Saturday, November 28, 2015

FERMAT'S LAST THEOREM PART NINE: APPENDICES D - F

APPENDIX D
FERMAT’S LAST THEOREM (FLT)
(A BRIEF PRESENTATION OF THE CLOSE 1965 PROOF
WITH WELL-DEFINED STANDARD NOTATION [FLT65C]
EXPANDED AND REWRITTEN IN 2013)

Notation: In this discussion, three types of numbers and two types of factors are defined precisely 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.
·        Consistent with , meaning “is not equal to”, the oblique strike through a symbol will indicate the negation of the symbol; e.g.: g(x) ɇ f(x) means g(x) is not an algebraic factor of the polynomial f(x) and A B means A is not an integer factor of B.

Consider the equation zp – xp = yp, equivalent to Equation (1) in the 1965 proof. Since it is sufficient (Appendix C) to consider p is a prime number > 2, and thus an odd prime, we can factor the left side of the equation to obtain:
(z-x)( zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = y, equivalent to Equation (2) of FLT65C.
For variable integer values of x, represented by X, let zp-1 + zp-2X + zp-3X2 +•••+ zXp-2 + Xp-1 = f(z), and z-X = g(z). Then g(z)f(z) = Yp, for all integer values X and Y. {Equation (3) of FLT65C}

For Fermat’s last theorem to be falsified, X, Y and z must be integers, so we will replace X and Y with X1 and Y1, representing specific integers. But, since we do yet not know whether z can actually be an integer if x and y are integers in an FLT solution, we must continue to represent it by z, a variable over the field of real numbers.
By Corollary I of the DIVISION ALGORITHM, since f(z) and g(z) are polynomials in z of degree n = p -1 and m = 1, respectively, when f(z) is divided by g(z), the remainder, r(z), will contain any and all algebraic factors common to both.
And by COROLLARY II, the remainder when f(z) is divided by g(z) will be f(X1). And so:
r(z) = f(X1) = X1p-1 + X1p-2X1 + X1p-3X12 +•••+ X1p-1 = pX1p-1, a unique constant made up of integer factors for any p > 2. …                                                        Equation (4).

Since for any solution of Equation (1), X1, Y1 and z, if an integer, may be considered to be relatively prime, no factor of X1, and, therefore, of X1p-1, may be contained in g(z) = z –X1. Therefore, if f(z) and g(z) have a common factor, it must be p. It also follows that for their product to be equal to the perfect p-power, (Y1)p, either p f(z), p g(z), or they must both be perfect p-powers of integers.
By exactly the same reasoning as above, we can write the FLT equation as zp – yp = xp and factor as:
(z - Y)(zp-1 + zp-2Y + zp-3Y2 +•••+ Yp-1) = Xp. …                                                Equation (5)
Let zp-1 – zp-2Y1 + zp-3Y1 2 -•••+ Y1 p-1 = f1(z) and z - Y1 = g1(z).
Then, dividing f1(z) by g1(z), analogous to Equation (4), we obtain r1(z) = pX1p-1Equation (6)
Then, either p f1(z) and p g1(z), or they are perfect p-powers of integers.
For a given case of zp = X1p + Y1p, it is possible that either p f(z) or p f1(z). But, if p is a factor of one, the other has to be a perfect p-power, since X1 and Y1, and therefore, f(z) and f1(z) are relatively prime. For a given z = Z1, either p f(Z1) p Y1, or p f1(Z1) p X1. If neither X1.nor Y1 contains p, because they are relatively prime, both must be perfect p-powers. Therefore, in any event, one of them at least, must be a perfect p-power, not containing p as a factor.

Since f(z) and f1(z) are both of the same form, we may choose either one or the other as not containing p. So we may choose f(z) = zp-1 + zp-2X1 + zp-3X12 +•••+ X1p-1 = Ap, with A p.

By the Division Algorithm, for the two polynomials, g(z) ≠ 0 and f(z), over the field of real numbers, with degrees m and n respectively, and n > m > 1, there exist unique polynomials q(z) and r(z) such that f(z) = q(z)g(z) + r(z), where r(z) is either zero or of degree smaller than m.
If n = m = 1, as in the case when p = 2, or if f(z) is not equal to an integer raised to the pth power, as in that case when A is not an integer, but is the pth root of a prime number, q(z) and r(z) are not unique and COROLLARY II does not hold. Thus, this proof does not apply to the case n = 2, or to non-integer solutions of the FLT equation. But when p > 2, and we assume there is an integer solution of equation (1), because the set of real numbers is closed with respect to addition, for any value of z, there is some real number s, such that z – s = A, and COROLLARY II tells us that if g(z) = z – s, a polynomial of degree 1 in z, q(z) and r(z) are unique and the remainder, r(z) will be of degree m < 1 = zero degree, and of the form f(s).
Therefore: f(z) = (z-s)q(z) + f(s) over the field of real numbers, and by COROLLARY III, if q(z) and f(s) are unique, f(z) e (z – s), IF AND ONLY IF, f(s) = 0. Thus when p >2, we have:

(z – s) e f(z) f(s) = sp-1 + sp-2X1 + sp-3X12 +•••+ X1p-1 = 0….                    Equation (6).

Since both s and z can take on any real number value, if they are not integers, f(s) = 0 is not a contradiction and there are an infinite number of real number solutions for the FLT equation, with exactly p solutions for any value of p. But the Division Algorithm and Corollaries hold over the field of real numbers, including integers, so for specific integer values z = Z1, s = S1 and A = A1, (z – s) e f(z) (Z1 – S1) f(Z1) f(S1) = 0.

Note that if q(z) and f(s) are not unique, corollary III does not apply and multiple non-zero values of  f(s) may be found. See Appendix E.
But f(S1) = S1p-1 + S1p-2X1 + S1p-3X12 +•••+ X1p-1 = 0 is an impossibility because X1 and S1 are positive integers, and the sum of p positive integers cannot equal zero.

Therefore, for specific integer values z = Z1, s = S1 and A = A1, (Z1 – S1) f(Z1) f(S1) 0, and thus
g(z) = (Z – S) f(z) = zp-1 + zp-2X1 + zp-3X12 +•••+ X1p-1.

But, for there to be a triple integer solution for the FLT equation, falsifying FLT, there must be specific integers X1, Y1, Z1, A1, and S1, such that (Z1 – S1) = A1, and f(Z1) = A1p, but if q(z) and f(s) are unique,

(Z1 – S1) f(Z1) A1 A1p and therefore we have a clear contradiction, proving Fermat’s Last Theorem (FLT65).

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FURTHER DISCUSSION AND CLARIFICATION
OF THE CLOSE FLT65C PROOF OF FERMAT’S LAST THEOREM

Using the standard notation for real variables, integer variables and specific integer values outlined above and in the discussion of Objection #4, we can write: zp – xp = xp. This is equation (1) in the 1965 proof, without the restriction of x, y and z to integer values.

Since it is sufficient to consider p as prime numbers greater than 2, we can factor the left side of the equation to obtain:

(z - x)( zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = yp, equation (2) of the FLT65C version of the proof.

