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Thread: Fermat's Theorem, revisited

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    Default Fermat's Theorem, The Binomial Theorem, and its relation to Physics

    To the reader: I finally organized my perspective in the following .pdf

    Fermat and Physics, (Updated 9/20/2016)

    Fermat Matrices (Added 12/06/2016

    Fermat's Theorem and the Lorentz Transform (added 12/13/2016)

    Quick and Dirty Proof of Fermat's Theorem (added 12/15/2016)

    which is my take on the relationship between Fermat's Theorem, the Binomial Theorem, and particle physics. As I produce more documents, I will provide links to them here. Previous work (including this one) are available on my website (in my signature below), on the page I devote to my work: My Take on Relativity

    My approach depends on the underlying assumption that the relation c = a + b makes sense for positive integers a,b, and c, and therefore the relation also makes sense.. (If neither of these relationships makes sense to you, then the proof will be invalid for you, but you probably should be institutionalized before you do any real damage to yourself or others...

    I show how the theorem is true (almost by inspection) by relating it to the Binomial Theorem, and the difference in the Binomial Expansion between , and the rest of the expansion, which I call rem(a,b).

    Note: My proof is also invalid if the Binomial Theorem is false. Good luck with proving that (although it seems to me at a superficial glance that Dr. Wiles is trying to prove it from the real numbers, similar to Einstein's unsuccessful quest for global covariance by applying tensor analysis and differential geometry to cosmology to be verified by our observations from the surface of the earth ....

    (Some in this thread have a lot of difficulty with more than one dimension and the subsequent relationships to curvature.)

    In the form of a vector space over the spaces of integers (a,b,;c) the independent integer sets {a} and {b} are related to the dependent (resultant) set {c} by the Binomial Expansion:

    ,

    where the outer product between independent (but interacting) sets of positive integers is represented for the additional dimension on the right hand side (a pseudo vector, since the resultant set now the resultant of the vectors in three dimensional vector space with each vector a unique power of n (or for rem(a,b) a sum of multiple powers of positive integers) .

    I then show that if a and b are independent positive integers (i.e., number lines) in a linear integer space (a,b), rem(a,b) must be positive for n > 3, thus proving Fermat's Theorem. This arises from the obsevation that only the positive cross products are relevant for the interaction terms in rem(a,b)

    (Note: In the case of the Binomial Expansion for (n = 2), c2 = a2 + b2 only for Pythagorean triplets, where the dot product between the legs of the right triangle eliminates the term represented by 2ab in the expansion (a + b)2 (triangle equality, conservation of particles). If the triangle is not Pythagorean, the case is represented by the full expansion: (a + b)2 = a2 + 2ab + b2 where the term 2ab represents the interaction between the elements, and for higher powers of n is analogous to the rem(a,b) in the full Binomial Expansion.

    I then show its relation to the Pauli Spin matrices (where the fermion is a model of an electron with intrinsic spin, which removes the degeneracy. This "spin" effect corresponds to the case v > c in the Lorentz transform (the off-diagonal elements in the matrix representation of the transform (I am working on a paper to specifically address this - stay tuned).

    If any of you would like to discuss this elsewhere, pm me and we'll start a conversation. (Let me know about any typos in the pdf; I think I have most of them out, but I'm never sure......
    -------------------------------------------------------------
    This list has degenerated into heated and incoherent noise noise, particularly from grapes (who did point out a couple of typos) and neverfly; who either haven't taken the time to read my proof (or, in the case of grapes, to understand it) or are ignorant of the basic concepts of linear independence.

    Let me know about any typos remaining. (There are some other threads about cross products forming the bases vectors of linear spaces of polynomials. Pm me if interested.

    For this thread, Start at Post #255, page 26, unless you want to observe a comedy in technical discourse - there will be enough after that page to keep you still laughing....

    Note to reader: (This thread was moved and edited without my permission; the physics thread has been appended at #255, page 26 where I clean up my discussion and relate Fermat's theorem, the Binomial theorem, and conserved particle counting in Physics to the forum.

    I consider this grossly unethical behavior on the part of the moderator (probably neverfly)

    I've been trying to delete my previous posts where my original post was attacked with total misunderstanding and ignorance immediately after, but have stopped, since it is not worth the effort. Just read my ignore list to understand where I'm coming from. Am I angry? You bet. These threads have been a lot of work on my part, and I have tried to answer questions honestly (but I admit, that I have been repelled by the ignorance and dthe purposeful misdirection of some of the questions).

