Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fermat Last Theorem: a^n+b^n<c^n

  1. Jul 8, 2008 #1
    Designations:
    a_(i) - the i-th digit from the end in the number a in the prime base n> 2.
    a_(i, j) - the number, comprised of the digits from i to j of the number a.
    9 – designation of digit n-1,
    8 – designation of digit n-2.

    Proof of FLT

    (01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.

    Let us lead the number u=a+b-c to the form:
    (02°) u = (n^k)(n^p-1)
    with the aid of the multiplication of the equality of 01° by the sufficiently large number d^n (which, as is known, in the prime base n exists), as a result of which the equality of 01° is converted into the equality
    (1°) a^n+b^n-c^n = 0, where c> a> b and a+b> c.

    Let us break all digits of the number u into three intervals:
    1) [p, q+1],
    2) [q, r+1],
    3) [r, 1],
    where p is greatest rank of the number c, q is the greatest rank of the significant part of the number u, r is greatest rank of the zero end of the number u.
    I.e. the number u = u_(p, q+1)n^q+u_(q, r+1)n^r+u_(r, 1) = /00… 00//99… 99//00… 00/.

    Since the end u_(r, 1) (=0) does not change from the multiplication of the number u on d and the number of a^n+b^n-c^n (as we will see more lately) is considerably less than zero, then the endings a_(r, 1), b_(r, 1), c_(r, 1) can be rejected as infinitesimal quanitiies in comparison with the number of a^n+b^n-c^n.

    And the now approximate values of the numbers a, b, c with their MAXIMUM value take the form:
    (2°)
    a=/99… 98//99… 99/+ 1,
    b=/00… 00//99… 99/+ 1,
    ==============
    c=/99… 99//99… 99/+ 1.

    Or:
    (3°)
    a=n^{p-r}-n^{q-r},
    b=n^{q-r},
    ========
    c=n^{p-r}.

    But with these values of the numbers a, b, c we have a equality:
    (4°) a+b = c, and therefore
    (5°) a^n+b^n-c^n<<0.

    It is easy to see that an increase (decrease) in the numbers a, b, c by 1 does not can change inequality sign of 5° (with the sufficiently large number d from 02°).

    It remains to add that the inequality of 5° remains true and for ANY positive numbers
    (6°)
    a=n^{pr} - n^{q- R} of+a',
    b=n^{q- R}+b' ,
    ========
    c=n^{pr}+b',

    with the condition of a'+b'- c'=0 (and with the sufficiently large number d from 02°).

    Thus, the expression 1° (or 01°) in FLT is inequality a^n+b^n-c^n<<0.

    FLT is proven.

    (On July 6, 2008. [Mezos])
     
  2. jcsd
  3. Jul 8, 2008 #2
    Mezos,

    Can you show me why (for all integers "u" and for all primes n>2) (there exists integers d, k, and p such that)

    u d^n = n^k ((3^p)-1)?

    DJ
     
  4. Jul 9, 2008 #3
    Let u=3^ku', where u'_(1)≠0.
    From u'+1 of the numbers 3^t/u' (t=1, 2... u'+1) exist two numbers with the equal remainders. A difference in these two numbers is divided into u', i.e., to u' is divided a certain number 3^p-1.
     
  5. Jul 13, 2008 #4
    Re: Fermat Last Theorem: Final text.

    Proof FLT. Final text.

    Designations:
    a_(i) – the i-th digit from the end in the number a in the prime base n>2.
    a_(p, r) – the number, comprised of the digits of the number a from the rank p to the rank r.
    9 – designation of digit n-1.

    (01°) let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.

    (02°) Then it is easy to show, A+B>C>A>B>U>0, where U=A+B-C.

    Let us lead the number U (in the prime base n) to the form
    (03°) u =Ud=(n^k)(n^p-1) with the aid of the multiplication of the equality 01° by the appropriate number d^n (which, as is known, exists), as a result of which the equality of 01° is converted into the equality
    (1°) a^n+b^n-c^n = 0,
    (2°) where a+b>c>a>b>u>0,
    (3°) u=a+b-c, where the number u has r digits (but the highest digit of the number c let us designate by letter p) and, let us note, that
    (4°) the number u is EVEN.

    Let us try to find the digital solution of equation (3°) lower the r-digit endings are isolated from the high-range digits by two skew lines. Let
    (5°)
    a*=…?v//x…,
    +
    b*=…?v//y…,
    -
    c*=…?w//z…
    ==========
    u = …00//9…

    It is easy to see that the sum of the numbers
    1. x+y is as the minimum equal to 9
    2. Consequently, x is as the minimum equal to 9/2.
    3. Consequently, z is as the minimum equal to 9/2.
    4. Consequently, x+y is as the minimum equal to 9+9/2…
    And so on to the eventual result:
    5. x=y=z=9.
    6. Consequently, v is as the minimum equal to 1.
    However, on this the calculation of numbers to end does not can, since in this case u_(r+1) is not equal to zero. In order to remove this contradiction, is necessary number w_ (r+1) to make equal to one.
    Of course other numbers, except these two, can enter into the discharges of higher than the r-th; however, when they satisfy the requirements: their sum is even.