In the 1965 proof (FLT65), the FLT equation is expressed as an N-degree polynomial in the variable Z. This translates to a p-degree polynomial in the variable z in the current standardized notation. The rationale for this approach to proving FLT by focusing on one of the variables as an independent unknown is based on the fact that any equation in three unknowns, including the FLT equation, has an infinite number of solutions, but, if we assume specific values for two of the variables, we can solve for the third. In a similar manner, if two of the variables of the FLT equation are restricted to the ring of integers, we will be able to determine whether any values of the third variable can be integers.

In FLT65, the author focused on z as the independent unknown by setting X = X1 and Y = Y1, intending only to imply that they were integer variables. Later on, he indicated that they were specific integer values, without changing the way they were represented. This lack of clear definition of notation may be the cause of the confusion that gave rise to the concern #ii.

The thinking was that taking the trouble to distinguish between variables and specific values of the variables was unnecessary because the Division Algorithm and corollaries apply to both, as ultimately, whatever their values, they are elements of the field of real numbers. Reviewers, however, have pointed out that this may not necessarily true for integer polynomials of the form of f(Z). It turns out to be the case in this instance, however, because of the unique form of the FLT equation and its factors, and the requirement that the only solutions being considered are those for which all three variables have integer values satisfying the Fermat equation.

Because at least three of the most qualified reviewers of FLT65 raised the concern about the applicability of the algorithm and corollaries to integer factors, it is clear that the proof may be difficult to follow unless one proceeds through the whole process with clearly defined and justified notation. When this is done below, we see that the legitimate application of the algorithm and corollaries to all real number variable polynomials derived from the FLT equation leads to an unavoidable contradiction that proves FLT.

In this more detailed explanation, we will adhere to the notation defined above, viz. we will use x, y and z for unrestricted variables over the field of real numbers, X, Y and Z for variables restricted to the sub-set of real numbers that make up the ring of integers, and X1, Y1, and Z1 for specific integers; and for clarity, 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. Also, consistent with ≠, meaning “is not equal to”, the oblique strike through a symbol will indicate the negation of the symbol; e.g.: g(x) ɇ f(x) means g(x) is not an algebraic factor of the polynomial f(x) and A B means A is not a factor of B.

For variable integer values of x, represented by X, let zp-1 + zp-2X + zp-3X2 +•••+ zXp-2 + Xp-1 = f(z)
And z -X = g(z). Then
(3.)     g(z)f(z) = Yp, for all integer values X and Y.
For Fermat’s last theorem to be falsified, X and Y must be specific integers, call them X1 and Y1, and z must also be an integer. But, since we do not yet know whether z can actually be an integer in the FLT equation, we must continue to represent it by z, a variable over the field of real numbers.
By Corollary I, since f(z) and g(z) are polynomials of degree n = p -1 and m = 1, respectively, over the field of real numbers, when f(z) is divided by g(z), the remainder, r(z), will contain any and all factors common to both. See Appendix A and B for the proof of this.
And by COROLLARY II, the remainder when f(z) is divided by g(z) = z - X1, will be f(X1). And so:
(4.)     r(z) = f(X1) = X1p-1 + X1p-2X1 + X1p-3X12 +•••+ X1p-1 = pX1p-1, a unique integer for any p > 2. Interestingly, it is this unique integer remainder in the case of the FLT equation that allows us to extend the application of the Division Algorithm from variable polynomials over the field of real numbers to integers, a subset ring of the field of real numbers.
Since for any integer solution of equation (1), X1 and Y1 may be considered to be relatively prime, no factor of X1, or, therefore, of X1p-1, may be contained in g(z) = z –X1, because, if z is to be an integer, z –X1 must contain a factor of Y1. Therefore, if f(z) and g(z) have a common factor, it must be p. It also follows that for their product to be equal to the perfect p power, (Y1)p, one of them, i.e. either f(z) or g(z), must contain p or they must be both perfect p-powers of integers.
Similarly, we can factor the FLT equation as
(5.)     (z-Y)( zp-1 + zp-2Y + zp-3Y2 +•••+ Yp-1) = Xp. And by exactly the same reasoning as above, for any particular Y = Y1, we can have
zp-1 – zp-2Y1 + zp-3Y1 2 -•••+ Y1 p-1 = f1(z) and z - Y1 = g1(z).
Then, either f1(z) and g1(z) contain p as a single common factor, or they are perfect p-powers of integers. So for a given case of zp = X1p + Y1p , if either f(z) or f1(z) contains p, the other has to be a perfect p-power, since we have concluded that we only have to consider relatively prime X, Y and Z, implying both cannot contain p. Now, p e f(z) p Y1 and p e f1(z) p X1. But X1.and Y1 are relatively prime. If neither X1.nor Y1 contains p, both, being relatively prime, must be perfect p-powers. Therefore, in any event, one of them at least, must be a perfect p-power, not containing p as a factor.
Therefore, we may choose f(z) = zp-1 + zp-2X1 + zp-3X12 +•••+ X1p-1 = Ap, and/or
f1(z) = zp-1 – zp-2Y1 + zp-3Y1 2 -•••+ Y1p-1 = Bp, A and B integers, and at least one, A or B, does not contain p. Also note that since g(z)f(z) = Y1p, and f(z) = Ap, A Y1 and since g1(z)f1(z) = X1p, f1(z) = Bp, B X1.
Since f(z) and f1(z) are both of the same form, we may choose either of them as the one not containing p. So we may choose p ɇ f(z) = zp-1 + zp-2X1 + zp-3X12 +•••+ X1p-1 = Ap, and p ɇ A, meaning that A will not contain p.
We cannot, at this point, assume that z = Z1, a specific integer, for some x = X1 and y = Y1, because this cannot be justified unless we can show that the requirement that f(z) = Ap does not lead to a contradiction.
For any value of X1, the conditions f(z) = Ap, and A f(z) are necessary conditions for an integer solution of the FLT equation to exist. And since A is a positive integer variable, with any specific X1 < Y1 < any z = Z1 that will satisfy the FLT equation, we can set A = z – S, where z is an unrestricted variable, and S is a positive integer variable. Note that S is a variable over the ring of integers. Because we don’t know whether z can be an integer, the specific value of S, e.g. S1, when X = X1, is dependent on the specific integer value, A1, of A.
The Division Algorithm tells us that for two polynomials, g(z) ≠ 0 and f(z), over the field of real numbers, with degrees m and n respectively, and n > m > 1, there exist unique polynomials q(z) and r(z) such that f(z) = q(z)g(z) + r(z), where r(z) is either zero or of degree smaller than m.
Notice that if n = m = 1, as in the case when p = 2, or if f(z) is not equal to an integer raised to the pth power, as in that case when A is not an integer, but is the non-integer pth root of a prime number, q(z) and r(z) are not unique and COROLLARY II does not hold, and thus this proof does not apply to the case n = 2, or to non-integer solutions of the FLT equation. But when p > 2, and we assume there is at least one solution of equation (1) where x, y and z are equal to integers, if there FLT is false, and COROLLARY II tells us that if g(z) = z – s, which is a polynomial of degree 1 in z, q(z) and r(z) are unique and the remainder, r(z) will be of degree m < 1 = zero degree, i.e., a constant, of the form f(s).
Therefore: f(z) = (z - s)q(z) + f(s) over the field of real numbers, and by COROLLARY III, when q(Z) and r(z) are unique,  f(z) is divisible by z – s, IF AND ONLY IF, r(z) = f(s) = 0. Thus when p >2, we have:

(6.)     (z – s) e f(z) f(s) = sp-1 + sp-2X1 + sp-3X12 +•••+ X1p-1 = 0.

