## Monday, November 23, 2015

### FLT65 PROOF PART 5

FERMAT'S LAST THEOREM PART 5

FERMAT’S LAST THEOREM AND THE CLOSE FLT65 PROOF
Before explaining in detail how some reviewers have missed certain critical details of FLT65, and demonstrating why their concerns are unwarranted, and why their ‘counterexamples’ are invalid, a brief description of FLT and FLT65 is in order.

Fermat’s Last Theorem (FLT) states that xn + yn ≠ zn, when n is an integer greater than 2 and x, y, and z are integers. FLT is stated here in the negative, as an inequality (“it cannot equal”), to emphasize the fact that the FLT65 proof consists of the negation of the existence of the equation xn + yn = zn when x, y and z are integers with n >2. (The cases n = 1 and n = 2 were known to have infinitely many solutions and are thus not part of FLT.)

FLT65 is based on, and relies upon application of the division algorithm to algebraic polynomials of real variables, and in the case of FLT, to integer polynomials because only integer solutions to the FLT equation are of interest. This reliance is of such basic importance that complete proofs of the algorithm and three corollaries, with requirements for their application were included in the 1965 proof to help clarify the proof of FLT. Reviewers who skipped over the division algorithm proof in FLT65, might have missed the importance of the uniqueness requirements which are very relevant to the division algorithm as applied to the FLT equation. It is important to note that they are clearly stated and their importance is explained in detail in the first three pages of FLT65. See Appendix A.

Brief statements of the division algorithm and its three relevant corollaries, with examples to show how they apply, are provided below. For convenience and quick reference, proofs of the division algorithm and its corollaries are also provided separately in Appendix B of this presentation.

The examples the division algorithm and corollaries provided below are specifically designed to show the relevance to FLT65.

FLT65 focuses on n = prime numbers greater than 2 because non-prime exponents can always be factored into prime numbers and the case of n = 4 had already been proved by many. It is, therefore, clear that it is sufficient for FLT65 to focus purely on n = prime numbers greater than 2. FLT65 proves FLT by demonstrating that assuming an integer solution of the FLT equation produces a contradictory inequality when n is prime greater than 2.
See Appendix C for formal proof that it is sufficient to prove FLT for n equal to prime numbers greater than 2.

The Division Algorithm
The division algorithm is an algebraic expression of the basic relationship between the
·       dividend, f,
·       divisor, g,
·       quotient, q, and
·       remainder, r,
in the process of division, where f(X), g(X), q(X), and r(X) are algebraic functions of some variable, represented here by the symbol X.
The division algorithm is stated as follows: For any two polynomials, g(X) ≠ 0 and f(X) over the field of real numbers, of degree m and n, respectively, and n > m > 0, there exist unique polynomials such that f(X) = q(X)g(X) + r(X) where r(X) is either zero or of degree smaller than m.
The Division Algorithm is true for any real value of X, rational, irrational or integer.
(Note that in  the Division Algorithm, n can be any positive integer, 1, 2, 3,… , while, in the FLT65 proof, n is restricted to integer values > 2.)

Example: Let f(X) = X4 + AX3 + A2X2 + A3 X + A4, and g(X) =X - a. Then, for this polynomial (a polynomial of the same form as f(Z) in FLT65)the mathematical operation of division, represented by f/g = q + r, becomes:
(X4 + AX3 + A2X2 + A3 X + A4)/(X-a) = q(X) + r(X)/(X-a), where, by the division algorithm, r(X) is either zero or of degree less than 4. Carrying out the polynomial long division to determine the algebraic polynomial form of q and r, we get:
q(X)= X3 + (A+a)X2 +(A2 +aA+a2)X+(A+aA2 + a2A2 + a3) and r(X)= A+aA3 + a2A2 +  a3A +a4. And if X and a are positive, r(X) cannot be zero and is not of degree smaller than m, so by the division algorithm stated above, the polynomial (X4 + AX3 + A2X2 + A3 X + A4) is not divisible by the polynomial (X-a). But what about the integer values of these algebraic polynomials?
We can assign A, X and a integer values to illustrate the transition from algebraic polynomials to integers: Let A = 1, a = 2, and X = 5, and substitute these integer values into f(X) = q(X)(X-a) + r(X). This gives us:
(54 + 53 + 52 + 5 + 1)= (53 + 3x52 +7x5+15)(5-2)+31 781 = 250x3+31 = 781.
The values of the integer polynomials in this example, reduced to single integers, are f(X)=781, the divisor, X-a =3, q(X)=250, and r(X)=31.

