**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

**, when***x*+ y^{n}*≠*^{n}*z*^{n}*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*(The cases**x**when^{n}+ y^{n}= z^{n}**x, y**and**z**are integers with n >2.*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

**,**, and*q*
·
remainder,

**,***r*
in the process of division, where

**are algebraic functions of some variable, represented here by the symbol***f(X), g(X), q(X), and r(X)***.***X***The division algorithm is stated as follows:**For any two polynomials

*,*and

**g(X) ≠ 0****over the field of real numbers, of degree**

*f(X)*

*m**and*

*, respectively, and*

**n****, there exist**

*n > m*__>__0**polynomials such that**

*unique***where**

*f(X) = q(X)g(X) + r(X)***is either zero or of degree smaller than**

*r(X)***.**

*m*
The Division Algorithm is true
for any real value of

**, rational, irrational or***X**integer*.
(Note that in the
Division Algorithm,

**can be any positive integer, 1, 2, 3,…***n**, while, in the FLT65 proof, n is restricted to integer values > 2*.)**Example**:

*Let f(X) =***, and**

*X*^{4 }+ AX^{3}+ A^{2}X^{2}+ A^{3 }X + A^{4}**. Then, for this polynomial (a polynomial of the same form as**

*g(X) =X - a***in FLT65)the**

*f(Z)**mathematical*

*operation of division*, represented by

**, becomes:**

*f/g = q + r***where, by the division algorithm,**

*(X*^{4 }+ AX^{3}+ A^{2}X^{2}+ A^{3 }X + A^{4})/(X-a) = q(X) + r(X)/(X-a),

*r(X**)*is either zero or of degree less than

**4**. Carrying out the polynomial long division to determine the algebraic polynomial form of

**and**

*q***, we get:**

*r***and**

*q(X)= X*^{3 }+ (A+a)X^{2}+(A^{2 }+aA+a^{2})X+(A^{3 }+aA^{2}+ a^{2}A^{2}+ a^{3})**And if**

*r(X)= A*^{4 }+aA^{3}+ a^{2}A^{2}+ a^{3}A +a^{4}.*and*

**X***are positive,*

**a****cannot be zero and is not of degree smaller than**

*r(X)***, so by**

*m***the polynomial**

*the division algorithm stated above,***is not divisible by the polynomial**

*(X*^{4 }+ AX^{3}+ A^{2}X^{2}+ A^{3 }X + A^{4})**But what about the integer values of these algebraic polynomials?**

*(X-a).*
We can assign

**and***A, X***integer values to illustrate the transition from algebraic polynomials to integers: Let***a***and***A = 1, a = 2,***, and substitute these integer values into***X = 5**f(X) =***This gives us:***q(X)(X-a) + r(X).*

*(5*^{4 }+ 5^{3}+ 5^{2}+ 5 + 1)= (5^{3}+ 3x5^{2}+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*. This example demonstrates that

**r(X) is 31**, which is not divisible by the integer value of X-a = 3*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

*and*

**f(X)**

*g(X)**contain a common factor*

*,*contains it also. It is important to note tha

**r(X)***t this corollary is true for*algebraic polynomials over the field of real numbers,

*integer polynomials and the integer values of those integer polynomials.*

If

**and***f(X)***are integer polynomials, i.e., all terms and coefficients are integers and***g(X)***is an integer variable, and they contain a common factor,***X***, then the remainder integer variable polynomial***X-a***contains***r(X)***, and the integer value of the remainder also contains the integer value of***X-a***.***X-a***Example**: the polynomials

**and**

*f(X) = X*^{4}+4X^{3}-X^{2 }-16X-12**contain the common factor**

*g(X) = X*^{2}-4**. Corollary II says that because**

*X–2*

*f(X) = X*^{4}+4X^{3}-X^{2 }-16X-12 = q(X)(X^{2}-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