Since both s and z can take on any real number value, if they are not integers, f(s) = 0 is not a contradiction and there are an infinite number of real number solutions for the FLT equation, with exactly p -1 non-integer solutions and one trivial integer solution: X1 p +(0) p = Z1p. But, if s is to be from the ring of integers, and q(s) and r = f(s) are unique, which they must be for an integer solution of the FLT equation, f(S) = 0 is a contradiction.

Since the Division Algorithm and Corollaries hold over the field of real numbers, including integers, for specific integer values z = Z1, s = S1 and A = A1, (z – s) e f(z) (Z1 – S1) f(Z1) f(S1) = 0.

But f(S1) = S1p-1 + S1p-2X1 + S1p-3X12 +•••+ X1p-1 = 0 is an impossibility because X1 and S1 are positive integers, and the sum of positive integers cannot equal zero. Therefore:

g(z) = (z – S) cannot be a factor of f(z) = zp-1 + zp-2X1 + zp-3X12 +•••+ X1p-1. Furthermore, because the Division Algorithm and Corollaries apply across the field of real numbers, including the integers, it follows that, for any real values of z, S and A, for there to be a triple integer solution for the FLT equation, falsifying FLT, there must be specific integers X1, Y1, Z1, A1, and S1, such that z = Z1 , S = S1 and A = A1, q(s) and f(s) are unique integers, and for the FLT equation, (z – S1) = A and f(z) = Ap, therefore:

(z – S) ɇ f(z) (Z1 – S1) f(Z1) A1 A1p and we have a clear contradiction, proving FLT.

Note that f(z) and g(z) are polynomials in the variable z throughout the entire discussion, up to the contradiction S1p-1 + S1 p-2X1 + S1p-3X12 +•••+ X1p-1 = 0; and also note that because the division algorithm and corollaries apply across the field of real numbers, including the integers, we are justified in substituting unique integers, that have to exist in order for FLT to be falsified, into the algebraic forms to obtain the integer factors for the hypothetical triple integer solution of the FLT equation. It is the uniqueness of these algebraic forms derived from the FLT equation, and the fact that any integer solution of the FLT equation must also satisfy the equation S1p-1 + S1 p-2X1 + S1p-3X12 +•••+ X1p-1 = 0 that assures that the fact that f(s) ≠ 0 implies that there are no integer solutions for equation (1).
With the substitution of integers into the algebraic forms of f(z) and f(s) we see that the ‘counterexamples’ provided by some reviewers did not work, because they produce non-unique q(z) and f(s) and thus do not apply. And we see that the contradiction obtained by assuming that for given integer values of x and y, an integer z was possible must apply to the integer factors as well as the algebraic factors. Therefore the concern that the factorization of XN + YN = ZN, as a polynomial in Z, might not provide a contradiction in the numerical factorization of the polynomial for some specific integer solution, i.e., concern #ii is not relevant, and the proof of FLT is complete.
When a mathematical statement is true, it can usually be proved in more than one way. The proof that in the case of FLT, the division algorithm applies to integer polynomials and the single integer value they can be reduced to is no exception. If the reader is not fully convinced by the argument presented above, we can address concern #ii in another way, as follows:
Given f(S) 0, the fact that the integers form a ring, which is a sub-set of the field of real numbers, but technically not a field because the integers are not closed with respect to division, might lead us to believe that for some Z1 that might satisfy the FLT equation along with X1 and Y1, the remainder r(S1) = f(S1) = S1p-1 + S1p-2X1 + S1p-3X12 +•••+ X1p-1, not being zero, might contain g(z) = (z – S1) = A1 as a factor. Assuming that there is such an actual integer triple solution for the FLT equation, the integers X1, Y1, Z1, and p might be so large that it would take a million years for the fastest computer available to search for and find this integer solution. The point is that it is possible that there could be an integer solution that we could never find by trying endless combinations of integers from the infinite ring of relatively prime integers. But we can set up a process of infinite descent4 by referring to the simple process of long division.

In the process of long division, first we estimate the quotient. Next we multiply our estimate by the divisor, and then subtract the result from the dividend. If the remainder is greater than the divisor, we increase the quotient estimate and repeat the process again until the remainder is either zero or less than the divisor. We can follow this same simple procedure with f(S) as the dividend, g(S) as the divisor and r(S) as the remainder, where S is an integer variable over the ring of integers from which the specific integers of a specific solution of the FLT equation must come if FLT is to be falsified.

Because the ring of integers is closed with respect to addition, for every value of S, there is some integer value of K, such that g(S) = S – K = A. Since the remainder, r(z) = f(S) ≠ 0, let’s assume it contains the divisor, A = S - K, i.e., (S – K) e f(S), as it must be for concern #ii to have any validity. Remembering that A, Z and S must all be integers for FLT to be falsified, and due to the well-ordered nature of the ring of integers, allowing for the basic operations of addition and subtraction, there is some (X,Y,Z) = (Xi,Yi,Zi), specific integers, such that A = Zi – S and there is also some specific integer K, such that A = Zi – S = S – K. Then dividing f(S) by S – K, in accordance with the Division Algorithm, we get a remainder equal to f(K):

(7.)     f(K) = Kp-1 + Kp-2Xi + Kp-3Xi2 + ••• + Xip-1

Since S – K = A, and A has to be a positive integer in order for FLT to be falsified, K < S, and f(K) < f(S). Just as in simple long division, we can repeat this process with A = K – K1, K1 < K, and with smaller and smaller Ki until the remainder, f(Ki) obtained is either zero or smaller than A. No matter how large or small the integers of the FLT falsifying solution are, and how large the remainder f(S) may be, the process of infinite descent  obtained by successively dividing by A will eventually reduce f(Ki) for that solution to its smallest possible integer value. Now, in order for FLT to be falsified, f(S) must contain A, and for FLT to be falsified, the smallest integer value of f(Ki) must be equal to zero. In our infinite descent, the smallest possible f(Ki) will occur when Ki =1.

But, the smallest possible f(Ki) = f(1) = 1 + Xi + Xi2 + ••• + Xip- 1, which is still a sum of positive integers, and thus cannot equal zero, or contain A, as it must for FLT to be falsified. This constitutes an infinite descent resulting in a contradiction. Assuming that for xp + yp = zp, x = X1 and y = Y1, X1 and Y1 specific integers, we reach the contradiction of an infinite descent, proving that z ≠ Z, Z ≠ Z1, and f(z) ≠ Ap, a perfect pth power of an integer, and all sufficient and necessary conditions for the definitive proof of FLT have been met. See Appendix E for application of the method of infinite descent to a proposed counterexample to FLT65.

