# Thread: Fermat's Theorem, Relativity, Quantum Field Theory

1. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Originally Posted by BuleriaChk
Once more into the breach - yet another shot at the algebra in the PDF. I really think it is ok now. (the Proof is still valid)

As an exercise, set a = b = 1 in the Expansion

for a model of the basis set (1,1) in two dimensions...
What do you mean STILL valid? It never was valid and still isn't!

2. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

From the perspective of Quantum mechanics, Planck's constant is a model of the photon energy necessary to eject an electron from a flat surface. It is wave-length independent, so the surface is not a model of an atom. Atomic physics models the atom itself, in which case differences in photon energies ("colors") are detectable in the hyperfine structure, in which photons only have spin polarization, but not "rest" masses of their own (That is, the mass is modeled as interactive spin polarization; equal and opposite via the null vector characterization, and so relevant to the Binomial Theorem representation).

The relativistic circle is the limit as the interaction -> 0 for a single photon, where the energy is modeled as the final state of particle creation for a single photon of invariant wavelength (or "period")). This photon energy can be multiplied for a model of non-interacting bosons, and the context for probability is Bose-Einstein statistics.

The fine structure constant (as opposed to the hyperfine) models the interaction between charged particles. This is modeled by orbital angular momentum, which changes as electrons change shells within the atom; the neutrino is equivalent to the binding energy or the electron within (e.g.) a hydrogen atom, where h is an approximation that ignores the curved geometry of the atom but does introduce the continuum as the limit of . This introduces the concept of charge (as independent of photon polarization), so the probabilities are modeled as Bessel functions via the Schrodinger equation to model the changes in state for different configurations of the nucleus (as opposed to n(n+1) from a flat surface), resulting in the Periodic Table and classical (non-relativistic) quantum mechanics. The inclusion of spin is then the interaction between the electron spin and the photon polarization.

Fermi-Dirac statistics model this with invariant electron charge/mass ratio. It gets complicated - that's why they got the big bucks...

Nevertheless, the relativistic unit circle and the binomial theorem is the foundation of any mathematical model that includes interactions, expressible by comparing metrics - including gravity. It there are no interactions, there are only individual "affine" particles (ok, integers) and it is impossible to apply any relationship (except imaginary) because there is no common origin for any metric (all that exists is "counting" mentally, which was Wittgenstein's point in "Remarks on the Foundations of Mathematics"; he just hadn't made the mathematical jump to relativity and the Binomial Theorem.

The application of probabilistic analysis inn classical quantum mechanics (i.e., no hyperfine splitting) removes the states of the photon from the context (via the conjugate transpose of the wave equation relative to h, so the only observable is the difference between the initial and final states independent of Planck's constant (and thus light energy))

There is more to this story (neutrons, gravity), but I don't have the space to write it here...

3. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Pythagorean Triples

Pythagorean Triples exist because all arithmetic operations are carried out within the confines of each coefficient number line of the individual vectors.

The point is that the arithmetic operation of multiplication (interaction) of terms within the coefficients of each vector (e.g. ab) does not apply to Pythagorean Triples, even though the it appears that way in a triangle diagram; the equation for Pythagorean Triples, because , and because the products with exponents only are calculated independently for each vector.

(In Fermat's theorem, that means the terms and are the only terms that are non-interactive, but all the other terms in the Binomial expansion are products of a and b).

Pythagorean triples then exist because all arithmetic operations are carried out within the coefficients of their respective vectors, with no interactive operations between coefficients of the vectors as specific instances of counting.

For example,

Pythagorean Triples are the reason that Fermat specified n>2 in his theorem.

(From the perspective of STR, it means that light-on-light interactions are not included in the analysis.)

4. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

THis is mathematics, Relaitivty and quantum mechanics is wholy irrelevant to the field.

5. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

For example, for Pythagorean Triples, , where the factor of is a result of the metric of the relativistic unit circle being one dimension () for positive integers.

6. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Originally Posted by BuleriaChk
Pythagorean Triples

Pythagorean Triples exist because all arithmetic operations are carried out within the confines of each coefficient number line of the individual vectors.

The point is that the arithmetic operation of multiplication (interaction) of terms within the coefficients of each vector (e.g. ab) does not apply to Pythagorean Triples, even though the it appears that way in a triangle diagram; the equation for Pythagorean Triples, because , and because the products with exponents only are calculated independently for each vector.

(In Fermat's theorem, that means the terms and are the only terms that are non-interactive, but all the other terms in the Binomial expansion are products of a and b).

Pythagorean triples then exist because all arithmetic operations are carried out within the coefficients of their respective vectors, with no interactive operations between coefficients of the vectors as specific instances of counting.

For example,

Pythagorean Triples are the reason that Fermat specified n>2 in his theorem.

(From the perspective of STR, it means that light-on-light interactions are not included in the analysis.)
Your entire argument is bogus, because we know that there are numbers c, a, and b where .

In fact, for any positive integer n, and any a and any b, there is always a c where .

7. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Originally Posted by grapes
Your entire argument is bogus, because we know that there are numbers c, a, and b where .

In fact, for any positive integer n, and any a and any b, there is always a c where .
I edited post#15.

You keep braying "bogus" like a donkey without any idea WTF what you are talking about.

