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})Ïµ 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 Ïµ 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.