APPENDIX D
FERMAT’S LAST THEOREM (FLT)
(A BRIEF PRESENTATION OF THE CLOSE 1965 PROOF
WITH WELLDEFINED STANDARD NOTATION [FLT65C]
EXPANDED AND REWRITTEN IN 2013)
Notation:
In this discussion, three types of numbers and two types of factors are defined
precisely for clarity:
·
Lowercase letters, like x, y and z represent variables with
no numerical restrictions.
·
Uppercase letters like X represent variables restricted to
integers.
·
Uppercase letters with subscripts like X_{1}
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 an
integer factor of B.
Consider the equation z^{p} – x^{p} = y^{p}, equivalent to Equation (1) in the 1965 proof. Since it is sufficient (Appendix C)
to consider p is a prime number > 2, and thus an odd prime, we can factor
the left side of the equation to obtain:
(zx)( z^{p1} + z^{p2}x + z^{p3}x^{2}
+•••+ x^{p1}) = y, equivalent to Equation
(2) of FLT65C.[1]
For variable integer values of x, represented by X, let z^{p1}
+ z^{p2}X + z^{p3}X^{2} +•••+ zX^{p2} + X^{p1}
= f(z), and zX = g(z). Then g(z)f(z) = Y^{p}, 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 X_{1} and Y_{1}, 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(X_{1}). And so:
r(z) = f(X_{1}) = X_{1}^{p1} + X_{1}^{p2}X_{1}
+ X_{1}^{p3}X_{1}^{2} +•••+ X_{1}^{p1}
= pX_{1}^{p1}, a unique constant made up of integer factors for any p > 2. … Equation (4).
Since for any solution
of Equation (1), X_{1}, Y_{1}
and z, if an integer, may be considered to be relatively prime, no factor of X_{1},
and, therefore, of X_{1}^{p1}, may be contained in g(z) = z –X_{1}.
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 ppower, (Y_{1})^{p},
either p ∈ f(z), p ∈
g(z), or they must both be perfect ppowers of integers.
By exactly the same
reasoning as above, we can write the FLT equation as z^{p} – y^{p}
= x^{p} and factor as:
(z  Y)(z^{p1}
+ z^{p2}Y + z^{p3}Y^{2} +•••+ Y^{p1}) = X^{p}.
… Equation
(5)
Let z^{p1} – z^{p2}Y_{1}
+ z^{p3}Y_{1} ^{2} •••+ Y_{1} ^{p1}
= f_{1}(z) and z  Y_{1} = g_{1}(z).
Then, dividing f_{1}(z)
by g_{1}(z), analogous to Equation (4), we obtain r_{1}(z) = pX_{1}^{p1}…
Equation (6)
Then,
either p ∈ f_{1}(z) and p ∈
g_{1}(z), or they are perfect ppowers of integers.
For a given case of z^{p} = X_{1}^{p}
+ Y_{1}^{p}, it is possible that either p ∈
f(z) or p ∈ f_{1}(z). But, if p is a factor of
one, the other has to be a perfect ppower, since X_{1} and Y_{1},
and therefore, f(z) and f_{1}(z) are relatively prime. For a given z =
Z_{1}, either p ∈ f(Z_{1}) →
p ∈ Y_{1}, or p ∈
f_{1}(Z_{1}) → p ∈
X_{1}. If neither X_{1}.nor Y_{1} contains p, because
they are relatively prime, both must be perfect ppowers. Therefore, in any
event, one of them at least, must be a perfect ppower, not containing p as a
factor.
Since f(z) and f_{1}(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) = z^{p1} + z^{p2}X_{1} + z^{p3}X_{1}^{2}
+•••+ X_{1}^{p1} = A^{p}, 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 p^{th} power, as in that case when A
is not an integer, but is the p^{th} 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
noninteger 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) = (zs)q(z) + f(s) over the field of real
numbers, and by COROLLARY III, if
q(z) and f(s) are unique, 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) = s^{p1} + s^{p2}X_{1} + s^{p3}X_{1}^{2}
+•••+ X_{1}^{p1} = 0…. Equation
(6).
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, with exactly p solutions
for any value of p. But the Division Algorithm and Corollaries hold over the
field of real numbers, including integers, so for specific integer values z = Z_{1},
s = S_{1} and A = A_{1}, (z – s) e f(z) →
(Z_{1} – S_{1}) ∈
f(Z_{1}) → f(S_{1}) = 0.
Note that if q(z)
and f(s) are not unique, corollary III does not apply and multiple
nonzero values of f(s) may be found.
See Appendix E.
But f(S_{1}) = S_{1}^{p1} + S_{1}^{p2}X_{1}
+ S_{1}^{p3}X_{1}^{2} +•••+ X_{1}^{p1}
= 0 is an impossibility because X_{1} and S_{1}
are positive integers, and the sum of p positive integers cannot equal zero.
Therefore, for specific integer values z = Z_{1}, s =
S_{1} and A = A_{1}, (Z_{1} – S_{1}) ∈
f(Z_{1}) → f(S_{1}) ≠ 0, and thus
g(z) = (Z – S) ∉ f(z)
= z^{p1} + z^{p2}X_{1} + z^{p3}X_{1}^{2}
+•••+ X_{1}^{p1}.
But, for there to be a triple integer solution for the FLT
equation, falsifying FLT, there must be specific integers X_{1}, Y_{1},
Z_{1}, A_{1}, and S_{1}, such that (Z_{1} – S_{1})
= A_{1}, and f(Z_{1}) = A_{1}^{p}, but if q(z) and f(s) are unique,
(Z_{1} – S_{1}) ∉ f(Z_{1}) →
A_{1} ∉ A_{1}^{p} and therefore we
have a clear contradiction, proving Fermat’s Last Theorem (FLT65).
__________
FURTHER
DISCUSSION AND CLARIFICATION
OF
THE CLOSE FLT65C PROOF OF FERMAT’S LAST THEOREM
Using the standard notation for real variables, integer
variables and specific integer values outlined above and in the discussion of
Objection #4, we can write: z^{p}
– x^{p} = x^{p}. This is equation (1) in the 1965 proof, without the restriction of x, y and
z to integer values.
Since it is sufficient to consider p as prime numbers greater
than 2, we can factor the left side of the equation to obtain:
(z  x)( z^{p1} + z^{p2}x + z^{p3}x^{2}
+•••+ x^{p1}) = y^{p}, equation
(2) of the FLT65C version of the proof.
In the 1965 proof (FLT65), the FLT equation is expressed as an
Ndegree polynomial in the variable Z. This translates to a pdegree 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 assume specific 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 ring of
integers, we will be able to determine whether any values of the third variable
can be integers.
In FLT65, the author focused on z as the independent unknown
