THOUGHTS ON A SUNDAY AFTERNOON

My purpose in sharing these thoughts is not to spare you the effort of thinking for yourself, but to stimulate your thinking.

THOUGHTS

Beyond surviving crossing the street, the highest goals of awareness and thinking, whether in science, religion or individual thought, are:

1. To describe the indescribable,

2. Explain the inexplicable, and

3. Know the unknowable.

Ludwig Wittgenstein said “Whereof one cannot speak, thereof one must be silent.” And then, ignoring his own advice, he went on to write many volumes of philosophical ponderings about what is knowable and what is not.

Philosophers are wordy and philosophy is an endless use of words to try to think about the meaning of words. But words obscure as much as they elucidate. As finite distinctions, they cannot describe Reality, which is infinite.

TRIADIC REALITY

Nikola Tesla said “If you only knew the magnificence of the 3, 6 and 9, then you would have the key to the universe.”

Why would he say this? I think it is because we actually live in a ‘3, 6, 9’ reality, in three (3) types of dimensional domains, with six (6) degrees of freedom, and a total of nine (9) rotational dimensions. The three dimensional domains are space, time and consciousness, all of which have measurable extent. Within these domains, there are three dimensions, each with two directions of movement making six degrees of freedom: Space has width, with movement to the left and right, height, with movement up and down, and depth, with movement forward and back. As individualized consciousness, we inhabit finite physical bodies limited to a five dimensional domain: three dimensions of space, one of time and one of consciousness. There are three aspects of substance or content of reality: matter, energy and consciousness.

But Max Planck, the father of quantum physics said “There is no matter as such!”

Why would he say that? Because the measure of mass is inertia, the resistance to motion, and inertia is caused by spinning in three or more dimensions.

Planck went on to say: “All matter originates and exists only by virtue of a force. We must assume behind this force the existence of a conscious and intelligent Mind. This Mind is the matrix of all matter."

So, if we agree with Planck, we can say that mass is concentrated energy, and energy is concentrated consciousness, consciousness is the insubstantial substance that makes up all reality, and in its infinite form it is God.

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 z^p – x^p = y^p, 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)( z^p-1 + z^p-2x + z^p-3x² +•••+ x^p-1) = y, equivalent to Equation (2) of FLT65C.[1]

For variable integer values of x, represented by X, let z^p-1 + z^p-2X + z^p-3X² +•••+ zX^p-2 + X^p-1 = f(z), and z-X = g(z). Then g(z)f(z) = Y^p, 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) = X₁^p-1 + X₁^p-2X₁ + X₁^p-3X₁² +•••+ X₁^p-1 = pX₁^p-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 X₁^p-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 z^p – y^p = x^p and factor as:

(z - Y)(z^p-1 + z^p-2Y + z^p-3Y² +•••+ Y^p-1) = X^p. …                                                Equation (5)

Let z^p-1 – z^p-2Y₁ + z^p-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) = pX₁^p-1… Equation (6)

Then, either p ∈ f₁(z) and p ∈ g₁(z), or they are perfect p-powers of integers.

For a given case of z^p = X₁^p + Y₁^p, 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) = z^p-1 + z^p-2X₁ + z^p-3X₁² +•••+ X₁^p-1 = A^p, 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 p^th power, as in that case when A is not an integer, but is the p^th 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) = s^p-1 + s^p-2X₁ + s^p-3X₁² +•••+ X₁^p-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) = S₁^p-1 + S₁^p-2X₁ + S₁^p-3X₁² +•••+ X₁^p-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) = z^p-1 + z^p-2X₁ + z^p-3X₁² +•••+ X₁^p-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) = A₁^p, but if q(z) and f(s) are unique,

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

__________

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: z^p – x^p = x^p. 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)( z^p-1 + z^p-2x + z^p-3x² +•••+ x^p-1) = y^p, 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[2] 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 z^p-1 + z^p-2X + z^p-3X² +•••+ zX^p-2 + X^p-1 = f(z)

And z -X = g(z). Then

(3.)     g(z)f(z) = Y^p, 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) = X₁^p-1 + X₁^p-2X₁ + X₁^p-3X₁² +•••+ X₁^p-1 = pX₁^p-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 X₁^p-1, may be contained in g(z) = z –X_1, 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)( z^p-1 + z^p-2Y + z^p-3Y² +•••+ Y^p-1) = X^p. And by exactly the same reasoning as above, for any particular Y = Y₁, we can have

z^p-1 – z^p-2Y₁ + z^p-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 z^p = X₁^p + Y₁^p , 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) = z^p-1 + z^p-2X₁ + z^p-3X₁² +•••+ X₁^p-1 = A^p, and/or

f₁(z) = z^p-1 – z^p-2Y₁ + z^p-3Y₁ ² -•••+ Y₁^p-1 = B^p, A and B integers, and at least one, A or B, does not contain p. Also note that since g(z)f(z) = Y₁^p, and f(z) = A^p, A ∈ Y₁ and since g₁(z)f₁(z) = X₁^p, f₁(z) = B^p, 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) = z^p-1 + z^p-2X₁ + z^p-3X₁² +•••+ X₁^p-1 = A^p, 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) = A^p does not lead to a contradiction.

