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April 18th, 2019, 09:51 PM   #11
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Outside the fact that all your formulas are incorrect, as I pointed above, your drawings are not correct either: from what I see, your red triangles seem to keep the same angles as the original one (but with a slight rotation). Either you keep the same angles but with the larger base, your side is not $x^{\frac{n}{2}}$ anymore, either you keep the same side and your angles do not follow anymore.

Also, you say $z<y+1$ but your drawing shows $z^{\frac{n}{2}}<(y^n+1)^{\frac{1}{2}}$, or $z^n<y^n+1$, not the same.
By the way, $z^{\frac{n}{2}}=(y^n+x^n)^{\frac{1}{2}}<(y^n+1)^{ \frac{1}{2}}$ means $x<1$. Is it?

Last edited by skipjack; April 26th, 2019 at 11:06 AM.
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April 18th, 2019, 10:35 PM   #12
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@michaelcweir, ps -- Re my technical question in post #8, I see what you're doing. You keep dropping a leg of length $x^\frac{n}{2}$ at whatever angle makes the new right angle, and then for example the other leg of the second triangle is $\sqrt{y^\frac{n}{2} - x^\frac{n}{2}}$. Perfectly sensible. Got it. The earlier isolating of the $x$-term is a red herring.

But there's no proof of FLT in there. You simply have a whole bunch of triangles with irrational sides. Doesn't mean anything.
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April 26th, 2019, 10:32 AM   #13
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The big idea behind this proof is that you can construct a right triangle based on on the equation

X^n + Y^n = Z^n

As you correctly stated, the next iteration of the triangles uses the smaller Y leg of previous triangle as
the hypotenuse of the new one. The triangle is completed with a new shorter Y leg than the hypotenuse.
(The new triangles is also a right triangle.)

As you increase the number of iterations, the Y leg gets shorter and shorter. As the number of iterations goes to infinity the Y leg approaches length Y, but never gets there.

At some point,, the length of Y leg will be less than (Y^n +1 )^1/n. At this point, the process of introducing new triangles stops. Now a new series of triangles is constructed starting with Y leg
(Y^n + 1)^1/n , X^n/2, calculated hypotenuse leg ( (Y^n + 1)^2/n + X^n)^n/2 ) This leg is longer than the Y leg of the next hypotenuse leg of the descending triangles.

AS we back up each step on the descending triangles there is a corresponding hypotenuse of the ascending (Y^n +1 ) that is longer than the descending hypotenuse ( call it the Z leg). Eventually when you get back to the triangle with hypotenuse of length Z^n/2. The hypotenuse of triangle with hypotenuse (Y^n +1)^n/2 is longer than that. Thus Z is less than Y+1.

If z is less than Y+1, it is also less than any other integer greater than Y +1. This then proves FLT
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April 26th, 2019, 10:38 AM   #14
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The big idea behind this proof is that you can construct a right triangle based on on the equation

X^n + Y^n = Z^n

As you correctly stated, the next iteration of the triangles uses the smaller Y leg of previous triangle as
the hypotenuse of the new one. The triangle is completed with a new shorter Y leg than the hypotenuse.
(The new triangles is also a right triangle.)

As you increase the number of iterations, the Y leg gets shorter and shorter. As the number of iterations goes to infinity the Y leg approaches length Y, but never gets there.

At some point,, the length of Y leg will be less than (Y^n +1 )^1/n. At this point, the process of introducing new triangles stops. Now a new series of triangles is constructed starting with Y leg
(Y^n + 1)^1/n , X^n/2, calculated hypotenuse leg ( (Y^n + 1)^2/n + X^n)^n/2 ) This leg is longer than the Y leg of the next hypotenuse leg of the descending triangles.

AS we back up each step on the descending triangles there is a corresponding hypotenuse of the ascending (Y^n +1 ) that is longer than the descending hypotenuse ( call it the Z leg). Eventually when you get back to the triangle with hypotenuse of length Z^n/2. The hypotenuse of triangle with hypotenuse (Y^n +1)^n/2 is longer than that. Thus Z is less than Y+1.

