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

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

>>>
http://en.wikipedia.org/wiki/Fermat's_Last_Theorem

People who see final picture of my evidence - think that it is very easy

At the end

1 000 000 USD prize for person who will solve similar equation .

I need small help I can offer 10 % of award details later .

IF 1 AND 2 not exist

we can't solve main equation

X^n +Y^n =Z^n (see line 3 page 1 ) !!!

THE END OF EVIDENCE

more details about BEAL AWARD 1 000 000 USD

Beal's Conjecture -- from Wolfram MathWorld

I'm looking for people who can take award with me ( I need small help
)

details >>> Fermat's last theorem

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

Yes Tesla. It's me again.

You may offer money to individual members via PM, but not on the open Forum.

-Dan

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

This is award fight proposition not solary proposition

I would like to cooperate with somone who like math

first info in my topic

I solved problem 360 years old unsolved !!! Mathematica

my 5 lines will be in all books physics also because Femat's equation = key to solve one general equation for physics

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

Some thoughts:

1) Why are you using $W^+$? By definition g and t belong to $\mathbb{Z} ^+$. As g must be positive and greater than 1 it would be much simpler to use use $\mathbb{Z}$.

2) You haven't proven that $\left ( g^n - 1 \right )^{1/n}$ is not rational (much less an integer) for some values of g and n. It isn't (as far as I know) but you need to provide proof of this. The proof shouldn't be that hard to come up with but it points out that you haven't been rigorous in presenting a full solution to the problem.

3) Did you even look at that wikipedia link you provided? I couldn't even get close to an understanding of the related theorems provided in the links.

You say that you are looking for someone to help you with finishing your proof. You'll need a good one because I don't think this is going anywhere any time soon. And just to mention "the 300 pound gorilla in the room"...You do realize that this problem was finally proven late last century (in the 90's?) after 400 years of a lot of effort? I sincerely doubt that a 10 line proof will ever be found. I'm not trying to be mean, I'm just saying to not get your hopes up on this derivation.

If it makes you feel any better I once came up with a "proof" of the four color theorem. But it wasn't a proof. Convincing? Yes, or at least I think so... but a convincing argument is not the same as a proven argument.

-Dan

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

Originally Posted by tesla2
This is award fight proposition not solary proposition

I would like to cooperate with somone who like math

first info in my topic

I solved problem 360 years old unsolved !!! Mathematica

my 5 lines will be in all books physics also because Femat's equation = key to solve one general equation for physics
We cross posted here. Let me address this also.
This is award fight proposition not solary proposition
I'm sorry but I don't understand this sentence. I'll take a stab at it: You seem to be saying that this is simply a part of the award that you will share and not a paid position. This is still a problem. No offers of monetary (or other) values will be tolerated on the open Forum.

I didn't see any science related comment over at PHF. Are you saying you found a link between FLT and Physics?

You are unlikely to find a proof of anything using Mathematica. Evidence, yes, but no proofs.

-Dan

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

Fermat's Last theorem was proved in 1994 by Andrew Wiles..

Wiles's proof of Fermat's Last Theorem - Wikipedia, the free encyclopedia

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

Andrew Wiles evidence 200 pages A4

My evidence 2 pages

"Why are you using ? By definition g and t belong to " . "As g must be positive and greater than 1 it would be much simpler to use use "

20 x 1/4 = 5

2) You haven't proven that is not rational (g^n - 1 )^1/n

please see ( page 2, line 1 ) please use g from condition 1

MAIN EQUATION ( page 1 line 3 and 4 ) CAN BE SOLVE ONLY IF CONDITION 1 !!! AND !!! CONDITION 2 from page 2 will be

if exist condition 1 =====> condition 2 not exist

functions can not be solve . Graphs will not cross ( are parallel )

g = W+ why ? z=g*x ..... x= 20 g= 1/4 z=5

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

There are several problems with this "proof":

1. Regarding whether $(g^n-1)^{1/n}$ is rational or not: I find it helpful to plug in some actual numbers and see how it works out. Nothing in this presentation precludes the use of n=2, so let's see what happens if x=3, y=4, z=5 and n=2. For these numbers you have t=4/3 and g = 5/3 At the bottom of page 1 you end up with:

$(g^n-1)^{1/n} = ((\frac 5 3)^2 -1)^{1/2} = \frac 4 3$

which is the value of t and is rational. So you still need to prove that for n>2 this is irrational.