by setting X = X_{1} and Y = Y_{1}, intending only to imply
that they were integer variables. Later on, he indicated that they were
specific integer values, without changing the way they were represented. This
lack of clear definition of notation may be the cause of the confusion that
gave rise to the concern #ii.
The 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 may not necessarily true for integer
polynomials of the form of f(Z). It turns out to be the case in this instance,
however, because of the unique form
of the FLT equation and its factors, and the requirement that the only
solutions being considered are those for which all three variables have integer
values satisfying the Fermat equation.
Because at least three of the most qualified reviewers of
FLT65 raised the concern about the applicability of the algorithm and corollaries
to integer factors, it is clear that the proof may be 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 all real number 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 subset of real numbers that make up the ring[2] of integers, and X_{1},
Y_{1}, and Z_{1} 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 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.
For variable integer
values of x, represented by X, let z^{p1} + z^{p2}X + z^{p3}X^{2}
+•••+ zX^{p2} + X^{p1} = f(z)
And z X = g(z). Then
(3.) g(z)f(z)
= Y^{p}, for all integer values X and Y.
For Fermat’s last
theorem to be falsified, X and Y must be specific integers, call them X_{1}
and Y_{1}, 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 and B for the
proof of this.
And by COROLLARY II, the remainder when f(z)
is divided by g(z) = z  X_{1}, will be f(X_{1}). And so:
(4.) r(z)
= f(X_{1}) = X_{1}^{p1} + X_{1}^{p2}X_{1}
+ X_{1}^{p3}X_{1}^{2} +•••+ X_{1}^{p1}
= pX_{1}^{p1}, a unique integer for any p > 2.
Interestingly, it is this unique integer remainder in the case
of the FLT equation that allows 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), X_{1}
and Y_{1} may be considered to be relatively prime, no factor of X_{1},
or, therefore, of X_{1}^{p1}, may be contained in g(z) = z –X_{1,
}because, if z is to be an integer, z –X_{1 }must contain a factor
of Y_{1}. 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,
(Y_{1})^{p}, one of them, i.e. either f(z) or g(z), must
contain p or they must be both perfect ppowers of integers.
Similarly, we can factor
the FLT equation as
(5.) (zY)(
z^{p1} + z^{p2}Y + z^{p3}Y^{2} +•••+ Y^{p1})
= X^{p}. And by exactly the same reasoning as above, for any particular
Y = Y_{1}, we can have
z^{p1}
– z^{p2}Y_{1} + z^{p3}Y_{1} ^{2}
•••+ Y_{1} ^{p1} = f_{1}(z) and z  Y_{1} = g_{1}(z).
Then,
either f_{1}(z) and g_{1}(z) contain p as a single common
factor, or they are perfect ppowers of integers. So for a given case of z^{p}
= X_{1}^{p} + Y_{1}^{p }, if either f(z) or f_{1}(z)
contains p, the other has to be a perfect ppower, 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 ∈
Y_{1} and p e f_{1}(z) →
p ∈ X_{1}. But X_{1}.and Y_{1}
are relatively prime. If neither X_{1}.nor Y_{1} contains p,
both, being relatively prime, must be perfect ppowers. Therefore, in any
event, one of them at least, must be a perfect ppower, not containing p as a
factor.
Therefore, we may choose
f(z) = z^{p1} + z^{p2}X_{1} + z^{p3}X_{1}^{2}
+•••+ X_{1}^{p1} = A^{p}, and/or
f_{1}(z) = z^{p1}
– z^{p2}Y_{1} + z^{p3}Y_{1} ^{2}
•••+ Y_{1}^{p1} = B^{p}, A and B integers, and at
least one, A or B, does not contain p. Also note that since g(z)f(z) = Y_{1}^{p},
and f(z) = A^{p}, A ∈ Y_{1} and since
g_{1}(z)f_{1}(z) = X_{1}^{p}, f_{1}(z)
= B^{p}, B
∈ X_{1}.
Since f(z) and f_{1}(z)
are both of the same form, we may choose either of them as the one not
containing p. So we may choose p ɇ
f(z) = z^{p1} + z^{p2}X_{1} + z^{p3}X_{1}^{2}
+•••+ X_{1}^{p1} = A^{p}, and p ɇ
A, meaning that A will not contain p.
We cannot, at this
point, assume that z = Z_{1}, a specific integer, for some x = X_{1}
and y = Y_{1}, because this cannot be justified unless we can show that
the requirement that f(z) = A^{p} does not lead to a contradiction.
For any value of X_{1},
the conditions f(z) = A^{p}, 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 X_{1}
< Y_{1 }< any z = Z_{1} 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, e.g. S_{1}, when
X = X_{1}, is dependent on the specific integer value, A_{1},
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 p^{th} power, as in that case when A is not an integer, but is the
noninteger p^{th} root of a prime number, q(z) and r(z) are not unique
and COROLLARY II does not hold, and
thus this
proof does not apply to the case n = 2, or to noninteger 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, if there FLT is false, and 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, when q(Z) and r(z) are unique, 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) = s^{p1} + s^{p2}X_{1} + s^{p3}X_{1}^{2}
+•••+ X_{1}^{p1} = 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, with exactly p 1
noninteger solutions and one trivial integer solution: X_{1} ^{p} +(0) ^{p} = Z_{1}^{p}.
But, if s is to be from the ring of integers, and q(s) and r = f(s) are unique, which they must be for an integer
solution of the FLT equation, 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 = Z_{1},
s = S_{1} and A = A_{1}, (z – s) e f(z) →
(Z_{1} – S_{1}) ∈
f(Z_{1}) → f(S_{1}) = 0.
But f(S_{1}) = S_{1}^{p1} + S_{1}^{p2}X_{1}
+ S_{1}^{p3}X_{1}^{2} +•••+ X_{1}^{p1}
= 0 is an impossibility because X_{1} and S_{1} are positive
integers, and the sum of positive integers cannot equal zero. Therefore:
g(z) = (z – S) cannot be
a factor of f(z) = z^{p1} + z^{p2}X_{1} + z^{p3}X_{1}^{2}
