Saturday, November 28, 2015


Over the past 50 years, the only serious question raised in regard to the FLT65 proof has been the concern that the division algorithm for algebraic polynomials might not be applicable to a polynomial factor of the Fermat equation for specific integer values of x, y and z as stated in FLT65. This concern prompted a few mathematicians reviewing FLT65 (3 out of more than 50) to produce what they considered to be counterexamples. However, as we will see, even when mathematically correct, none of the examples offered by reviewers are applicable to the FLT equation because they apply to functions that are not related to the Fermat equation, or because they produce non-unique quotients and remainders in the division process, producing an unresolvable contradiction and thus are not valid counterexamples.

Why it has taken so long?
Because FLT65 is a ‘simple’ proof, using only concepts that were available to Pierre Fermat in 1637, a mathematician’s first reaction to it is likely to be: “It can’t be true, because anything simple would have been found by any one of the many world-class mathematicians who have tried to prove FLT over the years, like Descartes, Euler, Derichlet, Dadekind, Gauss, and Hilbert, just to name a few.”

So most professional mathematicians won’t bother to look at a simple proof of FLT because they consider it a waste of time. That is apparently why many, even most, of the mathematicians to whom the author submitted FLT65 never even looked at it. Those who did look at it, instead of taking it seriously as a potential proof, looked for the first hint of a mistake – any mistake, however minor - to dispose of it as quickly as possible.

The author’s first experience with this kind of intellectual reticence came in 1966 when he submitted the proof to a to a university professor of mathematics who happened to be the Chairman of the Iowa Academy of Sciences, at the time, hoping to get it accepted and published.
It was very disappointing that the first professional mathematician to see FLT65 rejected it on the erroneous belief that it would also apply when n = 2 and thus would be disproved by the existence of the well-known Pythagorean triples. His hasty rejection of FLT65 revealed the fact that he hadn’t read very far. If he had, he would have seen that the case n = 2 is excluded near the bottom of the first page, in the proof of the division algorithm, where it is shown that, when the degree of the divisor and dividend are the same, which is the case when n = 2, the division algorithm, of key importance in the proof, doesn’t apply. The division algorithm does, of course, apply for all values of x, y and z when n > 2.  See Appendix A.

Due to acute disappointment with the unwarranted rejection, and his circumstances as a young professional starting a career in applied mathematics and physics, the author didn’t respond at the time, but waited for a chance to submit it to another mathematician for review. When he did, he quickly learned that most professional mathematicians would not take purported FLT proofs seriously based on the fact that every well-known mathematician living between 1637 and 1965 had tried to prove or disprove FLT and failed. Undaunted, the author continued to submit the proof to mathematicians for review over the years.
Most of the submissions and responses to FLT65 were documented, particularly those from mathematicians who bothered to reply. Of those, a small number, even though they did not identify an actual error, offered the opinion that FLT65 was not a valid proof. It is likely that this opinion was based on the reasoning cited above about all the great mathematicians who had tried, because they gave no mathematical argument supporting their opinions.

This discussion will concentrate on the few who did provide some mathematic arguments that they thought revealed a flaw in FLT65 that at least brought it into question.

The responses from those to whom the proof was submitted fall into one of the following categories:

1.    Some declined to look at it either because they didn’t have time or they didn’t want to take the time to review it.
2.    Some refused to look at it because number theory was not their field of expertise and/or interest.
3.    Some refused to look at it because they did not believe a simple proof of FLT possible.
4.    Several of the reviewers, including all the members of a graduate class in number theory, found nothing wrong with the proof.
5.    Some of the reviewers believed the methodology of FLT65 to be incomplete and/or invalid.