    En Fin: I am under no illusion my work is original in the world; however, I haven't ever seen it before, and it is the result of my personal investigation (it all started with the "time dilation" equation.... those who know will understand

    I am publishing it in the hope that it may save some time for those who don't have the big bucks for a university, but have an interest in this (and related) topics and so enrich your lives..
    Last edited by BuleriaChk; 12-15-2016 at 09:16 PM.
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  2. #2
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    I just realized why FLT is correct, using the Binomial Theorem.

    Fermat's Last Theorem

    The question really boils down to "why does the equation

    (an + bn cn ) for a, b, c positive integers

    is true for n > 2, but works for n = 2.


    My take is that one uses complex numbers for the proof, but not for the theorem itself. The reason is that i2 can be thought of as the additive "destructor" of positive integers (e.g., a), since a2 + (ia)2 = a2 - a2 = 0 .

    Now consider the equation (a + b)2 = a2 + 2ab + b2. Then 2ab = a2 + b2 , so that the l.h.s. can be destroyed by multiplying each of the positive integers by i so that -2ab = 2(ia)(ib) = 2i2(ab) so that a2 + b2 +2ab - 2ab = 0 if the destructor is included, and the equation is valid (is a circle - since eliminating a and b is equivalent to eliminating the ab translation at the origin).

    But the Binomial Expansion for n > 2 will always have elements a3 and b3 (eg like a3b2) which means that a corresponding destructor cannot be found, since (e.g.) (ia)3 = i(i2)a= -ia which are not only negative, but also complex, and therefore cannot be eliminated.

    QED

    (This has very interesting consequences for STR and QFT, which is why I got to thinking about it. There are many more consequences; I'm absolutely sure they have been published somewhere or are covered in a course I missed somewhere, but I haven't seen it, which is why I'm posting this.) I think it is why Einstein failed in his goal to provide a set of laws that were globally covariant, which is what I'm working (slowly, painfully) towards.. (lots of indices...

    That said, so far I have only been wrong once, and that was when I thought I made a mistake...
    One of your mistakes:
    My take is that one uses complex numbers for the proof, but not for the theorem itself. The reason is that i2 can be thought of as the additive "destructor" of positive integers (e.g., a), since a2 + (ia)2 = a2 - a2 = 0 .

    Now consider the equation (a + b)2 = a2 + 2ab + b2. Then 2ab = a2 + b2 , so that the l.h.s. can be destroyed by multiplying each of the positive integers by i so that -2ab = 2(ia)(ib) = 2i2(ab) so that a2 + b2 +2ab - 2ab = 0 if the destructor is included, and the equation is valid
    That equation is never valid, for positive integers.

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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    I just realized why FLT is correct, using the Binomial Theorem.

    Fermat's Last Theorem

    The question really boils down to "why does the equation

    (an + bn cn ) for a, b, c positive integers

    is true for n > 2, but works for n = 2.


    My take is that one uses complex numbers for the proof, but not for the theorem itself. The reason is that i2 can be thought of as the additive "destructor" of positive integers (e.g., a), since a2 + (ia)2 = a2 - a2 = 0 .

    Now consider the equation (a + b)2 = a2 + 2ab + b2. Then 2ab = a2 + b2 , so that the l.h.s. can be destroyed by multiplying each of the positive integers by i so that -2ab = 2(ia)(ib) = 2i2(ab) so that a2 + b2 +2ab - 2ab = 0 if the destructor is included, and the equation is valid (is a circle - since eliminating a and b is equivalent to eliminating the ab translation at the origin).

    But the Binomial Expansion for n > 2 will always have elements a3 and b3 (eg like a3b2) which means that a corresponding destructor cannot be found, since (e.g.) (ia)3 = i(i2)a= -ia which are not only negative, but also complex, and therefore cannot be eliminated.

    QED

    (This has very interesting consequences for STR and QFT, which is why I got to thinking about it. There are many more consequences; I'm absolutely sure they have been published somewhere or are covered in a course I missed somewhere, but I haven't seen it, which is why I'm posting this.) I think it is why Einstein failed in his goal to provide a set of laws that were globally covariant, which is what I'm working (slowly, painfully) towards.. (lots of indices...

    That said, so far I have only been wrong once, and that was when I thought I made a mistake...
    Regarding grape's point, did you try any numerical examples for a and b? That equation is obviously never equal to 0.

    I don't see what you're saying when you just say 2ab=a^2+b^2. Let a=2 and b=3. It's clearly wrong.
    Last edited by mathnerd; 08-09-2016 at 06:22 PM.

