Tuesday, June 10, 2014
Fermat's Last Theorem 1965 Proof Validated
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 zp – xp = yp, (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) = Yp, where g(z)=z–X and f(z)=zp-1+zp-2X+zp-3X2+…+Xp-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) = Ap and g(z) = Bp, 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) = zp-1+zp-2X+zp-3X2+…+Xp-1 = (z-s)p = Ap.
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 Ap 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 Ap, as a factor of Y, is an integer.
Example: Let X = 2, Y = 3, and p = 3: solving the Fermat equation, z3 = 23 + 33 = 8 + 27 → z = (35)1/3, and f(z) = (35)2/3+2(35)1/3+4 = A3 → 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 Ap, where A is a specific integer factor of Y, if it does not contain the algebraic factor z-s?
Clearly, existence of an integer Ap, as a factor of Yp, 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 1st 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, Ap, and (2.) Integer solutions are a subset of real number solutions, implying by Corollary III that no integer values of X and s (X1 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)= Ap = (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 =Ap, with Y =AB, and the division Algorithm and corollaries apply to them. But f(z) is also equal to zp-1+zp-2X+zp-3X2+…+Xp-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)= zp-1+zp-2X+zp-3X2+…+Xp-1, let’s assume there are integer solutions for the Fermat equation so that A is an integer factor of Y. Let A= z-s1, then (z-s1)ϵ f(s1), where s1 is the first in a series of integers, si, such that z-s1= s1-s2 = s2-s3 = s3-s4 =… si-si+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 si, there is a unique si+1 such that si-si+1= A, an integer factor of Y. Since A, as an integer factor of Y, is a positive integer: z > s1 > s2 > s3 > s4 > … >si.
Note that the infinite series of integer polynomials z-s1, s1-s2, s2-s3, s3-s4, … is exhaustive, containing all possible integer representations of A for any hypothetical integer solution of the Fermat equation.
Dividing f(z) by z-s1 we have: f(z)/(z-s1)= q1(z)+f(s1)/(z-s1)= q1(z)+ f(s1)/(s1-s2). We may substitute s1-s2 for z-s1 here, because z-s1= s1-s2 = s2-s3= s3-s4 = … si-si+1.
Dividing f(s1) by (s1-s2), we have:
f(z)/(z-s1)= q1(z)+f(s1)/(s1-s2)= q1(z)+q2(z)+f(s2)/(s1-s2)
We may continue in this manner:
f(z)/(z-s1)= q1(z)+q2(z)+…+qp-2(s1)+…+qi(si)+ f(si)/(si-si+1), ad infinitum, dividing the successive remainders, f(si) by A= z-s1=… si-si+1, purging f(s) of all “hidden” factors equal to the integer A.
If z and the si are integers, dividing f(si) by A represented by increasingly smaller values of si-1 and si, 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 pth division, if the si are integers, as shown in the example below, exactly when this smallest f(si) 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-s1 =A → 13-s1 = 2 → s1=11, and z-s1 = s1-s2 = s2-s3=…=A→ s2=9, s3=7, s4=5, s5=3, s6=1, s7=-3, s8=-5, s9=-7, … ad infinitum.
For p=3, f(si) = (si)2+ (si)X+ X2. Evaluating the remainder f(si) for successive integer values of si, we have:
1.) f(s1) = (11)2+ (11)(7)+ 72 = 121+77+49=247
2.) f(s2) = (9)2+ (9)(7)+ (7)2 = 81+63+49=193
3.) f(s3) = (7)2+ (7)(7)+ (7)2 = 49+40+49=138
4.) f(s4) = (5)2+ (5)(7)+ (7)2 = 25+35+49=129
5.) f(s5) = (3)2+ (3)(7)+ (7)2 = 9+21+49=79
6.) f(s6) = (1)2+ (1)(7)+ (7)2 = 1+7+49=53
7.) f(s7) = (-3)2+ (-3)(7)+ (7)2 = 9+(-21)+49=37, the minimum value of f(si).
8.) f(s8) = (-5)2+ (-5)(7)+ (7)2 = 25+(-35)+49=39
9.) f(s9) = (-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 ϵ Y, A cannot contain p.
As we divide the remainder, f(si), repeatedly by (si-si+1), we see that the successive remainders are all of the form: f(si)= sip-1+sip-2X+sip-3X2+…+Xp-1. As long as si is positive, f(si) is obviously non-zero. But, if the si 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(si) is of the (p-1)th degree in si and X, for all prime exponents, p, f(si) will always be positive, even as si becomes increasingly negative as in the example above.
Returning to f(z)/(z-s1)= q1(z)+q2(z)+…+qp-2(s1)+…+qi(si)+ f(si)/(si-si+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-s1)=(z-s1)p-1=Ap-1, is an integer, and all the of terms on the right-hand side: the qi, f(s1) and s1-s1+1, are integers, and therefore, f(z)/(z-s1)=(z-s1)p-1 an integer, equals a sum of integers plus an integer fraction that is greater than zero. If z and si 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 ith division, to a sum of integers and an irreducible rational fraction. This contradiction is not obtained if z, si 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, s1, s2, and s3 are integers, i.e., FLT is falsified, for any positive integer value of z, f(z) = z2+zX+X2, a positive integer, and the divisor, z-s and thus the quotients q1(z), q2(s1) and q3(s2), obtained as we divide successively by (z-s1), (s1-s2) and (s2-s3), are positive integers:
f(z)/(z-s1)= q1(z)+f(s1)/(s1-s2)= q1(z)+q2(s1)+f(s2)/(s2-s3)=q1(z)+q2(s1)+q3(s2)+f(s3)/(s3-s4)
and the remainder f(s1)= s12+s1X+X2 by (s1-s2), we have s1+s2+X+f(s2)/(s1-s2)→ q2(s1)= s1+s2+X, and dividing the remainder f(s2)= s22+s2X+X2 by (s2-s3), we have s2+s3+X+f(s3)/(s2-s2) → q3(s2)= s2+s3+X. Substituting these values of the qi(si) into the equation for f(z)/(z-s1), we have:
f(z)/(z-s1)= z+s1+X+ s1+s2+X+ s2+s3+X+ f(s3)/(s3-s4)= z+2s1+2s2+3X+ f(s3)/(s3-s4).
Clearly, the left-hand side of the equation, f(z)/(z-s1), is equal to an integer if z and s1 are integers and f(z) = (z-s)3 (necessary conditions for integer solutions for p=3). Since f(z)=A3, z, s1, 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 si are positive. This looks like the signature of an irrational number, just what you would expect if z is irrational. Since f(si) can never equal zero, even when si becomes negative, there are no integers, z and si, 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(si), being of degree p-1, will always have p terms, and when the si 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, q1(z) is the (p-2)th quotient, while for p=5, q3(z) is the (p-2)th quotient, and the remainder is a function of z and is divided by z-s1 until the (p-1)th quotient. Thereafter, divisors are (si-si+1), with i= 1, 2, 3,… ad infinitum. For integer solutions, however, the qi 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(si) occurs when the first si becomes negative. If additional divisions are carried out, the negative si does not make f(si) 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(si), are shown to contain no general equation integer factors equal to A = z-s1= s1-s2 = s2-s3=…(si-si+1). Because of the contradiction in the general equation, f(z)/(z-s1)=q1(z)+q2(z)+…+qp-2(s1)+…+qi(si)+f(si)/(si-si+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-s1, validates FLT65.