Those who believed the method was incomplete and/or invalid cited one or more of the following reasons to consolidate their argument:
i.       They thought it contradicted the existence of the Pythagorean triples (n = 2)
ii.     The case n = 4 was not addressed
iii.   Conclusions drawn from the factorization of the FLT equation, xn + yn = zn, as an algebraic polynomial might not apply to the numerical factorization of the polynomial for some specific integer solution (X1, Y1, Z1). (See Example #1 below.)
iv.   Concern that the algebraic remainder r(Z) of the FLT65 proof might be non-zero for specific integer solutions to FLT even if Z-a divides f(Z). (Example #2 below)
v.     Concern that no contradiction is obtained by applying the method of infinite descent [9] to the non-zero remainders obtained by repeatedly dividing f(z) by z - a. (Example #3 below).
vi.   They found the notation to be unclear or confusing (Objection #4 below)

Resolution of Concern #i: It contradicted the existence of the Pythagorean triples (n = 2)”
The first professional mathematician to whom FLT65 was submitted rejected it because he thought that, if correct, it would imply no integer solutions for X2 + Y2 = Z2, which would of course be a problem, because the Pythagorean integer triples are well known integer solutions to this equation. But when the degree of the FLT equation is 2, the degree of f(Z) and g(Z), m = n = 1, and the division algorithm does not apply. The fact that the algorithm applies only if n > m is clearly stated on the first page of FLT65, and thus X2 + Y2 = Z2 is excluded from the proof and FLT65 cannot be falsified on this basis because it does not apply to FLT.
Resolution of Concern #ii: The case n = 4 was not addressed”  
A few reviewers correctly noted that FLT65 does not specifically address the case n = 4. Technically, proof of FLT for n = 4 is needed for completeness because 4 is neither prime nor the product of primes greater than 2. Proof of FLT for n = 4 covers all values of n equal to powers of 2, which are not multiples of primes >2.  With FLT proved for n = 4, since all n equal to non-prime odd integers are factorable into prime numbers, it is sufficient to consider only n = p, odd primes, in any proof of FLT. Proof of FLT for n = 4 was not included in FLT65 because it had already been proved by a number of mathematicians, including the author (1962), Legendre (1930), Hilbert (1910), Kronecker (1900), Euler (1750), and Fermat himself (1640). Fermat published a proof by infinite descent [10] that the area of a right triangle cannot be equal to the area of a square, which implies no integer solution is possible for the case n = 4.
So the fact that the case n = 4 is not specifically mentioned in FLT65 does not invalidate the proof for n = p, prime numbers > 2, or FLT65 in any way. For completeness in this discussion, the author’s proof by infinite descent of the case n = 4 is included in Appendix B of this paper.

Concerns #iii – #v: In general, critical reviewers who offered what they considered to be counterexamples disproving FLT65, questioned whether the division algorithm, dealing with algebraic polynomials, can be applied to integers, and pointed out that for some n - 1 degree algebraic polynomials that are  not factorable into polynomials by virtue of a non-zero remainder, there may be specific integers that produce a value for f(Z) that is divisible by the integer value of Z – a, in spite of the non-zero remainder. They jumped to the conclusion that this represented a flaw in FLT65 and a few provided examples to try to make the point. Those examples are presented below, but it is quite easy to show that, whereas the examples may be independently mathematically correct, they are not applicable to the FLT equation. And, of course, an example that does not apply to the FLT equation obviously cannot be a valid counterexample of the FLT65 proof.

Concern #vi: There is some legitimacy to the criticism that the original notation used in FLT65 can be confusing. But that does not refute the FLT65 proof.  In the proof of the division algorithm in FLT65, e.g, N was used for the degree of the FLT equation and n was used for the degree of one of the factors. Later, n was used instead of N. The switch from N to n for the degree of the FLT equation may have been what led to the summary rejection by the first reviewer in 1966. And the fact that FLT65 notation did not clearly distinguish between integer and real number variables may have caused confusion contributing to the suspicion that the division algorithm might not apply to hypothetical integer solutions, but the proof should not be dismissed on the basis of unconventional notation. This concern is addressed in Appendix D of this presentation, where standard notation distinguishing between real number variables, integer variables, specific integer values, and rational and irrational solutions is used.
Of the fifty-plus professional mathematicians to whom FLT65 has been submitted, only three have offered arguments of refutation, and of those three, two used examples like Examples #1 and #2 below; only one (Example #3) commented on the method of infinite descent, implied by the uniqueness requirement in FLT65, but not mentioned specifically.
All three invoked Concern vi (Objection #4), but effectively that is semantic but does not invalidate the proof: The revised nomenclature as used in the Appendix D is easier to understand.
As indicated above and demonstrated below, none of these arguments actually refute FLT65. The actual examples provided by reviewers trying to disprove FLT65 are now presented below.

