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March 16th, 2019, 08:37 AM   #1
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one page proof of Fermat's Last Theorem

I am sending you an an invitation to see the one page proof of Fermat's Last Theorem. This proof uses Fermat's Right Triangle Theorem to prove the Theorem. Also included in the proof is the the diagram Fermat could not include in the margin of the book. It is an amazing proof

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March 16th, 2019, 09:01 AM   #2
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Math Focus: Wibbly wobbly timey-wimey stuff.
If it could be done by simple Trigonometry, it would have been found out loooooong ago. Ain't buyin' it.

-Dan
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March 16th, 2019, 10:17 AM   #3
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Originally Posted by topsquark View Post
If it could be done by simple Trigonometry, it would have been found out loooooong ago. Ain't buyin' it.
AND...IF a proof is ever found, what will have been gained?
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March 16th, 2019, 04:13 PM   #4
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For what it's worth, Fermat lived 30 years after writing that note in his copy of Diophantus but never mentioned FLT again. He must have realized he didn't have it else he'd have written up his marvelous demonstration.

ps -- Today I learned!

https://en.wikipedia.org/wiki/Fermat...iangle_theorem

Last edited by Maschke; March 16th, 2019 at 05:11 PM.
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March 16th, 2019, 08:29 PM   #5
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Originally Posted by michaelcweir View Post
I am sending you an an invitation to see the one page proof of Fermat's Last Theorem.
Question. Where in your proof do you use that n > 2? The way it is right now, your proof goes through for n = 2, but then it's wrong since there are FLT solutions for n = 2.
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March 17th, 2019, 11:28 AM   #6
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Quote:
Originally Posted by michaelcweir View Post
I am sending you an an invitation to see the one page proof of Fermat's Last Theorem. [/youtube]
Can you explain the meaning of $\alpha$?

I'm walking through your proof using the standard Pythagorean triple $3^2 + 4^2 = 5^2$. As far as I can tell, your proof shows this triple to be impossible, since you never used the fact that $n > 2$. You need to explain this else your proof is cooked.

So in this example, $x = 3$, $y = 4$, $z = 5$, and $n = 2$.

Now you say that $W = x^{\frac{n}{2}} - \alpha$. In this case we have $W = x - \alpha$ or $W = 3 - \alpha$.

So what is the meaning of $\alpha$? You write that $0 < \alpha < 1$ but I don't see what $\alpha$ is supposed to be. Do you mean to simply take $W = x$ in this case?

I hope you will take the time to respond to these concerns. To sum up, my questions are:

1) Where do you use that $n > 2$? Without that fact you have disproved the well-known existence of Pythagorean triples.

2) What is $\alpha$? In the case of $x = 3$, $y = 4$, $z = 5$, and $n = 2$, when you write that $W = 3 - \alpha$, what is $\alpha$?

ps -- Fermat's right triangle theorem asserts that not all of the labelled sides can be integers. But in the case at hand, if $0 < \alpha < 1$ then $W = 3 - \alpha$ is already not an integer, so of course you haven't proved anything.

On the other hand if $\alpha = 0$, then all the $\alpha$ terms in your squaring operations are zero and you have no contradiction.

I would say at this point that this basic example of $9 + 16 = 25$ cooks your proof. I do await your response.

Last edited by Maschke; March 17th, 2019 at 11:48 AM.
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March 17th, 2019, 04:02 PM   #7
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0<a<1

This is a good question.
A quick note about notation. W = Z (n/2),- a, not X(n/2)-a, as you state.

1) case n=2 is implicit in using a right triangle.For the case when n=2 , a = 0

2) Let's take your example for 3,4,5 for n=3.
3(3) + 4(3) = 27 + 64= 91 < 125 =5(3)

In this case, the cube root of 91 is approximately 4.49. 5(3) is greater than 4.49 (3).

While X(n) is an integer, X(n/2) may not be integer example 3(3) is 27, an integer. But 27(1/2) is 5.38 approximately.

It is relatively easy to show that Z must be smaller than Y +1, whether Z,Y are rational or irrational, and whatever n is greater than 2.

We have constructed 2 right triangles joined together where 2 sides out of the three are chosen to be integers. The square of the third side contains a term with Z(n/2) in th esquare. If that number is not an integer, then the square cannot be qan integer, a contradiction with Pythagoras's Thm. Therefor the assumption that Z is an integer must be wrong.
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March 17th, 2019, 04:03 PM   #8
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ps -- Typo. You have $W = 5 = \alpha$, not $3 - \alpha$. Same objection. If $\alpha$ is strictly positive then $W$ is already a non-integer so you've proved nothing. But if $\alpha$ is zero, then all the $\alpha$ terms in your $W^2$ and similar expressions are zero, so again you've proved nothing.

Your basic problem is that you haven't said what $\alpha$ is. And of course your "proof" would go through for $n = 2$, denying the existence of Pythagorean triples.
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March 18th, 2019, 09:51 AM   #9
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You really don't know what you are taking about, do you?

If you let a = 0 you have every case of a triplet for n=2.

(Z-1) (n) + (Z-1) (n) < Z(n) for every n>2

for n = 3, 3(3) + 4(3) < 5(3) for example

other examples are 4,5,6 and 5,6,7.

So the right triangle formed by with legs 3(3/2) and 4(3/2) must have a hypotenuse that is shorter than 5(3/2). a designates that difference.
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March 18th, 2019, 10:00 AM   #10
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sorry (Z-2)(n) + (Z-1)(n) < Z(n)
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