+•••+ X_{1}^{p1}. 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 X_{1}, Y_{1}, Z_{1}, A_{1}, and S_{1},
such that z = Z_{1} , S = S_{1} and A = A_{1}, q(s) and
f(s) are unique integers, and for the
FLT equation, (z – S_{1}) = A and f(z) = A^{p}, therefore:
(z – S) ɇ
f(z) → (Z_{1} – S_{1}) ∉
f(Z_{1}) → A_{1} ∉
A_{1}^{p} 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 S_{1}^{p1} + S_{1} ^{p2}X_{1}
+ S_{1}^{p3}X_{1}^{2} +•••+ X_{1}^{p1}
= 0; and also note that because the division algorithm and corollaries apply
across the field of real numbers, including the integers, we are justified in
substituting unique 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. It is the uniqueness of
these algebraic forms derived from the FLT equation, and the fact that any
integer solution of the FLT equation must also satisfy the equation S_{1}^{p1}
+ S_{1} ^{p2}X_{1} + S_{1}^{p3}X_{1}^{2}
+•••+ X_{1}^{p1} = 0 that assures that the fact that f(s) ≠ 0
implies that there are no integer solutions for equation (1).
With the substitution of
integers into the algebraic forms of f(z) and f(s) we see that the
‘counterexamples’ provided by some reviewers did not work, because they produce
nonunique
q(z) and f(s) and thus do not apply. And we see that the contradiction
obtained by assuming that for given integer values of x and y, an integer z was
possible must apply to the integer factors as well as the algebraic factors. Therefore the concern that the
factorization of X^{N} + Y^{N} = Z^{N}, 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.
When a mathematical
statement is true, it can usually be proved in more than one way. The proof
that in the case of FLT, the division algorithm applies to integer polynomials
and the single integer value they can be reduced to is no exception. If the
reader is not fully convinced by the argument presented above, we can address
concern #ii in another way, as follows:
Given f(S) ≠ 0, the
fact that the integers form a ring,
which is a subset 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 Z_{1} that might satisfy the_{ }FLT equation
along with X_{1} and Y_{1}, the remainder r(S_{1}) =
f(S_{1}) = S_{1}^{p1} + S_{1}^{p2}X_{1}
+ S_{1}^{p3}X_{1}^{2} +•••+ X_{1}^{p1},
not being zero, might contain g(z) = (z – S_{1}) = A_{1} as a
factor. Assuming that there is such
an actual integer triple solution for the FLT equation, the integers X_{1},
Y1, Z_{1}, 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 descent^{4} 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 again 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 wellordered nature of the ring of integers, allowing
for the basic operations of addition and subtraction, there is some (X,Y,Z) =
(X_{i},Y_{i},Z_{i}), specific integers, such that A = Z_{i}
– 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) = K^{p1} + K^{p2}X_{i}
+ K^{p3}X_{i}^{2} + ••• + X_{i}^{p1}
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 – K_{1}, K_{1} < K, and with smaller and smaller K_{i}
until the remainder, f(K_{i}) 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 [3] obtained by
successively dividing by A will eventually reduce f(K_{i}) 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(K_{i}) must be equal to zero. In our infinite
descent, the smallest possible f(K_{i}) will occur when K_{i}
=1.
But, the smallest possible f(K_{i}) = f(1) = 1 + X_{i}
+ X_{i}^{2} + ••• + X_{i}^{p 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 x^{p} + y^{p} = z^{p},
x = X_{1} and y = Y_{1}, X_{1} and Y_{1}
specific integers, we reach the contradiction of an infinite descent, proving
that z ≠ Z, Z ≠ Z_{1}, and f(z) ≠ A^{p}, a perfect p^{th}
power of an integer, and all sufficient and necessary conditions for the
definitive proof of FLT have been met. See
Appendix E for application of
the method of infinite descent to a proposed counterexample to FLT65.
__________
APPENDIX E ^{9}
Infinite Descent
INTRODUCTION
This
appendix presents the complete details of an infinite descent analysis of the
best example proposed by a mathematician specialist in number theory as a
counterexample invalidating FLT. The example consists of producing a set of
integer values for the FLT equation that appear to invalidate the application
of corollary III of the division algorithm to the integer values of factors of
the FLT equation in FLT65. But the
example ignores the uniqueness requirements for application of the corollary.
As a result of this omission, the example provides us with a way to elucidate
the contradiction validating the conclusion drawn in FLT65, that the nonzero remainder
when the algebraic polynomial factor f(Z)
= Z^{n1}+XZ^{n2}+…+
X^{n2}Z+X^{n1} of the FLT Diophantine equation Y^{n} = Z^{n
} X^{n} is divided by Z – a, is proof of Fermat’s Last
Theorem.
FACTORS OF THE FLT EQUATION
The FLT
equation, y^{n} = z^{n
} x^{n}, defined over the field of real numbers,
with n
> 2, is factorable, in the following manner:
Y^{n
}= (z – x)(z^{n1}+xz^{n2}+…+ x^{n2}z+x^{n1})
= g(z)f(z).
Assuming there may be integer solutions to the FLT
equation, we have shown in Appendix C
that we may let x = X and y = Y, integer
variables, such that
g(z)
= z – X = J^{n}^{ },
and f(z) =z^{n1}+Xz^{n2 }+…+X^{n1
}= K^{n}, with J and
K relatively prime integers for a hypothetical integer solution of
the FLT equation.
Because the integers are closed with respect to
addition and subtraction, we know that for every value of z and K, there is a real
number a, such that z – a = K. By the Principle of Designation of Variables we can designate x and y as independent integer variables, X and Y; but at this
point we do not know whether z and a can actually be integers or not, so