**is not equal to zero.**

*X-2*
For polynomials with

**variable,***X**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,

**must be greater than***X***2.**So, for this demonstration, let**, then***X =*5**=***X – a***5 – 2 = 3,**and factoring**and***f***, we have***g***.**

*(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) =***and***X*1008,^{4}+4X^{3}-X^{2 }-16X-12q(X) =*q(X)*=*X*3 = 882,^{2}+4X+*r*=**6X**.^{2}- 24 = 150 – 24 = 126 = 3x42
We can check our results for

**by substituting the integer value of the integer polynomials***r***and***f, q***into***g**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

**is divided by**

*f(X)***(X–a)**is

*f(a)**.*

**Division Algorithm Corollary III:**Given a polynomial,

**, of degree greater than one, if**

*f(X)*

*q**and*

*are unique, then*

**r****is divisible by**

*f(X)***(**.

*X – a*)*IF, AND ONLY IF*,*f(a)*= 0
Using
the example presented for the division algorithm above:

*f(X) =***,**

*X*^{4 }+ AX^{3}+ A^{2}X^{2}+ A^{3 }X + A^{4}

*A =1**and*, so that

**a = 2**

*f(X) =***.**

*X*^{4 }+ X^{3}+ X^{2}+ X + 1
Evaluating

**f(a)***for this example, f(a) =**2*^{4 }+ 2^{3}+ 2^{2}+ 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,

**and***f(X) = X*^{4}+4X^{3}-X^{2 }-16X-12**. Therefore,***a = 2**f(a) = 2*= 16+32-4-32-12 = 0, confirming that^{4}+4x2^{3}-2^{2 }-16x2-12*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 (**X**) such that_{1}, Y_{1}, Z_{1}**X**and_{1}, Y_{1},**Z**are integers. It is_{1}*necessary and sufficien*t to show this for**n = p**, for**p**a prime number and for**X**,_{1}**Y**and_{1}**Z**relatively prime._{1}
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
·

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 **It is necessary and sufficient to prove that there is no solution (X**_{1}, Y_{1}, Z_{1}) such that X_{1}, Y_{1}, and Z_{1}are relatively prime integers**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:

**X**^{N}+ Y^{N}≠ Z^{N }
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) = a

_{n}X^{n}+ a_{n-1}X^{n-1}+•••+ a_{1}X+ a_{0}
And g(X) = b

_{m}X^{m}+ b_{m-1}X^{m-1}+•••+ b_{1}X+ b_{0, }b_{m }≠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 f

_{1}(X), of lower degree than f(X), by writing
(B) f

_{1}(X) = f(X) - a_{n/}a_{m}X^{n-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 f

_{1}(X) is such a polynomial, we may write:
(C)
f(X)
= q

_{1}(X)g_{1}(X) + r_{1}(X),
where r

_{1}(X) is either zero or of a degree less than m. So from (B.) and (C.):
f(X)
= | a

_{n}/a_{m}X^{n-m}+ q_{1}(X) | g_{1}(X) + r_{1}(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.) X

^{N}+ Y^{N }= Z^{N}, 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 Z

^{N}= (Mx)^{N}+ (My)^{N}= M^{N}(X^{N}+ Y^{N}). And thus Z = Mz, i.e. Z contains M. Also, X^{N}+ Y^{N }= Z^{N}→ (Mx)^{N}+ (My)^{N}= (Mz)^{N}. Factoring M^{N}out, we have x^{N}+ y^{N }= z^{N}
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

X

^{ab}+ Y^{ab }= Z^{ab}→ (X^{a})^{b}+ (Y^{a})^{b }= (Z^{a})^{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, X

^{N}+ Y^{N }= Z^{N}→ Z^{N}– X^{N}= Y^{N }. Factoring, we have:
(2.) (Z-X)(Z

^{N-1}+ Z^{N-2}X + Z^{N-3}X^{2}+•••+ X^{N-1}) = Y^{N.}
For any given X, say X
= X

