## Wednesday, February 12, 2014

### Fermat's Last Theorem 1965 proof and Update

CLARIFICATION AND EXPLANATION OF CLOSE’S 1965 PROOF OF
FERMAT’S LAST THEOREM
By Edward R. Close PhD

Concerning whole numbers, while certain squares can be separated into two squares, it is impossible to separate a cube into two cubes or a fourth power into two fourth powers or, in general any power greater than the second into two powers of like degree. I have discovered a truly marvelous demonstration, which this margin is too narrow to contain.” - Pierre de Fermat, circa 1637 

ABSTRACT:
Fermat’s last theorem (FLT) states that no three positive integers X1, Y1, and Z1 can satisfy the equation xn +yn = zn for any integer value of n greater than two. Though a proof was claimed by Pierre de Fermat in 1637, it was never found and Fermat’s Last Theorem remained officially without proof until Andrew Wiles published an extensive solution in 1995. Prior to that, however, Edward Close in 1965 provided a solution and submitted a proof, which he then published in 1977 (FLT65). The Close proof though never refuted presented difficulties for reviewers because of the unconventional notation used and three reviewers have suggested that the difference between algebraic polynomial factors and integer factors as used made the proof unclear and possibly incorrect. In this paper, Close provides FLT65C as a clearer rewrite of FLT65. He also demonstrates that FLT65 provided a true and adequate proof of FLT. The FLT65 proof remains unchanged in this presentation but adherence to conventional notation makes it clearer, and the polynomial critique is shown to be irrelevant in this rewrite (FLT65C). Additionally, the fact that the case of the exponent N=2 is excluded from the proof is shown more clearly. This paper provides a 2-page FLT65C solution, and then in the discussion amplifies and clarifies areas that may help reviewers understand the proof. Several pertinent appendices are attached. Appendix A demonstrates the proofs of the division algorithm and corollaries because they are central to the proof. Additionally, the application of primes and relative primes is an important part of the proof of FLT65: Appendix B shows the proof of sufficiency for X, Y and Z relatively prime and N = primes > 2, in any proof of FLT. And importantly, Appendix C provides the original FLT65 published proof.

INTRODUCTION
Sometime around 1637, Pierre de Fermat scribbled a brief statement in Latin in the margin of a book on Diophantine equations. This theorem came to be known as “Fermat’s Last Theorem” (FLT), because it remained without a known proof for more than three centuries. In modern representation, FLT is stated as follows: No three positive integers X1, Y1, and Z1 can satisfy the equation xn +yn = zn for any integer value of n greater than two. The first generally accepted proof has been attributed to Sir Andrew Wiles, who is said to have spent seven years completing it. Dr. Wiles’ proof was published in 1995.

In 1965, three decades before Wiles’ proof was announced, I (Edward R. Close) had produced a proof of FLT. At the time, I had completed my BA degree in mathematics with a minor in physics, one semester of a Master’s program in theoretical physics, and had been teaching mathematics for three years. The 1965 proof was submitted to a mathematician at Iowa State University. He returned it with a note saying that “It could not be a valid proof because it would apply to the case n = 2”. From his response, I knew he had not read past the first page, because the case n = 2 was eliminated on the second page. I learned that most professional mathematicians consider reviewing a purported proof of FLT a waste of time, and look for anything that appears to be an error and reject the “proof” as quickly as possible. I was very discouraged and went on with my life, but later published my proof anyway in an appendix of “The Book of Atma”.  Over the years, my 1965 proof of Fermat’s Last Theorem (FLT65) has been submitted to a number of professional and amateur mathematicians for review and comment. The responses from those to whom the proof was submitted fall into one of the following categories:

1.      Some declined to look at it either because they didn’t have the time or they didn’t want to take the time to review it.
2.      Some refused to look at it because number theory was not their field of expertise and/or interest.
3.      Some refused to look at it because they did not believe a simple proof of FLT possible.
4.      Several of the reviewers found nothing wrong with the proof.
5.      Some of the reviewers found the argument inconclusive.

Those who found the argument inconclusive cited one or both of the following reasons:

i.        They found the notation unclear or confusing.
ii.      They were concerned that conclusions drawn from the factorization of the FLT equation, xn + yn = zn, as a polynomial in n, might not apply to the numerical factorization of the polynomial for some specific integer solution (X1, Y1, Z1), providing a potential loop-hole in the proof. 