The following attempt to produce a counterexample is typical of several offered by reviewers.

Example #1:
“When the polynomial f(Z) = Z + X is divided by the polynomial Z - a, the remainder is f(a) = a + X, which is non zero for a and X equal to positive integers. However, if, for example, Z =5, X =3 and a = 3, we find that f(Z) = 8, which is divisible by Z- a = 2, even though the remainder is not zero. This invalidates the use of Corollary III, of central importance in the Close FLT65 proof.”

Refutation by ERC:
This example does not in fact invalidate the use of the Division Algorithm or any of its corollaries in the FLT 65 proof because the degree, n, of f(Z), and the degree, m, of Z – a in this example are the same. And m = n, is a condition excluded in application of the Division Algorithm. See Appendix B. Also, the polynomial f(Z) used in the example is not the f(Z) of the FLT equation used in the proof, so this example fails on at least two counts.
This explanation therefore refutes the reviewer’s attempted disproof of FLT65.

A more compelling example produced by one mathematician, is the following:
Example #2: “For n =3, f(Z) = Z2 + ZX + X2 = (Z – a)3, the nth power of the first order divisor of the proof. Let Z = 73, X = 17 and a = 54, giving Z – a = 73 -54 = 19; f(Z) = Z+ XZ + X2 = 6859, and (Z- a)3 = 193. Now, the algebraic polynomial f(Z) is not divisible by the algebraic polynomial Z – a because when f(Z) is divided by Z – a, the remainder f(a) is not zero. But, when the integer value of f (Z), obtained when the numbers given are substituted for the variables, is divided by the integer value of Z – a, we see that f(Z) is divisible by Z – a.”

This, the reviewer believed, contradicts the claim in FLT65 that there can be no integer solutions of the FLT equation if the remainder, f(a) ≠ 0 when f(Z) is divided by Z – a.

Refutation by ERC:
This example avoided one of the mistakes made in the first example by using the actual FLT polynomial f(Z) for n = 3. But, for integer solutions to the Fermat equation, both f(z) and the factor g(Z) = Z – X must be equal to integers raised to the nth power (See the Brief Description of FLT65 above and Appendix A). In the example, the factor Z – X = 73 – 17 = 56, which clearly is not the 3rd power of an integer.

This fact therefore refutes the reviewer’s attempted disproof of FLT65; but, while this example is not a valid counterexample, it is instructive, because it affords us an opportunity to go a step further and see why the non-zero remainder f(a), obtained when f(Z) is divided by z – a, is both necessary and sufficient to prove FLT, as stated in FLT65.

Further discussion by the author of the application of the Division Algorithm in FLT65:
Substituting integer variables A, X and Z into Equation (2.), f(z)/(z-a) =q(z)+f(a)/(z–a), the division algorithm equation for algebraic polynomials when f(z) is the p -1 degree factor of the FLT equation, we have f(Z)/(Z–A)=q(Z) + f(A)/(Z–A) f(Z) =q(Z)(Z–A) + f(A). Then, by the Theorem of Parallel Factorization, If f(Z) is the factor (Zp-1 + XZp-2 + …+ Xp-1) of the FLT equation, then f(Z) = (Z – A)p and, by Corollary I, f(A) must contain Z – A. If we also substitute specific integers A1, X1 and Z1, representing a hypothetical integer solution of the FLT equation, directly into the Division Algorithm equation, we get f(Z1)/(Z1 –A1)=q(Z1)+f(A1)/(Z1–A1), and by the Theorem of Parallel Factorization for Polynomials of Integer Variables, f(Z1) and f(A1) contain Z1 – A1. So we see that for all real number solutions of the FLT equation, integer or not, f(z) and f(a) must contain z – a; for integer variables, f(Z) and f(A) must contain Z – A; and for specific integers, f(Z1) and f(A1) must contain Z1–A1. In all three cases, the unique form of f(z), an algebraic factor of the FLT equation, is preserved.

