Tuesday, October 14, 2014

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 Zn-1+XZn-2 + X2Zn-3 ++ Xn-2Z + Xn-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 + sZ2 – (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.

Friday, September 19, 2014


“To everything there is a season, and a time to every purpose under heaven”
- Ecclesiastes 3:1

Yes, to everything there is a season; and I believe the time is ripe for a global quantum leap in human consciousness. Not just an increase in knowledge, it must be a triadic leap: a physical, mental and spiritual awakening. Anything less leads to serious problems: If enlightenment is just intellectual and physical, it fosters prideful ego and eventual disillusionment as dissolution of the physical body, i.e. physical death, approaches. Awareness of the triadic nature of reality, on the other hand, reveals a reality of which the observable physical universe is only a small part, and explains why there is something rather than nothing. Triadic enlightenment integrates the logic of science, the philosophy of religion and the expanded awareness of spirituality.
The number of people on this planet ready to make this leap to a comprehensive understanding of reality may finally be reaching critical mass, a necessary condition for the inevitable shift out of the limiting paradigmatic belief in mechanistic materialism that has characterized science, the limiting dogmatic beliefs that have characterized religions, and the unrealistic fantasies that have characterized “new-age” spiritualism. Gradually, a few individuals on the leading edge of the bell curve have begun to transcend the limitations of materialistic science, religious dogma and spiritual fantasy, into an expanded awareness. This book is the story of my personal journey from the confusion of fragmented belief systems to the certainty of triadic enlightenment.
An early version of this book was completed in 1997. It was intended to be a readable introduction to Transcendental Physics, the work I completed in 1996 and published in 1997. Presenting a new scientific paradigm, Transcendental Physics reversed the basic assumption of conventional science, the a priori assumption that consciousness is an epiphenomenon arising from the evolution of matter and energy, with the hypothesis that a primary form of consciousness is the ground from which all patterns of reality, including the physical universe, originate. Transcendental Physics, the book, contained specific, detailed interpretations of complex relativity and quantum mechanics experiments and introduced some new mathematical concepts developed for purpose of putting consciousness into the equations expressing the known Laws of Nature. The Search for Certainty manuscript, on the other hand, was written for readers with less technical training. It traced the development of the ideas behind Transcendental Physics as I had experienced them, and was thus at least partly autobiographical. The purpose was to present the paradigm-shifting ideas of Transcendental Physics in non-technical terms. Dr. David Stewart, who was familiar with and even part of many of the events reported in the 1997 version of the Search for Certainty, reviewed the manuscript, and had this to say:

“For the first time, the common basis for all sciences and all religions is revealed - not in vague philosophical terms, but in concrete ways you can understand and put into practice in your own life. You can take scriptures or the works of science and, by being selective, prove almost anything. But Ed Close, in this monumental work, did not do that. Taking into consideration the totality of physics, both modern and classical, dodging no part of it, Dr. Close has applied relentless and impeccable logic to produce an intellectual triumph of our time, a unified theory that makes science and religion one. This achievement has been claimed by others before, but always there was a flaw. There are no flaws in Close’s paradigm. The search for certainty ends here for those with the capability of comprehending what Close has done for us. Both scientists and theologians, centuries hence, will thank Dr. Close for what he has done for us. This is truly the first mathematically complete articulation of the relationship between human consciousness, divine consciousness, and material reality. This could well be the most important work of the 20th century. What Einstein and his contemporaries started a century ago, Close has finished. And what makes his achievement even more remarkable is that he was able to articulate it in terms the layman can understand.”

March, 1997
David Stewart, PhD, Geophysicist, Educator, and Author

Tuesday, September 16, 2014


I posted announcement of my autobiography on Facebook last night at about 3:50 am. Considering the late hour, there was a huge response. There were also some questions; so I am going to start posting answers and excepts from Search for Certainty here.

I am currently working on my autobiography. The title is “The Search for Certainty, a Personal Journey”, and it is about 90% finished. It tells my life’s story, from an idyllic childhood in the St. Francois Mountains, to solving problems that have stumped scientists for decades, to becoming an international speaker on a variety of topics including environmental engineering, hydro-geology, pure mathematics, relativity, quantum physics, consciousness studies, alternative healthcare, and longevity. It tells the story of how I discovered that I have an IQ of perhaps as much as 30 points higher than Einstein’s, and how I became a distinguished member of three of the most difficult high IQ and Creative Achievement Organizations to qualify for in the world. It relates my experiences on five continents, and unusual experiences in places like the Great Pyramid, the Middle East, Saudi Arabia, and the Puerto Rican Rain Forest. In the process, it gives my answers to the following questions: Who are we? Where did we come from? Where are we going? I will make excerpts available on my Transcendental Physics blog soon, and I am considering making pre-publication purchases available. Let me know if you are interested.

