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Math Help - Fermat's last theorem ( My own Evidence 5 lines one a4)

  1. #16
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    g must be W+

    Y= ( g^n - 1 )^1/n *X please back to first line my evidence t= ( g^n - 1 )^1/n t is W+ t*X = N+ ( 1/3 *30 = 10 ) ---- t=1/3 , X=30 Y = 10


    to have W+ and solve ( g^n - 1 )^1/n we must use this equation g= ( (p/q)^n +1 )^1/n

    after put g= ( (p/q)^n +1 )^1/n to equation ( g^n - 1 )^1/n we have Y= (p/q) * X p/q it is def. W+ , p,q are C+ and q not equal ZERO


    g= ( (p/q)^n +1 )^1/n / ^n


    we have g^n = (p/g)^n + 1

    we can not find g that will be able solve Y= ( g^n - 1 )^1/n *X ====> we can not find g that will solve main equation THE END OF EVIDENCE


    ************************************************** *********************

    ABOUT MAROSZ EVIDENCE BEAL PRIZE CONJECTURE ( new picture green points - thank You again that You help me find mistake )



    and words

    Line III we have two functions ( right ) and (left )
    We can make two graphs (right ) and (left ) :


    I zero common points ==> not exist A that can solve A^x +B^x =C^z
    success !!! to get Beal Prize

    parallel situation (LEFT) =(RIGHT) but exist +1 OR (right ) graph is more wide
    1- two the same parallel graph situation (d=e but not equal 1 and t=n and g=n ) graph can not cross
    2- (right equation - graph ) is more wide (d>e and t>g>n ) graph can not cross

    OR we have two lines situation( d=e=1)
    Two lines situation :
    1* Fermat's Last Theorem
    2* d=e =1 ( we have two lines )
    Equation III and point 1* and point 2* : 1+ (t/n)^x = (g/n)^x
    line III n^x + t^x =g^x Impossible to solve n,t,q = Natural +

    II only one common point success !!! to get Beal Prize
    we can have situation where exist only one A that can solve equation A^x +B^x =C^z .

    Two graphs can cross only in one point please imagine graphs shape if :

    e > d and g >n and t < n and g>n
    OR one line and one curve situation ( d=1 t=n and e > d and g>n)


    III all points are common ==> we have theorem that fit to all A ???
    We can not have got two ideal the same graphs (left ) curve and ( right ) curve are different
    We can't have two ideal the same graph ( left ) line and ( right ) line
    THE END OF EVIDENCE

    Author :
    I'm 32 years old Engineer and inventor

    my patents and vision - city can fly !!! > > SIMPLE PATENTS
    Last edited by tesla2; June 20th 2013 at 09:11 AM.
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  2. #17
    MHF Contributor ebaines's Avatar
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    Tesla2: you have not addressed the issues I raised in my previous post. So I ask you again - Please label the horizontal axis of your graph.

    And second, please explain what you mean by this:

    Quote Originally Posted by tesla2
    II only one common point success !!! to get Beal Prize we can have situation where exist only one A that can solve equation A^x +B^x =C^z .
    Beal is not about showing one possible value of A; it's about showing that for A, B and C to be integers A, B and C must all share a common factor.

    Also you have another error: the condition d = e leads to the equation A^x + B^y = Z^y, not  A^x + B^x = Z^y.
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  3. #18
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    This is more cleare explain Beal's problem

    BEAL's Prize they not request thar "z" can not be equal "y" ( please see start condition )

    ************************below TXT I copied from AMS page *************************

    Beal's Conjecture is related. If a, b, c, x, y, and z are all positive integers and x, y, z are greater than two, ( for me y=z can be !!!)

    a^x + b^y = c^z
    is only possible when a, b and c have a common prime factor.

    Read more: Beale Conjecture 1 Million Dollar Prize - Business Insider

    *****************************************

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  4. #19
    Forum Admin topsquark's Avatar
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    Tesla. I think ebaines has shown you a great deal of patience and persistence in trying to help you. It is clear to me that your ideas on FLT need a serious upgrade. I am going to ask you (nicely) to take a break and look at your ideas before you come back to this.

    -Dan
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  5. #20
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    Please confirm condition before start read my pdf
    x, y, z > 2 there is zero words that z=y or x=y ( x,y,z = variables )


    where A, B, C, x, y, and z are positive integers with x, y, z > 2,*** then A, B, and C have a common prime factor.
    >>> Beal's conjecture - Wikipedia, the free encyclopedia


    MAROSZ ( me ) *** - x, y, z > 2 it is very Important sentence ( FALSE !!! )

    Pdf version
    >>> https://docs.google.com/file/d/0B0BX...it?usp=sharing
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  6. #21
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    Quote Originally Posted by ebaines View Post
    Tesla2:

    1. Regarding your Fermat's Last Theorem page - back in post #8 I asked you to explain what the plot of the lower line is, and you still have not answered that. As I said before, it appears that the upper line is a plot of  g^n versus t, and the lower line is  g^n versus g. Specifically: please label the horizontal axis on your graph.

    Attachment 28651

    2. Regarding Beal's Conjecture: all you have shown is that under your condition 1 (d=e and g=t=n, so that z=y and A=B=C) there is no solution to A^x +A^y = A^y for positive values of A. This is obvious. Under condition 2 (d>e and t>g>n, hence y>z and B> C >A) you show that if B^y > C^z then there is no solution for the equation  A^x + B^y =C^Z. Again - this is obvious. It does not prove Beal's Conjecture.
    Y= (g^n -1)^1/n *x
    and
    Z= g*x

    (g^n -1)^1/n must be W+ and g must exist if not exist You will not solve

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  7. #22
    Forum Admin topsquark's Avatar
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    Enough.

    -Dan
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  8. #23
    Super Member Rebesques's Avatar
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    Re: Fermat's last theorem ( My own Evidence 5 lines one a4)

    One of the lesson to be had from this topic, is that there is a pretty good reason that Wiles' original proof was so long.
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