2. On page 2 for condition 1 you have: $g = \sqrt[n]{(\frac p q ) ^n + 1}$. Please explain how you got that. Since g=p/q, your equation is equivalent to: $g=\sqrt[n]{g ^n + 1}$. Clearly this is incorrect. I think what it should actually be is: $g=\sqrt[n]{t ^n + 1}$.

3. You then define $m = \sqrt[n]{(\frac p q ) + 1}$, and plot the function $(\frac p q )^n + 1$ versus p/q, or stated another way: $(g^n+1)$ versus g, and get the upper curve. That's fine. But I don't understand what the the lower curve is supposed to be - what is it a plot of? It says m^n, but it's not. Please clarify that.

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

please look what is on page 2

condition 1 Y= ( g^n - 1 ) ^1/n * X

http://2.bp.blogspot.com/-sMulRUVcSv...0/CIMG2350.JPG

see what is on page 1 ( line 1 and line 4 )

Y= t*x , t= ( g^n - 1 ) ^1/n Z = g*x

4/3 * 30 =40 Y = N+

t:? You have right but not to the end t = p/q ( p,q are C+ and q can not be zero - it is W+ definition t is inside W+ )

http://4.bp.blogspot.com/-t3RQQN3-Pk...0/CIMG2349.JPG

...please see in my blog Next Evidence 1 000 000 USD PRIZE BEAL function G(A ) and F( A ) last drawing ?

Fermat's last theorem

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

Originally Posted by tesla2
t:? You have right but not to the end t = p/q
No - you defined g = p/q, not t=p/q. g and t are not the same thing.

Originally Posted by tesla2
...please see in my blog Next Evidence 1 000 000 USD PRIZE BEAL function G(A ) and F( A ) last drawing ?

Fermat's last theorem
A few errors in the third page as well:

1. In equation III where you divide A^(sx) by A^x the result should be A^(sx-x), not A^(sx-1). Same error when dividing A^gx by Ax; the result is A^(gx-x). With this correction equation IV becomes:

$\frac 1 {t^{sx}} + A^{(s-1)x} = \frac {n^{gx}}{t^{sx}} A^{(g-1)x}$

2. Since t and n are functions of A, the correct expression for functions G(A) and F(A) are:

$G(A) = (\frac A B)^{sx} + A^{(s-1)x}$

$F(A) = \frac {C^{gx}}{B^{sx}}A^{(s-1)x}$. Note that if s = 1, meaning that x = y, then F(A) is a constant.

These two functions will cross once, which is pretty obvious given that you started with $A^x + B^y = C^z$: the solution is quite simply $A = (C^z - B^y)^{1/x}$. Here's an example - given:

B=3
C=3
x=3
y=3
z=5

With these values at A = 6 you have G(A) = F(A) = 9. This is because $6^3 + 3^3 = 3^5$. But how does this prove Beal's Conjecture? What you need to show is that if G(A) and F(A) cross at an integer value for A then A, B, and C must all share a prime factor. You are nowhere near proving that.

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

Just to add... the attached figure are plots of G(A) and F(A) for two different scenarios. The first is using the parameters from my previous post of $A^3 + 3^3 = 3^5$ which yields A=6 for F(A) = G(A) . The second is for a different set of parameters for the equation $A^3 + 2^4 = 5^4$, the solution of which is $A=607^{1/3}$. So again, the trick is to to show under what conditions is A an integer.

Attachment 28647

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

Thank You that You found mistake

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

THANK YOU VERY MUCH

TO EXPLAIN MY MISTAKE I marked my mistake in my blog ( I found old page in basket )

I also add G(A) and F( A) description ( drawing and other my words not change ) please try study

A^x +B^y = C^z A,B,C N+ and >0
x,y,z N+ and >2

One more thank You --->

You elimintated calculation error not a logical error >>> Fermat's last theorem

good that we have this forum

Thank You BR Maciej

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

Line III we have two functions ( right ) and (left )

We can make two graphs and we can have three situation :

I zero common points ==> not exist A that can solve A^x +B^x =C^z
success !!! to get Beal Prize - parallel situation or (right ) graph is more wide
1- I showed two the same parallel graph situation ( d=e and t=n and g=n )
2- (right equation - graph ) is more wide (d>e and t>g>n )
II only one common point success !!! to get Beal Prize
I showed that we can have situation where exist only one solution ( one golden A ) that can solve equation A^x +B^x =C^z .Two graphs can cross only in one point
III all points are common ==> we have theorem that fit to all A
always exist "+1" ! we will never be able IDEAL cover two different graphs
THE END OF EVIDENCE

Post's Author : > SIMPLE PATENTS

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

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.

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.

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