    But then the sum of all numbers, which figure in the Fermat’s equation, will be ODD. Consequently, and the number of a^n+b^n-c^n will be ODD, which is impossible

    Thus, FLT is proven completely and indisputably.
     
  6. Jul 13, 2008 #5

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Re: Fermat Last Theorem: Final text.

    Simple enough. In fact, it's easy to show that [itex]8|(a+b-c).[/itex]

    I have no idea what you're doing here. x, y, and z are the low-digits of a, b, and c respectively (that is, their residues mod some number) but why restrictions 2 and 3? And what does "9 – designation of digit n-1" mean, after all?
     
  7. Jul 13, 2008 #6
    Re: Fermat Last Theorem: Final text.

    1. a+b-c is even - IF we assume that a^n+b^n-c^n and IF u=c.
    OR: a+b-c is even FROM a^n+b^n-c^n=0, but a+b-c is odd FOR a^n+b^n-c^n=0!

    2. 9 is symbol of the digit n-1 in prime base n (n=/10!!!).
     
  8. Jul 13, 2008 #7

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    None of that made sense. And yu still haven't addressed why you know that both x and z are at least "9" / 2.
     
  9. Jul 14, 2008 #8
    - Since the numbers x, y, z are first in the numbers a, b, c.
    +
    Cf.:
    (2°) where a+b>c>a>b>u>0,...

    +++++++++++++

    Replacement of text after 5°:

    6°. We analogously find that also all the remaining digits - up to the rank r (where r - number of final zeros in the number u) are only nine. And now it becomes obvious (taking into account that the number d can be as as desired to large), that a^n+b^n-c^n> > 0.
     
  10. Jul 14, 2008 #9
    Alas, the idea of proof proved to be hopeless. Long time out.
    Thanks
     
  11. Jul 14, 2008 #10

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Ah, sorry to hear that. Can't say I'm surprised; it's quite a hard problem.
     
  12. Jul 15, 2008 #11
    Is it possible to prove that there is no elementary proof for FLT, or is it possible that Fermat himself did find a proof (as he claimed) which nobody since has been able to repeat?
     
  13. Jul 15, 2008 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What happened to Fermat was, almost certainly, what happens to mathematicians all the time- he thought he saw a general method of proving the theorem, then, when he tried to work out the details found that it did not generalize. Fermat did publish a proof that there are no integer solutions for n= 3. That was probably the method he was thinking would work for all n. But later he published a proof for n= 4 using a completely different method.
     
  14. Jul 16, 2008 #13
    I revealed surprising phenomenon in the Fermat's equality (and I prepare note):
    there is located the second (moreover fractional!) the solution with the same first (from p to q+1) digits.
     
  15. Jul 17, 2008 #14

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    I don't know what that means. Are you claiming that there are solutions with [itex]a,b,c\in\mathbb{Q}[/itex] or just [itex]a,b,c\in\mathbb{R}[/itex]? The first is equivalent to solutions with [itex]a,b,c\in\mathbb{Z},[/itex] of course.
     
  16. Jul 17, 2008 #15
    Equality-double in FLT. (Passed moment of the rejected idea)

    Equality-double in FLT. (Passed moment of the rejected idea)

    Beginning is identical:

    Designations:
    a_(i) – the i-th digit from the end in the number a in the prime base n> 2.
    a_(p, r) – the number, comprised of the digits from p to r of the number a.
    9 – designation of digit n-1,
    8 – designation of digit n-2.

    Proof of FLT

    (01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.
    (02°) Then, as it is easy to show, A+B>C>A>B>U>0, where U=A+B-C.

    Let us lead (in prime base n) the number u=a+b-c to the form:
    (03°) u =Ud= [n^(p-r)-1)](n^r)
    with the aid of the multiplication of the equality of 01° by the corresponding number d^n (which, as is known, in the prime base n exists), as a result of which the equality of 01° is converted into the equality
    (1°) a^n+b^n-c^n = 0, where a+b>c>a>b>u>0 and u=a+b-c.

    Let us break all digits of the number u into three intervals:
    1) [p, q+1],
    2) [q, r+1],
    3) [r, 1], where
    p is greatest rank of the number c,
    q is the greatest rank of the significant part of the number u,
    r is greatest rank of the zero ending of the number u.
    I.e. the number
    (04°) u = u_(p, q+1)n^q+u_(q, r+1)n^r+u_(r, 1) = /00… 00//99… 99//00… 00/.