_{1}let Z^{N-1}+ Z^{N-2}X_{1}+ Z^{N-3}X_{1}^{2}+•••+ X_{1}^{N-1}= f(Z)
And Z-X

_{1}= g(Z). Then
(3.) g(Z)f(Z)
= Y

^{N}.
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 X

^{N}+ Y^{N }= Z^{N}, 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-X_{1}, will be f(X_{1}). And
(4.) f(X

_{1}) = X_{1}^{N-1}+ X_{1}^{N-2}X_{1}+ X_{1}^{N-3}X_{1}^{2}+•••+ X_{1}^{N-1}= NX_{1}^{N-1}.
Since X, Y and Z are
relatively prime, no factor of X

_{1}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 X

^{N}+ Y^{N }= Z^{N},
(5.) (Z-Y)(Z

^{N-1}+ Z^{N-2}Y + Z^{N-3}Y^{2}+•••+ Y^{N-1}) = X^{N}.
And by exactly the same
reasoning as above, for any particular Y = Y

_{1}, if we let
Z

^{N-1}– Z^{N-2}Y_{1}+ Z^{N-3}Y_{1}^{2}-•••+ Y_{1}^{N-1}= f_{1}(Z) and
Z - Y

_{1}= g_{1}(Z), then either f_{1}(Z) and g_{1}(Z) contain N as a common factor, or they are perfect N-powers. So for a given case of Z^{N}= X_{1}^{N}+ Y_{1}^{N }, if either f(Z) or f_{1}(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 Îµ f(Z) → N
Îµ Y

_{1}and N Îµ f_{1}(Z) → N Îµ X_{1}. But X_{1}.and Y_{1}are relatively prime. If neither X_{1}.nor Y_{1}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) = Z

^{N-1}+ Z^{N-2}X_{1}+ Z^{N-3}X_{1}^{2}+•••+ X_{1}^{N-1}= A^{N}, and/or
f

_{1}(Z) = Z^{N-1}– Z^{N-2}Y_{1}+ Z^{N-3}Y_{1}^{2}-•••+ Y_{1}^{N-1}= B^{N}, A and B integers <Z. and at least one does not contain N.
Since they are both of
the same form, set f(Z) = Z

^{N-1}+ Z^{N-2}X_{1}+ Z^{N-3}X_{1}^{2}+•••+ X_{1}^{N-1}= A^{N}. 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 N

^{th}power, as in that case when A is the N^{th}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) Îµ f(Z) → f(a) = 0.

But f(a) = a

^{N-1}+ a^{N-2}X_{1}+ a^{N-3}X_{1}^{2}+•••+ X_{1}^{N-1}= 0 is an__impossibility__because if Fermat’s last theorem is falsified, X_{1}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**X**has no solutions in positive integers when N is an integer > 2. And so the proof of Fermat’s last theorem is complete.^{N}+ Y^{N}= Z^{N}**Edward R. Close December 18, 1965**

**APPENDIX B**

^{[18]}

**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

**polynomials q(x) and r(x) such that**

*unique*
(A) f(x) = q(x)g(x) + r(x)

Where
r(x) is either zero or of degree

*smaller than m*.
Let
f(x) = a

_{n}x^{n}+ a_{n-1}x^{n-1}+•••+ a_{1}x + a_{0}
And g(x) = b

_{m}x^{m}+ b_{m-1}x^{m-1}+•••+ b_{1}x + b_{0, }b_{m }≠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**in this case.***not unique*
Now we
can form f

_{1}(x), of lower degree than f(x), by writing
(B) f

_{1}(x) = f(x) - a_{n}/a_{m}x^{n-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 f

_{1}(x) is such a polynomial, we may write:
(C)
f(x)
= q

_{1}(x)g_{1}(x) + r_{1}(x),
where r

_{1}(x) is either zero or of a degree less than m. So from (B) and (C):
f(x) =
| a

_{n}/a_{m}x^{n-m}+ q_{1}(x) | g_{1}(x) + r_{1}(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)

**when m>n.***are unique*
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

**, and the Division Algorithm is proved over the field of real numbers.**

*q(x) and r(x) are 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).