The notation in the 1965 proof may have confused readers because it is non-standard and not as well defined as it should have been. At the time, I used unconventional notation deliberately because I thought it made it easier to distinguish when the arguments were being considered as variables from when they were being considered as specific integers. The subscript identified an algebraic symbol as an integer, as opposed to a variable. I failed, however, to provide a notation that distinguished between integer variables and specific integer constants, and this was problematic because confusion may arise if a specific notation is not clearly defined and its use properly justified. Therefore, in this discussion, I will use three types of notation precisely defined for greater 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 a factor of B.

My position has been, and remains, that FLT65, my 1965 proof, is conclusive and complete, even though the notation may be confusing. I maintain this position for two simple reasons: (A) The division algorithm and corollaries applied in the proof are valid over the entire field  of real numbers, which includes the integers, and (B) Application of the division algorithm to the FLT equation produces a unique remainder that allows us to determine the numerical type of the third variable when two of them are assumed to be integers.

Reviewers in category 5 above, with concern number ii, thought that conclusions drawn from the factorization of the FLT equation, xn + yn = zn, as a polynomial in z, might not apply to the numerical factorization of the polynomial for some specific integer solution (X1, Y1, Z1). While this is a legitimate concern, it is dispelled by detailed clarification of notation and step-by-step explanation of the reasoning.
After this introduction, I will present the mathematics of FLT65 with no more discussion than is necessary to be complete. Then I will present a more detailed discussion and explanation of the proof, showing how unique features of the FLT equation provide proof that X, Y and Z cannot all three be integers, and how this eliminates concern number ii above.
The Division Algorithm and three Corollaries, presented briefly below, are central to this proof, and their proofs were included in FLT65. In order to make this presentation as complete as possible, these supplementary proofs are included in Appendix A.
The division algorithm states that if g(x) 0 and f(x) are any two polynomials over a field, of degree m and n, respectively, and n > m, then there exist unique polynomials q(x) and r(x) such that f(x) = q(x)g(x) + r(x), where r(x) is either zero or of degree smaller than m.
Corollary I states that if f(x) and g(x) contain a common factor, r(x) contains it also.
Corollary II states that the remainder, r(x), when f(x) is divided by (x-a) is f(a).
Corollary III states that a polynomial, f(x), of degree greater than one x-a, IF AND ONLY IF, f(a) = 0.
To prove FLT, it is necessary to show that for n 3, there is no solution (X1,Y1, Z1) such that X1, Y1 , and Z1 are integers. It is necessary and sufficient to show this for n = p, where p is a prime number and X1, Y1 and Z1 are relatively prime. Proofs of these two statements are given in the 1965 proof. They are relatively simple and well-known. For efficiency they are not included in the proof below, but for completeness and clarity, they are included in Appendix B. The method employed in the 1965 proof of FLT, is to show that if we restrict x and y to the field of integers, z cannot be an integer. Clearly, when this is shown, FLT is proved.

FERMAT’S LAST THEOREM (FLT)
(A TWO PAGE PRESENTATION OF THE CLOSE 1965 PROOF [FLT65C])

Consider the equation zp – xp = yp. …… equivalent to Equation (1) in the 1965 proof. Since 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) = yp…… 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, 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 (7).

Since both s and z can take on any real number value, if they are not integers, f(s) = 0 is not a contradiction because there are an infinite number of real number solutions for the FLT equation. 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.

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, 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, therefore:

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

DISCUSSION AND EXPLANATIONS
OF THIS PROOF OF FERMAT’S LAST THEOREM (CLOSE FLT65C)

This is the same proof FLT65C but with detailed discussions and explanations:
Using the notation outlined in the introduction, 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 p is a prime number, 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 up-dated 1965 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 arbitrarily choose 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 field of integers, we will be able to determine whether any values of the third variable can be integers.

In FLT65, Close 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, I indicated that they were specific integer values, without changing the way they were represented. This lack of clear definition of notation is at least part of the confusion that gave rise to the concern #ii.