designating z as the dependent
variable and z – a as the divisor,
where a will be an integer if
z is an integer, and dividing the factor f(z) by z  a, we have:
f(z)/(z
– a) = q(z) + r(z)/(z – a) where, by corollary II of the division
algorithm, r(z) = f(a).
Therefore, f(z)/(z – a) = q(z) + f(a)/(z – a) Equation
(E1)
INFINITE DESCENT VALIDATION OF FLT65
We can generalize the proof for all n = p > 2, as follows:
f(z)/(z
– a) = q(z) + f(a)/(z – a), where f(z) = (z^{p1}+Xz^{n2}+…+
X^{n2}z + X^{n1})
Multiplying both sides of Equation (E1) by (z – a)
gives us f(z) = q(z)(z – a) + f(a).
Transposing f(a) yields f(z) – f(a) = q(z)(z – a), and from
this, we can derive the algebraic polynomial form of q(z), as follows:
q(z)(z
– a) = f(z) – f(a) = (z^{p1}+Xz^{p2}+…+ X^{p2}z + X^{p1})
 (a^{p1}+Xa^{p2}+…+X^{p1}).
Then, collecting like powers of X,
we have q(z)(z – a) = z^{p1 }–
a^{p1} +(z^{p2} – a^{p2})X +…+ (z – a) X^{p2}
+ X^{p1 }– X^{p1}.
Simplifying and dividing by (z – a), we have:
q(z)
=(z^{p2}+az^{p3}+…+ a^{p3}z + a^{p2}) + (z^{p3}+az^{p4}+…+
a^{p4}z + a^{p3})X +… + (z + a)X^{p3} + X^{p2}
From this, we can see that for n = 3, the polynomial form of q(z) is q(z) = z + a + X;
For n =
4, q(z) = z^{2} + az + a^{2}
+ (z + a)X + X^{2}
For n =
5, q(Z) = z^{3 }+ az^{2}
+ a^{2}z + a^{3 }+(z^{2} + az + a^{2})X +(z +
a) X^{2} + X^{3}; ּ
And for n
= p:
q(z)
=(z^{p2}+az^{p3}+…+ a^{p2})+(z^{p3}+az^{p4}+…+
a^{p3})X+…+(z + a)X^{p3}+X^{p2}^{ }Eq. (E2)
Given X = X_{1}
and Y = Y_{1}, specific integers, and assuming that z can be an integer, let z = Z, then f (z) = f(Z) = Z^{p1 }+ X_{1}Z^{p2} ^{+}
… + X^{p1}, an integer polynomial of n1 degree in Z. But,
since at this point, we don’t know whether a
can be an integer or not for a specific integer value of Z, Eq. (E2) becomes q(Z)=(Z^{p2}+aZ^{p3}+…+
a^{p2})+(Z^{p3}+aZ^{p4}+…+ a^{p3})X+…+(Z +
a)X^{p3}+X^{p2}, an algebraic polynomial in Z
and a, and Eq. (E1) becomes:
f(Z,a)/(Z
– a) = q(Z,a)_{ }+ f(a)/(Z – a) Eq. (E3)
For a specific integer solution, X = X_{1}, Y_{1} = J_{1}K_{1}, Z = Z_{1} and Y_{1}^{p
}= J_{1}^{p}K_{1}^{p}, f(a) = a^{p1 }+ a^{p2}X_{1}^{ }+ …+ aX_{1}^{p2
}+ X_{1}^{p1}, and Eq. (E3) becomes Eq. (E4),
an equation in the variable a:
f(Z_{1})/(Z_{1} – a) = q(a)
+ f(a)/(Z_{1} – a) Eq. (E4)
If there is an integer solution, then z and a will be integers, but, since we do not know whether there can be
an integer solution for the FLT equation, we must continue to represent z as an integer variable, Z, and a as a real number variable, which may be integer,
rational fraction or irrational, depending upon whether or not the hypothesis
that Z can be an integer is
validated.
With z
as an integer variable, Z, and a as a real variable, f(Z) = Z^{p1}+ X_{1}Z^{p2
}+ … + X_{1}^{p1}_{ }= (K_{1})^{p}, q(Z) = (Z ^{n2}+aZ_{1}^{n3}+…+ a^{n2})
+ (Z^{n3}+aZ ^{n4}+…+ a^{n3})X+…+(Z + a)X^{n3}+X^{n2},
and
f(a)
= a^{p1}+X_{1}a^{p2}+…+X_{1}^{p1}. Substituting
these hypothetical integer values into Eq.
(E4), we have:
(K_{1})^{p}/K_{1}^{ }= (K_{1})^{p1}
= (Z^{p2}+aZ^{p3}+…+ a^{p2}) + (Z^{p3}+aZ^{p4}+…+
a^{p3})X+…+(Z + a)X^{p3}+X^{p1} +
+
(a^{p1}+X_{1}a^{p2}+…+X_{1}^{p1})/K_{1}^{ }(An algebraic polynomial in a and
Z.) Eq. (E5)
IMPORTANT
NOTE:
Note that the polynomials making up the equations in this
discussion are all algebraic polynomials in one or more of the real variables,
z, Z and a, to which the division algorithm and its corollaries definitely
apply.
By inspection of Eq. (E4), we see that, if Z
is an integer, the left side of the equation, f(Z)/K_{1} = (K_{1})^{p1}, is an integer
and a and q(a) are integers. Since
integers are closed with respect to addition, we see that f(a) must contain Z – a = K_{1} as an integer factor, for any integer value of a.
Eq.
(E5) represents the first step in the process of long
division of the integer polynomial f(Z)
= Z^{p1 }+ X_{1}Z^{p2 }+ X_{1}^{2}Z^{p3
}+…+ X_{1}^{p2}Z^{ }+X_{1}^{p1 }by
the integer polynomial Z_{ }– a = K_{1}.
The quotient is q(Z)=(Z^{p2}+aZ^{p3}+…+a^{p2}) + (Z^{p3}+aZ^{p4}+…+a^{p3})X_{1}+…+(Z
+ a)X_{1}^{p3}+X^{p2}, and_{ }the remainder is f(a) = a^{p1}+X_{1}a^{p2}+…+X_{1}^{p1}.
In long division, if the remainder is larger than
the divisor, it indicates that the quotient is too small, a new quotient is
used, and the process is repeated with a larger and larger quotient until the
remainder is either zero or smaller than the divisor. If the remainder is zero,
we know that the dividend is divisible by the divisor, if the remainder is not
zero, then the dividend is not divisible by the divisor.
Because the integers, a subset of the field of real numbers, are closed with respect to
addition and subtraction, for any integer, K_{1}, there are an infinite number of pairs
of integers, a_{i}, a_{i+1}, such that
z
– a = K_{1} ≡ a – a_{1} ≡ a_{1} – a_{2} ≡
a_{2} – a_{3} ≡…≡
a_{k1} – a_{k}, with a >a_{1} >a_{2} >…
>a_{k}, k = 1, 2, 3 … ∞.
If there actually is an integer solution, X_{1}, Y_{1}, Z_{1},
for the FLT equation, then z – a ≡ Z_{1} – A_{1}
≡ K_{1} expressed
as the difference between a specific pair of integers. The problem becomes how
to find the unique pair of integers, a_{i1} and a_{i}, if they exist, that will
produce the minimum remainder, r = f(a_{i}),
consistent with integer values of Z and a that will satisfy the FLT equation.
We can approach this problem by carrying out the division indicated in the last
term of Equation (E1), i.e. (a^{p1}+X_{1} a^{p2
}+… +X_{1}^{n1})/(Z – a). To carry out the indicated
division, we replace Z – a by its
equivalent a_{ } a_{1} with a as the variable, which we can do by virtue of the Principle of Designation of Variables,
and repeatedly divide by Z  a in
