I think we agree that for z

^{p}– x^{p}= y^{p}, (the Fermat equation) with x =X and y=Y (integers), we may assume X and Y relatively prime and choose Y not containing p (as demonstrated in FLT65). There are many solutions with X & Y relatively prime integers. The question is: can any of them yield z = Z, an integer?
Factoring the Fermat equation, we have:

**g(z)f(z) = Y**where

^{p},**g(z)=z–X**and

**f(z)=z**

^{p-1}+z^{p-2}X+z^{p-3}X^{2}+…+X^{p-1}.
For integer solutions, we may assume
that the integer values of

**g(z)**and**f(z)**are relatively prime (also demonstrated in FLT65). Thus, since**Y**is an integer, we may set**f(z) = A**and^{p}**g(z) = B**,^{p}**A**and**B**relatively prime integer factors of**Y**, and for any real**z**,*integer or not*, there is a real number,**s**_{,}such that**z-s = A**. Thus**f(z) = z**^{p-1}+z^{p-2}X+z^{p-3}X^{2}+…+X^{p-1}= (z-s)^{p}= A^{p}.**Here we begin to see the**

*connection*between A, an integer factor of Y, and z-s_{, }a first-degree polynomial factor_{ }of the polynomial f(z) of degree p-1. For integer solutions of the Fermat equation, if there are any, z-s and f(z) must become integer polynomials equal to A and A^{p}respectively.
For hypothetical integer solutions, falsifying
FLT,

**z**,**s**and**A**will have to be integers. For any of the infinite number of non-integer solutions,**z**will be irrational, making it impossible for**A**to be an integer, even though**A**, as a factor of^{p}**Y**,*is*an integer.
Example: Let X = 2, Y = 3, and p = 3:
solving the Fermat equation, z

^{3}= 2^{3}+ 3^{3 }= 8 + 27 → z = (35)^{1/3}, and f(z) = (35)^{2/3}+2(35)^{1/3}+4 = A^{3}→ A = [(35)^{2/3}+2(35)^{1/3}+4]^{1/3}, clearly not an integer.
Returning to my 1965 proof (FLT65),
which depends on application of Corollary III of the Division Algorithm to the
result of dividing

**f(z)**by**z-s**: The Division Algorithm tells us that for*all*real number solutions,**f(z)/(z-s)**_{=}_{ }**q(z) + R(z)/(z-s)**, and Corollary I tells us that if**f(z)**contains**z-s**, a non-zero**R(z)**must also contain**z-s**. Corollary II tells us that the value of**R(z)**, when**f(z)**is divided by**z-s**, is**f(s)**and Corollary III tells us that the**p-1**degree polynomial**f(z)**is divisible by the one degree polynomial**z-s**, if and only if,**f(s)=0**. In FLT65, the fact that**f(s)****≠****0**for any positive integer value of**s**is seen as a conclusive contradiction of the assumption of integer solutions, and therefore proof of FLT. Some reviewers have questioned whether Corollary III applies to integer polynomials. I believe this is adequately answered in FLT65 with the observation that integer solutions are a subset of the real number solutions to which the Division Algorithm and corollaries__do__apply.
A
secondary question, raised by Dr. Stanly Ogilvy, a professional mathematician who
reviewed FLT65, and echoed in some form by at least two other reviewers, is the
following: Even though a non-zero remainder results when

**f(z)**is divided by**A= z-s**, indicating that**z-s**cannot be an algebraic factor of**f(z)**, if there is an integer solution, that remainder will still contain integer values of**z-s**as integer factors, hidden from detection by algebraic division. He used modulus algebra to demonstrate the point. The question then becomes: How can we determine whether the integer value of the polynomial**f(z)**can contain**A**, where^{p}**A**is a specific integer factor of**Y**, if it does not contain the algebraic factor**z-s**?
Clearly, existence of an integer

**A**, as a factor of^{p}**Y**, implies that, if there are integer values of^{p}**z**and**s**, the integer value**z-s**has to be a factor of the integer value of**f(z)**. With x and y as integers (X and Y) we know that**z**can be irrational, as demonstrated above, but can it be an integer? This seems at first a very difficult question to answer because we can produce examples of integer polynomials of degree >1 that, when divided by a 1st degree polynomial, have remainders that contain an integer factor equal to the value of the dividing 1^{st}degree polynomial, that with repeated divisions produce a zero remainder, and other examples that don’t. But none of these example polynomials are of the exact same algebraic form as**f(z)**, a factor of the Fermat equation.
As an aside, in my opinion this question
is also covered in the FLT65 proof by two statements: (1.) For integer
solutions,