We see that the integer value of r(X) is 31, which is not divisible by the integer value of X-a = 3. This example demonstrates that f(X), which is of the same form as f(Z) in FLT65 with n = 5, is not divisible by X-a, of the same form as Z-a, and that this is true for both the algebraic polynomials, and for  their reduced integer values.
This example illustrates the relationship between the polynomial remainder and integer remainder when the f(Z) factor of the FLT equation is divided by Z -a, which is especially important in the application of  corollary III of the division algorithm in FLT65 and as discussed below.
Corollary I: If f(X) and g(X) contain a common factor, r(X) contains it also. It is important to note that this corollary is true for algebraic polynomials over the field of real numbers, integer polynomials and the integer values of those integer polynomials.
If f(X) and g(X) are integer polynomials, i.e., all terms and coefficients are integers and X is an integer variable, and they contain a common factor, X-a, then the remainder integer variable polynomial r(X) contains X-a, and the integer value of the remainder also contains the integer value of X-a.
Example: the polynomials f(X) = X4+4X3-X2 -16X-12 and g(X) = X2-4 contain the common factor X–2. Corollary II says that because f(X) = X4+4X3-X2 -16X-12 = q(X)(X2-2) + r(X), r(X) contains X – 2, and we  can see that this is true for both algebraic polynomials and their integer values, as long as X-2 is not equal to zero.
For polynomials with X variable, f(X) = (X+1)(X+2)(X +3)(X-2) = q(X)(X+2)(X-2) + r(X), which gives us r(X)=(X+1)(X+2)(X +3)(X-2)-q(X)(X+2)(X-2)= {(X+1)(X+2)(X +3)-q(X)(X+2}(X-2), so we can see by simple inspection of this equation that r(X) contains X-2.
Since the divisor cannot equal zero, X must be greater than 2. So, for this demonstration, let X = 5, then X – a = 5 – 2 = 3, and factoring f and g, we have
(X+1)(X+2)(X +3)(X-2)=q(X)(X+2)(X-2)+r. substituting X=5.this becomes
6x7x8x3 =7x3xq + r, and even without going farther, we can see by inspecting this equation that r contains 3, the integer value of X-2.
Carrying out the algebraic and substituting X = 5, we get f(X) = X4+4X3-X2 -16X-12q(X) = 1008, q(X) = X2+4X+3 = 882, and r = 6X2- 24 = 150 – 24 = 126 = 3x42.
We can check our results for r by substituting the integer value of the integer polynomials f, q and g into f(X) = q(X)g(Z) + r(X): 1008 = 882x21 + r r = 1008 – 882 = 126.
This shows r = 126 = 3x42, verifying the fact that r contains the integer value of X–a.
Corollary II: The remainder when a polynomial f(X) is divided by (X–a) is f(a).
Division Algorithm Corollary III: Given a polynomial, f(X), of degree greater than one, if q and r are unique, then f(X) is divisible by (X – a) IF, AND ONLY IF, f(a) = 0.
Using the example presented for the division algorithm above:
f(X) = X4 + AX3 + A2X2 + A3 X + A4, A =1 and a = 2, so that f(X) = X4 + X3 + X2 + X + 1.
Evaluating f(a) for this example, f(a) = 24 + 23 + 22 + 2 + 1=31, which is clearly non-zero, confirming what we found by polynomial long division and integer substitution above.
Using the example for corollary I, presented above, f(X) = X4+4X3-X2 -16X-12 and a = 2. Therefore, f(a) = 24+4x23-22 -16x2-12 = 16+32-4-32-12 = 0, confirming that f(X) is divisible by X–2, which we found above by factoring f(X) and substituting integer values into the algebraic polynomial forms of the division algorithm equation.
_______
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, for p a prime number and for X1, Y1 and Z1 relatively prime.
Proofs that these two conditions, namely
·       It is necessary and sufficient to prove FLT for the exponent n equal to primes greater than 3 (n = p 3), and
·       It is necessary and sufficient to prove that  there is no solution (X1, Y1, Z1) such that X1, Y1, and Z1 are relatively prime integers
Proofs that these conditions are necessary and sufficient are given in FLT65. For efficiency, they are not included here, but for easy reference, completeness and clarity, they are presented in Appendix C.

Appendices

APPENDIX A:
The original published proof of FLT65: Close’s Fermat’s Last Theorem
Including proof of the Division Algorithm and its corollaries
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:
(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.
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) = 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 1 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 B 
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 C:
PROOF OF SUFFICIENCY FOR X, Y AND Z RELATIVELY PRIME
AND N = PRIMES > 2, IN ANY PROOF OF FLT
Proofs of the sufficiency of considering X,Y and Z relatively prime, and N restricted to primes greater than 2 were included in FLT65 as published in the Book of Atma in 1977. The author developed the proof of FLT for N = 4 while finishing his degree in mathematics in 1962, but never published it because it had already been published by others. In retrospect, it may have been worthy of publication because of its utter simplicity. It is a simple, straight-forward extension of the Derivation of the Ratio Formula as published in Appendix A of the Book of Atma.
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). This proof was included in FLT65. N= 4 is a special case since 4 is neither prime nor a multiple of primes > 2.

PROOF OF FLT FOR N = 4 BY INFINITE DESCENT
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, the answer is no, because we can eliminate all even powers of 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: Derivation of the Pythagorean triples formula from properties of rational numbers was published by the author in 1977.

Using the formulae derived for the Pythagorean triples, 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, and
Z2 = P2 + Q2

Now with this application of the Pythagorean triple formulae, we have obtained another Pythagorean triple: Y2 + Q2 = P2. By comparison with the equation (X2)2 + (Y2)2 = (Z2)2 and the formulae above, we see that P < Z2, Q< X2, and Y < Y2.

Thus, by assuming that a triple integer non-zero solution exists for the equation (X2)2 + (Y2)2 = (Z2)2 (2PQ)2 + (P2 - Q2)2 = (P2 + Q2)2 we can produce another integer triple solution with smaller integers: Y2 + Q2 = P2.

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 two 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 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 other non-prime odd integers are factorable into prime numbers, it is sufficient to consider N = p, a prime in any proof of FLT. QED.