My 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 is not necessarily true for polynomials in general. It turns out to be the case in this instance, however, because of the unique form of the FLT equation and the requirement that the only solutions being considered are those for which all three variables have integer values.

Because at least three of the most qualified reviewers of FLT65 raised the concern about the applicability of the algorithm and corollaries to factors of integers, it is clear that the proof is 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 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 will indicate the negation of 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 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 will allow 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 f(z) = zp-1 + zp-2X1 + zp-3X12 +•••+ X1p-1 = Ap, and 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, 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 pth root of a prime number, q(z) and r(z) are not unique and COROLLARY II does not hold, and 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, 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, 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. But, if s is to be from the ring of integers, 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, 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 justified substituting 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.
When the author did this, he found that the contradiction obtained by assuming that for given integer values of x and y, an integer z was possible also applied 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. However, if the reader is not convinced by this argument, we can address concern #ii in additional detail 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 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.

Therefore, we may state unequivocally that the equation xp + yp = zp has no solutions in positive integers when p is an integer > 2, and the proof of Fermat’s last theorem is complete.
APPENDIX A:
PROOFS OF The Division Algorithm and Corollaries
The Division Algorithm
THEOREM: (The division algorithm) If g(x) ≠ 0 and f(x) are any two polynomials over a field, of degree m and n, respectively, and n>m, then there exist unique polynomials q(x) and r(x) such that
(A)       f(x) = q(x)g(x) + r(x)
Where r(x) is either zero or of degree smaller than m.
Let f(x) = anxn + an-1xn-1 +•••+ a1x + a0
And g(x) = bmxm + bm-1xm-1 +•••+ b1x + b0, bm ≠0.
If q(x) is zero or of a degree smaller than m we have
f(x) = 0•g(x) + f(x)
and if n = m, q(x) becomes a constant, Q, and r(x) may be of any degree from zero to n, depending on the value of the constant Q. Thus q(x) and r(x) are not unique in this case.
Now we can form f1(x), of lower degree than f(x), by writing
(B)       f1(x) = f(x) - an/am xn-m •g(x).
We may now complete the proof of this theorem by induction: Assume the algorithm true for all polynomials over the field. Since f1(x) is such a polynomial, we may write:
(C)   f(x) = q1(x)g1(x) + r1(x),

where r1(x) is either zero or of a degree less than m. So from (B) and (C):
f(x) = | an/am xn-m + q1(x) | g1(x) + r1(x).
or f(x) = Q(x)g(x) + R(x), the desired form of f(x). All that remains is to prove that Q(x) = q(x) and R(x) = r(x), that is, that q(x) and r(x) are unique when m>n.
Now, f(x) = q(x)g(x) + r(x)
And f(x) = Q(x)g(x) + R(x).
This implies Q(x)g(x) + R(x) = q(x)g(x) + r(x), or [Q(x) – q(x)]g(x) = r(x) – R(x).
If m = n, the right member of this equation is either zero or of degree less than m, the degree of g(x). Hence, unless Q(x)– q(x) = 0 → Q(x) = q(x) and r(x) – R(x) = 0 → r(x) = R(x), the left member of the equation will be of a higher degree than m, and we have a contradiction. Thus Q(x) = q(x) and r(x) = R(x), which means that q(x) and r(x) are unique, and the division algorithm is proved over the field of real numbers.
Corollary I
If f(x) and g(x) contain a common factor, r(x) contains it also.
This follows by inspection of equation (A):
f(x) = q(x)g(x) + r(x) → r(x) = f(x) - q(x)g(x).
Let the common factor be represented by M. Then f(x) = Mf1(x) and g(x) = Mg1(x), so that r(x) = Mf1(x) - q(x)Mg1(x) = M∙[f1(x) - q(x)g1(x)] r(x) = M(a function of x), QED.
Note that this proof holds for M constant or variable, integer or polynomial factor.
COROLLORY II:
The remainder when a polynomial is divided by (x-a) is f(a):.
This follows at once from the division algorithm:
Let g(x) = x-a. Then (A) becomes
f(x) = (x-a)q(x) + r(x).
By substitution,
f(a) = (a-a)q(a) + r, where r is an element of the field
or, f(a) = r,
so that f(x) = (x-a)q(x) + f(a), QED.
Notice that if q(x) and r(x) are unique, f(a) cannot contain x-a, and so it follows that
COROLLORY III: A polynomial, f(x), of degree greater than one is divisible by x-a IF AND ONLY IF, f(a) = 0.
Note that the division algorithm and all three corollaries hold when f(x) = f(X)and g(x) = g(X), expressions involving only integers, since it is valid over the entire field of real numbers, and integers are elements in the field of real numbers.