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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by mathnerd View Post
    Regarding grape's point, did you try any numerical examples for a and b? That equation is obviously never equal to 0.

    I don't see what you're saying when you just say 2ab=a^2+b^2. Let a=2 and b=3. It's clearly wrong.
    Case n = 2, a=2, b=3, 2ab=12

    f(a,b) = a2 + 2ab + b2 = 4 + 12 + 9

    g(a,b) = 2(i2)(i3) = = 2(-1)2*3 = -12


    f((a,b) - g(a,b) = a2 + b2 + 12 - 12 = a2 + b2 = 13 which is a positive integer.

    QED

    =========================================

    Case n 3

    Example n = 3

    NOW try it with n > 3 for the binomial theorem for (a + b)3 and compare with terms in a3 + b3 - g3(a,b) , any positive integers a and b

    Hint: in the expansion, try sticking an i = in front of the a's and b's in g3(a,b) and see if you can come up with integers....

    (If you're confused about what g3(a,b) is, let me know...)

    Ok, I'll make it easier: (a + b)3 = a3 + 3a2b + 3 ab2 + b3

    g3(a,b) = 3a2b + 3 ab2 = 12 + 12 = 24 and cannot be eliminated by subtraction - that is, multiplication of a positive integer by (i2)

    g'(ia,ib) = 3(ia)2(b) + 3(ia)(ib)2 = 3(-4*3i) +3(2*-9i) = -36i -54i = -90i (if my arithmetic is correct; I'll know more after the beer wears off....)

    so g3(a,b) cannot be eliminated by subtracting -g3(a,b), and therefore (a + b)3 cannot be a positive integer. QED

    That is, there will always be odd terms in the expansion for higher orders n > 2, which means that there will always be factors of i (as opposed to i2= ()2 = -1 in the inner terms in the expansion, so the expansion cannot be eliminated by subtraction for n > 2, a and b positive integers.

    (Sigh) Oh Lord, but it's hard to be humble when you're perfect in every way....

    (Does anyone know the toll-free number of the Abel committee?)
    Last edited by BuleriaChk; 08-09-2016 at 09:10 PM.
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    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    a=2, b=3, 2ab=12

    f(a,b) = a2 + 2ab + b2 = 4 + 12 + 9


    g(a,b) = 2(i2)(i3) = = 2(-1)2*3 = -12


    f((a,b) - g(a,b) = a2 + b2 + 12 - 12 = a2 + b2 = 13 which is a positive integer.

    QED

    NOW try it with n > 3 for the binomial theorem for (a + b)3 and compare with terms in a3 + b3 - g3(a,b) , any positive integers a and b

    Hint: in the expansion, try sticking an i = in front of the a's and b's in g3(a,b) and see if you can come up with integers....

    (If you're confused about what g3(a,b) is, let me know...)

    Ok, I'll make it easier: (a + b)3 = a3 + 3a2b + 3 ab2 + b3

    g3(a,b) = 3a2b + 3 ab2 = 12 + 12 = 24 and cannot be eliminated by subtraction - that is, multiplication of a positive integer by (i2)

    g'(ia,ib) = 3(ia)2(b) + 3(ia)(ib)2 = -12i -12i = -24i

    so g3(a,b) cannot be eliminated by subtracting -g3(a,b), and therefore (a + b)3 cannot be a positive integer.

    That is, there will always be odd terms in the expansion for higher orders n > 2, which means that there will always be factors of i (as opposed to i2= ()2 = -1 in the inner terms in the expansion, so the expansion cannot be eliminated by subtraction for n > 2, a and b positive integers.

    (Sigh) Oh Lord, but it's hard to be humble when you're perfect in every way....

    (Does anyone know the toll-free number of the Abel committee?)
    QED? You're up next for the Abel prize right after John Gabriel, last I heard.

    So what with sticking an i in front? What does it show? If you do this for degree 4, you can get integers too.

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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by mathnerd View Post
    QED? You're up next for the Abel prize right after John Gabriel, last I heard.

    So what with sticking an i in front? What does it show? If you do this for degree 4, you can get integers too.
    Write out the expansion for n = 4 and see if you can eliminate the terms by using multiplicative complex numbers (which either put a (+ or -) 1 (axby term even) or an i (axby term odd) in front of each positive integer a or b).

    the term 4a3b in the expansion (a + b)4 contains an odd term: (ai)3bi = (ia3)bi= (a3b) so the term is positive, and the 6a2b2 term still a positive integer, but I would still need an additional term -6a2b2 to eliminate it so that c4 = (a + b)4. (i.e., the expansion terms vanish), which was my original "proof".