the form of the difference between successively smaller and smaller values of a_{i }and a_{i+1}. Each division produces an addition to q(Z) and a diminishing f(a_{i}) until a minimum
remainder is reached. Using the process of long division of polynomials we can
determine whether the integer value of the minimum f(a_{i}) is actually divisible by a_{i1 } a_{i }=
K_{1} or not if f(a) ≠ 0. If it is not, we have a
contradiction, since for an integer solution, if f(a) is not zero, it must
contain Z – a.
Returning to Eq.
(E1), for n = p,
any prime >2, we can set K_{1}
= a – a_{1} and divide
f(a)=
a^{p1 }+ X_{1}a^{p2}+…+X_{1}^{p1},
an algebraic polynomial in a, by a – a_{1}, to obtain:
f(Z)/(a
– a_{1})=(Z^{p2}+aZ^{p3}+…+a^{p2})+(Z^{p3}+aZ^{p4}+…+a^{p3})X_{1}+…+(Z
+ a)X_{1}^{p3}+X_{1}^{p2}
+ (a_{1}^{p1 }+ X_{1}a_{1}^{p2
}+ … + X_{1}^{p1})/(a  a_{1}). Eq.
(E6)
This process can be repeated indefinitely with
smaller and smaller values of a_{i1}
and a_{i}. Each time the
last term is expanded by dividing f(a_{i})
by a_{i1} – a_{i},
an additional quotient term, q(a_{i}),
and a new remainder, f(a_{i}),
are produced. Each additional
quotient term added to previous quotient is smaller than the one before, and
each new remainder is smaller than the one before, because Z > a, each a_{i}
is smaller than the one before, and X_{1}
is a constant. This descent will continue until f(a_{i}) reaches a minimum. At that point, a_{i} must be_{ }either
zero, positive, or negative. Looking at the three possible outcomes, we see
that:
1.) If
a_{i} for the minimum f(a_{i}) is zero, then a_{i1} – a_{i} = a_{i1 }= K_{1}, and f(a_{i}) = a_{1} ^{p1}+X_{1}a_{1}^{p2
}+…+X_{1}^{p1 }= 0 + 0 +…+X_{1}^{p1 }= X_{1}^{p1},
which we see by inspection of Eq. (E6),
^{ }must contain a_{i1}
because of closure of integers with respect to addition. This implies that X_{1}^{p1 }contains K_{1} which is a factor of Y_{1}, and this is a contradiction of the relative prime
sufficiency condition for the FLT equation (See Appendix C). Therefore, a
cannot equal zero for minimum f(a_{i}).
2.) The
smallest a_{i} cannot be
positive: A positive a_{i}
will not produce the smallest f(a_{i}),
because f(a_{i}), which is always positive, will be
smallest with the first negative a_{i}.
After the first negative a_{i},
successive negative values of a_{i}
will only produce larger values for f(a_{i}), because odd powers of
negative numbers are negative and even powers of negative numbers are positive.
With f(a_{i}) = a_{1} ^{p1}+X_{1}a_{1}^{p2
}+…+X_{1}^{p1}, the even powers of a_{i} and X_{1} will always outnumber and be greater than the odd
powers of a_{i} and X_{1}.
3.) Finally,
because the a_{i} producing
the smallest f(a_{i}) cannot
be zero or positive, it will have to
be negative, and q(a_{i}) and
f(a_{i}) will be unique,
implying by corollary II and III that the smallest f(a_{i}) must be a positive integer not containing Z – a = a_{i1} – a_{i}.
But, if f(Z) is an integer, and q(a_{i}) is an integer, by
inspection of Eq. (E6) and closure of integers with respect to
addition, f(a_{i}) must
contain K_{1}, the integer
value of Z – a. This is a
contradiction unless the a_{i}’s,
and consequently, Z, are nonintegers.
Therefore, this contradiction tells us that, while Z – a must be an integer
factor of the integer value of f(Z), Z and a cannot be integers, proving
FLT for all p > 2.
These infinite descent contradictions prove
that because f(a) ≠ 0, a and Z cannot be integers, and therefore, FLT is
proved for all n > 2.
THE
CASE p = 3
This demonstration may be easier to follow by
using p = 3, the smallest and least
complicated case as an example. Returning to Eq. (E1), for p = 3, we
can let K_{1} =a – a_{1}
and divide a^{2}+ aX_{1}^{
}+X_{1}^{2}, an algebraic polynomial in a, by a – a_{1}, to obtain: f(a)/(a
– a_{1})= a+a_{1}+X_{1}
+(a_{1}^{2}+a_{1}X_{1}+X_{1}^{2})/K_{1}.
So we have:
f(Z_{1})/(Z_{1}
– a) = Z_{1} + a + X_{1} +(a + a_{1} + X_{1}) + (a_{1} ^{2}+ a_{1}X_{1}^{ }+X_{1}^{2})/(a_{1}
–a_{2}). Eq. (E7)
This process can be repeated indefinitely with
smaller and smaller values of a_{i}.
Each time the last term is expanded by dividing f(a_{i1}) by a_{i1}
– a_{i}, an additional quotient term, q(a_{i}) = (a_{i1}
+ a_{i} + X_{1}) and a new remainder, f(a_{i}) are produced.
Each additional quotient term added to the previous quotient is smaller than
the one before, and each new remainder is smaller than the one before, because Z_{1} > a, each a_{i} is smaller than the one
before, and X_{1} is a
constant. This descent will be continued until f(a_{i}) reaches a minimum. At that point, a_{i} must be_{ }either zero, positive, or negative.
Looking at the possible outcomes with n
= 3, as in the general case, we see that,
1.) If
a_{i} for the minimum f(a_{i}) is zero, then a_{i1} – a_{i} = a_{i1 }= K_{1}, and f(a_{i}) = a_{i}^{2}+
a_{i}X_{1}^{ }+X_{1}^{2 }= 0 + 0 + X_{1}^{2
}= X_{1}^{2}, which we see by inspection of Eq. (E7) ^{ }must
contain a_{i1} because of closure of
integers with respect to addition. This implies that X_{1}^{2 }contains K_{1} which is a
factor of Y_{1}, and this is
a contradiction of the relative prime sufficiency condition for the FLT
equation (See Appendix C). Therefore, a
cannot equal zero for minimum f(a_{i}).
2.) The
smallest a_{i} cannot be
positive. If a_{i} is
positive, it will not produce the smallest f(a_{i}),
because f(a_{i}) is always
positive, and it will be smallest with the first negative a_{i}. After the first negative a_{i}, successive negative values of a_{i} will only produce larger
values for f(a_{i}), because
f(a_{i}) = a_{i}^{2 }_{ }+aX_{1}^{ }+ X_{1}^{2},
and the even powers of a_{i}
and X_{1} outnumber and are
greater than the odd powers (negative) of
a_{i} and X_{1}.
3.) Finally,
since the a_{i} producing
