLT65. I also believe that it becomes even clearer
when illustrated with a numerical example, and I will use one provided by the critics.

While
there are a few competent mathematicians who agree with me that FLT65 is a
valid proof of FLT, more of them agree with you. For example, while reviewing
my work, a Nobel Prize physicist and a very competent Israeli number-theory
professor of mathematics, responded with what, in their opinions are counter
examples that call FLT65 into question and, they believed, might even refute
it.

They both correctly noted that my argument in FLT65 is that when the factor
of the Fermat equation f(z) = z

^{p}+xz^{p-1}+… + x^{p-1 }is divided by z – a, the remainder, f(a) cannot be zero, while, for any integer solution, f(z) is definitely divisible by z – a. In fact, f(z) = A^{p}divided by z - a = A is A^{p-1}, where, if there is an integer solution to the Fermat equation, A is an integer, and this produces an inescapable contradiction. They argued that this is, or may be, incorrect because they could produce examples for the equation when p = 3 with the remainder f(a) non-zero even though f(z) is clearly divisible by z - a when certain integers are chosen for z, x and a.
Here is one such example
offered by the math professor:

Let z=7 and x=4. Thus 3
divides z

^{2}+xz+x^{2}, because f(z) = 49 + 28 + 16 = 93 = 3x31. So for a=4, the integer z - a = 3 divides the integer z^{2}+xz+x^{2}. However, in the polynomial ring R[Z], the polynomial z - a does not divide the polynomial z^{2}+xz+x^{2}=z^{2}+4z+16. Indeed, the remainder is a^{2}+xa+x^{2}> 0. Thus, he reasoned, the non-zero remainder when dividing polynomials does not prove that f(z) is not divisible by z – a = A if x, y, z, and a are integers.
There is however, a
serious error in this argument. The error lies in the fact that, after choosing
z = 7 and a = 4, the value for x is arbitrarily, and

*incorrectly*chosen to make f(z) divisible by 3, allowing the production of a spurious “counter example”. The error is compounded by assuming that this supposed disparity in divisibility between the polynomial f(z) and its integer value may exist for the Fermat equation.
In fact, if z = 7 and a
= 4 in the Fermat equation, then x

*cannot*be equal to 4. This is easily and clearly demonstrated as follows:
The division algorithm
expresses the essence of the fundamental operation of division for all real
numbers, including integers. Corollary III of the division algorithm says that
f(z) is divisible by z – a IF AND ONLY IF f(a) = 0. Therefore, in this example fabricated
by the math professor, in order to see what x must be to satisfy the equation
when z = 7 and a = 4, we must set f(a) = 4

^{2}+4x+x^{2}= 0. When this equation is solved for x, we see that x cannot be equal to 4. In fact, solving this equation for x, we see that the two values of x satisfying the equation with z = 7 and a = 4 are – 2 + 2Ö3i and – 2 - 2Ö3i, which are complex numbers, and definitely not integers.
This
is easily generalized for all integer values of z and a, and for all p>2
because all values of p are odd allowing the factorization into z – x and z

^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+ ••• + x^{p-1}, a polynomial of p terms; and that is exactly what FLT65 does. The polynomial f(a) can never equal zero if x, z and a are integers, which they must be for an all-integer solution of the Fermat equation z^{p}– x^{p}= (z-x)(z^{p-1}+ z^{p-2}x + z^{p-3}x^{2}+ ••• + x^{p-1}) = y^{p}. This proves FLT.
Now
one must ask: Why has this simple proof, which I believe is, in essence,
Fermat’s “marvelous proof”, been overlooked for more than 300 years, even by
the world’s most brilliant mathematicians???

It appears to go back to Leonhard
Euler and Carl Friedrich Gauss, arguably two of the most brilliant
mathematicians of all time. Euler used complex numbers to prove FLT for p = 3,
and Gauss developed modular algebra in an effort to prove or disprove the
solvability of Diophantine equations including FLT. Unfortunately, like many
mathematical procedures, modular algebra obscures as much about integer and
non-integer polynomials as it reveals. When Gauss was unable to produce a proof
using this method, he famously announced that he could produce any number of
such theoretical propositions that could be neither proved nor disproved, and
thus would waste no more time on it. This set the tone for many professional
mathematicians in the years to follow.

Kurt
Gӧdel’s incompleteness theorem proved that there
are always logical propositions that cannot be proved or disproved within the
mathematical system giving rise to them. This strengthened Gauss’s speculation
that FLT might not be provable using basic mathematics. Add to this the increasingly
extreme specialization encouraged by academia in the last 200 years, and you
have a general attitude that Fermat must have been mistaken about having a
proof.

Especially
after Andrew Wiles and Richard Taylor produced a torturously complex proof of hundreds
of pages in 1995, it was thought probable that Fermat had fooled himself into
believing that he had a proof, when in fact he had not, because the complex
theorems used in Taylor and Wiles’ proofs were not available to Fermat in 1637.
This line of reasoning, while convincing, of course does not prove there can be
no simple proof.

After
many years of trying to get professional mathematicians to take my 1965 proof
of FLT seriously, I had given up. When I discovered in about 1989 or 1990 that
FLT had an important application in quantum physics, I revisited FLT65. In
2010, even though the quantum physics application only required FLT to be true
for p £ 9, I mentioned my 1965
proof to Dr. Neppe, who was intrigued, and after studying it and proving it
correct for himself, urged me to resume efforts to get it recognized and
accepted.

To
date, only a few competent mathematicians have agreed with me that FLT65 is a
valid proof, but, importantly, no one has shown me any real proof that FLT65 is
not valid. The proclivity of professional mathematicians to dismiss it because
of the belief that no simple proof is possible has led even otherwise competent
mathematicians to think erroneously that examples like the one presented above
disprove FLT65.

Even those who have acknowledged that such examples are not
counter examples because they have no relevance to actual solutions of the
Fermat equation, apparently are loathe to think that FLT65 could be valid.

I
believe that the simplified FLT65 approach presented above should convince some
skeptics, perhaps including you, my friend, of the truth of FLT65, if it is
carefully and thoroughly considered.

With
Regards,

Ed
Close June 16, 2017

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