For any value of X₁, the conditions f(z) = A^p, 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 p^th power, as in that case when A is not an integer, but is the non-integer p^th 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) = s^p-1 + s^p-2X₁ + s^p-3X₁² +•••+ X₁^p-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 = Z₁^p. 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) = S₁^p-1 + S₁^p-2X₁ + S₁^p-3X₁² +•••+ X₁^p-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) = z^p-1 + z^p-2X₁ + z^p-3X₁² +•••+ X₁^p-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) = A^p, therefore:

(z – S) ɇ f(z) → (Z₁ – S₁) ∉ f(Z₁) → A₁ ∉ A₁^p 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 S₁^p-1 + S₁ ^p-2X₁ + S₁^p-3X₁² +•••+ X₁^p-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 S₁^p-1 + S₁ ^p-2X₁ + S₁^p-3X₁² +•••+ X₁^p-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 X^N + Y^N = Z^N, 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) = S₁^p-1 + S₁^p-2X₁ + S₁^p-3X₁² +•••+ X₁^p-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) = (X_i,Y_i,Z_i), specific integers, such that A = Z_i – 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) = K^p-1 + K^p-2X_i + K^p-3X_i² + ••• + X_i^p-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 K_i until the remainder, f(K_i) 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 [3] obtained by successively dividing by A will eventually reduce f(K_i) 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(K_i) must be equal to zero. In our infinite descent, the smallest possible f(K_i) will occur when K_i =1.

But, the smallest possible f(K_i) = f(1) = 1 + X_i + X_i² + ••• + X_i^{p- 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 x^p + y^p = z^p, 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) ≠ A^p, a perfect p^th 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) = Z^n-1+XZ^n-2+…+ X^n-2Z+X^n-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)(z^n-1+xz^n-2+…+ x^n-2z+x^n-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) =z^n-1+Xz^n-2 +…+X^n-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) = (z^p-1+Xz^n-2+…+ X^n-2z + X^n-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) = (z^p-1+Xz^p-2+…+ X^p-2z + X^p-1) - (a^p-1+Xa^p-2+…+X^p-1). Then, collecting like powers of X, we have q(z)(z – a) = z^p-1 – a^p-1 +(z^p-2 – a^p-2)X +…+ (z – a) X^p-2 + X^p-1 – X^p-1.

Simplifying and dividing by (z – a), we have:

q(z) =(z^p-2+az^p-3+…+ a^p-3z + a^p-2) + (z^p-3+az^p-4+…+ a^p-4z + a^p-3)X +… + (z + a)X^p-3 + X^p-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) =(z^p-2+az^p-3+…+ a^p-2)+(z^p-3+az^p-4+…+ a^p-3)X+…+(z + a)X^p-3+X^p-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) = Z^p-1 + X₁Z^{p-2 +} … + X^p-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)=(Z^p-2+aZ^p-3+…+ a^p-2)+(Z^p-3+aZ^p-4+…+ a^p-3)X+…+(Z + a)X^p-3+X^p-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 Y₁^p = J₁^pK₁^p, f(a) = a^p-1 + a^p-2X₁ + …+ aX₁^p-2 + X₁^p-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) = Z^p-1+ X₁Z^p-2 + … + X₁^p-1 = (K₁)^p, q(Z) = (Z ^n-2+aZ₁^n-3+…+ a^n-2) + (Z^n-3+aZ ^n-4+…+ a^n-3)X+…+(Z + a)X^n-3+X^n-2, and

f(a) = a^p-1+X₁a^p-2+…+X₁^p-1. Substituting these hypothetical integer values into Eq. (E-4), we have:

(K₁)^p/K₁ = (K₁)^p-1 = (Z^p-2+aZ^p-3+…+ a^p-2) + (Z^p-3+aZ^p-4+…+ a^p-3)X+…+(Z + a)X^p-3+X^p-1 +

+ (a^p-1+X₁a^p-2+…+X₁^p-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) = Z^p-1 + X₁Z^p-2 + X₁²Z^p-3 +…+ X₁^p-2Z +X₁^p-1 by the integer polynomial  Z – a = K₁.

The quotient is q(Z)=(Z^p-2+aZ^p-3+…+a^p-2) + (Z^p-3+aZ^p-4+…+a^p-3)X₁+…+(Z + a)X₁^p-3+X^p-2, and the remainder is f(a)  = a^p-1+X₁a^p-2+…+X₁^p-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, a_i, a_i+1, such that

z – a = K₁ ≡ a – a₁ ≡ a₁ – a₂ ≡ a₂ – a₃ ≡…≡ a_k-1 – a_k, with a >a₁ >a₂ >… >a_k, 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, a_i-1 and a_i, if they exist, that will produce the minimum remainder, r = f(a_i), 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. (a^p-1+X₁ a^p-2 +… +X₁^n-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 a_i and a_i+1. Each division produces an addition to q(Z) and a diminishing f(a_i) 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(a_i) is actually divisible by a_i-1 - a_i = 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)= a^p-1 + X₁a^p-2+…+X₁^p-1, an algebraic polynomial in a, by a – a₁, to obtain:

f(Z)/(a – a₁)=(Z^p-2+aZ^p-3+…+a^p-2)+(Z^p-3+aZ^p-4+…+a^p-3)X₁+…+(Z + a)X₁^p-3+X₁^p-2 + (a₁^p-1 + X₁a₁^p-2 + … + X₁^p-1)/(a - a₁).                                                                                            Eq. (E-6)

