Proof of Fermat's Last Theorem for n = 3 and n = 5 using the logic of my 1965 proof

This proof that there is no co-prime integer solution (X,Y,Z) of Xⁿ + Yⁿ = Zⁿ 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^n-1+XZ^n-2 + X²Z^n-3 +…+ X^n-2Z + X^n-1 = Aⁿ = (Z – s)ⁿ with A, X, Z and a co-prime 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) = Z² + XZ + X²  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 s² + sX + X², 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 3s² as stated by a recent reviewer. When f(Z) is divided by Z – s, R = f(s) = s² + sX + X², not 3s². The only way f(s) can equal 3s², is for s = X; in which case Z – s = Z – X, which clearly is not a factor of f(Z) = Z² + XZ + X², for X and Z co-prime. We can get a remainder equal to 3s² by dividing s² + sX + X² 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 Z³ – X³ represented by f(Z) = Z² + XZ + X² is equal to A³ 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) = s² + sX + X².

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:

(Z² + XZ + X²)/(Z – s) = Z + X + s + f(s)/(Z – s) →

(Z² + XZ + X²) = (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 co-prime 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 Z² + XZ + X² = A³, an integer factor of Y, and by inspection,  Z² + XZ + X² 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 = a² + b². Since f(Z) = Z² + XZ + X² = (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:

Z² – 2sZ + s² = Z + s → Z² – (2s + 1)Z  + s² – s = 0. We can solve this equation for Z using the quadratic formula and get Z = [(2s + 1) + √2s]/2, a non-integer 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 s² – s = s(s – 1), and s and Z must be co-prime for Y and Z to be co-prime, 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) = s² + sX + X², 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 Z² + XZ + X² = 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 co-prime 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:

(Z⁴ + XZ³ + X² Z² + X³Z + X⁴)/(Z – s) = Q(Z) + f(s)/(Z – s), where Q(Z) = Z³ + (s + X) Z² + (s² + X² )Z + s³ + X³, and f(s) = (s⁴ + Xs³ + X² s² + X³s + X⁴).

So we have f(Z) = (Z⁴ + XZ³ + X² Z² + X³Z + X⁴) = Q(Z)(Z – s) + f(s).

And, since f(Z) = A³ = (Z – s)⁵, f(s) must contain Z – s as a factor, so we have: f(s) = (s⁴ + Xs³ + X² s² + X³s + X⁴) = 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) = {Z³ + (s + X) Z² + (s² + X² )Z + s³ + X³} contains (Z – s)⁴. Note: determining what this means in terms of co-prime 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 = Z⁴ + (s + X) Z³ + (s² + X² )Z² + (s³ + X³)Z

- F(Z) = - Z⁴        - XZ³            - X² Z²             - X³Z - X⁴. Subtracting term by term,

Q(Z)Z – f(Z) = sZ³ + s²Z² + s³Z – X⁴ = {sZ(Z² + sZ + s²) – X⁴}which contains Z – s.

Subtracting sZ(Z – s)² = sZ(Z² - 2sZ + s²) from {sZ(Z² + sZ + s²) – X⁴}→ X⁴ contains Z – s.

So Q(Z) contains (Z – s)⁴ and Z + s = (Z – s)⁴ implies that X⁴ contains Z – s as an integer factor, which contradicts co-prime 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 non-zero implies X, Y and Z cannot be co-prime 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 non-integer remainder for F(Z)/(Z - s), whether containing Z - s or not, insures no integer solution for the Fermat equation.