APPENDIX B:
PROOF OF SUFFICIENCY FOR X, Y AND Z RELATIVELY PRIME
AND N = PRIMES > 2, IN ANY PROOF OF FLT
X, Y AND Z RELATIVELY PRIME:
When developing a proof of FLT, in the FLT equation XN + YN = ZN, X, Y and Z may be considered as three relatively prime integers.
X, Y and Z may be considered to be relatively prime because if two of them, say X and Y, contain a common factor or factors, M, then Z contains M also. Proof:
If X = MX1 and Y = MY1, then ZN = (MX1)N + (MY1 )N = MN(X1N + Y1N). And thus Z = MZ1 , i.e. Z contains M. Also, XN + YN = ZN → (MX1)N + (MY1)N = (MZ1)N. Factoring MN out, we have X1N + Y1 N = Z1N, with X1, Y1 and Z1 relatively prime.
This demonstration proves that any case of the FLT equation with X, Y and Z not relatively prime can be reduced to a case wherein they are relatively prime, and thus if we prove the theorem for X, Y and Z relatively prime, no non-relatively prime case can exist.
N RESTRICTED TO PRIME NUMBERS > 2 IN FLT PROOF:
Definition: a prime number is any integer that is only divisible by itself and 1. The first prime number after unity, and the only even prime number, is 2. But when N = 2, the equation XN + YN = ZN is known as the Pythagorean Theorem equation, which has an infinite number of integer solutions known as the Pythagorean triples. For example: 32 + 42 = 52. This is why Fermat stated the theorem in the way he did. Translated from Latin, it reads:
Concerning whole numbers, while certain squares can be separated into two squares, it is impossible to separate a cube into two cubes or a fourth power into two fourth powers or, in general any power greater than the second into two powers of like degree. I have discovered a truly marvelous demonstration, which this margin is too narrow to contain.” Pierre de Fermat, circa 1637 

If we can prove FLT for N equal to prime numbers greater than 2, i.e. N = p > 2, we will have proved FLT for all N. Proof:
Let N = ab, with B prime and A any other prime or composite of primes. Then
Xab + Yab = Zab → (Xa)b + (Ya)b =(Za)b
This is clearly a case of the FLT equation with N prime. We can assume XYZ 0, to eliminate trivial solutions that are obtained when one of the triples equals zero, and as shown above, we can assume that x,y,z are relatively prime (sometimes called co-prime).
Since we know that the equation has integer solutions when N = 2, we must consider the cases N = 2a:
X2a + Y2a = Z2a → (Xa)2 + (Ya)2 =(Za)2, which is a Pythagorean equation. Since we know that the Pythagorean equation has integer solutions, the question here becomes: can all three members of a Pythagorean triple be powers of integers. Fortunately, we can eliminate all even powered N from a proof of FLT as follows:
If a = 2, we have: (X2)2 + (Y2)2 = (Z2)2.

Solution triples for this Pythagorean Theorem equation may be obtained using the well-known formulas for Pythagorean triples. Note:
Close published his derivation of the ratio formulas for the Pythagorean triples in 1977. The formulas were derived from the FLT equation when n = 2 using the properties of real numbers. 