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APPENDIX E 9
Infinite Descent

INTRODUCTION
This appendix presents the complete details of an infinite descent analysis of the best example proposed by a mathematician specialist in number theory as a counterexample invalidating FLT. The example consists of producing a set of integer values for the FLT equation that appear to invalidate the application of corollary III of the division algorithm to the integer values of factors of the FLT equation in FLT65.  But the example ignores the uniqueness requirements for application of the corollary. As a result of this omission, the example provides us with a way to elucidate the contradiction validating the conclusion drawn in FLT65, that the non-zero remainder when the algebraic polynomial factor f(Z) = Zn-1+XZn-2+…+ Xn-2Z+Xn-1 of the FLT Diophantine equation Yn = Zn - Xn is divided by Z – a, is proof of Fermat’s Last Theorem.

FACTORS OF THE FLT EQUATION
The FLT equation, yn = zn - xn, defined over the field of real numbers, with n > 2, is factorable, in the following manner:
Yn = (z – x)(zn-1+xzn-2+…+ xn-2z+xn-1) = g(z)f(z).

Assuming there may be integer solutions to the FLT equation, we have shown in Appendix C that we may let x = X and y = Y, integer variables, such that

g(z) = z – X = Jn , and f(z) =zn-1+Xzn-2 +…+Xn-1 = Kn, with J and K relatively prime integers for a hypothetical integer solution of the FLT equation.

Because the integers are closed with respect to addition and subtraction, we know that for every value of z and K, there is a real number a, such that z – a = K. By the Principle of Designation of Variables we can designate x and y as independent integer variables, X and Y; but at this point we do not know whether z and a can actually be integers or not, so designating z as the dependent variable and z – a as the divisor, where a will be an integer if z is an integer, and dividing the factor f(z) by z - a, we have:

f(z)/(z – a) = q(z) + r(z)/(z – a) where, by corollary II of the division algorithm, r(z) = f(a).

Therefore,              f(z)/(z – a) = q(z) + f(a)/(z – a)                                 Equation (E-1)

INFINITE DESCENT VALIDATION OF FLT65
We can generalize the proof for all n = p > 2, as follows:

f(z)/(z – a) = q(z) + f(a)/(z – a), where f(z) = (zp-1+Xzn-2+…+ Xn-2z + Xn-1)

Multiplying both sides of Equation (E-1) by (z – a) gives us f(z) = q(z)(z – a) + f(a). Transposing f(a) yields f(z) – f(a) = q(z)(z – a), and from this, we can derive the algebraic polynomial form of q(z), as follows:

q(z)(z – a) = f(z) – f(a) = (zp-1+Xzp-2+…+ Xp-2z + Xp-1) - (ap-1+Xap-2+…+Xp-1). Then, collecting like powers of X, we have q(z)(z – a) = zp-1 – ap-1 +(zp-2 – ap-2)X +…+ (z – a) Xp-2 + Xp-1 – Xp-1.
Simplifying and dividing by (z – a), we have:

q(z) =(zp-2+azp-3+…+ ap-3z + ap-2) + (zp-3+azp-4+…+ ap-4z + ap-3)X +… + (z + a)Xp-3 + Xp-2
From this, we can see that for n = 3, the polynomial form of q(z) is q(z) = z + a + X;
For n = 4, q(z) = z2 + az + a2 + (z + a)X + X2
For n = 5, q(Z) = z3 + az2 + a2z + a3 +(z2 + az + a2)X +(z + a) X2 + X3; ּ
And for n = p:
q(z) =(zp-2+azp-3+…+ ap-2)+(zp-3+azp-4+…+ ap-3)X+…+(z + a)Xp-3+Xp-2                      Eq. (E-2)

Given X = X1 and Y = Y1, specific integers, and assuming that z can be an integer, let z = Z, then f (z) = f(Z) = Zp-1 + X1Zp-2 + … + Xp-1, an integer polynomial of n-1 degree in Z. But, since at this point, we don’t know whether a can be an integer or not for a specific integer value of Z, Eq. (E-2) becomes q(Z)=(Zp-2+aZp-3+…+ ap-2)+(Zp-3+aZp-4+…+ ap-3)X+…+(Z + a)Xp-3+Xp-2, an algebraic polynomial in  Z and a, and Eq. (E-1) becomes:

f(Z,a)/(Z – a) = q(Z,a) + f(a)/(Z – a)                                                 Eq. (E-3)

For a specific integer solution, X = X1, Y1 = J1K1, Z = Z1 and Y1p = J1pK1p, f(a) = ap-1 + ap-2X1 + …+ aX1p-2 + X1p-1, and Eq. (E-3) becomes Eq. (E-4), an equation in the variable a:

f(Z1)/(Z1 – a) = q(a) + f(a)/(Z1 – a)                                      Eq. (E-4)

If there is an integer solution, then z and a will be integers, but, since we do not know whether there can be an integer solution for the FLT equation, we must continue to represent z as an integer variable, Z, and a as a real number variable, which may be integer, rational fraction or irrational, depending upon whether or not the hypothesis that Z can be an integer is validated.

With z as an integer variable, Z, and a as a real variable, f(Z) = Zp-1+ X1Zp-2 + … + X1p-1 = (K1)p, q(Z) = (Z n-2+aZ1n-3+…+ an-2) + (Zn-3+aZ n-4+…+ an-3)X+…+(Z + a)Xn-3+Xn-2, and
f(a) = ap-1+X1ap-2+…+X1p-1. Substituting these hypothetical integer values into Eq. (E-4), we have:

(K1)p/K1 = (K1)p-1 = (Zp-2+aZp-3+…+ ap-2) + (Zp-3+aZp-4+…+ ap-3)X+…+(Z + a)Xp-3+Xp-1 +
+ (ap-1+X1ap-2+…+X1p-1)/K1         (An algebraic polynomial in a and Z.)                        Eq. (E-5)

IMPORTANT NOTE:
Note that the polynomials making up the equations in this discussion are all algebraic polynomials in one or more of the real variables, z, Z and a, to which the division algorithm and its corollaries definitely apply.

By inspection of Eq. (E-4), we see that, if Z is an integer, the left side of the equation, f(Z)/K1 = (K1)p-1, is an integer and a and q(a) are integers.  Since integers are closed with respect to addition, we see that f(a) must contain Z – a = K1 as an integer factor, for any integer value of a.

Eq. (E-5) represents the first step in the process of long division of the integer polynomial f(Z) = Zp-1 + X1Zp-2 + X12Zp-3 +…+ X1p-2Z +X1p-1 by the integer polynomial  Z – a = K1.