    So can I find positive integers c and such that c4 = d4 + 6a2b2 for some positive integer d on the l.h.s. ? So I would need

    c = So there may be a buried in there somewhere....


    That said, you may have a point about the even terms - there may also be an issue about the way the coefficients are generated, so I'll have to think about it a bit more.... but not too much more....

    (I'm beginning to see that my first proof was correct, actually. The only issue here was why n=2 works. 2ab is a positive integer, but not a power of either integer. so c = , so that may not make sense either.... )

    I'm starting to think that the proof may have something to do with hyper-areas (i.e., topology - factoring out the products of primes and then looking at what's left in each term), which is way above my pay grade....

    Well, ok, hold off on that call to the Abel committee - the line may be busy since they're probably calling John anyway...)

    Update: I think it is that the even terms on the r.h.s. will either be positive (the middle terms for even expansions, so can't use i to eliminate it, Not sure, though). Whatever the case, the issue is eliminating the expansion terms by subtraction and still retain the positive integer characterization for n > 2 to disprove the theorem. At least if the Binomial theorem is relevant... (which it may not be, after all)
    Last edited by BuleriaChk; 08-09-2016 at 10:36 PM.
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    The question really boils down to "why the equation

    (an + bn cn ) for a, b, c positive integers

    is true for n > 2, but works for n = 2.
    Wiles proof explains why that is the case.

    Now consider the equation (a + b)2 = a2 + 2ab + b2. Then 2ab = a2 + b2[/QUOTE]
    This is not true in general, you either need a negative sign on the last and the nIF AND ONLY IF
    otherwise that equality is false.


    Quote Originally Posted by BuleriaChk View Post
    so that the l.h.s. can be destroyed by multiplying each of the positive integers by i so that -2ab = 2(ia)(ib) = 2i2(ab) so that a2 + b2 +2ab - 2ab = 0 if the destructor is included, and the equation is valid (is a circle - since eliminating a and b is equivalent to eliminating the ab translation at the origin).
    WHAT origin? Going nuts on your fucking geometry crap again?

    Quote Originally Posted by BuleriaChk View Post
    But the Binomial Expansion for n > 2 will always have odd elements like a3 and b3 (eg like a3b2) which means that a corresponding destructor cannot be found, since (e.g.) (ia)3 = i(i2)a= -ia which are not only negative, but also complex, and therefore cannot be eliminated by subtracting the expansion, since at least one of the terms in the expansion will be imaginary for the terms containing both odd and even powers.

    QED
    No Q.E.D because you have not established the relation between your "destructor" and the solution. It's a non-sequitor here.

    Quote Originally Posted by BuleriaChk View Post
    That said, so far I have only been wrong once, and that was when I thought I made a mistake...
    Actually we've seen you being wrong many times so this statement is false.

    Your thought is at best cute but it does not in anyway established logically any form of connection.

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    Default Re: Fermat's Theorem, revisited

    Actually, I just realized that the case for n=2 is true because the set of integers for which this is true (a,b,c) are Pythagorean triples. If Wiles has beat me to the punch, I congratulate him....

    Since Pythagorean triples satisfy the triangle inequality, I think that higher order powers (and the real numbers) might be analogous to "non-linear elements" in the number system, similar to the concept of curvature in relation to the derivative. Not sure how that would work, but it would probably have something to do with traces and determinants in relation to prime numbers (conjecture on my part).

    That said, I'm out of here; those on my ignore list are now free to continue to masturbate here without my interruption; my best wishes for a positive and speedy climax....

    Update: I found Fermat's proof! It really is very simple, but I don't have time to write it here yet.

    Stay Tuned...
    Last edited by BuleriaChk; 08-10-2016 at 10:40 AM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  9. #9
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    That said, I'm out of here; those on my ignore list are now free to continue to masturbate here without my interruption; my best wishes for a positive and speedy climax....

    Update: I found Fermat's proof! It really is very simple, but I don't have time to write it here yet.

    Stay Tuned...
    OK, I won't bother with the rest of his errors

    Remember the last time he thought he found a short proof of Fermat's Last Theorem?

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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by grapes View Post
    OK, I won't bother with the rest of his errors

    Remember the last time he thought he found a short proof of Fermat's Last Theorem?
    What was the other 'proof'?

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