flt

  1. M

    Fermat's Last Theorem FLT

    Are there any resources which describe FLT in a very tangible way which will motivate students to be interested in this subject?
  2. M

    Six functions related to FLT

    Since Fermat's last theorem has been proved, can it be that with powers greater than 2, the following functions can't all be simultaneously equal to zero with odd powers and all but one of them similarly with even powers? The functions are A^n Mod(B + C), B^n Mod(A - B), C^n Mod (A - B), (B^n...
  3. M

    A Novel attempt to find a simple proof of F.L.T

    A NOVEL ATTEMPT TO FIND FERMAT’S PROOF I have realized something that seems to have not been considered before, in the attempts to find a simple proof of Fermat’s “LAST” theorem and which could lead experts finding a simple proof. My attempt is based on the fact that; unlike the power...
  4. C

    FLT from a system of infinite linear equations

    FLT from a system of infinite linear equations As told from the Power properties they can be written as Sum: $$A^n= \sum_{x=1}^{A} M_n$$ $$B^n= \sum_{x=1}^{B} M_n$$ $$C^n= \sum_{x=1}^{C} M_n$$ Where: $$M_n=[x^n-(x-1)^n]$$ Than FLT state: $$C^n=A^n+B^n$$ Can be rewritten...
  5. M

    An extension to FLT

    A THEORETICAL EXTENSION TO FLT 1. The smallest sum of a set of four different positive integers A B C D that are related by A = INT(( Bp + Cp )1/p +0.5) and D = INT(( Ap - Bp - Cp )1/p + 0.5), A B and C being relatively prime, IS WHEN B - C = 1 2. The value of A is an expression in any...
  6. W

    Vaguely related to FLT...

    It's well known that no integer solution exists for A^3 + B^3 = C^3. But is anything known about (A^3)*(I^2) + (B^3)*(J^2) = (C^3)*(K^2) where I,J,K are also integers? Is there a way of finding solution(s) to an equation of this form?
  7. C

    Still on FLT END...

    (EDIT 1) Hi, Sorry to bore all you ...still on FLT END... I hope someone wanna still follow me on my complicate modulus solution... I several time poste the fact that FLT imply: (1a) C^n=2 A^n + Delta (1b) C^n=2 B^n + Delta This is due to the simple facat that: C^n= \sum_{1}^{C}Mn...
  8. C

    The end of FLT in math

    Hi friends: FLT "symmetric" condition: (0a) C^n = 2A^n + \sum_{A+1}^{B} Mn (0b) C^n = 2B^n - \sum_{A+1}^{B} Mn With $Mn=(X^n-(X-1)^n)$ Calling $Delta= \sum_{A+1}^{B}Mn $ It's also clear that deriving the (0a) and the (0b) we have the same derivate: y'= 2nX^{(n-1)} So: C^3=...
  9. C

    FLT n=3 means A divisible by (C-B) ?

    FLT n=3 means A divisible by C-B ? FLT rewritten with Sum is: A^3 = \sum_{X=B+1}^{C} (3X^2-3X+1) or A^3 = \sum_{X=1}^{C-B} (3(X+B)^2-3(X+B)+1) or A^3 = \sum_{X=1}^{C-B} (3X^2+6BX+3B^2-3X-3B +1) or A^3 = \sum_{X=1}^{C-B} ((3X^2-3X +1) +6BX-3B +3B^2) or A^3 = (C-B)^3 + 3B...
  10. C

    FLT (end of the story fo all n)

    I'm not so happy I've not received any comments on FLT for n=2 and n=3... So I post here what happen for all n just using words: - I already explain why I use X instead of i,m,n or else as index of the Sum: is because I wanna stick what I'm doing on a X,Y plane, and show via Natural or...
  11. C

    FLT (n=2, n=3) the correct end of the story:

    I hope this time all works fine: Original link deleted - see later post(s). The conclusion is: mixed product, after FLT application, from n>2 leaves at least a grain of sand in the Binomial Develope Gears. For n=3 the Rest is: 6ABC=0 The PDF if much longer since explain my Complicate...
  12. C

    FLT (hope the end of the story...)

    I post lot on, but I hope this time all thinks will be at the right place: We all know Fermat Theorem C^n<>A^n+B^n c1- for A,B,C integers, c2- for A,B,C coprimes, c3- for n>=3. Knowing Mr. Wiles give us a proof (that many people can't understand) The goal is to find a simple proof...
  13. U

    § flt §

    An Elementary Proof of Fermat's 'last theorem' {Notation: '<>' - 'not equal'. '¤=' - 'congruent'.} The argument depends on the following: If a ¤= b (mod r) and c ¤= d (mod r) then a +/- c ¤= b +/- d (mod r). If a ¤= b (mod r) and c ¤= d (mod r) then ac ¤= bd (mod r). If...
  14. M

    what's up with FlT?

    So much was written and discussed here about Fermats Last theorem. But no one talks about Fermats little theorem! What do we know about pseudoprimes? What do we know about their distribution and characteristics? Say if we test for Fermat pseudoprimes to base 2 we get: A001567 [341, 561, 645...
  15. M

    A Visual "Proof" of FLT

    The attachment provides convincing visual evidence in favor of FLT, buy not a 100% proof. Perhaps it could lead someone to a simple 100% proof. Another attachment to follow.
  16. C

    FLT and Beal

    Remembering that a power of an integer or a rational "A", can be expressed as: A^n = \sum_{1/k}^{A} Mnk with n>=2 and: Mnk = (x^n-(x-1)^n)/k^n FLT (example n=3) can be reduced to the following StepSum step 1/k equation: \sum_{x=1/k}^{(C-B)/B} 3(x+1)^2/k -3(x+1)/k^2 +1/k^3...
  17. M

    Proof of FLT

    I'd like to present one of the versions (the last one) of my proof of FLT. It would be great if there is somebody brave enough to examine it in hope to reveal fatal flaws. I will be grateful even if he or she succeeds though not surprised if no comments appear at all. This is not an easy prey...
  18. J

    Proving FLT with Congruences?

    New member, happy to be here. Greetings to everyone, including Greathouse and our friend Stefano. I think I've got just such a proof ready. Yet there are rumors that it can't be done with congruences. Some sort of algebraic structure that always balances the congruences, perhaps? I figure...
  19. C

    FLT as parametric equation

    FLT for n=3 is equal to say: (-2 B^3 k^3+3 B^2 k^3-3 B^2 k^2-B k^3+3 B k^2-2 B k+2 C^3 k^3+3 C^2 k^3-3 C^2 k^2+C k^3-3 C k^2+2 C k) - (2 A k-3 A k^2-3 A^2 k^2+A k^3+3 A^2 k^3+2 A^3 k^3)<> 0 for A,B,C,k INTEGERS It comes from my step sum: \sum_{X=1/k}^{A}{( 3X^2/k...
  20. C

    FLT proof

    Sorry I re-post cleaning as I can, hopeing someone agree with me that this is the FLT proof or what he, probably, call infinite descent (or he is looking for): Example for n=3 but nothing change for n>3 A^3 = C^3-B^3 From FLT reduced to a sum: \sum_{x =1}^{A} {(3x^2 -3x +1)} =...