The quotient is q(Z)=(Zp-2+aZp-3+…+ap-2) + (Zp-3+aZp-4+…+ap-3)X1+…+(Z + a)X1p-3+Xp-2, and the remainder is f(a)  = ap-1+X1ap-2+…+X1p-1.
In long division, if the remainder is larger than the divisor, it indicates that the quotient is too small, a new quotient is used, and the process is repeated with a larger and larger quotient until the remainder is either zero or smaller than the divisor. If the remainder is zero, we know that the dividend is divisible by the divisor, if the remainder is not zero, then the dividend is not divisible by the divisor.
Because the integers, a subset of the field of real numbers, are closed with respect to addition and subtraction, for any integer, K1, there are an infinite number of pairs of integers, ai, ai+1, such that

z – a = K1 a – a1 a1 – a2 a2 – a3 ak-1 – ak, with a >a1 >a2 >… >ak, k = 1, 2, 3 … .

If there actually is an integer solution, X1, Y1, Z1, for the FLT equation, then z – a Z1 – A1 K1 expressed as the difference between a specific pair of integers. The problem becomes how to find the unique pair of integers, ai-1 and ai, if they exist, that will produce the minimum remainder, r = f(ai), consistent with integer values of Z and a that will satisfy the FLT equation. We can approach this problem by carrying out the division indicated in the last term of Equation (E-1), i.e. (ap-1+X1 ap-2 +… +X1n-1)/(Z – a). To carry out the indicated division, we replace Z – a by its equivalent a - a1 with a as the variable, which we can do by virtue of the Principle of Designation of Variables, and repeatedly divide by Z - a in the form of the difference between successively smaller and smaller values of ai and ai+1. Each division produces an addition to q(Z) and a diminishing f(ai) until a minimum remainder is reached. Using the process of long division of polynomials we can determine whether the integer value of the minimum f(ai) is actually divisible by ai-1 - ai = K1 or not if f(a) ≠ 0. If it is not, we have a contradiction, since for an integer solution, if f(a) is not zero, it must contain Z – a.

Returning to Eq. (E-1), for n = p, any prime >2, we can set K1 = a – a1 and divide
f(a)= ap-1 + X1ap-2+…+X1p-1, an algebraic polynomial in a, by a – a1, to obtain:

f(Z)/(a – a1)=(Zp-2+aZp-3+…+ap-2)+(Zp-3+aZp-4+…+ap-3)X1+…+(Z + a)X1p-3+X1p-2 + (a1p-1 + X1a1p-2 + … + X1p-1)/(a - a1).                                                                                            Eq. (E-6)

This process can be repeated indefinitely with smaller and smaller values of ai-1 and ai. Each time the last term is expanded by dividing f(ai) by ai-1 – ai, an additional quotient term, q(ai), and a new remainder, f(ai), are produced. Each additional quotient term added to previous quotient is smaller than the one before, and each new remainder is smaller than the one before, because Z > a, each ai is smaller than the one before, and X1 is a constant. This descent will continue until f(ai) reaches a minimum. At that point, ai must be either zero, positive, or negative. Looking at the three possible outcomes, we see that:

1.) If ai for the minimum f(ai) is zero, then ai-1 – ai = ai-1 = K1, and f(ai) = a1 p-1+X1a1p-2 +…+X1p-1 = 0 + 0 +…+X1p-1 = X1p-1, which we see by inspection of Eq. (E-6),  must contain ai-1 because of closure of integers with respect to addition. This implies that X1p-1 contains K1 which is a factor of Y1, and this is a contradiction of the relative prime sufficiency condition for the FLT equation (See Appendix C). Therefore, a cannot equal zero for minimum f(ai).

2.) The smallest ai cannot be positive: A positive ai will not produce the smallest f(ai), because f(ai), which is always positive, will be smallest with the first negative ai. After the first negative ai, successive negative values of ai will only produce larger values for f(ai), because odd powers of negative numbers are negative and even powers of negative numbers are positive. With f(ai) = a1 p-1+X1a1p-2 +…+X1p-1, the even powers of ai and X1 will always outnumber and be greater than the odd powers of ai and X1.

3.) Finally, because the ai producing the smallest f(ai) cannot be zero or positive, it will have to be negative, and q(ai) and f(ai) will be unique, implying by corollary II and III that the smallest f(ai) must be a positive integer not containing Z – a = ai-1 – ai. But, if f(Z) is an integer, and q(ai) is an integer, by inspection of Eq. (E-6) and closure of integers with respect to addition, f(ai) must contain K1, the integer value of Z – a. This is a contradiction unless the ai’s, and consequently, Z, are non-integers. Therefore, this contradiction tells us that, while Z – a must be an integer factor of the integer value of f(Z), Z and a cannot be integers, proving FLT for all p  > 2.

These infinite descent contradictions prove that because f(a) ≠ 0, a and Z cannot be integers, and therefore, FLT is proved for all n  > 2.
THE CASE p = 3
This demonstration may be easier to follow by using p = 3, the smallest and least complicated case as an example. Returning to Eq. (E-1), for p = 3, we can let K1 =a – a1 and divide a2+ aX1 +X12, an algebraic polynomial in a, by a – a1, to obtain: f(a)/(a – a1)= a+a1+X1 +(a12+a1X1+X12)/K1. So we have:

f(Z1)/(Z1 – a) = Z1 + a + X1 +(a + a1 + X1) + (a1 2+ a1X1 +X12)/(a1 –a2).            Eq. (E-7)

This process can be repeated indefinitely with smaller and smaller values of ai. Each time the last term is expanded by dividing f(ai-1) by ai-1 – ai, an additional quotient term, q(ai) = (ai-1 + ai + X1) and a new remainder, f(ai) are produced. Each additional quotient term added to the previous quotient is smaller than the one before, and each new remainder is smaller than the one before, because Z1 > a, each ai is smaller than the one before, and X1 is a constant. This descent will be continued until f(ai) reaches a minimum. At that point, ai must be either zero, positive, or negative. Looking at the possible outcomes with n = 3, as in the general case, we see that,

1.) If ai for the minimum f(ai) is zero, then ai-1 – ai = ai-1 = K1, and f(ai) = ai2+ aiX1 +X12 = 0 + 0 + X12 = X12, which we see by inspection of Eq. (E-7)  must contain ai-1 because of closure of integers with respect to addition. This implies that X12 contains K1 which is a factor of Y1, and this is a contradiction of the relative prime sufficiency condition for the FLT equation (See Appendix C). Therefore, a cannot equal zero for minimum f(ai).

2.) The smallest ai cannot be positive. If ai is positive, it will not produce the smallest f(ai), because f(ai) is always positive, and it will be smallest with the first negative ai. After the first negative ai, successive negative values of ai will only produce larger values for f(ai), because f(ai) = ai2  +aX1 + X12, and the even powers of ai and X1 outnumber and are greater than the odd powers (negative) of ai and X1.