the smallest, and therefore unique f(a_{i}) cannot be
zero or positive, it will have to be negative, and q(a_{i}) and f(a_{i})
will be unique, implying that the smallest f(a_{i}) must be a
positive integer that cannot contain
K_{1 }because it must be smaller than K_{1}. But, if f(Z)
is an integer, and q(a_{i})
is an integer, by inspection of Eq.
(E7) and closure of integers
with respect to addition, we see that f(a_{i})
must
contain the integer value of Z – a.
This is a contradiction, unless the a_{i}’s,
and consequently, Z, are not
integers. Therefore, while the integer value of Z – a is a factor of the integer value of f(Z), Z and a cannot be integers, proving FLT for n =
3.
EXAMPLE
This process of descent is somewhat easier to follow using
actual integers, and we can use the integer values from Example #3 provided by the number theorist reviewer: Z = 73, X_{1} = 17 and a = 54. To start with. Using these
integer values, infinite descent reveals the fact that the nonzero remainder
obtained by dividing the algebraic polynomial f(z) = Z^{2 }+ X_{1}Z+ X_{1}^{2} by
Z – a, leads to contradictions
proving that these values do not constitute a counterexample to the FLT65
proof.
To invalidate FLT65, specific values of a, X and Z that
produce an integer value for f(Z) that is divisible by the integer value Z  a,
even when the algebraic polynomial f(Z) divided by Z – a produces a nonzero
remainder, must produce a unique
maximum quotient and minimum remainder; otherwise, corollary III of the
division algorithm does not apply. On the other hand, if the quotient and
remainder are unique, which they must be for an integer solution to the FLT
equation, corollary III does apply,
and the nonzero remainder proves FLT.
Note that for any
numerical example, the number of steps are finite, but the method is properly
called infinite descent because the process can descend from any hypothetical
integer value of Z, however large.
Step one of Descent:
Evaluation of the First Remainder
Using the integer values X_{1} = 17, z = 73 and
a = 54, yields: z
– a = 73 – 54 =
19;
f(Z) = 73^{2 }+ 73x17 ^{+}
17^{2}
= 5329 + 1241 + 289 = 6859; q(Z) = (Z + a + X_{1 }) = 73
+ 54 + 17 = 144; and r_{1 }= f(a) = a^{2}+Xa +X^{2}
= 54^{2 }+ 54x17 ^{+} 17^{2}
= 4123. Substituting these results into Eq. (E7), we have:
f(Z)/(Z –a) =19^{3}/19 =19^{2}
=144+4123/19 = (2736+4123)/19^{ }=6859/19^{ }=361 = 19^{2}.
This indicates that f(Z) is divisible by Z – a for
these integer values of a, X and Z, and the remainder is not zero, but the remainder, f(a) =
4,123, is much larger than the divisor,
Z_{ } a _{ }= 19, indicating
that the quotient and remainder are not
unique and we must divide again, increasing the quotient and decreasing the
remainder, to find the unique integer values, which can only be obtained when
the remainder is either equal to zero or has the minimum possible integer
value.
Note that the process of
repeatedly dividing the remainder by Z 
a = a_{i1 } a_{i }=
19 for decreasing values of a_{i1
}and a_{i}, does
not change the value of f(Z)/( Z –a) =
q(Z) + f(a_{i}) because as f(a_{i})
decreases, q(Z) increases preserving
the total value of f(Z)/( Z –a),
which is 19^{2}.
Step Two of Descent:
Evaluation of the Second Remainder
We have all the integer values needed to evaluate Eq. (E7), except a_{1} and we can determine a_{1} as follows: K_{1}
= z – a = 73 – 54 = 19, and z – a = a – a_{1} = 19 → a_{1 }= a – 19 = 54 – 19 = 35.
Substituting a_{1 }= 35
into the last two terms of Eq. (E7), we
have:
q(a_{1})
= 144 + (a + a_{1} + X_{1}) = 144
+ (54 + 35 + 17) = 250, and
f(a_{1})/(a
–a_{1}) ={(35)^{2} +35x17^{ }+ (17)^{2}}/19 = 2109/19^{ }
Substituting these integer values into Eq. (E6), we have:
f(Z)/(Z –a) = 19^{2}
= 144 + 106 + 2109/19^{ }= (4750 +2109)/19= 6859/19^{ }= 19^{2}
Verifying that, with the new remainder
f(Z) may be divisible by Z – a for
these integer values of a, X and Z.
Step Three of Descent:
Evaluation of the Third Remainder
Next, a_{1 } a_{2
}= 35 – a_{2 }=19, yielding a_{2 }= 16
Substituting a_{2 }=
16 into the last two terms of Eq.
(E7)
expanded, we have:
q(a_{2})
= 250 + (a_{1} + a_{2} + X_{1}) = 250 + (35
+ 16 + 17) = 318 and
f(a_{2})/(a_{1}
–a_{2}) ={(16)^{2} +16x17^{ }+
(17)^{2}}/19 = 817/19^{}
Thus, f(Z)/(Z –a) = 19^{2}
= 318 + 817/19 = (6042 + 817)/19 =6859/19 = 19^{2}
Step Four of Descent:
Evaluation of the Fourth Remainder
Next, a_{2 }– a_{3
}= 16 – a_{3 }=19, yielding a_{3 }=  3
Substituting a_{3 }=
 3 into the last two terms of the expanded equation, we
have:
q(a_{3})
= 318 + (a_{2} + a_{3} + X_{1}) =318 + (16 + (3) + 17) = 348, and
f(a_{3})/(a_{2}
–a_{3}) ={(3)^{2} + (3x17)^{ }+
(17)^{2}}/19 = 247/19^{}
Then: f(Z)/(Z –a) =144
+ 106 + 68 + 30 + 247/19 =(6612 + 247)/19 =6859/19 = 19^{2}
Step Five of Descent:
Evaluation of the Fifth Remainder
Next, a_{3 }– a_{4
}=  3 – a_{4 }= 19, yielding a_{4 }=  22
Substituting a_{4 }=
 22 into the last two terms of the equation, we have:
q(a_{4})
= 348 + (a_{3} + a_{4} + X_{1}) = 348+ (
3 + ( 22) + 17) = 340, and
f(a_{4})/(a_{3}
–a_{4}) ={(22)^{2} + (22x17)^{ }+
(17)^{2}}/19 = 399/19^{}
Then we have: f(Z)/(Z
–a) = 348 + ( 8) + 399/19 = (6460 + 399)/19 =6859/19 = 19^{2}
Step Six of Descent:
Evaluation of the Sixth Remainder
And, a_{4 }– a_{5
}=  22 – a_{5 }= 19, yielding a_{5 }=  41
Substituting a_{5 }=
 41 into the last two terms of the equation, we have:
q(a_{5})
= 340 + (a_{4} + a_{5} + X_{1}) = 340 + (
22 + ( 41) + 17) = 294, and
f(a_{5})/(a_{4} – a_{5})
={(41)^{2} + ( 41x17)^{ }+ (17)^{2}}/19 = 399/19, and a_{3 }
a_{4 }= 19 → a_{4 }=  22, yielding r_{5 }= f(a_{4}) = a_{4}^{2}+a_{4}X+X^{2
}= 299.
TABLE
E1 Summary of
Remainder Descent with Search Interval = 19
Descent Step #