Using the formulae, we know that there must be two relatively prime integers, P and Q, with PQ > 0, such that:

X2 = 2PQ
Y2 = P2 - Q2 Y2 + Q2 = P2
Z2 = P2 + Q2

Now with this application of the formulae, we have obtained another
Pythagorean triple:
Y2 + Q2 = P2. By inspection of the equations for X2, Y2 and Z2 above, we see that P < Z, Q < X, and Y < Y2. Thus, by assuming that a triple integer non-zero solution exists for the equation (X2)2 + (Y2)2 = (Z2)2 we can produce another integer triple solution with smaller integers. Repeated applications of the formulae will produce smaller and smaller triples, leading to contradiction by infinite descent: Since any integer solution will lead to a smaller integer solution, the smallest integer of the smallest triple must eventually equal the smallest non-zero integer, 1, yielding a triple (1,C,D), where C and D are positive integers, such that (12)2 + (B2)2 = (C2)2→ (C2)2 - (B2)2 = 1. Given that the smallest integer of the successive non-zero triples in the infinite descent will eventually reach unity, and B > A > 1, we can demonstrate the universality of the contradiction as follows: Let C = 3, and D = 2, the smallest positive integers larger than 1. Then (32)2 - (22)2 = 1 → 81 – 16 = 1, which is a clear contradiction, and any pairs of larger integers C1 > C and D1 > D, will lead to larger discrepancies. So we have to conclude that there can be no integer solution triples for the equation (X2)2 + (Y2)2 = (Z2)2.

This, of course, is proof of FLT for N = 4, which was first proved by Fermat3. However, this has greater significance than just a proof for N = 4, because any case of X2a + Y2a = Z2a when a is an even number, is a case of N = 4: If a = 2m, m a positive integer, X2a + Y2a = Z2a → X4m + Y4m = Z4m → (Xm)4 + (Ym)4 = (Zm)4. Thus the proof of FLT for N = 4 is proof of FLT for all even N, and since all non-prime odd integers are factorable into prime numbers, it is sufficient to consider N = p, a prime in any proof of FLT. QED

APPENDIX C:
The original proof of 1965: Close’s Fermat’s Last Theorem (FLT65)
PROOF OF FERMAT’S LAST THEOREM
E. R. CLOSE
The theorem:                 XN + YN ≠ ZN
When X, Y and Z are integers > 0
And N is any prime integer >2.
The following theorem and corollaries are proved here because their proof makes their application clearer and because certain aspects, not ordinarily mentioned, are of particular interest in the proof of Fermat’s last theorem.
THEOREM: (The division algorithm) If g(X) ≠ 0 and f(X) are any two polynomials over a field, of degree m and n, respectively, and n>m, then there exist unique polynomials q(X) and r(X) such that
(A)        f(X) = q(X)g(X) + r(X)
Where r(X) is either zero or of degree smaller than m.
Let f(X) = anXn + an-1X n-1 +•••+ a1X+ a0
And g(X) = bmXm + bm-1X m-1 +•••+ b1X+ b0, bm ≠0.
If q(X) is zero or of a degree smaller than m we have
f(X) = 0•g(X) + r(X)
and if n = m, q(X) becomes a constant, Q, and r(X) may be of any degree from zero to n, depending on the value of the constant Q. Thus q(X) and r(X) are not unique in this case.
Now we can form f1(X), of lower degree than f(X), by writing
(B)        f1(X) = f(X) - an/am Xn-m •g(X).
We may now complete the proof of this theorem by induction: Assume the algorithm true for all polynomials over the field. Since f1(X) is such a polynomial, we may write:
(D)    f(X) = q1(X)g1(X) + r1(X),
where r1(X) is either zero or of a degree less than m. So from (B.) and (C.):
f(X) = | an/am Xn-m + q1(X) | g1(X) + r1(X).
or f(X) = Q(X)g(X) + R(X), the desired form of f(X). All that remains is to prove that Q(X) = q(X) and R(X) = r(X), that is, that q(X) and r(X) are unique when m>n.
Now, f(X) = q(X)g(X) + r(X)
And f(X) = Q(X)g(X) + R(X). This implies Q(X)g(X) + R(X) = q(X)g(X) + r(X), or [Q(X) – q(X)]g(X) = r(X) – R(X).
If m = n, the right member of this equation is either zero or of degree less than m, the degree of g(X). Hence, unless Q(X) – q(X) = 0 → Q(X) = q(X) and r(X) – R(X).= 0 → r(X) = R(X)., the left member of the equation will be of a higher degree than m, and we have a contradiction. Thus Q(X) = q(X) and r(X) = R(X), which means that q(X) and r(X) are unique, and the division algorithm is proved.
Again notice the important fact that if m = n, i.e. f(X) and g(X) are of the same degree, Q(X) and q(X) must be of zero degree in X. So R(X) and r(X) are now of the same degree from zero to n, if f(X)X = g(X). The degree of R(X) and r(X) now depend upon the values Q(X) and q(X), so that Q(X) and q(X) are not necessarily equal, as R(X) and r(X)are not, and q(X) and r(X) are not unique.
COROLLARY I: If f(X) and g(X) contain a common factor, r(X) contains it also.
This follows by inspection of equation (A):
f(X) = q(X)g(X) + r(X) → r(X) = f(X) - q(X)g(X).
COROLLARY II:
The remainder when a polynomial is divided by (X-a) is f(a):.
This follows at once from the division algorithm:
Let g(X) = X-a. Then (A) becomes
f(X) = (X-a)q(X) + r(X).
By substitution,
f(a) = (a-a)q(a) + r, where r is an element of the field
or, f(a) = r,
so that f(X) = (X-a)q(X) + f(a).
Notice that if q(X) and r(X) are unique, f(a) cannot contain X-a, and so it follows that
COROLLARY III: A polynomial, f(X), of degree greater than one is divisible by X-a IF AND ONLY IF, f(a) = 0.
It may be remarked that the division algorithm and all three corollaries hold when f(X) and g(X) are expressions involving only integers, since integers are elements in the field of real numbers.