**f(z)**must be equal to a perfect**p**-power integer,**A**, and (2.) Integer solutions are a subset of real number solutions, implying by Corollary III that no integer values of^{p}**X**and s (X_{1}and a in FLT65) can produce an**f(s)**zero remainder. For integer solutions, the integer polynomial**z-s**must be a factor of the integer polynomial**f(z)**, and integer value of the integer polynomial**f(s)**for some**z**and**s**must contain**A**, which by Corollary III, is possible,__if and only if__the remainder is zero, and for positive integer values of**s**, the remainder,**f(s)**, is never zero. However, since not everyone agrees with me, let’s proceed to analyze the non-zero remainder produced in the process of dividing**f(z)**by**z-s**, by dividing it (the non-zero remainder) by**z-s**at least an additional**p-1**times.
I start by noting that for

*all*solutions of the Fermat equation,**f(z)= A**, and^{p}= (z-s)^{p}**f(z)**will be divisible by**z-s**__exactly__**p**times. This is not a problem if**z-s**is irrational. For all solutions to the Fermat equation, including hypothesized integer solutions,**f(z)**and**z-s**are polynomials in**z**whose terms involve values of**s**and**X**. If**z**can be an integer, there is a subset of polynomials for which the sums of those terms are integer factors of**Y: f(z) = (z-s)**, with^{p }=A^{p}**Y =AB**, and the division Algorithm and corollaries apply to them. But**f(z)**is also equal to**z**, and by Corollary III of the Division Algorithm, the remainder,^{p-1}+z^{p-2}X+z^{p-3}X^{2}+…+X^{p-1}**f(s)**, obtained when the polynomial**f(z)**is divided by**z-s**must be zero if**z-s**is a factor of**f(z)**. Can the remainder be zero for a subset of the polynomials with integer variables and constants, and not for all others with non-integer terms? If so, the integer value of the non-zero remainder obtained by dividing**f(z)**by**z-s**must contain the integer value of**z-s****≡****A**.
So, given that

**f(z)=****z**, let’s assume there are integer solutions for the Fermat equation so that^{p-1}+z^{p-2}X+z^{p-3}X^{2}+…+X^{p-1}**A**is an integer factor of**Y**. Let**A= z-s**, then_{1}**(z-s**_{1})**ϵ****f(s**, where_{1})**s**is the first in a series of integers,_{1}**s**, such that_{i}**z-s**_{1}= s_{1}-s_{2 }= s_{2}-s_{3 }= s_{3}-s_{4 }=**…**_{ }**s**. And by definition of the integers as a ring, closure with respect to addition and subtraction insures that for any integer value of_{i}-s_{i+1 }= A**s**, there is a unique_{i}**s**such that_{i+1}**s**, an integer factor of_{i}-s_{i+1}= A**Y**. Since**A**, as an integer factor of**Y**, is a positive integer:**z > s**_{1}> s_{2}> s_{3}> s_{4 }>**…****>s**._{i}**Note that the infinite series of integer polynomials z-s**

_{1,}s_{1}-s_{2}

_{,}**s**

_{2}-s_{3}

_{,}**s**

_{3}-s_{4},**…**

**is exhaustive, containing**

*all possible**integer*representations of*A for any hypothetical integer solution of the Fermat equation.*
Dividing

**f(z)**by**z-s**we have:_{1}**f(z)/(z-s**._{1})= q_{1}(z)+f(s_{1})/(z-s_{1})= q_{1}(z)+ f(s_{1})/(s_{1}-s_{2})**We may substitute****s**for_{1}-s_{2}**z-s**here_{1 }**because**_{,}**z-s**_{1}= s_{1}-s_{2 }= s_{2}-s_{3}= s_{3}-s_{4 }=**…****s**._{i}-s_{i+1}
Dividing

**f(s**by_{1})**(s**, we have:_{1}-s_{2})**f(z)/(z-s**

_{1})= q_{1}(z)+f(s_{1})/(s_{1}-s_{2})= q_{1}(z)+q_{2}(z)+f(s_{2})/(s_{1}-s_{2})
We
may continue in this manner:

**f(z)/(z-s**

_{1})= q_{1}(z)+q_{2}(z)+**…**

**+q**

_{p-2}(s_{1})+**…**

**+q**

_{i}(s_{i})+ f(s_{i})/(s_{i}-s_{i+1})_{,}*ad infinitum*, dividing the successive remainders,

**f(s**by

_{i})**A= z-s**…

_{1}=**s**

_{i}-s_{i+1}_{,}purging

**f(s)**of all “hidden” factors equal to the integer

**A**.

If

**z**and the**s**are integers, dividing_{i}**f(s**by_{i})**A**represented by increasingly smaller values of**s**and_{i-1}**s**, must eventually, in a finite number of steps, produce the smallest possible integer polynomial remainder. Even though, assuming there are integer solutions, we would expect this smallest remainder to be zero, occurring with the_{i}**p**division, if the^{th}**s**are integers, as shown in the example below, exactly when this smallest_{i}**f(s**will occur depends upon the size of the integer_{i})**A***relative*to the value of the hypothetical integer**z**.
If
X=7, z=13 and A=2, then z-s