    Let us examine first the case a_(R, 1)+b_(R, 1)-c_(R, 1)=0

    Is attained the following process:

    Let us replace all digits in the numbers a, b, c before (i.e. it is less) the range p by zero. We will obtain the numbers a*_(p, p), b*_(p, p), c*_(p, p)
    (here it is below for simplicity of formulas zero endings equal for all numbers we do not write! Either it is convertible all numbers into n-mal fractions with the integer parts of a_p, b_p, c_p)
    …and value of function F (a*, b*, C*), or F_ (p).

    Then let us enter in the numbers a*, b*, c* the digits of the previous (i.e. the smaller) range p -1 and let us calculate value F_ (p, p -1) with the values of the numbers a_(p, p -1), b_(p, p -1), c_(p, p -1).

    Then let us enter in the numbers a*, b*, c* the digits of range p -2 let us calculate value F_(p, p-2) with the values of the numbers a_(p, p-2), b_(p, p-2), c_(p, p-2).

    And so on - up to the restoration of all digits of the numbers a, b, c.

    ***
    Taking into account the formula of 4°, it is easy to see that the value
    (5°) F_(p, q+1)= a_(p, q+1)^n+b_(p, q+1)^n-c_(p, q+1)^n<0.

    (6°) In the interval from (p, q) to (p, r+1) function F(a*, b*, C*) it will be strictly increasing.

    (7°) A in the interval from (p, r+1) to (p, 1) function F (a*, b*, C*) it will be as a whole of that diminishing.

    ***
    Thus, function F (a*, b*, C*) in the closed interval [p, 1] in the first subinterval appears as a whole of that diminishing, in the second subinterval - strictly increasing, on the third - as a whole of that diminishing.

    But the hence it follows that extrapolated diagram of the function F (a*, b*, C*) from (a*_(p, p), b*_(p, p), c*_ (p, p)) to (a, b, c) has also THE FRACTIONAL solution (a' , b' , c'), moreover 0<a'<a,0<b'<b, 0<c'<c!!!

    There is above how to think over!

    Continuation follows.
     
  17. Jul 18, 2008 #16
    Re: Equality-double in FLT. (Passed moment of the rejected idea)

    This is confusing to me. Assume that n = 97, then would "9" = 96 and "8" = 95 in base n. If not then why are "9" and "8" so important in base n for all n?
     
    Last edited: Jul 18, 2008
  18. Jul 18, 2008 #17
    Re: Fermat Last Theorem: Final text.

    Per my understanding in base n = 97 "9" = 96. Thus since x+y-z = "/9" , x +y must be at a minimum = 96
    2. follows when you make the assumptiom that x>= y which is always a proper assumption since a and b are interchangable.
    3. is not at all clear since x = 49, y =47, z = 0 looks like a possibility to me. So as CRGreathouse pointed out, Victor, you need to address this.
     
    Last edited: Jul 18, 2008
  19. Jul 18, 2008 #18
    Re: Equality-double in FLT. (Passed moment of the rejected idea)

    Interpretation and the conclusion:
    Either the numbers a, b, c are not given to the canonical form
    or (a^n)_(1) = -a_(1), (b^n)_(1) = -b_(1), (c^n)_(1) = -c_(1),
    that contradicts Little Fermat's theorem.
     
  20. Jul 18, 2008 #19
    Re: Fermat Last Theorem: Final text.

    1. For the proof are important numbers 1, n-1 and n-2.

    2. For the ranks in the interval [q, r+1] and digits x, y, z of the numbers a, b, c is observed the equality x+y- z = 9, i.e., 96 (in the base 10).
    Exception can be only for the numbers of the rank of r+1: x+y- z = 8, i.e., 95.

    3. For the ranks in the intervals [p, q+1] and [r, 1] and digits x, y, z of the numbers a, b, c is observed the equality (x+y- z) _ (1) = 0, in particular (in base 97): (49+47-0) _ (1)≠0.
     
  21. Jul 18, 2008 #20
    Re: Fermat Last Theorem: Final text.

    Victor We all seem to find your statements somewhat confusing. Yet you are giving still more cryptic statements as you attempt to clarify that which we find confusing. Why not show a simple example of your proof using a simple and concrete example by applying your method to a proof for say n = 7 so at least some of us can get a better idea of what you are saying and doing?
     
    Last edited: Jul 18, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fermat Last Theorem: a^n+b^n<c^n
  1. Su(n) to sl(n,C) (Replies: 6)

  2. Inverse of b mod n? (Replies: 1)

  3. Planes in C^n (Replies: 14)

Loading...