The fact that there is a non-zero remainder when f(z) is divided by z – a proves FLT because if there is an integer z when x and y are specific integers X1 and Y1, then the quotient, q(Z), and the remainder, f(a) are unique, and Z – a divides f(Z) if and only if f(a) = 0. It is this uniqueness requirement that produces unavoidable contradictions when x, y and z are assumed to be integers. These unavoidable contradictions are demonstrated below and in detail in Appendix E.
The importance of this uniqueness in the FLT65 proof, apparently overlooked by most reviewers, is illustrated in the example above by the fact that the ‘counterexample’ integers do not produce a first-degree factor Z – X that can be equal to B3, where B a positive integer, as must be the case for integer solutions of FLT. This eliminated Z = 73, X =17 and a = 19 as a counterexample. Now let’s see how the failure of this specific-integer example can shed some light on the general case.

As shown in FLT65 and above: The FLT equation, yn = zn - xn, with n equal to a prime >2, is factorable, yielding yn = (z – x)(zn-1+xzn-2+…+ xn-2z+xn-1) = g(z)f(z). The factor f(z) is a unique form of the general polynomial form Azn +Bzn-1x +Czn-2x2 +…+Mxn with all coefficients, A, B, C, …, M  = 1.

If there are integer solutions to the FLT equation, then we can let X and Z be integer variables such that g(Z) = Z – X = J1n , and f(Z) =Zn-1+XZn-2+…+Xn-1 = K1n, with J1 and K1 relatively prime integers, so that y = J1K1 and yn = J1nK1n. Since integers are closed with respect to addition and subtraction, for any Z, there will always be an ‘a’ such that K1 = Z – a, for any value of Z, and for an integer solution, f(Z1) = (Z1 – a)n. Thus, we have:

 f(Z1)/(Z1 – a) = q(Z1) + (an-1+X1an-2+…+X1n-1)/(Z1 – a)                      

In the terms of long division, Z1 – a is the divisor, q(Z1) is the quotient,
and r = (an-1+X1an-2+…+X1n-1) = f(a) is the remainder.

Because n = 3 is the simplest case for FLT, and because n = 3 in the reviewers’ examples, let’s have a closer look at this equation when n = 3.

When n = 3, for a specific value of Z = Z1, a becomes the unknown, and Equation (2.) becomes:

f(Z)/(Z – a) =  q(a) + f(a)/(Z1 – a) = Z1 + X1 + a + (a2+aX1 +X12)/(Z1 – a)                    (3.)

If all terms of this equation are integers, as they must be if there are integer solutions to the FLT equation, then, because the integers are closed with respect to addition and subtraction, and because f(Z) = (Z – a)3,  by simple inspection of this equation, we can see that f(a) must be divisible by Z – a. But Z – a = K1, and for any integer, K1, there is an a1, such that a – a1 = K.

Dividing f(a) by a – a1 = Z – a, we get: f(a)/(a – a1) = a + X + a1 + (a12+Xa1 + X2)/(a – a1), implying, again that, because integers are closed with respect to addition and subtraction, f(a1)= a12+Xa1 + X2 is also divisible by a – a1 = a1 – a2. Just like in simple long division, with each division, the remainder, f(ai) gets smaller and smaller, until with some unique value of ai and ai-1, the remainder, f(ai), reaches a unique minimum, with a unique quotient, q(ai). Corollary III of the division algorithm says that if the quotient and remainder are unique, f(Z) is divisible by Z – ai IF AND ONLY IF f(ai) = 0. But f(ai) can never equal zero if the ai’s are integers.

Thus, f(Z) cannot be equal to (Z – a)3 if a is an integer; and if a is not an integer, y cannot be an integer, and FLT is proved for n =3.

To see exactly how this works with specific integers, we can insert the integers from the ‘counterexample’ given in Example #2 above, namely: Z = 73, X = 17, a = 54. These specific values were selected by a mathematician reviewing FLT65 for the express purpose of creating an f(Z) equal to Z2 + XZ + X2  that, when divided by Z – a using algebraic long division, produces a remainder f(a) that is non-zero while, nevertheless, the integer value of that remainder is divisible by the integer value of Z – a, ostensibly showing that corollary III of the division algorithm does not necessarily apply to integers in the same way it applies to algebraic polynomials. However, this is a case of subtle, albeit unintentional misdirection, because, as we have seen, the Integers that accomplish this, e.g. Z = 73, X = 17, a = 54, do not apply to the Fermat equation.