_{1 }=A → 13-s_{1}= 2 → s_{1}=11, and z-s_{1}= s_{1}-s_{2 = }s_{2}-s_{3}=…=A→ s_{2}=9, s_{3}=7, s_{4}=5, s_{5}=3, s_{6}=1, s_{7}=-3, s_{8}=-5, s_{9}=-7, …*ad infinitum*.
For
p=3, f(s

_{i}) = (s_{i})^{2}+ (s_{i})X+ X^{2}. Evaluating the remainder f(s_{i}) for successive integer values of**s**, we have:_{i}
1.)
f(s

_{1}) = (11)^{2}+ (11)(7)+ 7^{2 }= 121+77+49=247
2.)
f(s

_{2}) = (9)^{2}+ (9)(7)+ (7)^{2}= 81+63+49=193
3.)
f(s

_{3}) = (7)^{2}+ (7)(7)+ (7)^{2}= 49+40+49=138
4.)
f(s

_{4}) = (5)^{2}+ (5)(7)+ (7)^{2}= 25+35+49=129
5.)
f(s

_{5}) = (3)^{2}+ (3)(7)+ (7)^{2}= 9+21+49=79
6.)
f(s

_{6}) = (1)^{2}+ (1)(7)+ (7)^{2}= 1+7+49=53
7.)

**f(s**_{7}) = (-3)^{2}+ (-3)(7)+ (7)^{2}= 9+(-21)+49=37, the minimum value of f(s_{i}).
8.)
f(s

_{8}) = (-5)^{2}+ (-5)(7)+ (7)^{2}= 25+(-35)+49=39
9.)
f(s

_{9}) = (-7)^{2}+ (-7)(7)+ (7)^{2}= 49+(-40)+49=58
Since we have no integer values from actual
integer solutions to work with in this example, integer values for X,

**z**and**A**are arbitrarily chosen, with the exception that**z**must be larger than**A**(**A**is a factor of Y and Y<**z**), and**A**cannot contain**p**. This is established in the setup of FLT65 where, since X,Y and Z are relatively prime, only one of them, at most, can contain**p**as a factor, and the Fermat equation is symmetric with respect to X and Y, so we could choose either X or Y as not containing**p**. I chose**Y**, and, since**A****ϵ****Y**,**A**cannot contain**p**.
As we divide the remainder,

**f(s**, repeatedly by_{i})**(s**we see that the successive remainders are all of the form:_{i}-s_{i+1})_{,}**f(s**. As long as_{i})= s_{i}^{p-1}+s_{i}^{p-2}X+s_{i}^{p-3}X^{2}+…+X^{p-1}**s**is positive,_{i}**f(s**is obviously non-zero. But, if the_{i})**s**are integers, as they must be if there are integer solutions, if we repeat the division enough times, they will eventually become negative. Because the remainder_{i}**f(s**is of the (_{i})**p-1)**degree in^{th}**s**and X, for all prime exponents,_{i}**p**,**f(s**will always be positive, even as_{i})**s**becomes increasingly negative as in the example above._{i}
Returning to

**f(z)/(z-s**_{1})= q_{1}(z)+q_{2}(z)+**…****+q**_{p-2}(s_{1})+**…****+q**the general equation derived above, inspection of this equation at once reveals a very basic numerical contradiction: Under the assumption that there are integer solutions for the Fermat equation, the left-hand side of the equation,_{i}(s_{i})+ f(s_{i})/(s_{i}-s_{i+1})_{,}**f(z)/(z-s**, is an integer, and all the of terms on the right-hand side: the_{1})=(z-s_{1})^{p-1}=A^{p-1}**q**,_{i}**f(s**and_{1})**s**, are integers, and therefore,_{1}-s_{1+1}**f(z)/(z-s**an integer, equals_{1})=(z-s_{1})^{p-1}**a sum of integers plus an integer fraction that is greater than zero. If****z**and**s**are integers, we have an integer on the left-hand side of the equation, equal at any given point in the succession of divisions, call it the_{i}**i**division, to a sum of integers and an irreducible rational fraction. This contradiction is not obtained if^{th}**z**,**s**and_{i}**A**are not integers.
It may be helpful to look at this
process with actual prime number values of