3.) Finally, since the ai producing the smallest, and therefore unique f(ai) cannot be zero or positive, it will have to be negative, and q(ai) and f(ai) will be unique, implying that the smallest f(ai) must be a positive integer that cannot contain K1 because it must be smaller than K1. But, if f(Z) is an integer, and q(ai) is an integer, by inspection of Eq. (E-7) and closure of integers with respect to addition, we see that f(ai) must contain the integer value of Z – a. This is a contradiction, unless the ai’s, and consequently, Z, are not integers. Therefore, while the integer value of Z – a is a factor of the integer value of f(Z), Z and a cannot be integers, proving FLT for n = 3.

EXAMPLE
This process of descent is somewhat easier to follow using actual integers, and we can use the integer values from Example #3 provided by the number theorist reviewer: Z = 73, X1 = 17 and a = 54. To start with. Using these integer values, infinite descent reveals the fact that the non-zero remainder obtained by dividing the algebraic polynomial f(z) = Z2 + X1Z+ X12 by Z – a, leads to contradictions proving that these values do not constitute a counterexample to the FLT65 proof.

To invalidate FLT65, specific values of a, X and Z that produce an integer value for f(Z) that is divisible by the integer value Z - a, even when the algebraic polynomial f(Z) divided by Z – a produces a non-zero remainder, must produce a unique maximum quotient and minimum remainder; otherwise, corollary III of the division algorithm does not apply. On the other hand, if the quotient and remainder are unique, which they must be for an integer solution to the FLT equation, corollary III does apply, and the non-zero remainder proves FLT.

Note that for any numerical example, the number of steps are finite, but the method is properly called infinite descent because the process can descend from any hypothetical integer value of Z, however large.

Step one of Descent: Evaluation of the First Remainder
Using the integer values X1 = 17, z = 73 and a = 54, yields: z – a = 73 – 54 = 19;
f(Z) = 732 + 73x17 + 172 = 5329 + 1241 + 289 = 6859; q(Z) = (Z + a + X1 ) = 73 + 54 + 17 = 144; and r1 = f(a) = a2+Xa +X2 = 542 + 54x17 + 172 = 4123. Substituting these results into Eq. (E-7), we have:
f(Z)/(Z –a) =193/19 =192 =144+4123/19 = (2736+4123)/19 =6859/19 =361 = 192.

This indicates that f(Z) is divisible by Z – a for these integer values of a, X and Z, and the remainder is not zero, but the remainder, f(a) = 4,123, is much larger than the divisor, Z - a  = 19, indicating that the quotient and remainder are not unique and we must divide again, increasing the quotient and decreasing the remainder, to find the unique integer values, which can only be obtained when the remainder is either equal to zero or has the minimum possible integer value.
Note that the process of repeatedly dividing the remainder by Z - a = ai-1 - ai = 19 for decreasing values of ai-1 and ai, does not change the value of f(Z)/( Z –a) = q(Z) + f(ai) because as f(ai) decreases, q(Z) increases preserving the total value of f(Z)/( Z –a), which is 192.
Step Two of Descent: Evaluation of the Second Remainder
We have all the integer values needed to evaluate Eq. (E-7), except a1 and we can determine a1 as follows: K1 = z – a = 73 – 54 = 19, and z – a = a – a1 = 19 a1 = a – 19 = 54 – 19 = 35.  Substituting a1 = 35 into the last two terms of Eq. (E-7), we have:

q(a1) = 144 + (a + a1 + X1) = 144 + (54 + 35 + 17) = 250, and
f(a1)/(a –a1) ={(35)2 +35x17 + (17)2}/19 = 2109/19

Substituting these integer values into Eq. (E-6), we have:
f(Z)/(Z –a) = 192 = 144 + 106 + 2109/19 = (4750 +2109)/19= 6859/19 = 192 Verifying that, with the new remainder f(Z) may be divisible by Z – a for these integer values of a, X and Z.

Step Three of Descent: Evaluation of the Third Remainder
Next, a1 - a2 = 35 – a2 =19, yielding a2 = 16
Substituting a2 = 16 into the last two terms of Eq. (E-7) expanded, we have:

q(a2) = 250 + (a1 + a2 + X1) = 250 + (35 + 16 + 17) = 318 and
f(a2)/(a1 –a2) ={(16)2 +16x17 + (17)2}/19 = 817/19

Thus, f(Z)/(Z –a) = 192 = 318 + 817/19 = (6042 + 817)/19 =6859/19 = 192

Step Four of Descent: Evaluation of the Fourth Remainder
Next, a2 – a3 = 16 – a3 =19, yielding a3 = - 3
Substituting a3 = - 3 into the last two terms of the expanded equation, we have:

q(a3) = 318 + (a2 + a3 + X1) =318 +  (16 + (-3) + 17) = 348, and
f(a3)/(a2 –a3) ={(-3)2 + (-3x17) + (17)2}/19 = 247/19

Then: f(Z)/(Z –a) =144 + 106 + 68 + 30 + 247/19 =(6612 + 247)/19 =6859/19 = 192

Step Five of Descent: Evaluation of the Fifth Remainder
Next, a3 – a4 = - 3 – a4 = 19, yielding a4 = - 22
Substituting a4 = - 22 into the last two terms of the equation, we have:

q(a4) = 348 + (a3 + a4 + X1) = 348+ (- 3 + (- 22) + 17) = 340, and
f(a4)/(a3 –a4) ={(-22)2 + (-22x17) + (17)2}/19 = 399/19

Then we have: f(Z)/(Z –a) = 348 + (- 8) + 399/19 = (6460 + 399)/19 =6859/19 = 192

Step Six of Descent: Evaluation of the Sixth Remainder
And, a4 – a5 = - 22 – a5 = 19, yielding a5 = - 41
Substituting a5 = - 41 into the last two terms of the equation, we have:

q(a5) = 340 + (a4 + a5 + X1) = 340 + (- 22 + (- 41) + 17) = 294, and
f(a5)/(a4 – a5) ={(-41)2 + (- 41x17) + (17)2}/19 = 399/19, and a3 - a4 = 19 a4 = - 22, yielding r5 = f(a4) =  a42+a4X+X2 = 299.

TABLE E-1 Summary of Remainder Descent with Search Interval = 19
 Descent Step # Z – a = ai-1 - ai= K1 = 19 Quotient  q(ai) Remainder ri  = f(ai) RESULT: FLT65 Invalidated? 1 72 - 54 144 4123 = 217(Z – a) NO, q and r Not Unique 2 54 - 35 250 2109 =111(Z – a) NO, q and r Not Unique 3 35 - 16 318 817 = 43(Z – a) NO, q and r Not Unique 4 16  - 3 348 247 = 13(Z – a) NO, r is minimum for this search interval, but still larger than the divisor, Z- a 5 - 3  - 22 340 399 = 21(Z – a) NO, q smaller than maximum, And r larger than minimum 6 - 22  - 41 294 1273 = 67(Z – a) NO, r larger than minimum

Notice that the pairs of integers producing quotient additions and remainders before or after the smallest remainder found with this search interval do not produce unique quotients and remainders.  All of the pairs produce remainders divisible by Z – a = 19 and satisfy the equation. If the integer values, Z1 = 73, X1 = 17 and a = 54, actually comprise a valid counterexample to FLT65, there must be a pair, ai-1 and ai, that will produce the maximum quotient and minimum remainder, but none of these pairs produce unique maximum q and minimum r. This descent using step intervals equal to K1 = 19, shows us that the pairs (ai-1, ai) that produce a unique q(ai) and minimum f(ai), must lie between (16 ,- 3) and (- 3, -22).

