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April 17th, 2019, 06:18 PM   #1
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Easy proof of Fermat's last Theorem

Here is a video I made of a proof of FLT.

I have trouble with notation, so I made a YouTube video with handwritten equations and diagrams.



Enjoy the read.

Last edited by skipjack; May 23rd, 2019 at 12:03 AM.
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April 17th, 2019, 07:02 PM   #2
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Can you do me a favor? I don't watch your videos for the most part. I just look at your page of equations. I watch the video just long enough to find the clearest view of your page, then I take a screenshot.

Can you just post a clear photo of your page of exposition? With maybe a paragraph or two of explanation? I hate watching videos in general when I could just skim-read an article instead.

ps -- I watched the first ten seconds or so. That's quite a diagram! I'll have to see if I can figure out what you're doing.

I admit that I clicked off when you said that you wanted to show "how easy this proof is." I think that some appropriate level of introspection, self-awareness, and humility would be helpful in trying to tackle this problem. It's a difficult mathematical problem and even Wiles spent seven years up in his attic. Then when he finally published, an error was found that took him another year and a half to fix, with someone else's help. In fact the New York Times published a fascinating article about how Wiles had to find someone smart enough to help him; but not so senior that he'd get half the credit! The politics of math. The article ruffled some feathers I believe. Let me see if I can find it. Yes here it is.

https://www.nytimes.com/1995/01/31/s...s-bridged.html

Last edited by Maschke; April 17th, 2019 at 07:09 PM.
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April 18th, 2019, 11:29 AM   #3
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What is correct in your drawing (none of what you wrote match anything)?
You start with $z^{\frac{n}{2}}=(y^n+x^n)^{\frac{1}{2}}$, next you get $y^{\frac{n}{2}}$, then you should get $(y^n-x^n)^{\frac{1}{2}}$ and not what you wrote, then $(y^n-2\cdot x^n)^{\frac{1}{2}}$, $(y^n-3\cdot x^n)^{\frac{1}{2}}$, ....until you get smaller than $x^{\frac{n}{2}}$ (your triangle has another shape).
It means that you must have $y^n\geq (a+1)x^n$ to be able to make $a$ steps

I(?) am a bit confused
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April 18th, 2019, 03:13 PM   #4
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Quote:
Originally Posted by michaelcweir View Post
Here is a video I made of a proof of FLT.
Say that $x^5 + y^5 = z^5$.

Then you have a right triangle with sides $x^\frac{5}{2}$, $y^\frac{5}{2}$, and $z^\frac{5}{2}$.

But that triangle no longer has integer sides. So whatever point you are making can no longer be operative. I can see that if given any Fermat triple you can produce a smaller one, that is indeed an infinite descent argument. Even so it would not be a proof of FLT since it applies equally well to n = 2. Since your proof does not distinguish the cases n = 2 and n > 2, it can't be a proof of FLT, whatever else it may be.

But we don't even get that far. It is NOT an infinite descent argument, because given a triangle with integer sides, the next smaller triangle does NOT have integer sides if n is odd. So you no longer have a Fermat triple. You have no infinite descent.

Remember earlier I used the example of n = 10 to show that you'd get a smaller integer triangle with n = 5 and alpha = 0. I used 10 because it's an even number. Your n/2 always works if n is even. But if n is odd, it never works.

Now I do recall that in your very first attempt, you used the greatest integer function; so that starting with n = 5, the next smaller triangle would have exponent 2. But that doesn't make any algebraic sense. The greatest integer function loses information. There's no relationship based on the exponent laws between n = 5 and n = 2. It's not the same argument as when you use the exponent laws to go from 5 to 5/2.

I also recall asking you a couple of times about your use of the greatest integer function but you haven't replied. I ask again. But lately you've abandoned it. The problem is then that your smaller triangle no longer has integer sides.

What do you say?
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April 18th, 2019, 05:50 PM   #5
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IMG_6037.jpg

IMG_6036.jpg

I finally figured out how to upload images on this forum. Youtube would not upload just pictures, nor add them to an existing video.

The proof is rather easy, and you will probably agree once you understand it.

You can construct a right triangle with Zn/2, Xn/2, Yn/2.

What the proof shows is that for every n greater than 2, Z must be smaller than Y+1. That being the case, Z must be irrational.. thus proving FLT.

Using the first triangle, you can construct a new triangle from Yn/2, which is the hypotenuse of the new triangle, and Xn/2. The length of the Y leg becomes shorter than Yn/2.

You then create another triangle with the shorter Y leg being the hypotenuse of the new triangle.