There may be a number c, but it is not an integer unless it is a member of a Pythagorean Triple if a and b are integers. All three must be positive integers (and n>2) for Fermat's theorem to apply.

The Binomial Theorem is valid even for positive real numbers that are not integers, if there is scaling (i.e. multiplicative products between independent numbers (a,b) or (x, y) or (c,v) STR provides a linear scaling (ct,vt') between the numbers... which emphasizes their roles as vectors (hence, a "space-time" diagram with orthogonal axes..) That's why the coefficients of vectors are called scalars; the multiplicative product ab (or ct) "scales" a in terms of b, so it is a relation between metrics.

The equation is the equation of a circle, but it could also be a relation between the areas of three circles or three squares .... But there are no interactive terms such as xy in the equation.

There is no metric scaling because the metric is the same on a single number line (in a single dimension), and the values on each side of the equality refer to the same unique integer always.

(One can eliminate the interactive terms in two dimensions by which adds and subtracts an imaginary interaction from the equation)... This is possible because the dot product applies, so that (there is no scaling since the axes are orthogonal).

One cannot eliminate terms like in the Binomial Expansion by that method for n > 2, p < n, q < n so rem(a,b,n)>0. (Some of the terms will be imaginary and/or negative in rem(a,b,n) because of the different degrees of a and b in the terms.)

8. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Originally Posted by BuleriaChk
I edited post#15.

You keep braying "bogus" like a donkey without any idea WTF what you are talking about.

There may be a number c, but it is not an integer unless it is a member of a Pythagorean Triple if a and b are integers. All three must be positive integers (and n>2) for Fermat's theorem to apply.

My point is that your so-called proofs do not even use the restriction that the numbers must be integers.

You just say that there are no interaction terms...or there are interaction terms. But if your logic held, that should be true whether or not the numbers were integers. Essentially you've "proven" that algebra doesn't work. Sad.
The Binomial Theorem is valid even for positive real numbers that are not integers, if there is scaling (i.e. multiplicative products between independent numbers (a,b) or (x, y) or (c,v) STR provides a linear scaling (ct,vt') between the numbers... which emphasizes their roles as vectors (hence, a "space-time" diagram with orthogonal axes..) That's why the coefficients of vectors are called scalars; the multiplicative product ab (or ct) "scales" a in terms of b, so it is a relation between metrics.

The equation is the equation of a circle, but it could also be a relation between the areas of three circles or three squares .... But there are no interactive terms such as xy in the equation.

There is no metric scaling because the metric is the same on a single number line (in a single dimension), and the values on each side of the equality refer to the same unique integer always.

(One can eliminate the interactive terms in two dimensions by which adds and subtracts an imaginary interaction from the equation)... This is possible because the dot product applies, so that (there is no scaling since the axes are orthogonal).

One cannot eliminate terms like in the Binomial Expansion by that method for n > 2, p < n, q < n so rem(a,b,n)>0. (Some of the terms will be imaginary and/or negative in rem(a,b,n) because of the different degrees of a and b in the terms.)

9. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Originally Posted by grapes
My point is that your so-called proofs do not even use the restriction that the numbers must be integers.

You just say that there are no interaction terms...or there are interaction terms. But if your logic held, that should be true whether or not the numbers were integers. Essentially you've "proven" that algebra doesn't work. Sad.
This is absurd. Of course, for Fermat's Theorem I restrict the numbers to positive integers. The Binomial Theorem removes this restriction for real numbers, but the integers are real numbers (included in the set of real numbers). The Binomial Theorem for real numbers was proved by Newton. If the triple is NOT Pythagorean, then Fermat's Theorem cannot apply if n=2 (the case where the Binomial Theorem is irrelevant if the interaction terms are eliminated by

My application of application of the Binomial Theorem simply means that c cannot be an INTEGER if a and b are INTEGERS since . The reason is because there are non vanishing products in the Binomial Expansion for the case n > 2 (real positive numbers or real positive integers that cannot be eliminated by complex numbers. c can, of course, be a positive real number that is NOT an integer for n > 2.

Of course, it would be helpful if you could clarify whether you are speaking in terms of one or two dimensions (or higher e.g. n > 2) (an important distinction since Descartes..)

10. ## Re: Fermat's Theorem, Relativity, Quantum Field Theory

Originally Posted by BuleriaChk
This is absurd. Of course, for Fermat's Theorem I restrict the numbers to positive integers. The Binomial Theorem removes this restriction for real numbers, but the integers are real numbers (included in the set of real numbers). The Binomial Theorem for real numbers was proved by Newton. If the triple is NOT Pythagorean, then Fermat's Theorem cannot apply if n=2 (the case where the Binomial Theorem is irrelevant if the interaction terms are eliminated by

My application of application of the Binomial Theorem simply means that c cannot be an INTEGER if a and b are INTEGERS since . The reason is because there are non vanishing products in the Binomial Expansion for the case n > 2 (real positive numbers or real positive integers that cannot be eliminated by complex numbers. [B]c can, of course, be a positive real number that is NOT an integer for n > 2.
My point exactly. The binomial theorem is valid for nonintegers as well, so when you use it to "show" that cannot equal , you are "showing" it for all numbers, not just integers.

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