Memories of Other Lives, Fermat’s Last Theorem 
and Particle Physics

The title of this chapter may appear to be a mixture of unrelated subjects. But it is not. I have learned that, in my life, as it must also be in a true theory of everything, there are no unrelated parts. The resolution of the EPR paradox in quantum physics revealed that the consciousness of the observer is part of the observation, and in Transcendental Physics I proposed that consciousness, not matter, is the ground of all being. Consciousness is the only thing we experience directly, and that is why it is such a hard thing to define. We are immersed in consciousness, and we are consciousness. Thus, when we attempt to define consciousness, we are trying to describe the very essence of our own being. As individual sentient beings we contain limited amounts of consciousness, like bottles of water submerged in an ocean of water, except that, in our case, even the bottles are made of water. In this book I will show that the containers of our consciousness, the physical body and brain, which are composed of cells that are composed of molecules, composed of atoms, composed of sub-atomic particles, ultimately are composed of consciousness.

Wednesday, June 11, 2014

Understanding the math of Fermat

A few words about a subject that often gets a bad rap: MATHEMATICS. I believe that most people who “don’t like math” or say “it’s Greek to me” do so primarily because of inexcusably poor math teachers. Anyone who can speak and understand a language (like English, Spanish, etc.) can learn math. You learned your native tongue as a child, and that’s the best time to learn math, because learning math is just like learning a language, but it’s much, much simpler than French or German. An equation is just a simplified sentence: the left-hand side is the subject, the equal sign is the verb, and the right-hand side is the direct object. Mathematics properly understood is the language of the universe. This is why Albert Einstein could say “Ich will Gottes Gedanken zu wissen, alles andere sind nur unwichtige Einzelheit!” (I want to know God’s thoughts, all the rest is just unimportant detail!) Math is NOT just a left-brain thing. Music and poetry are mathematical, and real math is creative, especially if you realize that it works the way your mind (and the infinite mind of God) works!

Fermat's Last Theorem is like a poem or sonata, and it describes a beautiful part of reality. Translated into English, it says: While two perfect square areas can be added to produce a third perfect square area, no two perfect cubes, or whole numbers raised to any other  power, can be added to produce a whole number result of the same power. Expressed in part in the language of mathematics, this is:

A, B, C = (1, 2, 3, ...):  A2+B2=C 32+42=52 , 52+122=132 ...  (called Pythagorean triples)
But A3+B3C3 , A4+B4C4  ... An+BnCn  (where ≠ means “cannot be equal to”)

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==As2=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.

Wednesday, April 2, 2014


Notes on Fermat's Last Theorem
The Fermat equation is a Diophantine equation which we may try to solve for Z as I did in FLT65. As we saw before, when p = 3, we have a quadratic equation which we may solve using the quadratic formula, and we can show relatively easily that the solutions are non-integer, proving FLT for p = 3. Also I should mention that proof for the Fermat equation when n = 4, the only non-prime that cannot be reduced to a prime 3, was not included in FLT65 because, as you know, even prior to 1965, mathematicians had proved that there are no integer solutions for exponents up to a large number, certainly past nine, which is all that is necessary for application of FLT to the work Vernon Neppe and I are doing in TDVP.
So, just for the record, since FLT was proved for all cases from n = 3 to a very large number before Wiles’ proof in 1994, and even for n = 3 and 4 much before 1965, the question of the validity of my FLT65 proof has no bearing on the validity of TDVP, which is supported by some original mathematics developed by me, including an important application of FLT to elementary particle combination.

FLT states that there are no integer solutions for any p 3. If any of the solutions to the p – 1 degree Diophantine equation (1) are integers, FLT is falsified. If we could solve this Diophantine equation generalized for all p, we would see that none of the p – 1 solutions is an integer. But this is not a trivial problem. Number ten on David Hilbert’s famous list of important unsolved mathematical problems, published in 1900, was to find a general algorithm for solving Diophantine equations.

