New Short proof FLT Is there a catch?

[onylokram]

It may be easier on http://www.onylokram.wordpress.com



December 2008

Heads you win, tails I lose!



Yes, one of you is bound to win 1000 US dollars.



If you are the first one to find a valid error in the proof of FLT given in the margin of this page, O. K. Kram would give you an award of $1000.

Or, if you are the first one to approve and validate this proof, the symbolic award of 1000$ will go to you.



Terms & Conditions:

1) Any errors in, or Objections to this proof of FLT must be sent to me directly by e-mail [onylo_kram@yahoo.com]. Any validations or Approval of the proof, to qualify for the award must be sent in the individual’s name but through the Math faculty of a recognized University with full contact details of the authorized person at the University.

2) The award offer is made in good faith, with the objective of making progress. O. K. Kram will not be held responsible or liable for any damages of any kind whatsoever arising, directly or indirectly, out of this program. O. K. Kram will be the sole judge to determine the validity of an Entry and its qualification for the award in every respect – technical, administrative and interpretation/arbitration.

3) The award offer will expire on January 26 2009.

4) In case of validation/approval, 30 days will be allowed for anyone to come forward with an objection before the Award is confirmed and delivered. In the same manner, in case of Objection or Error found in the proof, 30 days will be allowed for any rectification or completion of the proof before an Award is confirmed and delivered.

5) This proof and comments have been published amongst others, in the Blog, Forums and Web site. These may not be copied or communicated directly or indirectly, in part, or as a whole, to a third party without the prior written consent of O. K Kram and upon specifying the ownership of Intellectual property held by Onylo K Kram. Under those terms, the complete text may be put on your Math Forum, if you judge appropriate.



…………..the foot note below implies that all rational numbers are objects in one dimension with n=1 in x2/n.x(1-2/n) and any change in ‘n’ is not visible in Arithmetic due to the understood dimension ‘n’ as always inexistent. That’s why FLT leads to a Truly Marvelous finding in Physics.




If X, Y, Z & N are integral solutions to

xn + yn = zn where n>2, then





( ZN/YN)1/2 will have to be of the form +/- b.c1/2 where ‘b’ and ‘c’ are rational values and never of the ‘complex’ form +/- a. d1/2 +/- b .c1/2 where a0 and d is rational(may or may not be a perfect square of a rational value) and c is definitely not a perfect square of a rational value..Statement (1)



The equation xn + yn = zn can be put as:

x(n-2).x2= x(n-2) . (x2+k2) 2 _ (x2-k2) 2…Eq2

(2k)² (2k)²

‘where k= x .

xn/2/(zn/2-yn/2)

i.e. zn will always equal x(n-2) . (x2+k2) 2

(2k)²

and yn will always equal x(n-2) . (x2-k2) 2

(2k)²

Hence ( zn/yn )1/2 = x2+k2

x2-k2

Let’s say X, ZN and YN are rational:

‘k’ by definition is of the complex form

+/- a. d1/2 +/- b .c1/2 (rational value divided by complex form)

Hence k2 will be of the complex form

+/- a +/- b .c1/2 with one rational part + a and one irrational part +/- b. c1/2

Therefore = x2+k2

x2-k2 will definitely* be of complex form +/-a +/- b .c1/2 Statement(2)

which is in contradiction with Statement (1)

Hence X, Y, Z & N cannot be integers to satisfy the equality xn + yn = zn… Q. F. D.





*since if k2 = -x2 +/- b .c1/2 , the numerator will become of the non-complex form+/- b .c1/2 but the denominator will become -2x2 +/- b .c1/2 which is of the complex form. Other objections received and invalidated:

Objection 1 : If Zn/2 and Yn/2 are both rational i.e. Z and Y were both perfect squares of rational values, k and thus k2 will not be of the complex form. True, in that case swap X and Y in the above demo so that k= y . (Let’s call “xn/2/(zn/2-yn/2)” as Xg)

yn/2/(zn/2-xn/2)

and the impossibility will reappear with k of a complex form.

Objection 2: What if x is also a perfect square of a rational value in addition to y and z! In that case, x, y and z are all perfect squares say of x1, y1 and z1, the x12n + y12n = z12n i.e. valid for n an even integer which has already been proved to be impossible. And finally, the trickiest one:

Objection 3: One can generate infinite sets of x, y, z and n by taking x as an integer (say 5) raising it to any power n (say 5). Taking any value of yn say 75 (i.e. 16807), we get zn = 19932. In this example (Zn/Yn)1/2 will be of the form c1/2 and not of a complex form

+/- a +/- b .c1/2 in spite of x, yn,and zn being rational! The above example has no error yet it is in contradiction to what has been shown in the above Proof of FLT. That does not invalidate the above proof. It only brings up a valid & mysterious contradiction where we know there is a catch. It’s like saying “How come?” the proof has no error but we have a real example that contradicts the proof. Yes, there is a catch – deep inside the example! Before we dissect that ‘catch’, we can also point out a mysterious catch in the Shimura Taniyama’s logic applied to prove FLT. If you apply that logic fully, it should not be possible to have rational/integral solutions to x2 + y2 = z2 but we know that there are infinite sets feasible. How Come? This does not mean that Wile & Taylor’s proof of Shimura-Taniyama conjencture is wrong in any manner. It will be interesting to see that the answer to the above two mysterious catches is one and the same and is directly linked to the ‘Truly Marvelous’ finding of Fermat (which some mathematicians, like Grisha may be able to derive from the above proof). I am eager to share that finding with you once this initial proof is validated to your satisfaction after any new objections have been successfully answered. It flows from the above that Xgn = Xgn.(Xg + 1/Xg)2 _ Xgn.(Xg - 1/Xg)2 covers ALL sets of solutions for xn + yn = zn simply for 1< Xg <= (1+ 21/2) Reason being:

As Xg increases, the component 1/Xg decreases and that x and y are interchangeable in the equation. For a given Xg, the corresponding value of Yg will be (Xg + 1)/ (Xg - 1) which is interestingly a closed loop and returns the value Xg for (Yg + 1)/ (Yg - 1)

[Doraki]

Or, if you are the first one to approve and validate this proof, the symbolic award of 1000$ will go to you.

I approve of and validate this proof.


Just kidding.

[ffpower]

lol...
Ya évidemment une erreur,faut juste trouver ou..Cela dit j ai la flemme de lire cette demo..

[Imod]

Sans compter que je ne vois pas la raison pour laquelle on devrait supporter de l'english sur un site francophone ! Si ça continue j'appelle Super Dupond et là , fini le pudding et la conduite à gauche et bienvenue au camembert et au gros rouge qui tache !!!

Non mais :hum:

Imod

[Timothé Lefebvre]

C'est vrai que c'est difficilement acceptable ...

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