Let the common factor be
represented by M. Then f(x) = M

**∙**f_{1}(x) and g(x) = M**∙**g_{1}(x), so that r(x) = M**∙**f_{1}(x) - q(x)**∙**M**∙**g_{1}(x) = M**∙[**f_{1}(x) - q(x)g_{1}(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

**, f(a) cannot contain x-a, and so it follows that***if q(x) and r(x) are unique*

*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 X

^{N}+ Y^{N}= Z^{N}, 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 = MX

_{1}and Y = MY_{1}, then Z^{N}= (MX_{1})^{N}+ (MY_{1})^{N}= M^{N}(X_{1}^{N}+ Y_{1}^{N}). And thus Z = MZ_{1}, i.e. Z contains M. Also, X^{N}+ Y^{N}= Z^{N}→ (MX_{1})^{N}+ (MY_{1})^{N}= (MZ_{1})^{N}. Factoring M^{N}out, we have X_{1}^{N}+ Y_{1}^{N }= Z_{1}^{N}, with X_{1}, Y_{1}and Z_{1}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 X

^{N}+ Y

^{N}= Z

^{N}is known as the Pythagorean Theorem equation, which has an infinite number of integer solutions known as the Pythagorean triples. For example: 3

^{2}+ 4

^{2}= 5

^{2}. 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*^{[1]}
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

X

^{ab}+ Y^{ab}= Z^{ab}→ (X^{a})^{b}+ (Y^{a})^{b}^{ }= (Z^{a})^{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:

X

^{2a}+ Y^{2a}= Z^{2a}→ (X^{a})^{2}+ (Y^{a})^{2 }= (Z^{a})^{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: (X

^{2})^{2}+ (Y^{2})^{2}= (Z^{2})^{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

^{[3]}.

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:

X

^{2}= 2PQ,

Y

Z

Now with this application of the Pythagorean triple formulae, we have obtained another Pythagorean triple:

^{2}= P^{2}- Q^{2}**→**Y^{2}+ Q^{2}= P^{2}, andZ

^{2}= P^{2}+ Q^{2}Now with this application of the Pythagorean triple formulae, we have obtained another Pythagorean triple:

**Y**^{2}+ Q^{2}= P^{2}. By comparison with the equation (X^{2})^{2}+ (Y^{2})^{2}= (Z^{2})^{2}and the formulae above, we see that P < Z^{2}, Q< X^{2}, and Y < Y^{2}.
Thus, by assuming that a triple integer
non-zero solution exists for the equation (X

^{2})^{2}+ (Y^{2})^{2}= (Z^{2})^{2}→ (2PQ)^{2}+ (P^{2}- Q^{2}**)**P^{2}= (^{2}+ Q^{2}**)**we can produce another integer triple solution with smaller integers: Y^{2}^{2}+ Q^{2}= P^{2}.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 (1

^{2})

^{2}+ (B

^{2})

^{2}= (C

^{2})

^{2}→ (C

^{2})

^{2}- (B

^{2})

^{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 (3

^{2})

^{2}- (2

^{2})

^{2}= 1 → 81 – 16 = 1, which is a clear contradiction, and any pairs of larger integers C

_{1}> C and D

_{1}> D, will lead to larger discrepancies. So we have to conclude that there can be no integer solution triples for the equation (X

^{2})

^{2}+ (Y

^{2})

^{2}= (Z

^{2})

^{2}.

This, of course, is

**proof of FLT for N = 4**, which was proved by Fermat^{3}. However, this has greater significance than just a proof for N = 4, because any case of X^{2a}+ Y^{2a}= Z^{2a}when a is an even number, is a case of N = 4: If a = 2m, m a positive integer, X^{2a}+ Y^{2a}= Z^{2a}→ X^{4m}+ Y^{4m}= Z^{4m }→ (X^{m})^{4}+ (Y^{m})^{4}= (Z^{m})^{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**.
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