Z – a =
a_{i1 } a_{i}=
K_{1} = 19_{}

Quotient
q(a_{i})

Remainder
r_{i } = f(a_{i})

RESULT:
FLT65
Invalidated?

1

72  54

144

4123 = 217(Z – a)

NO, q and r Not Unique^{}

2

54  35

250

2109 =111(Z – a)

NO, q and r Not Unique

3

35  16

318

817 = 43(Z – a)

NO, q and r Not Unique

4

16  3

348

247 = 13(Z – a)

NO, r is minimum for
this search interval, but still larger than the divisor, Z a_{}

5

 3  22

340

399 = 21(Z – a)

NO, q_{ }smaller
than maximum,
And r larger than
minimum

6

 22  41

294

1273 = 67(Z – a)

NO, r_{ }larger
than minimum

Notice that the pairs of integers producing quotient additions
and remainders before or after the smallest remainder found with this search
interval do not produce unique quotients and
remainders. All of the pairs produce remainders divisible by Z – a = 19 and satisfy the equation.
If the integer values, Z_{1} =
73, X_{1} = 17 and a = 54,
actually comprise a valid counterexample to FLT65, there must be a pair, a_{i1} and a_{i}, that will produce the maximum quotient and minimum remainder, but none of these
pairs produce unique maximum q and
minimum r. This descent using step
intervals equal to K_{1}_{
}= 19, shows us that the pairs (a_{i1},
a_{i}) that produce a unique q(a_{i}) and minimum f(a_{i}),
must lie between (16 , 3) and ( 3, 22).
With this process using descent
steps equal to19, the fourth step pair produces a quotient larger than the
pairs before and after, and the remainder is larger than those before and
after, but the remainder, 247, is
still much larger than the divisor, 19,
indicating that further divisions are required to identify the pair that will produce the unique minimum f(a_{i}).
Searching for the integer pair
that will produce the minimum remainder using an integer search interval of 19 is like trying to catch
a fish that is 19 centimeters long and a few centimeters in diameter with a net
with a mesh size of 19 centimeters. Since we are dealing with a Diophantine
equation, i.e., an equation with integer variables, the ultimately smallest net
we can use is one with an integer mesh size of one. To do this, we must start
with the first pair, Z and a, and proceed with descent steps
defined by an integer search interval of 1, with each successive a_{i 1 }and_{ }a_{i }smaller by 1, as follows: 73–54 = 72–53 = 71–52 =…= a_{i1}
 a_{i} = 19.
Table E2 displays the results of
the search refined to the minimum search interval of 1.
TABLE
E2
Descent Step #

Z – a =
a_{i1 } a_{i}=
K_{1} = 19_{}

Quotient
q(a_{i})

Remainder
r_{i } = f(a_{i})

RESULT:
FLT65
Invalidated?

1

73  54

144

4123 = 217x19

NO, r_{i} is not minimum

2

72  53

286

3999

NO, q(a_{i}) not maximum
and f(a_{i}) not minimum

3

71 – 52

426

3877

NO, (same reason as above)^{}

4

70 – 51

564

3757

NO, (same reason as above)

5

69 – 50

700

3639

NO, (same reason as above)

6

68 – 49

834

3523

NO, (same reason as above)

7

67 – 48

966

3409

NO, (same reason as above)

8

66 – 47

1096

3297

NO, (same reason as above)

9

65 – 46

1224

3187

NO, (same reason as above)

10

64 – 45

1350

3079

NO, (same reason as above)

11

63 – 44

1474

2973

NO, (same reason as above)

12

62 – 43

1596

2869=151x19

NO, r_{i} is neither minimum nor
unique

13

61 – 42

1716

2767

NO, q(a_{i}) not maximum
and f(a_{i}) not minimum

14

60 – 41

1834

2667

NO, (same reason as above)

15

59 – 40

1950

2569

NO, (same reason as above)

16

58 – 39

2064

2473

NO, (same reason as above)

17

57 – 38

2176

2379

NO, (same reason as above)

18

56 – 37

2286

2287

NO, (same reason as above)

19

55 – 36

2394

2197

NO, (same reason as above)

20

54 – 35

2500

2109 = 111x19

NO, f(a_{i}) divisible by Z – a,
but not Minimum

21

53 – 34

2604

2023

NO, q(a_{i}) not maximum
and f(a_{i}) not minimum

22

52 – 33

2706

1939

NO, (same reason as above)

23

51 – 32

2006

1857

NO, (same reason as above)

24

50 – 31

2904

1777

NO, (same reason as above)

24

49 – 30

3000

1699

NO, (same reason as above)

25

48 – 29

3094

1623

NO, (same reason as above)

26

47 – 28

3186

1549

NO, (same reason as above)

27

46 – 27

3276

1477

NO, (same reason as above)

28

45 – 26

3364

1407

NO, (same reason as above)

29

44 – 25

3450

1339

NO, (same reason as above)

30

43  24

3534

1273 = 67x19

NO, f(a_{i}) divisible by Z – a,
but not Minimum

32

42 – 23

3616

1209

NO, q(a_{i}) not maximum
and f(a_{i}) not minimum

33

41 – 22

3696

1147

NO, (same reason as above)

34

40  21

3774

1087

NO, (same reason as above)

35

39  20

3850

1029

NO, (same reason as above)

36

38 – 19

3924

973

NO, (same reason as above)

37

37 – 18

3996

919

NO, (same reason as above)

38

36  17

4066

867

NO, (same reason as above)

39

35  16

4134

817=43x19

NO, f(a_{i}) divisible by Z – a,
but not Minimum

40

34 – 15

4200

769

NO, q(a_{i}) not maximum
and f(a_{i}) not minimum

41

33 – 14

4264

723

NO, (same reason as above)

42

32  13

4326

679

NO, (same reason as above)

43

31  12

4386

637

NO, (same reason as above)

44

30  11

4444

597

NO, (same reason as above)

45

29 – 10

4500

559

NO, (same reason as above)

46

28 – 9

4554

523

NO, (same reason as above)

47

27 – 8

4606

489

NO, (same reason as above)

48

26 – 7

4656

457

NO, (same reason as above)

49

25  6

4704

427

NO, (same reason as above)

50

24  5

4750

399=21x19

NO, f(a_{i}) divisible by Z – a,
but not Minimum

51

23 – 4

4794

373

NO, q(a_{i}) not maximum
and f(a_{i}) not minimum

52

22 – 3

4836

349

NO, (same reason as above)

53

21 – 2

4876

327

NO, (same reason as above)

54

20  1

4914

307

NO, (same reason as above)

55

19  0

4950

289

NO, (same reason as above)

56

18 – (1)

4984

273

NO, (same reason as above)

57

17 – (2)

5016

259

NO, (same reason as above)

58

16 – (3)

5046

247 = 13x19

NO, f(a_{i}) is divisible by Z  a
but not minimum, and
q(a_{i}) is not maximum

59

15 – (4)

5074

237

NO, q(a_{i}) and f(a_{i})
not unique

60

14 – (5)

5100

229

NO, q(a_{i}) and f(a_{i})
not unique

61

13 – (6)

5124

223

NO, q(a_{i}) and f(a_{i})
not unique

62

12 – (7)

5146

219

NO, q(a_{i}) and f(a_{i})
not unique

63

11 – (8)

5166

217

NO, r = f(a_{i}) is minimum,
but q(a_{i}) is not maximum

64

10 – (9)

5184

217

NO, r = f(a_{i}) is minimum,
but q(a_{i}) is not maximum

65

9 – (10)

5200

219

NO, q(a_{i}) <max & f(a_{i})
>min

66

8 – (11)

5214

223

NO, q(a_{i}) <max & f(a_{i})
>min

67

7 – (12)

5226

229

NO, q(a_{i}) <max & f(a_{i})
>min

68

6 – (13)

5236

237

NO, q(a_{i}) <max & f(a_{i})
>min

70

5 – (14)

5244

247=13x19

NO, f(a_{i}) contains Z  a, but
q(a_{i}) & f(a_{i}) are not unique

71

4 – (15)

5250

259

NO, q(a_{i}) <max & f(a_{i})
>min

72

3 – (16)

5254

273

NO, q(a_{i}) <max & f(a_{i})
>min

73

2 – (17)

5256

289

NO, q(a_{i}) is maximum,
but r = f(a_{i}) not minimum

74

1 – (18)

5256

307

NO, q(a_{i}) is maximum, but not
unique, and r = f(a_{i}) is not minimum

75

0 – (19)

5254

327

NO, q(a_{i}) is not maximum,
And f(a_{i}) is not minimum

76, 77,…

1 – (20)
etc.