THE PROOF
Consider the equation
(1.)     XN + YN = ZN, where n is a prime number>0, and, X,Y and Z are relatively prime integers.>0. X,Y and Z may be considered to be relatively prime because if two of them, say X and Y, contain a common factor, M, then Z contains it also:
If X = Mx and Y = My, then ZN = (Mx)N + (My)N = MN(XN + YN). And thus Z = Mz, i.e. Z contains M. Also, XN + YN = ZN(Mx)N + (My)N = (Mz)N. Factoring MN out, we have xN + yN = zN
And so, any case of equation (1.) with X, Y and Z not relatively prime implies a case wherein they are relatively prime, and thus if we prove the theorem for X, Y and Z relatively prime, no non-relatively prime case can exist.
Furthermore, it is only necessary to consider N prime, since any non-prime case for N in (1.) implies a case wherein N is prime. For example, let N = ab, with b prime and a may be another prime or a composite of primes. Then
Xab + Yab = Zab → (Xa)b + (Ya)b =(Za)b Clearly a case of (1.) with n prime. Since all non-prime integers are factorable into prime numbers, it is sufficient to consider N as a prime in any proof.
Now, XN + YN = ZN → ZN – XN = YN . Factoring, we have:
(2.)     (Z-X)( ZN-1 + ZN-2X + ZN-3X2 +•••+ XN-1) = YN .
For any given X, say X = X1 let ZN-1 + ZN-2X1 + ZN-3X12 +•••+ X1N-1 = f(Z)
And Z-X1 = g(Z). Then
(3.)     g(Z)f(Z) = YN.
Remembering that for Fermat’s last theorem to be falsified, Y is an integer, as are X and Z, then for any particular case of XN + YN = ZN, it follows that either g(Z) and f(Z) contain a common factor, or they are both perfect N-powers of integers. By Corollary I, the remainder, when f(Z) is divided by g(Z), will contain any and all factors common to both. And by COR II, the remainder when f(Z) is divided by g(Z) = Z-X1, will be f(X1). And
(4.)     f(X1).= X1N-1 + X1N-2X1 + X1N-3X12 +•••+ X1N-1 = NX1N-1.
Since X, Y and Z are relatively prime, no factor of X1 may be contained in Z –X. Therefore, if f(Z) and g(Z) have a common factor, it must be N. It also follows that either f(Z) or g(Z) contains N or they are perfect N-powers.
Similarly, from XN + YN = ZN,
(5.)     (Z-Y)( ZN-1 + ZN-2Y + ZN-3Y2 +•••+ YN-1) = XN.
And by exactly the same reasoning as above, for any particular Y = Y1, if we let
ZN-1 – ZN-2Y1 + ZN-3Y1 2 -•••+ Y1 N-1 = f1(Z) and
Z - Y1 = g1(Z), then either f1(Z) and g1(Z) contain N as a common factor, or they are perfect N-powers. So for a given case of ZN = X1N + Y1N , if either f(Z) or f1(Z) contains N, the other has to be a perfect N-power, since we have concluded that we only have to consider relatively prime X, Y and Z, implying both cannot contain N.
Now, N ε f(Z) N ε Y1 and N ε f1(Z) N ε X1. But X1.and Y1 are relatively prime. If neither X1.nor Y1 contains N, both, being relatively prime, must be perfect N-powers. Therefore, one of them at least, must be a perfect N-power, not containing N as a factor.
Therefore, f(Z) = ZN-1 + ZN-2X1 + ZN-3X12 +•••+ X1N-1 = AN, and/or
f1(Z) = ZN-1 – ZN-2Y1 + ZN-3Y1 2 -•••+ Y1N-1 = BN, A and B integers <Z. and at least one does not contain N.
Since they are both of the same form, set f(Z) = ZN-1 + ZN-2X1 + ZN-3X12 +•••+ X1N-1 = AN. Then f(Z) is divisible by A, and since A is a positive integer < Z, we may write A = Z - a, where a is another integer smaller than Z.
The division algorithm tells us that for g(Z) ≠ 0 and f(Z), two polynomials 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 N = 2, or if f(Z) is not equal to an integer raised to the Nth power, as in that case when A is the Nth root of a prime number, q(Z) and r(Z) are not unique and COR. II does not hold. But when N>2, and X, Y and Z are integers, COR. II tells us that if g(Z) = Z - a, 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, i.e., zero in Z, a constant, of the form f(a). Therefore:
f(Z) = (Z-a)q(Z) + f(a) over the integer values of the field of real numbers, and by COR. III, f(Z) is divisible by Z - a IF AND ONLY IF f(a) = 0. The division algorithm and its corollaries apply over the field of real numbers, including integers. Thus when X, Y and Z are integers and N>2,
(6.)      (Z - a) ε f(Z) → f(a) = 0.
But f(a) = aN-1 + aN-2X1 + aN-3X12 +•••+ X1N-1 = 0 is an impossibility because if Fermat’s last theorem is falsified, X1 and a, are both positive integers. This implies that f(Z) cannot be a perfect integral N-power. Thus we have reached a complete contradiction by assuming X, Y and Z to be integers, and may state that the equation XN + YN ≠ ZN has no solutions in positive integers when N is an integer > 2. And so the proof of Fermat’s last theorem is complete.
Edward R. Close December 18, 1965

