The following is a detailed validation of my FLT proof of 1965 that came out of a discussion with Prof. Moshe Roitman of Haifa University in Israel.

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**^{p}, where** 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**^{p}
and **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**^{p}, as a
factor of **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**^{p}, where **A** is a specific integer factor of **Y**, if it does not contain the algebraic
factor **z-s**?

Clearly, existence of an integer **A**^{p}, as a factor of **Y**^{p}, implies that, if there
are integer values of **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**^{p}, and (2.) Integer
solutions are a subset of real number solutions, implying by Corollary III that
no integer values of **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**^{p} = (z-s)^{p},
and **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)**^{p }=A^{p},
with **Y =AB**, and the division
Algorithm and corollaries apply to them. But **f(z)** is also equal to **z**^{p-1}+z^{p-2}X+z^{p-3}X^{2}+…+X^{p-1},
and by Corollary III of the Division Algorithm, the remainder, **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**^{p-1}+z^{p-2}X+z^{p-3}X^{2}+…+X^{p-1},
let’s assume there are integer solutions for the Fermat equation so that **A** is an integer factor of **Y**. Let **A= z-s**_{1}, then **(z-s**_{1})**ϵ ****f(s**_{1}), where **s**_{1} is the first in a series
of integers, **s**_{i}, such
that **z-s**_{1}= s_{1}-s_{2
}= s_{2}-s_{3 }= s_{3}-s_{4 }=**…**_{ }**s**_{i}-s_{i+1 }= A. And by
definition of the integers as a ring, closure with respect to addition and
subtraction insures that for any integer value of **s**_{i}, there is a unique **s**_{i+1} such that **s**_{i}-s_{i+1}=
A, an integer factor of **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**_{1}
we have: **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**_{1}-s_{2} for **z-s**_{1 }here_{,}
because **z-s**_{1}= s_{1}-s_{2
}= s_{2}-s_{3}= s_{3}-s_{4 }= **…**** s**_{i}-s_{i+1}.

Dividing
**f(s**_{1}) by **(s**_{1}-s_{2}), we have:

**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**_{i}) by **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**_{i} are integers,
dividing **f(s**_{i}) by **A** represented by increasingly smaller
values of **s**_{i-1} and** s**_{i}, 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 **p**^{th} division, if the **s**_{i} are integers, as shown in the example below, exactly
when this smallest **f(s**_{i})
will occur depends upon the size of the integer **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**_{i}, we have:

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**_{i}), repeatedly by **(s**_{i}-s_{i+1})_{,}
we see that the successive remainders are all of the form: **f(s**_{i})= s_{i}^{p-1}+s_{i}^{p-2}X+s_{i}^{p-3}X^{2}+…+X^{p-1}.
As long as **s**_{i} is
positive, **f(s**_{i}) is
obviously non-zero. But, if the** s**_{i}
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
**f(s**_{i}) is of the (**p-1)**^{th} degree in **s**_{i} and X, for all prime
exponents, **p**,** f(s**_{i}) will always be positive, even as** s**_{i} becomes increasingly
negative as in the example above.

Returning to **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})_{,} 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, **f(z)/(z-s**_{1})=(z-s_{1})^{p-1}=A^{p-1},
is an integer, and all the of terms on the right-hand side: the **q**_{i}, **f(s**_{1}) and **s**_{1}-s_{1+1},
are integers, and therefore, **f(z)/(z-s**_{1})=(z-s_{1})^{p-1}
an integer, equals** **a sum of integers
plus an integer fraction that is greater than zero. If **z** and **s**_{i} 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**^{th} division, to a sum of
integers and an irreducible rational fraction. This contradiction is not
obtained if **z**,** s**_{i} and **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**_{2},
and **s**_{3} are integers,
i.e., FLT is falsified, for any positive integer value of **z**, **f(z) = z**^{2}+zX+X^{2},
a positive integer, and the divisor, **z-s**
and thus the quotients **q**_{1}(z),
**q**_{2}(s_{1}) and **q**_{3}(s_{2}), obtained
as we divide successively by **(z-s**_{1}),** (s**_{1}-s_{2}) and **(s**_{2}-s_{3}), are
positive integers:

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

**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**_{1}), is
equal to an integer if **z **and** s**_{1} are integers and f**(z) = (z-s)**^{3} (necessary
conditions for integer solutions for **p=**3).
Since **f(z)=A**^{3}, **z**, **s**_{1,
}X_{ }and **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**_{i}
are positive. This looks like the signature of an irrational number, just what
you would expect if **z** is irrational.
Since **f(s**_{i}) can never
equal zero, even when **s**_{i}
becomes negative, there are no integers, **z
**and **s**_{i}, such that the
integer polynomial **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**_{i}), being of degree **p-1**, will always have **p**
terms, and when the **s**_{i}
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 **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**_{1}(z) is the **(p-2)**^{th}
quotient, while for **p=5**, **q**_{3}(z) is the **(p-2)**^{th} quotient, and the
remainder is a function of **z** and is
divided by **z-s**_{1} until the
**(p-1)**^{th} quotient. Thereafter,
divisors are **(s**_{i}-s_{i+1}),
with** 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**_{i})
occurs when the first **s**_{i}
becomes negative. If additional divisions are carried out, the negative **s**_{i} does not make **f(s**_{i}) negative, but does
reduce the value of the sum of integers on the right-hand side of the equation.
Since **f(z)** is of this form for all **p**, and all possible integer polynomial
remainders, **f(s**_{i}), are
shown to contain no general equation integer factors equal to **A = z-s**_{1}= s_{1}-s_{2 }=_{
}s_{2}-s_{3}=**…****(s**_{i}-s_{i+1}). Because of the
contradiction in the general equation, **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})_{,} the only
solutions for which **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.