A few other historical notes might be of interest here: The great German mathematician Johann Carl Friedrich Gauss attempted to solve this problem 100 years before Hilbert, around 1800. To Gauss, FLT was just a possible conquest as a subset of Diophantine equations in the larger quest for more general ways to solve Diophantine equations. As part of this effort, he developed what is now known in number theory as modular algebra. Appearing almost incomprehensible and complex to the uninitiated, it is really quite simple: Developed specifically for application to Diophantine equations, and thus applying only to integers, it states that an integer A is said to be “congruent” to another integer, B, modulo N, when the difference of the two integers contains N as a factor.  It is written A B(Mod N), For example, 5 3(Mod 2) simply means 5 – 3 is divisible by 2. In case you might wonder why the identity symbol is used here, recall that it is used in geometry to indicate the congruency of angles, triangles and other geometric forms.

Why would Gauss, or anyone for that matter, want to express this simple mathematical relationship in such a convoluted way? The answer lies in the fact that trinomial Diophantine equations can be reduced to binomials: For example, notice that the Pythagorean Theorem equation X2 + Y2 = Z2 may be written Z2 – X2 = Y2 and factored to obtain (Z – X)(Z + X) = Y2. For the Pythagorean integer triple (Diophantine solution) X,Y,Z = 3,4,5: (Z – X) Y(Mod Y), and (Z2 – X2) Y(Mod Y2), i.e. (25 – 9) (Mod 16).  Gauss saw that this method could also be applied to the Fermat equation because it is in the same family of trinomial Diophantine polynomials as the Pythagorean Theorem equation.

By applying modular algebra to the first few prime integer exponents of Fermat’s equation, one can easily see why Gauss made the statement he did, when around 1805, he said:

“I confess that Fermat's Theorem as an isolated proposition has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of.”

The problem with modular algebra as an approach to solving Diophantine equations is that it only applies to binomial expressions or expressions whose terms can be re-grouped as sums of binomials. Proofs that there are no integer solutions for the Fermat equation when n = 3 and 4 using this method are relatively easy to obtain because the first factor is always a binomial and the second factor can be arranged to be treated as two binomials for n = 3, and 4 binomials when n = 4. But, when n 5 the task becomes increasingly more difficult, because the number of possible combinations of terms in the second factor increases exponentially from one prime to the next, so the task of checking all of them becomes onerous very quickly. The number of terms in the second factor of the equation for any prime, p, is equal to 2p -1, i.e. 16 for p = 5, 64 for p = 7, 1,024 for p = 11, etc. And there appears to be no pattern of combinations that would allow one to check for integer solutions by groups, just like there is no apparent pattern in the occurrence of primes. Gauss went on to prove FLT for n = 3 and 4, using what are called Gaussian numbers, complex numbers which when squared are integers, also known as complex conjugates: a + bi, which are, of course, binomials handily subject to modular algebra.
A number of professional mathematicians worked on the problem of trying to find a general algorithm for solving Diophantine equations until 1970, when Yuri Matiyasevich proved that that no such general algorithm is possible within the logic of mathematics as we know it. So there is no known straight-forward way to solve Diophantine equations for all p. I can tell you, however, that application of the calculus of dimensional distinctions (CoD), developed from G. Spencer Brown’s Laws of Form by yours truly, while it does not yield the numerical values of solutions, does indicate that the Fermat equation has no integer solutions. Of course, I really don’t expect you to accept this based on my statement alone, and I don’t expect you to want to learn a whole new system of mathematics to see the logic of it for yourself. So I will try to present the essence of the CoD argument in conventional concepts below.

Calculus of Dimensional Distinctions Visualization
First, think of algebraic polynomials as distinct geometric forms. Not in the conventional sense, where x, y and z are plotted on Cartesian coordinates; instead, think of the exponent of a term as an indicator of its dimensionality. Thus terms like X, AY, (a + b)Z, etc. are linear, terms like X2, XY, aY2, etc. are planar, terms like X3, XY2, XYZ, etc. are three dimensional, and terms with variables raised to the fourth power are four-dimensional, etc. The value of X, Y, etc. in each case is a multiple of some common unit. In the case of Diophantine equations, the unit is simply the unitary integer, 1. In this scheme of things, the degree of a polynomial indicates the dimensionality of the form it describes. An unrestricted Diophantine expression is, like the ring of integers, closed but infinite. If we set a polynomial equal to a finite constant, and limit its dimensionality, it is a closed and finite distinction, like a bubble.