APPENDIX D
ADDENDUM D to FLT65, November, 2011
For Fermat’s last theorem to be falsified, i.e., for the Fermat equation to have integer solutions, Z and a must be integers and f(Z) must equal (Z –a )N.
And we see that f(Z) = (Z –a )N f(Z) - (Z –a )N = 0 , and by substitution of the polynomial f(Z) and expansion of the second term,
ZN-1 + ZN-2X1 + ZN-3 X12 +•••+ X1N-1– [ZN - NaZN-1 + N(N – 1)/2!a2 ZN-2 +•••+ aN] = 0
Collecting like power-of-Z terms, we see that the left-hand side of this equation is an Nth degree polynomial in Z. Call it G(Z). This equation is an algebraic expression of the requirements that must be met for Fermat’s last theorem to be falsified.
Equation (1.) may also be considered to be an Nth degree polynomial in Z and can be written as follows:
F(Z) = ZN – X1N - Y1N = 0, and thus can be expanded to produce:
F(Z) = (Z-X1)( ZN-1 + ZN-2X1 + ZN-3 X12 +•••+ X1N-1) - (Z-X1)(Z –a )N = 0.
By inspection of this equation, we see that G(Z) is a polynomial factor of F(Z), and Thus the roots of G(Z) must also be roots of equation (1.).
The fundamental theorem of algebra tells us that any polynomial of degree N has exactly N solutions. Assuming there is a positive integral solution, Z1 = ri, (1 i N) of the Fermat equation for specific integers X1 and Y1, the N solutions (roots) of G(Z), represent all the N possible solutions of equation (1.), (X,Y,Z), Z = r1, r2, r3, … rN, for the specific values, X = X1, and Y1N = (Z –a)N f(Z). So any integer factor of f(a) that might represent an integer solution of (1.) must also be a factor of Z – a and G(Z).
In order for any integer factor of f(a) to represent an integral solution (X 1,Y1, Z1) of F(Z), it must be equal to some integer, r, where r is one of the roots, r1, r2, r3, … rN, of the polynomial F(Z). But, if r is a root of F(Z) = 0, it must also be one of the possible values of the one-degree polynomial Z – a, a polynomial factor of F(Z). If this is true, then by the division algorithm, F(Z)/(Z – a) = Q(Z) + F(a), and by Corollary III, F(a) must equal zero.
The equation F(a) = (a-X1)( aN-1 + aN-2X1 + aN-3 X12 +•••+ X1N-1) = 0 is satisfied by:
a - X1 = 0 and aN-1 + aN-2X1 + aN-3 X12 +•••+ X1N-1 = 0
So the N solutions of F(a) are a = X1 and N – 1 values of ai, none of which can be integers. The nature of the N – 1 values of ai, irrational, imaginary or complex, can be determined, but for the purpose of this proof, it is only necessary to observe that they cannot be integers since the sum of N positive integers cannot equal zero.
These N values of a yield the N solutions of F(Z) = 0 as follows:
Z-X1 = 0 Y1 = 0 X1N + (0)N = ZN , an integral solution, albeit trivial*, and
Z – ai = 0, i = 1,2,3,…N – 1 X1 N + Y1 N = (ai )N, where none of the ai are integers.
Therefore the N roots, of equation (1.): r1, r2, r3, … rN, are equal to the N roots of F(Z): X1, a1, a2 , a3 , … ain-1.
Since all of the N solutions of the Fermat equation are accounted for, and their numeric nature identified, for N equal to any prime number >2, the proof is complete.
• Note that Fermat, in his statement of the theorem, specified that there could be no positive solutions (X,Y,Z), excluding the solution (X,Y,0). However, X1N + (0)N = ZN, is, of course, one of the N legitimate solution of F(Z)