There are actually many combinations of integer values of X, Z and a, that can be chosen to produce non-zero remainders that are divisible by Z – a, but the most important question is: Can any of them produce the unique values of Z and a that will satisfy the FLT equation? With these values, dividing f(Z) by Z - a produces a remainder, f(a) = a2+Xa+…+X2 = 542+17x54+172= 4,123, which is divisible by Z – a = 19. It should come as no surprise that f(a) = 4,123 is divisible by Z – a = 19 (4,123/19 = 217), because, as shown above, it must be the case that f(a) is divisible by Z – a for integer solutions of the FLT equation.

We found, however, that this numerical example was not a counterexample to FLT65, because it did not satisfy the FLT equation. Unless the values chosen for X, Z and a satisfy the FLT equation, they produce a false q(Z), and in this case, q(Z) was much too small because it produces a remainder, R1=4,123, much larger than the divisor Z – a = 73 – 54 = 19. In long division, if the remainder is larger than the divisor, we increase the value of the quotient, multiply by the divisor and produce a new, smaller remainder; and we continue this procedure until we find the unique minimum remainder, which will be either zero or a positive number smaller than the divisor Z – a. If the minimum remainder is non-zero, then f(Z) is not divisible by Z – a, contradicting a necessary condition for the hypothesized integer solution of the FLT equation.

With the case at hand, we know that the remainder, r1=f(a), must be divisible by Z – a = 19, and it is: 4123/19 = 217. However, if there are integer solutions, X and Z are integers, and because integers are closed with respect to addition and subtraction, we can introduce another integer, a1, defined by a – a1 = Z – a = 19. In this example, for a = 54, a – a1= 19 a1= 35. With a new integer a1= 35, we can now treat a as an integer variable. Dividing the remainder r1= f(a) by a – a1, we get an additional quotient, q1(a1), and a new remainder, r2 = a12+a1X+X2 = 2,109. Thus

f(Z)/(Z – a) = q(Z) + q1(a) + 2,109/(Z – a).

But this is a contradiction because if Z and a are integers, we see that the integer pair (ai-1 and ai) that produces the smallest f(ai) are 16 and - 3. See Appendix E. But the fact that the minimum remainder is not zero and yet does contain Z – a = 19, tells us that if Z and a are integers, we have the contradiction that for the smallest f(ai), q(ai) and f(ai) are unique, implying that the smallest f(ai) must be a positive integer not containing Z – a; but, if f(Z) is an integer, and q(ai) is an integer, by inspection of Eq. (I/A-6) and closure of integers with respect to addition, f(ai) must contain the integer value of Z – a. This is an unresolvable contradiction unless the ai’s, and consequently, Z, are not integers, proving that the integers Z = 73, X = 17 and a = 54 do not provide a counterexample to FLT65 for n = 3. Notice that this conclusion is a direct result of the fact that r1 = f(a) is non-zero.
Thus we see that r1 = f(a) ≠ 0 leads to an unavoidable contradiction proving that when f(Z) is divided by Z – a and the remainder is non-zero, the integers Z = 73, X = 17 and a = 54 cannot satisfy the FLT equation, and thus do not provide a valid counterexample.

This further explains the refutation of the reviewer’s attempted disproof of FLT65.

Generalization of the procedure produces an infinite descent:
The validity of the application of corollary III of the division algorithm in FLT65 can be demonstrated by the evaluation of r1 = f(a), produced by dividing f(Z) by Z – a, while not applying specific values for a, X and Z, but treating them as variables and continuing in the manner described above, increasing the quotient with each division, in order to reduce the remainder as we evaluate f(a) with a as an integer variable. This constitutes an infinite descent [7] that ends when ri reaches its unique minimum, which it must, if the ai are integers, which they must be for integer solutions of the FLT equation. But this minimum f(a) is non-zero, creating the contradiction that because q(ai) and f(ai) are unique, the smallest f(ai) must be a positive integer not containing Z – a; but, if f(Z) is an integer, and q(ai) is an integer, then by inspection of Eq. (I/A-6) and closure of integers with respect to addition, f(ai) must contain the integer value of Z – a. This contradiction proves that, while Z – a may be equal to an integer, as in the case of Example #2, Z and a individually, cannot be integers, proving FLT.