Decreasing

Increasing

NO, q(a_{i}) < maximum,
and f(a_{i}) > minimum

Inspection of this table clearly shows that the minimum value
of the remainder, r_{i =} f(a_{i}), occurs between a_{i }=
8, and a_{i }= 9, which tells us that the a_{i} that produces
the minimum remainder is not an integer.
This contradicts the assumption that Z and a are integers: If a_{i }is
not an integer, then a_{i1 } a_{i} = 19 implies that a_{i1
}= 19 – a_{i} is also not an integer, and this inference passes
back up the descent all the way to a and Z_{1}. This contradiction
proves that the integers X_{1} = 17, a = 54, and Z_{1} = 73 do
not comprise a counterexample to FLT65.
Also, we see from this
table that the minimum integer value
of the remainder, r_{i =} f(a_{i})
= 217, is not unique, since it occurs
for two different integer pairs. Corollary III of the division algorithm says
that if q(Z) and r(Z) are unique,
f(a) cannot contain Za, and so it follows that a polynomial, f(Z), of degree greater than 1 is divisible by Za IF AND ONLY IF, f(a) = 0. This infers that corollary III does
not apply to the FLT equation for X_{1} = 17, a = 54, and
Z = 73. It does apply, however when f(a_{i})
and q(a_{i}) are unique. For unique f(a_{i}) and q(a_{i}), in this case,
nonintegers, f(a)
cannot contain Z – a, and f(Z) is divisible by Za, IF AND ONLY IF, f(a) = 0. But f(a) cannot equal zero for any
positive integer value of a, and if a is zero or negative, then the
hypothesis that Z can be an integer
for any specific integers, X_{1 }and Y_{1}, is proved false.
CONCLUSION: For there to be an integer solution of the FLT
equation, there must be an integer pair, a_{i1 } a_{i}, for
the integer values of X, a and Z that produces a unique q(a_{i})
maximum and a unique f(a_{i}) minimum. But by applying infinite descent
with the smallest possible integer search interval, one, with this exhaustive search, we find that there is no integer
pair a_{i1 } a_{i} that produces a unique q(a_{i})
and f(a_{i}) for X_{1} = 17, a = 54, and Z = 73, proving that
they do not comprise a counterexample to the conclusion of FLT65.
Finally, notice that the logical injunction “IF AND ONLY IF”
works in both directions: The statement ‘A is true IF AND ONLY IF B is true’,
implies that the converse: ‘B is true IF AND ONLY IF A is true’, is also true.
It follows that the proof of corollary III given in FLT65 also proves the
converse. Corollary III says:
“If q(Z) and r(Z) are
unique, then f(a) cannot contain Za, and f(Z) is divisible by Za, IF AND
ONLY IF, f(a) = 0.”
So the converse is also true: If f(a) = 0, f(a) cannot contain
Za, and f(Z) is divisible by Za IF AND ONLY IF q(Z) and r(Z) are unique.
It is also an established fact of logic that
the inverse of a statement has the same truth value as the converse of the
statement^{1}.
The truth table below shows the relationship between the conditional,
the converse, the inverse, and the contrapositive. Only the shaded row is
relevant to our discussion of ‘if and only if’ statements.
TABLE E3


Not p

Not q

Conditional
(if p, then q)

Converse

Inverse

Contrapositive








^{→}

T

T

F

F

T

T

T

T

T

F

F

T

F

T

T

F

F

T

T

F

T

F

F

T

F

F

T

T

T

T

T

T

This means that
the inverse and the converse of an ‘if and only if’
statement are logically equivalent. So, since the converse of the ‘if and only if’ statement of
corollary III is true, the inverse of corollary III is also true. Its inverse
is:
If q(Z) and r(Z) are not
unique, then f(a) contains Za, and f(Z) is not divisible by
Za, IF AND ONLY IF, f(a) ≠ 0.
Since these are ‘if and only if’
statements, the converse of this statement (also called the contrapositive) is
also true:
If f(a) ≠ 0, then f(a) contains Za, and f(Z) is not
divisible by Za, IF AND ONLY IF, q(Z) and r(Z) are not unique.
Thus the fact that f(a) cannot
equal zero for any integer value of Z
and a, (including the values they would have if there are integer solutions for
the FLT equation) implies that f(a) contains Z – a and f(Z) is not divisible by
Z–a, which is exactly what FLT65 says.
Eq. (E4): f(Z_{1})/(Z_{1} – a) = q(a) +
f(a)/(Z_{1} – a), with the remainder divided iteratively to the minimum
remainder becomes:
f(Z_{1})/(Z_{1} – a) = q(a_{i})
+ f(a_{i} )/(Z_{1} – a) Eq. (E8)
By inspection of this equation we
see that f(a_{i}) must
contain Z_{1} a, but if f(a_{i}) contains Z_{1}  a, r(Z) is not unique, and we must divide by Za again to produce a unique remainder, but if a is an integer, the remainder will
never be zero, implying that r(Z) is not
unique unless f(a) = 0. The fact that
f(a) cannot equal zero proves Z  a
cannot divide f(Z) because q(Z) and r(Z) are
not unique unless r(Z) is smaller than Z–a, in which case, Za does not divide
f(Z). Thus we have a contradiction that proves FLT, as stated in FLT65.
[1]
FLT65C is the 2013
rewrite of FLT65 distinguishing integer variables from real number variables
and constants using more conventional notation without modifying the
mathematical structure of any of the equations. This version of the proof is
designated as a clarification of FLT65, or FLT65C.
[2] 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.
[3] 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 ᵱ 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 ᵱ leads by a mathematical process of
one or more steps to the existence of a smaller
positive integer, N_{1} < N,
that also has or provides a function that has the property ᵱ, 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 N_{1}, will produce a
series of integers: N > N_{1} > N_{2} >…> N_{i}, 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 ᵱ 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.^{[9]}
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