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April 17th, 2019, 06:38 PM   #41
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I have not really investigated how to make the proof for n =3. Of more concern to me is why the case n =2 is able to be excluded. Perhaps you can shed some light on that.If it can't be excluded there may be something wrong with the proof.
If you can't prove it for n = 3, then what good is the method?

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April 17th, 2019, 06:50 PM   #42
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Originally Posted by michaelcweir View Post
I have not really investigated how to make the proof for n =3. Of more concern to me is why the case n =2 is able to be excluded. Perhaps you can shed some light on that.

Well that's a really deep question. It occupied some of the smartest mathematicians in the world for 350 years. It was finally solved by Wiles using advanced math developed only in recent decades. I'm afraid I can't even offer an intuitive or casual explanation of why there are Fermat triples for n = 2 but for no other positive integer. The buzzwords are modular forms and elliptic curves, but I can't even pretend to do justice to any of it. There are some excellent popularized books on FLT as I'm sure you know. Simon Singh's books are good. There are many references here ... https://en.wikipedia.org/wiki/Wiles%...s_Last_Theorem


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Originally Posted by michaelcweir View Post
I would like to extend an invitation to help me with the notation. i would like to include you as one of the authors of the proof.
I can help out with the notation on this board as we go. I appreciate your offer. Your approach is unlikely to bear fruit so it's not necessary for me to accept or decline. If there were an elementary proof using triangles it would have been found. There is by the way zero chance that Fermat himself actually had a proof. He lived if I recall some 30 years after writing down his remark yet never mentioned the subject again.
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Last edited by Maschke; April 17th, 2019 at 06:59 PM.
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April 17th, 2019, 07:23 PM   #43
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If you looked at the equations and the diagram you would understand that the proof is no different, no matter what the power is. For n=3, there are only 2 iterations of the infinite descent.,
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April 17th, 2019, 08:00 PM   #44
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If you looked at the equations and the diagram you would understand that the proof is no different, no matter what the power is. For n=3, there are only 2 iterations of the infinite descent.,
I'll make a run at your latest attempt in the next few days.
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April 25th, 2019, 02:41 PM   #45
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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 goe s 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.

Tf 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 25th, 2019, 03:09 PM   #46
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This then proves FLT
Even if I followed the rest of that, and I confess I didn't yet dive in and might not, I do not see how any of this proves FLT.

As I recall (I didn't review this thread before posting) you have an integer triangle, and you derive from it a triangle whose sides must all be irrational. So you're already out of the realm of Fermat triples. You have no infinite descent.

It's perfectly clear that you can keep drawing triangles and that the sides are getting smaller.

If at some point you show that your inequality must occur, what of it? All those sides are irrational. You are no longer talking about Fermat triples.

Does what I'm saying make sense to you? What am I missing? You have all these irrational triangles, they have nothing to do with Fermat triples. As far as I can see.

ps -- Ok I read your latest post a couple of times and it's extremely convoluted. You say that at some point you can't draw any new triangles? How can that be? You are just drawing another triangle whose hypotenuse is the previous triangle's lower leg (the horizontal one). That's your construction. That can go on forever.

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Originally Posted by michaelcweir View Post
At some point,, the length of Y leg will be less than (Y^n +1 )^1/n.
I don't doubt that, since the triangles all seem to be getting smaller. I wouldn't be surprised if the length of the sides goes to zero. I don't remember from before if that's obviously the case or not.

But why can't we keep drawing smaller triangles? I don't see this at all.

And what does any of this have to do with FLT? You haven't explained that. From the very first step you no longer have Fermat triples, all the sides are irrational.

Last edited by Maschke; April 25th, 2019 at 03:58 PM.
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