This process of infinite descent, which is simply an application of the long division process we all learn in elementary school, may be easier to see using actual numbers. So, using the example above, and continuing to increase the quotient in order to reduce the remainder as we evaluate f(a) as an algebraic polynomial in a, as variable Z is replaced by Z1, a specific integer value, and a takes on a  series of decreasing integer values, we have: Z - a = Z1 – a1  = a1 - a2 = 19 a2 = 16, yielding r3 = a22+a2X+X2 = 817; a2- a3 = 19 a3 = - 3, yielding r4 = a32+a3X+X2 = 147; a3 - a4 = 19 a4 = - 22, yielding r5 = a42+a4X+X2 = 299. So we see that, continuing in this way, ri reaches its minimum value as r4 = 147. For the detailed calculations, see Appendix E. The minimum ri is non-zero, producing unresolvable contradictions as shown above, proving that these integers do not produce a counterexample. This failure is a direct result of the fact that r1 = f(a) is non-zero.
The infinite descent procedure can be applied to any set of integer values of Z, X, a, a1, a2, a3,… ai, in the same way: If f(Z) is divisible by Z – a = a – a1 = a2 – a1 =… = ai-1 – ai, corollary II of the division algorithm says that the remainder of f(Z) divided by Z – a is f(a) for any real value of Z. For all primes p > 2, f(Z)/(Z – a)= q(Z) + f(a)/(a – a1)= q(Z) + q1(a) + f(a1)/(a1 – a2) = q(Z) + q1(a) + q2(a) + f(a2)/(a2 – a3) ... But no f(ai) can equal zero if Z, X and for integer solutions to the FLT equation, the ai are all integers, and the division process never terminates. In fact, after f(ai) reaches a minimum, it begins to increase again. (See Table I/A in Appendix E.) This is no problem for non-integer Z and a, but any set of integer values of Z, X and a, chosen so that f(a) contains Z – a, creates a contradiction as shown above.
Dividing f(Z) by (Z – a) produces an algebraic polynomial quotient, q(a) and a non-zero remainder, f(a). Repeated division of successive remainders, f(ai),  by ai-1 – ai, must eventually produce a unique minimum remainder, and since q(ai) and r(ai) for the minimum remainder are unique, the ‘If and only if’ condition of the division algorithm’s corollary III applies and the non-zero remainder proves that f(Z) cannot be divisible by Z – a, contradicting the condition f(Z) = (Z – a)p necessary for there to be integer solutions to the Fermat equation. Thus, f(Z)/(Z – a), which must be an integer if the FLT equation has integer solutions, cannot be an integer, and FLT is proved by assuming it to be false and showing that, because f(a) ≠ 0, the assumption that there are integer solutions leads to an unresolvable contradiction. For a detailed account of the application of this infinite descent method to FLT, see Appendix E.
This explanation clarifies the author’s refutation of the reviewer’s attempted disproof of FLT65. It also provides a demonstration of how Gödel’s incompleteness is overcome by expanding the mathematical paradigm of the reviewers with an additional axiomatic principle and theorems.

Example #3: This example is copied from the reviewer’s email with only minor non-substantial notational changes for consistency with previous examples.

Reviewer: “The algorithm that you describe does not terminate indeed, but this is not a contradiction since ai < 0 for sufficiently large i.”

ERC Response: It is not a contradiction only if a is a non-integer. This confusion may arise because p – 1 non-integer solutions to the FLT equation do exist: It can be shown that they are irrational nth roots of integers. However, if a is an integer, as must be the case for FLT to fail, it is a contradiction because the minimum remainder, some specific f(ai), must be less than Z – a or  equal to zero, which is an impossibility, since f(a) ≠ 0 f(a1) ≠ 0 f(a2) ≠ 0, … as stated in FLT65 and shown by infinite descent above and in Appendix E.