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

1.) If a_i for the minimum f(a_i) is zero, then a_i-1 – a_i = a_i-1 = K₁, and f(a_i) = a₁ ^p-1+X₁a₁^p-2 +…+X₁^p-1 = 0 + 0 +…+X₁^p-1 = X₁^p-1, which we see by inspection of Eq. (E-6),  must contain a_i-1 because of closure of integers with respect to addition. This implies that X₁^p-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(a_i).

2.) The smallest a_i cannot be positive: A positive a_i will not produce the smallest f(a_i), because f(a_i), which is always positive, will be smallest with the first negative a_i. After the first negative a_i, successive negative values of a_i will only produce larger values for f(a_i), because odd powers of negative numbers are negative and even powers of negative numbers are positive. With f(a_i) = a₁ ^p-1+X₁a₁^p-2 +…+X₁^p-1, the even powers of a_i and X₁ will always outnumber and be greater than the odd powers of a_i and X₁.

3.) Finally, because the a_i producing the smallest f(a_i) cannot be zero or positive, it will have to be negative, and q(a_i) and f(a_i) will be unique, implying by corollary II and III that the smallest f(a_i) must be a positive integer not containing Z – a = a_i-1 – a_i. But, if f(Z) is an integer, and q(a_i) is an integer, by inspection of Eq. (E-6) and closure of integers with respect to addition, f(a_i) must contain K₁, the integer value of Z – a. This is a contradiction unless the a_i’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 a_i. Each time the last term is expanded by dividing f(a_i-1) by a_i-1 – a_i, an additional quotient term, q(a_i) = (a_i-1 + a_i + X₁) and a new remainder, f(a_i) 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 a_i is smaller than the one before, and X₁ is a constant. This descent will be continued until f(a_i) reaches a minimum. At that point, a_i 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 a_i for the minimum f(a_i) is zero, then a_i-1 – a_i = a_i-1 = K₁, and f(a_i) = a_i²+ a_iX₁ +X₁² = 0 + 0 + X₁² = X₁², which we see by inspection of Eq. (E-7)  must contain a_i-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(a_i).

2.) The smallest a_i cannot be positive. If a_i is positive, it will not produce the smallest f(a_i), because f(a_i) is always positive, and it will be smallest with the first negative a_i. After the first negative a_i, successive negative values of a_i will only produce larger values for f(a_i), because f(a_i) = a_i²  +aX₁ + X₁², and the even powers of a_i and X₁ outnumber and are greater than the odd powers (negative) of a_i and X₁.

3.) Finally, since the a_i producing the smallest, and therefore unique f(a_i) cannot be zero or positive, it will have to be negative, and q(a_i) and f(a_i) will be unique, implying that the smallest f(a_i) 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(a_i) is an integer, by inspection of Eq. (E-7) and closure of integers with respect to addition, we see that f(a_i) must contain the integer value of Z – a. This is a contradiction, unless the a_i’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 = a_i-1 - a_i = 19 for decreasing values of a_i-1 and a_i, does not change the value of f(Z)/( Z –a) = q(Z) + f(a_i) because as f(a_i) 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, a_i-1 and a_i, 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 (a_i-1, a_i) that produce a unique q(a_i) and minimum f(a_i), 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(a_i).

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 a_{i -1} and a_i smaller by 1, as follows: 73–54 = 72–53 = 71–52 =…= a_i-1 - a_i = 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) <max & f(ai) >min
66	8 – (-11)	5214	223	NO, q(ai) <max & f(ai) >min

67	7 – (-12)	5226	229	NO, q(ai) <max & f(ai) >min
68	6 – (-13)	5236	237	NO, q(ai) <max & f(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) <max & f(ai) >min
72	3 – (-16)	5254	273	NO, q(ai) <max & f(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, r_{i =} f(a_i), occurs between a_i = -8, and a_i = -9, which tells us that the a_i that produces the minimum remainder is not an integer. This contradicts the assumption that Z and a are integers: If a_i is not an integer, then a_i-1 - a_i = 19 implies that a_i-1 = 19 – a_i 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, r_{i =} f(a_i) = 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(a_i) and q(a_i) are unique. For unique f(a_i) and q(a_i), 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, a_i-1 - a_i, for the integer values of X, a and Z that produces a unique q(a_i) maximum and a unique f(a_i) 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 a_i-1 - a_i that produces a unique q(a_i) and f(a_i) 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(a_i) + f(a_i )/(Z₁ – a)                                                                       Eq. (E-8)

By inspection of this equation we see that f(a_i) must contain Z₁ -a, but if f(a_i) 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.

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[1] 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.

[2] 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.

[3] 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 ᵱ 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 ᵱ 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 ᵱ, 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 >…> N_i, 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 ᵱ 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.^[9]