This proof that there is no coprime integer
solution (X,Y,Z) of X^{n} + Y^{n} = Z^{n} for n = 3 and
its generalization to n = p, primes > 2, provides validation of the method
of proof I used in FLT65. But, in my opinion, no such validation is needed because
the questions you have raised, and every valid question ever raised by any
reviewer of FLT65 are adequately answered by two simple statements
in FLT65: (1.) “A polynomial f(X), of degree greater than one, is divisible by
X – a IF, AND ONLY IF, f(a) = 0.” And, (2.), “… the integers are elements in
the field of rational numbers.” Application of statement (1.) to the integer
polynomial of the form Z^{n1}+XZ^{n2 }+ X^{2}Z^{n3}
+…+
X^{n2}Z + X^{n1} = A^{n }= (Z – s)^{n} with
A, X, Z and a coprime elements of the ring of integers, showing that the
remainder f(s) can never be zero, comprises a valid proof of FLT.
Questions raised by reviewers invariably arise from
the claim that: “While the Division Algorithm applies to algebraic polynomials,
it may not apply to the division of integers.” And some reviewers who make this
extraordinary claim, attempt to justify it with an example using integer values
of s, X and Z such that f(Z) = Z^{2} + XZ + X^{2} is divisible by Z – s. They assume that,
because f(Z) in their example is divisible by Z –s, the remainder f(s) is equal
to zero. It is easily demonstrated that this is not so. I have done so a number
of times in discussions with different reviewers.
It is a mistake to assume that the Division
Algorithm and corollaries do not apply to integer polynomials. To see this,
consider the fact that the remainder f(s) is
exactly the same, namely, it is equal to the integer s^{2} + sX + X^{2},
whether dividing the polynomial f(z) over the field of real numbers by z – s,
or dividing f(Z), an integer polynomial factor of Y for solutions of the FLT
equation, by the integer Z – s, with s and Z positive integers. In examples produced by a few reveiwers,
e.g., f(s) is not zero, it is equal to a multiple of Z – s. The remainder R = 0
is obtained and confused with f(s) by skipping the step involving f(s) obtained
by substituting the integer values of X and Z into f(Z). This step is easily
overlooked because the integers used in the example are small. Since f(s) ≠
0, it is easy to show that the integer Z in such example cannot be the Z in any X,Y,Z, integer solution
to the FLT equation.
(I have found several other examples of f(Z)/(Z – s)
where f(s) is a multiple of Z – s, but they also require an integer Z that cannot
be part of an integer solution of the Fermat equation, and, interestingly, all
of the examples I’ve found so far, including your example, involve integers X
and Z that are also members of Pythagorean triples. I see this as a basis for a conjecture that could become an important theorem if proven.)
Before proceeding with the proof for n = 3, there is
one other minor point I want to clear up: f(s)
is not equal to 3s^{2} as stated by a recent reviewer.
When f(Z) is divided by Z – s, R = f(s) = s^{2} + sX + X^{2},
not 3s^{2}. The only way f(s) can equal 3s^{2}, is for s = X;
in which case Z – s = Z – X, which clearly is not a factor of f(Z) = Z^{2}
+ XZ + X^{2}, for X and Z coprime. We can get
a remainder equal to 3s^{2} by dividing s^{2} + sX + X^{2}
by X – s, which yields X – s = Z – s →
X = Z and Y = 0, a trivial solution of the Fermat equation. However, f(s)
cannot equal zero for integer solutions of Fermat’s equation because s and X
are positive integers by definition. They are specific positive integer members
of a hypothetical integer solution of Fermat’s equation.
Proof
of FLT for n = 3:
For integer solutions of
the Fermat equation, the factor of Z^{3} – X^{3} represented by
f(Z) = Z^{2}
+ XZ + X^{2} is equal to A^{3}
and A = Z – s, A, Z and s integers. And, by Corollary II of the Division
Algorithm, f(Z) divided by Z – s produces a remainder equal to the integer f(s)
= s^{2} + sX + X^{2}.
Due to the fact that integers form a subset of the
real numbers, the integer polynomial f(Z) is a subset of the real number polynomial
f(z), and with X, Y and, dividing f(Z) by Z  s yields:
(Z^{2} + XZ + X^{2})/(Z – s) = Z + X + s + f(s)/(Z – s) →
(Z^{2} + XZ + X^{2}) = (Z + X + s)(Z
– s) + f(s) → f(s) = m(Z – s), m a positive integer.
By application of Fermat’s ‘Little’ Theorem, choosing Y
coprime with n = 3, ensures that f(Z) is an integer raised to the third power.
For a hypothetical primitive solution (X,Y,Z), f(Z) is equal to Z^{2} +
XZ + X^{2} = A^{3}, an integer factor of Y, and by inspection, Z^{2} + XZ + X^{2} is odd for all integer values of X and Z.
We may, therefore use Fermat’s
factorization method which says that for every odd integer N, there are two
relatively prime integers, a and b, such that N = a^{2} + b^{2}.
Since f(Z) = Z^{2} + XZ + X^{2} = (Z – s)^{3} = (Z –
s)(Z – s)^{2} = (a  b)(a + b) = (Z – s)(Z + s) →
(Z – s)^{2} = (Z + s).