With this process using descent steps equal to19, the fourth step pair produces a quotient larger than the pairs before and after, and the remainder is larger than those before and after, but the remainder, 247, is still much larger than the divisor, 19, indicating that further divisions are required to identify the pair that will produce the unique minimum f(ai).
Searching for the integer pair that will produce the minimum remainder using an integer search interval of 19 is like trying to catch a fish that is 19 centimeters long and a few centimeters in diameter with a net with a mesh size of 19 centimeters. Since we are dealing with a Diophantine equation, i.e., an equation with integer variables, the ultimately smallest net we can use is one with an integer mesh size of one. To do this, we must start with the first pair, Z and a, and proceed with descent steps defined by an integer search interval of 1, with each successive ai -1 and ai smaller by 1, as follows: 73–54 = 72–53 = 71–52 =…= ai-1 - ai = 19.

Table E-2 displays the results of the search refined to the minimum search interval of 1.

TABLE E-2
 Descent Step # Z – a = ai-1 - ai= K1 = 19 Quotient  q(ai) Remainder ri  = f(ai) RESULT: FLT65 Invalidated? 1 73 - 54 144 4123 = 217x19 NO, ri is not minimum 2 72 - 53 286 3999 NO, q(ai) not maximum and f(ai) not minimum 3 71 – 52 426 3877 NO, (same reason as above) 4 70 – 51 564 3757 NO, (same reason as above) 5 69 – 50 700 3639 NO, (same reason as above) 6 68 – 49 834 3523 NO, (same reason as above) 7 67 – 48 966 3409 NO, (same reason as above) 8 66 – 47 1096 3297 NO, (same reason as above) 9 65 – 46 1224 3187 NO, (same reason as above) 10 64 – 45 1350 3079 NO, (same reason as above) 11 63 – 44 1474 2973 NO, (same reason as above) 12 62 – 43 1596 2869=151x19 NO, ri is neither minimum nor unique 13 61 – 42 1716 2767 NO, q(ai) not maximum and f(ai) not minimum 14 60 – 41 1834 2667 NO, (same reason as above) 15 59 – 40 1950 2569 NO, (same reason as above) 16 58 – 39 2064 2473 NO, (same reason as above) 17 57 – 38 2176 2379 NO, (same reason as above) 18 56 – 37 2286 2287 NO, (same reason as above) 19 55 – 36 2394 2197 NO, (same reason as above) 20 54 – 35 2500 2109 = 111x19 NO, f(ai) divisible by Z – a, but  not  Minimum 21 53 – 34 2604 2023 NO, q(ai) not maximum and f(ai) not minimum 22 52 – 33 2706 1939 NO, (same reason as above) 23 51 – 32 2006 1857 NO, (same reason as above) 24 50 – 31 2904 1777 NO, (same reason as above) 24 49 – 30 3000 1699 NO, (same reason as above) 25 48 – 29 3094 1623 NO, (same reason as above) 26 47 – 28 3186 1549 NO, (same reason as above) 27 46 – 27 3276 1477 NO, (same reason as above) 28 45 – 26 3364 1407 NO, (same reason as above) 29 44 – 25 3450 1339 NO, (same reason as above) 30 43 - 24 3534 1273 = 67x19 NO, f(ai) divisible by Z – a, but  not  Minimum 32 42 – 23 3616 1209 NO, q(ai) not maximum and f(ai) not minimum 33 41 – 22 3696 1147 NO, (same reason as above) 34 40 - 21 3774 1087 NO, (same reason as above) 35 39 - 20 3850 1029 NO, (same reason as above) 36 38 – 19 3924 973 NO, (same reason as above) 37 37 – 18 3996 919 NO, (same reason as above) 38 36 - 17 4066 867 NO, (same reason as above) 39 35 - 16 4134 817=43x19 NO, f(ai) divisible by Z – a, but  not  Minimum 40 34 – 15 4200 769 NO, q(ai) not maximum and f(ai) not minimum 41 33 – 14 4264 723 NO, (same reason as above) 42 32 - 13 4326 679 NO, (same reason as above) 43 31 - 12 4386 637 NO, (same reason as above) 44 30 - 11 4444 597 NO, (same reason as above) 45 29 – 10 4500 559 NO, (same reason as above) 46 28 – 9 4554 523 NO, (same reason as above) 47 27 – 8 4606 489 NO, (same reason as above) 48 26 – 7 4656 457 NO, (same reason as above) 49 25 - 6 4704 427 NO, (same reason as above) 50 24 - 5 4750 399=21x19 NO, f(ai) divisible by Z – a, but not Minimum 51 23 – 4 4794 373 NO, q(ai) not maximum and f(ai) not minimum 52 22 – 3 4836 349 NO, (same reason as above) 53 21 – 2 4876 327 NO, (same reason as above) 54 20 - 1 4914 307 NO, (same reason as above) 55 19 - 0 4950 289 NO, (same reason as above) 56 18 – (-1) 4984 273 NO, (same reason as above) 57 17 – (-2) 5016 259 NO, (same reason as above) 58 16 – (-3) 5046 247 = 13x19 NO, f(ai) is divisible by Z - a but not minimum, and q(ai) is not maximum 59 15 – (-4) 5074 237 NO, q(ai) and f(ai) not unique 60 14 – (-5) 5100 229 NO, q(ai) and f(ai) not unique 61 13 – (-6) 5124 223 NO, q(ai) and f(ai) not unique 62 12 – (-7) 5146 219 NO, q(ai) and f(ai) not unique 63 11 – (-8) 5166 217 NO, r = f(ai) is minimum, but q(ai) is not maximum 64 10 – (-9) 5184 217 NO, r = f(ai) is minimum, but q(ai) is not maximum 65 9 – (-10) 5200 219 NO, q(ai) min 66 8 – (-11) 5214 223 NO, q(ai) min