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          L. Kronecker , “Vorlesungen über Zahlentheorie”, vol. I. Leipzig: Teubner pp. 35–38, 1901. Reprinted by New York: Springer-Verlag in 1978

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

In this “clarified” version (FLT65C), we will see that this problem never actually arises because z of the FLT equation is always a variable, either an unrestricted variable, z, or an integer variable, Z, never a constant, throughout the proof. The fact that this is not clearly indicated in the notation of the original FLT65 proof has caused confusion of the sort expressed in the example critique above. This confusion is fully clarified in this presentation. The key to understanding the proof is realizing that the legitimate application of the division algorithm and corollaries to the polynomial factors of the FLT equation, as polynomials in the variable z, leads to a unique remainder which reveals the fact that if x and y are integers, z cannot be an integer. Thus the question of factors of polynomials in the variable z, versus constant integer factors is resolved. Other than the need for clearer notation, this question is the only legitimate concern that reviewers of FLT65 have brought to my attention.
 In mathematics, a field is a group of ordered quantitative elements that contains a multiplicative inverse for every nonzero element. As such it is an arithmetical structure within which the operations of addition, subtraction, multiplication, and division are consistent and valid, and it supports an algebraic structure within which the same basic operations are valid. Geometrically, a field is a dimensional domain within which arithmetically and algebraically defined elements can be described and located. The three most basic mathematical fields are the field of real numbers, the field of rational numbers, and the field of complex numbers.
 FLT65C is the 2014 rewrite of FLT65 but without modifying any of the Equation statements. The proof above is 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.
 Around 1650, Pierre de Fermat used the technique of infinite descent to prove that the area of a right triangle with integer sides can never equal the square of an integer. Expressing this geometric proposition algebraically with the equation x4 + y4 = z2, he proved the proposition by proving that this equation has no relatively prime integer solutions. He noted that this also proves what we now know as FLT for the case n = 4, since the equation a4 + b4 =c4 can be written as c4  b4 = (a2)2. Later, a number of professional mathematicians produced proofs for the case n = 4, including Bernard de Bessy (1676), Leonhard Euler (1738), Adrien-Marie Legendre (1830), David Hilbert (1897), and Leopold Kronecker (1901),.

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