**p**:
For

**p=3**, if**z**,**s**,_{1}**s**, and_{2}**s**are integers, i.e., FLT is falsified, for any positive integer value of_{3}**z**,**f(z) = z**, a positive integer, and the divisor,^{2}+zX+X^{2}**z-s**and thus the quotients**q**,_{1}(z)**q**and_{2}(s_{1})**q**, obtained as we divide successively by_{3}(s_{2})**(z-s**,_{1})**(s**and_{1}-s_{2})**(s**, are positive integers:_{2}-s_{3})**f(z)/(z-s**

_{1})= q_{1}(z)+f(s_{1})/(s_{1}-s_{2})= q_{1}(z)+q_{2}(s_{1})+f(s_{2})/(s_{2}-s_{3})=q_{1}(z)+q_{2}(s_{1})+q_{3}(s_{2})+f(s_{3})/(s_{3}-s_{4})
and the remainder

**f(s**_{1})=**s**by_{1}^{2}+s_{1}X+X^{2}**(s**, we have_{1}-s_{2})**s**_{1}+s_{2}+X+f(s_{2})/(s_{1}-s_{2})**→****q**, and dividing the remainder_{2}(s_{1})= s_{1}+s_{2}+X**f(s**_{2})= s_{2}^{2}+s_{2}X+X^{2}^{ }by**(s**, we have_{2}-s_{3})**s**_{2}+s_{3}+X+f(s_{3})/(s_{2}-s_{2})**→****q**. Substituting these values of the_{3}(s_{2})= s_{2}+s_{3}+X**q**into the equation for_{i}(s_{i})**f(z)/(z-s**, we have:_{1})**f(z)/(z-s**.

_{1})= z+s_{1}+X+ s_{1}+s_{2}+X+ s_{2}+s_{3}+X+ f(s_{3})/(s_{3}-s_{4})= z+2s_{1}+2s_{2}+3X+ f(s_{3})/(s_{3}-s_{4})
Clearly, the left-hand side of the
equation,

**f(z)/(z-s**, is equal to an integer if_{1})**z**and**s**are integers and f_{1}**(z) = (z-s)**(necessary conditions for integer solutions for^{3}**p=**3). Since**f(z)=A**,^{3}**z**,**s**and_{1, }X_{ }**A**are positive integers, the right-hand side of the equation is an integer plus an irreducible rational fraction that becomes increasingly smaller with each division as long as the**s**are positive. This looks like the signature of an irrational number, just what you would expect if_{i}**z**is irrational. Since**f(s**can never equal zero, even when_{i})**s**becomes negative, there are no integers,_{i}**z**and**s**, such that the integer polynomial_{i}**z-s**divides the integer polynomial**f(z)**exactly**p**times.
Demonstrations
of this type produce the same conclusion for

**p**equal to any prime number, since**f(s**, being of degree_{i})**p-1**, will always have**p**terms, and when the**s**become negative, the negative terms, will always be out-weighed by larger positive terms. It may be a helpful exercise to construct this demonstration of few more with_{i}**p=**5, 7, … I have done it for**p=**5, but have not included it here to save space. The only difference between the case for**p=3**and**p=5**, is that for**p=3**,**q**is the_{1}(z)**(p-2)**quotient, while for^{th}**p=5**,**q**is the_{3}(z)**(p-2)**quotient, and the remainder is a function of^{th}**z**and is divided by**z-s**until the_{1}**(p-1)**quotient. Thereafter, divisors are^{th}**(s**, with_{i}-s_{i+1})**i= 1, 2, 3,****…***ad infinitum*. For integer solutions, however, the**q**_{i}_{ }are always integers, whether containing**z**or not, assuring the same contradiction of equating an integer with another integer plus an irreducible rational fraction for all**p**.
As illustrated
above, the smallest integer value of

**f(s**occurs when the first_{i})**s**becomes negative. If additional divisions are carried out, the negative_{i}**s**does not make_{i}**f(s**negative, but does reduce the value of the sum of integers on the right-hand side of the equation. Since_{i})**f(z)**is of this form for all**p**, and all possible integer polynomial remainders,**f(s**, are shown to contain no general equation integer factors equal to_{i})**A = z-s**_{1}= s_{1}-s_{2 }=_{ }s_{2}-s_{3}=**…****(s**. Because of the contradiction in the general equation,_{i}-s_{i+1})**f(z)/(z-s**_{1})=q_{1}(z)+q_{2}(z)+**…****+q**_{p-2}(s_{1})+**…****+q**the only solutions for which_{i}(s_{i})+f(s_{i})/(s_{i}-s_{i+1})_{,}**f(s)**= 0, require**s**and**z**to be non-integer. This means that there are no integer values of**z**and**s**that satisfy the Fermat equation.**The fact that the contradiction falsifying integer solutions of the Fermat equation is demonstrable with only one division of f(z) by z-s**._{1,}validates FLT65
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