 67 7 – (-12) 5226 229 NO, q(ai) min 68 6 – (-13) 5236 237 NO, q(ai) min 70 5 – (-14) 5244 247=13x19 NO, f(ai) contains Z - a, but q(ai) & f(ai) are not unique 71 4 – (-15) 5250 259 NO, q(ai) min 72 3 – (-16) 5254 273 NO, q(ai) min 73 2 – (-17) 5256 289 NO, q(ai) is maximum, but r = f(ai) not minimum 74 1 – (-18) 5256 307 NO, q(ai) is maximum, but not unique, and r = f(ai) is not minimum 75 0 – (-19) 5254 327 NO, q(ai) is not maximum, And f(ai) is not minimum 76, 77,… -1 – (-20) etc. Decreasing Increasing NO, q(ai) < maximum, and f(ai) > minimum
Inspection of this table clearly shows that the minimum value of the remainder, ri = f(ai), occurs between ai = -8, and ai = -9, which tells us that the ai that produces the minimum remainder is not an integer. This contradicts the assumption that Z and a are integers: If ai is not an integer, then ai-1 - ai = 19 implies that ai-1 = 19 – ai is also not an integer, and this inference passes back up the descent all the way to a and Z1. This contradiction proves that the integers X1 = 17, a = 54, and Z1 = 73 do not comprise a counterexample to FLT65.
Also, we see from this table that the minimum integer value of the remainder, ri = f(ai) = 217, is not unique, since it occurs for two different integer pairs. Corollary III of the division algorithm says that if q(Z) and r(Z) are unique, f(a) cannot contain Z-a, and so it follows that a polynomial, f(Z), of degree greater than 1 is divisible by Z-a IF AND ONLY IF, f(a) = 0. This infers that corollary III does not apply to the FLT equation for X1 = 17, a = 54, and Z = 73. It does apply, however when f(ai) and q(ai) are unique. For unique f(ai) and q(ai), in this case, non-integers, f(a) cannot contain Z – a, and f(Z) is divisible by Z-a, IF AND ONLY IF, f(a) = 0. But f(a) cannot equal zero for any positive integer value of a, and if a is zero or negative, then the hypothesis that Z can be an integer for any specific integers, X1 and Y1, is proved false.
CONCLUSION: For there to be an integer solution of the FLT equation, there must be an integer pair, ai-1 - ai, for the integer values of X, a and Z that produces a unique q(ai) maximum and a unique f(ai) minimum. But by applying infinite descent with the smallest possible integer search interval, one, with this exhaustive search, we find that there is no integer pair ai-1 - ai that produces a unique q(ai) and f(ai) for X1 = 17, a = 54, and Z = 73, proving that they do not comprise a counterexample to the conclusion of FLT65.
Finally, notice that the logical injunction “IF AND ONLY IF” works in both directions: The statement ‘A is true IF AND ONLY IF B is true’, implies that the converse: ‘B is true IF AND ONLY IF A is true’, is also true. It follows that the proof of corollary III given in FLT65 also proves the converse. Corollary III says:
“If q(Z) and r(Z) are unique, then f(a) cannot contain Z-a, and f(Z) is divisible by Z-a, IF AND ONLY IF, f(a) = 0.”
So the converse is also true: If f(a) = 0, f(a) cannot contain Z-a, and f(Z) is divisible by Z-a IF AND ONLY IF q(Z) and r(Z) are unique.
It is also an established fact of logic that the inverse of a statement has the same truth value as the converse of the statement1
The truth table below shows the relationship between the conditional, the converse, the inverse, and the contrapositive. Only the shaded row is relevant to our discussion of ‘if and only if’ statements.

TABLE E-3  Not p Not q Conditional (if p, then q) Converse Inverse Contrapositive        → T T F F T T T T T F F T F T T F F T T F T F F T F F T T T T T T

This means that the inverse and the converse of an ‘if and only if’ statement are logically equivalent. So, since the converse of the ‘if and only if’ statement of corollary III is true, the inverse of corollary III is also true. Its inverse is:
If q(Z) and r(Z) are not unique, then f(a) contains Z-a, and f(Z) is not divisible by Z-a, IF AND ONLY IF, f(a) ≠ 0.
Since these are ‘if and only if’ statements, the converse of this statement (also called the contrapositive) is also true:
If f(a) ≠ 0, then f(a) contains Z-a, and f(Z) is not divisible by Z-a, IF AND ONLY IF, q(Z) and r(Z) are not unique.
Thus the fact that f(a) cannot equal zero for any integer value of Z and a, (including the values they would have if there are integer solutions for the FLT equation) implies that f(a) contains Z – a and f(Z) is not divisible by Z–a, which is exactly what FLT65 says.
Eq. (E-4): f(Z1)/(Z1 – a) = q(a) + f(a)/(Z1 – a), with the remainder divided iteratively to the minimum remainder becomes:
f(Z1)/(Z1 – a) = q(ai) + f(ai )/(Z1 – a)                                                                       Eq. (E-8)
By inspection of this equation we see that f(ai) must contain Z1 -a, but if f(ai) contains Z1 - a, r(Z) is not unique, and we must divide by Z-a again to produce a unique remainder, but if a is an integer, the remainder will never be zero, implying that r(Z) is not unique unless f(a) = 0. The fact that f(a) cannot equal zero proves Z - a cannot divide f(Z) because q(Z) and r(Z) are not unique unless r(Z) is smaller than Z–a, in which case, Z-a does not divide f(Z). Thus we have a contradiction that proves FLT, as stated in FLT65.

 FLT65C is the 2013 rewrite of FLT65 distinguishing integer variables from real number variables and constants using more conventional notation without modifying the mathematical structure of any of the equations. This version of the proof is designated as a clarification of FLT65, or FLT65C.
 A ring is a set, S, of mathematical or algebraic elements for which the four basic operations of addition, subtraction, multiplication, and division apply. And if a, b and c represent elements of a ring, the four basic operations satisfy the following conditions:

1. Members of the set are additively associative: For all a, b and c, (a + b) + c = a + (b + c)
2. They are additively and multiplicatively commutative: For all a and b, a + b = b + a, and a x b is equal to b x a.
3. There exists a zero element, or additive identity, such that for all a, 0 + a = a+ 0 = a.
4. There exists an additive inverse: For every a there exists – a, such that a + (-a) = (-a) + = 0.
5. Added elements are multiplicatively distributive: For all a, b and c, ax(b + c) = axb + .axc and (b + c)xa = bxa + cxa.
6. Elements are multiplicatively associative: For all a, b and c, (a xb)xc = ax(bxc).
The simplest example of a ring is the set of integers, and the set of integer polynomials also forms a ring. The field of real numbers is also a ring, but, even though the ring of integers is a subset of the field of real numbers, it is not a field because its elements do not have multiplicative inverses.
 Infinite descent is a powerful method for proving or disproving propositions involving integers. In general, the method, which appears to have been one of Pierre de Fermat’s favorite methods of proof, may be described as follows: if is a property that integers or functions of integers may possess, and if the assumption that a given positive integer, N, or a function based on it has the property leads by a mathematical process of one or more steps to the existence of a smaller positive integer, N1 < N, that also has or provides a function that has the property , then no positive integer or form of the function involved can have that property. This conclusion is logically and mathematically valid because repeated applications of the same process that led from N to N1, will produce a series of integers: N > N1 > N2 >…> Ni, that also have the property. Since the process can be repeated again and again, leading to an infinitely decreasing sequence of positive integers - which is impossible - the assumption that is possessed by a given positive integer implies a contradiction and, hence, is false. This method may be applied to a set of integers, sums of integers, and any function that is reducible to an integer.
The method of infinite descent is commonly associated with the French mathematician Pierre de Fermat, probably because he was the first to state it explicitly.