With this equation, FLT
for n = 3 can be proved a number of ways. For example:
Z^{2} – 2sZ + s^{2} = Z + s → Z^{2}
– (2s + 1)Z^{ } + s^{2} –
s = 0. We can solve this equation for Z using the quadratic formula and get Z = [(2s + 1) + √2s]/2, a noninteger for all positive integer values
of s, proving FLT for n = 3. But this method of proof becomes problematic for n
> 3, since we only have a quadratic equation when n = 3. However, proof of FLT
for n = 3 can also be obtained by noticing that the equation implies that Z divides s^{2} – s = s(s – 1),
and s and Z must be coprime for Y and Z to be coprime, and s is a positive
integer < Z, so s(s – 1) cannot contain Z,
Similarly, since Z – s
is an integer factor of f(s), and f(s) = s^{2} + sX + X^{2}, an
odd integer, and f(s) = m(Z –s), applying Fermat’s
factorization method, we have:
f(s)
= m(Z – s) = (a + b)(a  b) = (Z +
s)(Z  s)
→ m = Z + s. But application of Fermat’s factorization to f(Z) above
gave us (Z – s)^{2} = Z + s.
These two results taken together imply that f(s) = (Z – s)^{3}. If f(s)
= (Z – s)^{3}, the equation Z^{2} +
XZ + X^{2} = Q(Z)(Z – s) + m(Z – s) = Q(Z)(Z – s) + (Z – s)^{3}
→ Q(Z) = (Z + X + s) contains (Z – s)^{2}. But (Z + X + s) contains (Z –
s)^{2 }and Z + s = (Z – s)^{2} implies X contains Z – s as an integer factor, which
contradicts coprime X, Y and Z, denying the existence of primitive solutions,
proving FLT for n = 3.
By combining these two
applications of Fermat’s factorization, we have a demonstration of the FLT65
method of proof in a form that can be extended to n = prime numbers > 3. To
see how this can be done, let’s also look at a proof for n = 5.
For n = 5, dividing f(Z) by Z  s yields:
(Z^{4} + XZ^{3} + X^{2 }Z^{2}
+ X^{3}Z + X^{4})/(Z
– s) = Q(Z) + f(s)/(Z – s), where
Q(Z) = Z^{3} + (s + X)^{ }Z^{2} + (s^{2} + X^{2}
)Z + s^{3} + X^{3}, and f(s) = (s^{4} + Xs^{3}
+ X^{2 }s^{2} + X^{3}s + X^{4}).
So we have f(Z) = (Z^{4} + XZ^{3} +
X^{2 }Z^{2} + X^{3}Z + X^{4}) = Q(Z)(Z – s) +
f(s).
And, since f(Z) = A^{3} = (Z – s)^{5},
f(s) must contain Z – s as a factor, so we have: f(s) = (s^{4} + Xs^{3}
+ X^{2 }s^{2} + X^{3}s + X^{4}) = m(Z – s), m a
positive integer.
By
inspection we see that f(Z) is an odd integer. So, applying Fermat’s factorization method, we also have:
f(Z) = (Z – s)^{5}
= (Z – s)(Z – s)^{4} = (a  b)(a + b) = (Z – s)(Z + s) → (Z – s)^{4} = (Z + s).
Similarly, since Z – s
is an integer factor of f(s), and f(s) is an odd integer, equal to m(Z –s),
applying Fermat’s factorization method
again, we have:
f(s)
= m(Z – s) = (a + b)(a  b) = (Z +
s)(Z  s)
→ m = Z + s. But application of Fermat’s factorization to f(Z) gave
us (Z – s)^{4} = Z + s.
These two results taken together, imply that f(s) = (Z – s)^{4} and from
the equation f(Z)
= Q(Z)(Z – s) + m(Z – s) = Q(Z)(Z – s) + (Z – s)^{3} , we see that Q(Z)
= {Z^{3} + (s + X)^{ }Z^{2} + (s^{2} + X^{2}
)Z + s^{3} + X^{3}} contains (Z – s)^{4}. Note:
determining what this means in terms of coprime X, Y and Z is a bit more
complicated than it was in the case n = 3, but it can be done as follows:
Since
f(Z) and Q(Z) contain Z – s as a common integer factor, the difference Q(Z)Z –
f(Z) must also contain Z – s as an
integer factor:
Q(Z)Z
= Z^{4} + (s + X)^{ }Z^{3} + (s^{2} + X^{2}
)Z^{2} + (s^{3} + X^{3})Z

F(Z) =  Z^{4}  XZ^{3}
 X^{2 }Z^{2}  X^{3}Z  X^{4}. Subtracting
term by term,
Q(Z)Z
– f(Z) = sZ^{3} + s^{2}Z^{2} + s^{3}Z – X^{4}
= {sZ(Z^{2} + sZ + s^{2}) – X^{4}}which contains Z – s.
Subtracting
sZ(Z – s)^{2} = sZ(Z^{2}  2sZ + s^{2}) from {sZ(Z^{2}
+ sZ + s^{2}) – X^{4}}→ X^{4} contains Z – s.
So
Q(Z) contains (Z – s)^{4 }and Z + s = (Z – s)^{4} implies that X^{4}
contains Z – s as an integer factor,
which contradicts coprime X, Y and Z, denying the existence of primitive solutions,
proving FLT for n = 5. The pattern we see emerging is: For n = p, any prime >2,
the fact that the remainder f(s) is nonzero implies X, Y and Z cannot be
coprime integers, proving FLT.
In conclusion: In my opinion, the proof of FLT for n = 3 and
n = 5, as presented above demonstrate the validity of FLT65. A noninteger
remainder for F(Z)/(Z  s), whether containing Z  s or not, insures no integer
solution for the Fermat equation.
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