Reviewer: “Consider the following example:
For n = 3, let a = 5 and X = 2.
Thus f(a) = a2 + aX + X2 = 39. Let Z – a =13 then Z – a = a – a1= 5 - a1 = 13.
We have a1 = - 8 < 0. So you cannot use this algorithm in connection with FLT.”

ERC Response: This does not follow. The point is that ri reaches a minimum without ever equaling zero, either before or after the minimum, implying that there will always be a non-zero remainder, which cannot be the case for integer solutions of the Fermat equation, but this poses no problem for non-integer solutions.

When the author pointed out to this reviewer that an actual counterexample to FLT65 would produce an integer solution for yn = zn - xn, which would also falsify Andrew Wiles’ proof, the reviewer did not respond and terminated our communications. Several months later, when asked by a third party to comment on FLT65, he replied that it is based on an incorrect argument. When asked to give a specific statement indicating why it is incorrect, he responded with the following (copied directly from his email):

The proof uses the following wrong argument:

Let f(X) and g(X) be monic polynomials over Z (the ring of integers), and
let r(X) be the remainder obtained when dividing f by g in Z[X].
Then, if m is an integer such that g(m) is not zero, the remainder obtained when dividing f(m) by g(m) is r(m).
[a polynomial is monic if its leading coefficient is 1].

For example, let f(X) = X+3, and g(X) =X. Then r(X) =3.
If X=3, the remainder obtained when dividing f(3) by g(3) is 0, while r(3)=3.”

ERC: An astonishing response, considering that it is easy to show that (1.) the ‘wrong argument’ presented as representative of the FLT65 proof is an incomplete and incorrect statement of the division algorithm. If it is to apply to the FLT equation factor f(Z), the monic polynomial f(Z) must be of degree > 2 and the degree of g(Z) must = 1. (2.) The example he provides, like that in example #1 above, has no relationship to f(Z), and thus no bearing on FLT or the FLT65 proof.

The author still has high regard for this reviewer, a respected professional number theorist. But, strangely, it is as if he completely ignored everything previously discussed and reverted to an easily refutable ‘counterexample’! I can only conclude that he answered hastily, without thinking it through.

Refuting these statements by the reviewer may seem unnecessary and redundant. However, for the sake of completeness, the following brief analysis of the reviewer’s latest statement is provided:

The polynomial factors of the FLT equation in FLT65 are, indeed monic polynomials over the ring of integers. It is important to note, however, that they are of degree n – 1 and 1, respectively, unequal, as required for application of the division algorithm, while the degree of f(X) and g(X) of the reviewer’s example are equal. And, of course, the f(X) of the example is not of degree n – 1, making it totally irrelevant to FLT and the FLT65 proof.

This further explains the refutation of the reviewer’s attempted disproof of FLT65.

Objection #4: confusion over the original notation, that may have contributed to objections and the invalid counterexamples discussed above:

The notation in the 1965 proof may have confused readers because it is non-standard and not as well defined as it could have been. At the time, the author used unconventional notation deliberately because he thought it made it easier to distinguish when the arguments were being considered as unrestricted real-number variables from when they were being considered as integer variables. The subscript identified an algebraic symbol as a specific integer, as opposed to an integer variable. For what he thought was a valid reason, the author did not provide a notation that distinguished between integer variables and specific integer constants, and this was problematic because confusion may arise if a specific notation is not clearly defined and its use properly justified. Therefore, in this discussion, the notation distinguishes three types of numbers precisely defined for clarity:

·       Lower-case letters, like x, y and z represent variables with no numerical restrictions.
·       Upper-case letters like X represent variables restricted to integers.
·       Upper-case letters with subscripts like X1 represent specific integer values of the variables.
In addition, we will distinguish between integer factors and algebraic factors as follows:
·       g(x) e f(x) means the polynomial g(x) is an algebraic factor of the polynomial f(x), or stated another way, g(x) is contained in f(x) as an algebraic factor; and
A B means A is an integer factor of the integer B, or A is contained in B as an integer factor.