The Y leg becomes shorter each time and as n goes to infinity, the Y leg approaches Y..

At some iteration of the triangle, the Y leg becomes shorter than (Yn +1)1/n.

This is represented by the bottom triangle at the bottom of the diagram. The red line represents the line (Yn +1)1/n.

So if you now back up the process, you will finally get this relationship

Yn +1 > Zn. So Zn must be irrational, and by extension Z is irrational. QED
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April 18th, 2019, 06:57 PM   #6
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Quote:
Originally Posted by michaelcweir View Post

You can construct a right triangle with Zn/2, Xn/2, Yn/2.

What the proof shows is that for every n greater than 2, Z must be smaller than Y+1. That being the case, Z must be irrational.. thus proving FLT.
I don't follow this at all. Take n = 5 and suppose you have integers $x$, $y$, and $z$ with $x^5 + y^5 = z^5$.

Then you do indeed have a right triangle with sides $x^\frac{5}{2}$, $y^\frac{5}{2}$, and $z^\frac{5}{2}$. But each of these sides is irrational unless $x$, $y$ and $z$ each happen to be perfect squares.

I don't see how that proves anything. You can keep dividing your exponents by 2 to get a smaller triangle, but that triangle also has irrational sides. I just don't see your point.

I wish you'd walk through a specific example. Do it for n = 5. Then maybe I'll understand your point.

Likewise the case n = 2 messes up your proof. You have a Fermat triple $3^2 + 4^2 = 5^2$. This gives a right triangle with sides 3, 4, and 5. In this case n is even so the n/2 triangle has integer sides.

What of it? I just can't follow your logic.

Last edited by Maschke; April 18th, 2019 at 07:07 PM.
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April 18th, 2019, 08:23 PM   #7
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Your video makes absolutely no sense. You should start by asking if your math is even right, not asking how you can publish already. One glaring thing is that Euclidean triangles are only defined for $\displaystyle x^n + y^n=z^n$ where $\displaystyle n=2$. So slapping a bunch of triangles together (though you didn't even do that right) doesn't show anything about large $\displaystyle n\epsilon N$. The geometry you should be looking into is that of Elliptic Curves.
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April 18th, 2019, 08:46 PM   #8
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@michaelcweir, Can you respond to a technical question? I'm confused about how you construct the second triangle.

In the picture in your video (but NOT the pic you posted just now) you have:

$x^n + y^n = z^n$ hence $z^n - y^n = x^n$. For whatever reason you are isolating the $x$ term.

This gives rise to a right triangle with sides $x^\frac{5}{2}$, $y^\frac{5}{2}$, and $z^\frac{5}{2}$. Now unless we are very unlucky and all three of $x$, $y$, and $z$ are perfect squares, at least one of those three sides is already irrational. You think showing some triangle's side is irrational proves FLT, I don't follow that, but let it pass. I have a different question.

So now working from the pic you just posted, I want to be able to draw the second triangle. By playing around with pencil and paper, one sees that the angle the new leg makes with the old leg (the one of length $y^\frac{5}{2}$ on the bottom of the first triangle) is a function of the length of the new leg you're drawing.

Now my question is, I'm confused about what is the length of that leg. In your picture you label it $y^\frac{5}{2} - x^\frac{5}{2}$. But that doesn't make any sense. Why are you subtracting these two numbers? Is this supposed to relate to your earlier isolating of the $x$-term? But that doesn't make sense either, you need a $z$ in there somewhere.

I hope this is clear. What is the length of the new leg you're drawing to make the second triangle?

And ... if it's possible ... can you give me the big picture? What is the purpose of all this? I can see that you're creating a bunch of triangles, all of which have at least one irrational side and usually three. But how does that prove FLT? I don't see the connection.

Last edited by Maschke; April 18th, 2019 at 09:00 PM.
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April 18th, 2019, 08:48 PM   #9
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Quote:
Originally Posted by Walter Rudin View Post
Your video makes absolutely no sense. You should start by asking if your math is even right, not asking how you can publish already. One glaring thing is that Euclidean triangles are only defined for $\displaystyle x^n + y^n=z^n$ where $\displaystyle n=2$. So slapping a bunch of triangles together (though you didn't even do that right) doesn't show anything about large $\displaystyle n\epsilon N$. The geometry you should be looking into is that of Elliptic Curves.
Welcome to the forum. That's an ambitious handle. Brings back memories.

Last edited by Maschke; April 18th, 2019 at 08:58 PM.
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April 18th, 2019, 09:01 PM   #10
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It is more to show respect to the great man.
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