If z is less than Y+1, it is also less than any other integer greater than Y +1. This then proves FLT
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April 26th, 2019, 10:38 AM   #15
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Quote:
Originally Posted by michaelcweir View Post
The big idea behind this proof is that you can construct a right triangle based on on the equation

X^n + Y^n = Z^n
Once again you're ignoring the question I've posed several times.

You claim that e/f > 1. But for n = 10 you get a right triangle in which e/f = 1. What do you say to that?

Quote:
Originally Posted by michaelcweir View Post

At some point,, the length of Y leg will be less than (Y^n +1 )^1/n. At this point, the process of introducing new triangles stops.
Why? Whatever length the leg is, it's greater than zero; and you can make a new right triangle with that leg as hypotenuse, just as with the earlier triangles. I already pointed this out and you didn't answer.


Quote:
Originally Posted by michaelcweir View Post
Now a new series of triangles is constructed starting with Y leg
(Y^n + 1)^1/n , X^n/2, calculated hypotenuse leg ( (Y^n + 1)^2/n + X^n)^n/2 ) This leg is longer than the Y leg of the next hypotenuse leg of the descending triangles.
So what?

Quote:
Originally Posted by michaelcweir View Post
AS we back up each step on the descending triangles there is a corresponding hypotenuse of the ascending (Y^n +1 ) that is longer than the descending hypotenuse ( call it the Z leg). Eventually when you get back to the triangle with hypotenuse of length Z^n/2. The hypotenuse of triangle with hypotenuse (Y^n +1)^n/2 is longer than that. Thus Z is less than Y+1.
So what? And why don't you respond directly to the questions I've posed?

Quote:
Originally Posted by michaelcweir View Post
If z is less than Y+1, it is also less than any other integer greater than Y +1. This then proves FLT
How?

If you do nothing else, please respond to my point about n = 10 and e/f = 1.
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April 26th, 2019, 12:18 PM   #16
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You are asking questions about the proof using Diophantine equations. I've answered those questions there. e/f is greater than 1 in the examples I've described and drawing I submitted. Of course when your example of e/f and n=10 doesn't apply to the point I'm making.
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April 26th, 2019, 01:23 PM   #17
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Quote:
Originally Posted by michaelcweir View Post
You are asking questions about the proof using Diophantine equations. I've answered those questions there. e/f is greater than 1 in the examples I've described and drawing I submitted. Of course when your example of e/f and n=10 doesn't apply to the point I'm making.
Why not? If n = 10 then e/f = 1 and your proof doesn't work.
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April 29th, 2019, 10:01 AM   #18
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I answered, or rather you answered your own questions in the other proof. I am asking what you now think of this proof. You have made no comment since April 26th.Do you need more time?
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April 29th, 2019, 10:42 AM   #19
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Quote:
Originally Posted by michaelcweir View Post
I answered, or rather you answered your own questions in the other proof. I am asking what you now think of this proof. You have made no comment since April 26th.Do you need more time?
I don't actually have an obligation to reply at all, I hope you understand that.

I have convinced myself that in fact $(x^n + y^n)^{\frac{1}{n}} - (x^2 + y^2) < 0$ for all cases of interest to FLT. This is not anything you were able to prove at all.

If I grant you this, I'd then have to dive into your algebraic manipulations. I confess I lack the interest but it's on my vague to-do list. Someone else found an error in your algebra that you did not address.

I may or may not get around to diving in to your algebraic argument. Meanwhile if you could rewrite it in light of the fact that you have your inequality backwards and for the wrong reason (ie you derived the wrong inequality using faulty logic) it might be helpful.
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April 29th, 2019, 07:41 PM   #20
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Oh trivial algebraic manipulations. Why didn't anyone else ever think of that! Great job on your proof.

Oh yeah here's a theorem which says your proof is wrong.
https://en.wikipedia.org/wiki/Hensel%27s_lemma

FLT can not have a proof by simple algebraic techniques since such a solution would also apply to p-adic solutions....of which there are infinitely many.....for every n.
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