The following is an example, copied here in its original format, from one of the reviewers of FLT65 who raised this concern:
“F(Z) = ZN – X1N - Y1N = 0, and thus can be expanded to produce:
F(Z) = (Z-X1)(ZN-1 + ZN-2X1 + ZN-3 X12 +•••+ X1N-1) - (Z-X1)(Z –a )N = 0.
By inspection of this equation, we see that G(Z) is a polynomial factor of F(Z) ... We can see the error here by using z for the variable and Z for the constant figuring in the Fermat equation….”

Response: In the “clarified” version (See FLT65C, Appendix D) we see that this problem never arises because the z of the FLT equation is always a variable, either an unrestricted real-number variable, z, or an integer variable, Z, throughout the proof except when it is replaced by a specific integer, indicated by Z1, in which case, a becomes a variable. The author acknowledges that this is not indicated as clearly as it could be in the original FLT65 proof, and may have caused some confusion. This confusion is clarified in this presentation.

The key to understanding the FLT65 proof is realizing that the legitimate application of the division algorithm and corollaries to the polynomial factors of the FLT equation, as polynomials in the variable z, leads to a unique  remainder which reveals the fact that if x and z are integers, y cannot be an integer. Thus the question of factors of polynomials in the variable z, versus constant integer factors is irrelevant.

This response by the author explains why the question of notation is not relevant in reviewers’ attempted disproofs of FLT65.

The FLT65 proof uses one of the most basic algorithms of mathematics, the division algorithm, and its three corollaries, to prove Fermat’s Last Theorem.
In the fifty years since the author completed the proof, it has been submitted to literally hundreds of people, among them, more than fifty professional mathematicians.
·       Many of them simply ignored it, because most mathematicians consider the probability of a simple proof of Fermat’s Last Theorem being valid to be statistically insignificant or nonexistent.
·       A few of them, however, found nothing wrong, and suggested a rewrite using standard notation and nomenclature, and submission to a mathematical journal for publication.
·       The first submission to a professional mathematician who was in a position to help get it published occurred in 1966, but it was summarily rejected without proper review. The author began the process several times after that, including a submission to Dr. Hans Zassenhaus, Editor of the Journal of Number Theory in April 1985, who was encouraging.
·       However, due to circumstances beyond anyone’s control, the process of review was never completed, and the author refrained from submitting the proof again until it was mentioned as a biographical aside in a publication by Neppe and Close in 2012 in the book:  Neppe VM, Close ER: Reality begins with consciousness: a paradigm shift that works (First Edition) 1 Edition. Seattle:, 2012. [16] It was picked up by a reader of this book who attempted to refute it.
·       Since then, it has been submitted to at least five more mathematicians for peer review. No reviewer has ever produced a valid mathematical argument disproving FLT65 to date.
·       It is shown in this presentation that the counterexamples offered are not applicable, and in the process of demonstrating this, the proof of FLT65 has been validated as never before.

In this presentation, FLT65 has been validated as follows:
·       It is shown that the concern that the division algorithm and its corollaries might not apply to integer factors derived from the algebraic polynomial factors of the FLT equation is unwarranted.
·       It is shown that application of corollary III of the division algorithm depends upon the requirement that q(a) and f(a) are unique for the minimum remainder, and that application to the non-zero remainder{r = f(a) ≠ 0} results of dividing the p-1 degree factor f(z) of the Fermat equation by z - a validates FLT65 by proving that f(Z) cannot contain Z – a, contradicting the fact that  it must contain Z – a, if there are integer solutions for the FLT equation.
·       It is also shown that while an integer solution of the FLT equation requires that the non-zero remainder, f(a), must contain Z – a, any set of integer values for Z, X and a, chosen so that f(Z) contains Z – a creates unresolvable contradictions.

Finally, the reason that FLT65 has been neither disproved nor widely accepted for fifty years, - apart from the fact that most mathematicians don’t believe a simple proof is possible – is a very subtle mistake made by reviewers that leads even very competent mathematicians to think that integer examples like Example #1 and Example #2 presented above invalidate the application of the division algorithm and corollary III to integer factors in FLT65 and prove FLT65 wrong.

The subtle mistake is caused by reviewers ignoring the uniqueness requirement for the application of corollary III of the division algorithm. Ignoring the uniqueness requirement leads to the erroneous conclusion that FLT65 is wrong. 

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