A NEW LOOK AT FERMAT'S LAST THEOREM

A NEW ANALYSIS OF MY 1965 PROOF

A friend pointed out to me that I spend a lot of time and effort these days defending my 1965 proof of Fermat’s Last Theorem (FLT65) against hypotheses and propositions that are designed to show flaws in FLT65’s logic. My friend suggested that instead of being on the defensive, perhaps it is time for me to take a different approach. The only legitimate questions raised by critics seem to involve, in one way or another, questioning the legitimacy of the application of the division algorithm and corollary III to integers in FLT65. So let’s have a look at how the division algorithm is applied in FLT65.

Division is one of the four fundamental operations of mathematics, and the division algorithm describes the operation of division for polynomials (algebraic summations with multiple terms). The following are examples of algebraic polynomials:

f(z) = z³ – a³; g(y) = y² + by + c; h(x) = x⁴ – 2x² + 5x + 13; and q(X) = 12X

A general expression representing an algebraic polynomial in x, of degree n, is given by;

f(x) = axⁿ + bx^n-1 + cx^n-2 + …+ sx + k, where there can be any number of terms, and a, b, c. …s, and k are constants that can be either positive, negative or zero.

The first part of FLT65 shows that for any polynomial f(x) over the field of real numbers, there exist unique polynomials, q(x) and g(x), such that f(x)/g(x) = q(x) + r(x)/g(x). This is nothing more than a statement of the division algorithm for polynomials. I also provided proof of the three corollaries of the division algorithm used in the FLT65 proof, including corollary I, which says that if f(x) and g(x) contain a common factor, r(x) contains it also; corollary II, which says that the remainder, when f(x) is divided by z-a is f(a); and corollary III that says that a polynomial f(x) of degree greater than 1, is divisible by the polynomial x-a, if and only if, f(a)=0. FLT65 also contains proof that if FLT is true for n = p, when p is a prime number greater than 2, then it is true for all n.

In FLT65, if x, y and z are integers, then, for a comprehensive proof, it is sufficient that they are relatively prime integers. For FLT65, y is chosen as a variable that does not contain p, and f(z) is defined as the larger factor of the right-hand side of the equation below, the Fermat equation:

y^p = z^p – x^p = (z-x)(z^p-1 + z^p-2x + z^p-3x² +•••+ x^p-1)

It is shown that z–x, and f(z), as factors of y^p, can be considered to be relatively prime. Thus f(z) = A^p, and A is an integer factor of y.

It is noted in FLT65 that the division algorithm and corollaries hold when the terms and coefficients of the polynomials are integers because the integers are elements of the field of real numbers. Also note that there are no restrictions on the application of the division algorithm and its corollaries with regard to the values of the coefficients, or the number of terms in the polynomial to which they are applied. This means that the statement of the division algorithm, f(x)/g(x) = q(x) + r(x)/g(x), is true whether f(x) and g(x) are polynomials of many terms or single terms.

Applying the algorithm and corollaries to the polynomial f(z), and defining g(z) as z-a, we have:

(z^p-1 + z^p-2x + z^p-3x² +•••+ x^p-1)/(z-a) = q(z) + f(a)/(z-a). Multiplying through by z-a, we have:

(z^p-1 + z^p-2x + z^p-3x² +•••+ x^p-1) = q(z)(z-a) + f(a).

From this it is clear that the polynomial f(z) is factorable into the two polynomial factors q(z) and z-a, if f(a) = 0.

We also know that, if there is an integer solution to the equation z^p – x^p = y^p, then y^p is an integer containing the integer factor f(Z) = A^p. = (Z-a)^p. This gives us two equations:

1)         (z^p-1 + z^p-2x + z^p-3x² +•••+ x^p-1) = q(z)(z-a) + f(a), an algebraic polynomial equation and

2)         (Z^p-1 + Z^p-2X + Z^p-3X² +•••+ X^p-1) = q(Z)(Z-a) + f(a), the same equation with all integer terms. Equation #2 exists if there is an integer solution (X,Y,Z) for z^p – x^p = y^p.

For an integer solution, if one exists, equation #2 can be reduced to an equation of single integer terms:

3)         A^p = Q·A + R, which implies that the integer f(Z) = A^p is equal to the product of the two factors Q and A, if and only if R = 0.

Critics have suggested that there is no correspondence between the polynomial factors of equation #1 and the integer factors of equation #3, in which case, f(a) ≠0 does not necessarily imply that R ≠0, and the fact that the polynomial f(z) cannot contain the polynomial z-a as a factor, has no bearing on whether the integer that f(Z) reduces to can contain the integer value that (Z-a) reduces to, or not.

So the real question whose answer will resolve the disagreement about the validity of FLT65 is:

What is the nature of the relationship between the polynomial factors of equation #1 and the integer factors of equation #3?

In FLT65, Z-a is defined as A, a factor of Y^p, and f(Z) = A^p, and A is an integer if there is an integer solution. By inspection of equations #2 & 3 above, we see that there is a one-to-one relationship between the four expressions of the two equations. By direct substitution of single-integer values for X, Z and a, assuming an integer solution of the FLT equation, for any prime degree, p, the polynomial factor (Z^p-1 + Z^p-2X + Z^p-3X² +•••+ X^p-1) yields A^p, the polynomial factor q(Z) yields Q, (Z-a) yields A, and f(a) yields R. This means that R = f(a) ≠ 0, and we have a contradiction:

For there to be an integer solution to the FLT equation, Y^p =A^pB^p =f(Z)B^p → f(Z) =A^p and this in light of equation #3 implies R has to be zero. But because of the form of f(z), with all coefficients equal to unity except the constant term, which is zero, f(a), which yields R, cannot be zero. And thus the fact that f(a) ≠ 0 for equation #1, implies that R ≠ 0 in equation #3, and this contradiction, featured in FLT65, is sufficient to prove there are no integer solutions for the FLT equation.

Edward R. Close August 3, 2017