The Fermat expression, Zp - Xp = Y1p, where Y1 is an integer constant and p = 3, is a closed distinction that can be visualized as a spherical bubble with radius equal to Y1. And f(Z) = Z2 + XZ + Z2 is a planar form, like an ellipse cut out of a plane. The other factor, g(Z) = Z – X, can be visualized as a line. Both factors are polynomials, and their dimensionometric forms, like that of their parent polynomial, are closed and finite, limited by the value of Y1. The ellipse, with major axis smaller than the diameter of the sphere and minor axis smaller than the radius of the sphere, is enclosed within the sphere. The line, of length equal to or shorter than the minor axis of the ellipse, is enclosed within the ellipse. Now visualize a three-dimensional integer grid originating from the center of the sphere. Each node of the 3D grid is located exactly one unit from each of the six nodes nearest to it. If there are any integer solutions with Y = Y1, (X,Z) = (X1,Z1), they occur where one point on the line and two points on the ellipse coincide with three points on the surface of the sphere, and then only if those three points coincide with nodes of the integer grid. It is not too hard to see, even in this simple visualization, that this is highly unlikely, if not an outright impossibility.

We can extend this visualization to all values of p, at least conceptually, by observing that for any prime, the factorization is the same, with f(Z) a hyper-ellipsoid of degree p -1 enclosed within the hyper-sphere of the Fermat Diophantine equation with Y = Y1, and g(Z) = Z – X is a line enclosed within the ellipsoid. A hyper-dimensional integer grid is still possible to visualize because it turns out that this p-dimensional hyperspace is still Euclidean, and the Euclidean theorem for division still applies!

Sunday, February 16, 2014


The human predicament is a Self-imposed enigma, born of circular reasoning: It is literally only a self-indulgent quirk of ego that leads serious minded people to the erroneous belief that the only thing that can actually be proved is unprovable, and that the illusion of that belief is provable!

The only thing that you can actually prove is the existence of consciousness. This is because consciousness is the only thing that you experience directly. All other knowledge is obtained very indirectly through the severely limited filter of the physical senses. The illusion of separate existence, on the other hand, is totally based on the fallacious belief that reality can exist without consciousness. This is the very basis of the current scientific paradigm, and yet it is totally un-scientific: A scientific hypothesis, by definition, must be open to proof or disproof. Thus the hypothesis that reality exists, or has ever existed, apart from consciousness, is literally un-provable, because no reality is observable without the existence of a conscious observer.

Once you allow yourself to think outside the self-imposed box of the ego-based illusion of separate existence, the clues of quantum physics and relativity become quite obvious: time and space, mass and energy, mind and matter, are not what you think they are! Their apparent existence as elements of a reality independent of consciousness depends upon the actual existence of your consciousness, and its willingness to indulge in the illusion of confinement to a separate existence.

It is, however, within your power, as a spark of the Reality of Primary Consciousness, to free yourself from identification with the illusion of a separate physical existence. It is time to grasp the key to the human dilemma; find the door in your cage, open it and allow the Eagle of your Spirit to escape the self-imposed confinement of material imprisonment and soar into the Infinite Reality beyond. The new sensation of impending freedom may be frightening, but there is nothing to fear but fear itself. The Way-Shower is always there, waiting for you in whatever form your heart requires, just outside the door, extending a helping hand.
The existence of a reality separate from consciousness is forever unprovable. On the other hand, the existence of Primary Consciousness, sometimes called God, is scientifically provable, once science is freed of the self-imposed constraints of materialism, and directly provable for each and every one of us through the direct experience of Infinity, once your consciousness is freed from the self-imposed illusion of identification with an imaginary ego. The place and time of your Enlightenment and Realization of Who you really are is very real and always available in the Here and Now. It only requires your willingness to drop the seemingly safe, comfortable and intoxicating illusion of ego confinement.

There is no available, or even possible, proof that life and consciousness have emerged from random arrangements of debris flying away from a great explosion. There are, however, many ways to prove the opposite, i.e. that the physical universe is a finite manifestation of Primary Consciousness. We can now prove scientifically that finite reality is an ever-expanding small part of an amazingly complex and beautiful Self-referential Reality, a conscious expression of the awe-inspiring grandeur of a glorious infinite existence, a blissful Reality without beginning or end. 

The idea that the reality we experience is part of an infinite conscious Reality is, of course, not new. What is new, is that we have proved it by putting consciousness into the equations of science for the very first time. See articles on the TRUE units of mass, energy and consciousness, and the e-book  "Reality Begins with Consciousness" on wwwBrainVoyage.com.