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April 19th, 2019, 05:31 PM   #1
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Fermat Last Theorem Diophantine Analysis

Analysis using Diopphantine Equations
Diophantine Equations are used to remove as many variables as possible and write the remaining unknowns in terms of the other unknowns. By analyzing the remaining terms as whole numbers, we can decided whether there are infinite number of solutions or zero solutions.

Because we have the given equation X^n + Y^n = Z^n.

From this equation we can derive to associated equations

1) X + Y = c/d * Z where c/d > 1. c,d are rational also c/d < 2^1/2

2) X^2 + Y^2 = e/f * Z^2 where e/f > 1 and c/d > e/f e,f are rational


from 1) we get d*X + d*Y = C*Z]

d*X = c*Z - d*Y

or finally X = c*Z/d - Y


If we substitute value for X into 2) we get

( c^2/d^2 * Z^2 - 2c/d * Z * Y + Y^2) + Y^2 = e/f * Z

rearranging

c^2/d^2 * Z^2 - 2c/d * Z * Y - e/f*Z = -2Y^2

or

Z *(c^2/d^2 * Z - 2c/d * Y - e/f) = -2Y^2

Now let Z = h *Y where 2 > h > 1 and substitute into equation

h * Y *((c^2/d^2 * h *Y) - 2c/d * Y - e/f) = -2Y^2

factor Y out of both sides of equation

and you get


h *((c^2/d^2 * h *Y) - 2c/d * Y - e/f) = -2Y

Now move -2Y to left hand side and extend h across members of brackets

c^2/d^2 * h^2 *Y - 2c/d * h * Y + 2Y - e/f * h = 0

move e/f * h to right hand side. Now left hand side has the Y come out of all terms on left hand side


(c^2/d^2 * h^2 - 2c/d * h + 2) * Y = e/f * h

or finally

Y = (e/f * h)/(c^2/d^2 * h^2 - 2 * c/d * h + 2)

We do not know what the exact value of y is, but we can determine the range

when Y is close to Z, e/f approximates 1, h approximates 1, c^2/d^2 approximates 1, but is always less than 1

In that case

Y is approximately (1 * 1)/((1 * 1^2) - 2 * 1 * 1 + 2) = 1/1 = 1, but less than 1




When Y is close to X, e/f approximates 1.3, h approximates 2, c^2/d^2 approximates 1.4

Then
Y is approximately (1.3 * 2)/(1.4^2 * 2^2 - 2 * 1.4 * 2 + 2)= 2.6/7.84 - 5.6 + 2

or Y is approximately 2.6/4.24 or approximately 0.6

These approximations could be made more defined, but the real issue has already been decided.

The range of values Y takes on is between 0.6 and 1. There is no way Y could be equal to 2, even if the numbers are refined. It should be noted that Y is the larger of the supposed values X,Y. This only leaves an integer value of less than 0 for X, which is not allowed.

This completes the proof for Fermat's Last Theorem using Diophantine Equations.
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April 19th, 2019, 10:55 PM   #2
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in this equation you give:
( c^2/d^2 * Z^2 - 2c/d * Z * Y + Y^2) + Y^2 = e/f * Z
you should write Z^2 instead of Z in right hand side
Check if this blows everything up or not
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April 20th, 2019, 11:33 AM   #3
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Quote:
Originally Posted by michaelcweir View Post

Because we have the given equation X^n + Y^n = Z^n.

From this equation we can derive to associated equations

1) X + Y = c/d * Z where c/d > 1. c,d are rational also c/d < 2^1/2

2) X^2 + Y^2 = e/f * Z^2 where e/f > 1 and c/d > e/f e,f are rational
This is already wrong. Take $3^2 + 4^2 = 5^2$. Then $\frac{e}{f} = 1$.

Proof busted. Your CLAIM that n > 2 (which you didn't bother to specify) must lead to $\frac{e}{f} > 1$ requires assuming FLT is true in the first place. This is the exact same error as in your first false proof. In that proof you CLAIM that alpha > 0 for n > 2 but it's perfectly clear that this assertion depends on the truth of FLT.

Indeed, my earlier counterexample works here as well. If $x^{10} + y^{10} = z^{10}$ then you have a Pythagorean triple $(x^5)^2 + (y^5)^2 = (z^5)^2$ and $\frac{e}{f} = 1$.

The only way you can prove that can't happen is by assuming FLT in the first place.

By the way I would appreciate your reply in your previous thread with the infinite descent of triangles. I pointed out that you do indeed have all those triangles, but they all have irrational sides and you haven't proved anything. I await your comment.
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Last edited by Maschke; April 20th, 2019 at 11:58 AM.
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April 24th, 2019, 02:19 PM   #4
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I did reply to your comment about the infinite descent of triangles. Let me know if you now have it.

There is no assumption that FLT is true for either equation 1 or 2

Equation 1 is derived from the fact that only the initial equation is true.There is the assumption that the values of X,Y,Z are rational. It really doesn't matter whether the numbers are rational or not. . This is purely a consideration of the number line and where this numbers are on the number line in relation to each other. That means that Z is larger than either X and Y. If Y is the larger of the two, than Z must be less than 2Y. ( I made a notation error and said c/d <2^1/2.The reasoning still holds. The range for Y is still is between 0.5 to 0.6 approximately.)

Equation 2 is derived from the fact that the squares of the numbers X,Y,Z have a a similar relationship to each other, the same as equation 1.
X^2 + Y^2 = e/f *Z^2 If X was as big as Y, then Y^2 +Y^2 = Z^2. 2Y^2 =Z^2 or

Z =2^1/2 Y

So Z = e/f Y means that the range of Z is between 1 to 1.4 (2^1/2).

As a side note, when n is 2, then e/f is 1 as you say. That's why Fermat specified that n be greater than 2.
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April 24th, 2019, 08:50 PM   #5
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Quote:
Originally Posted by michaelcweir View Post
I did reply to your comment about the infinite descent of triangles. Let me know if you now have it.
I don't believe so, not to my satisfaction. I'm referring to the end of this thread:

Easy proof of Fermat's last Theorem

In that thread you start with an integer right triangle and then keep dropping a new triangle underneath it whose hypotenuse is the lower leg of the previous triangle.

After doing this a number of times, you triumphantly say, "And see, some triangle has an irrational side." If I'm understanding you correctly. And then you say that proves FLT.

There are two problems. First, every side of every one of your triangles is irrational, except in exceptional cases where the sides of the first triangle are all squares. Even so you will soon get to the case where all your sides of all your triangles are irrational.

So far from proving that some side will eventually be irrational, all the sides of all the triangles are irrational. But regardless, I don't see how this proves FLT. You haven't explained that. I do await that explanation.



Quote:
Originally Posted by michaelcweir View Post
There is no assumption that FLT is true for either equation 1 or 2
Yes but you have the same problem for n = 10 and in fact for n being any even number. You will immediately have a Pythagorean triple in which e/f is exactly 1. You claimed that it must be greater than 1 but this is false as false can be.

Quote:
Originally Posted by michaelcweir View Post
Equation 1 is derived from the fact that only the initial equation is true.There is the assumption that the values of X,Y,Z are rational.
Integer, you mean.

Quote:
Originally Posted by michaelcweir View Post
It really doesn't matter whether the numbers are rational or not.
It matters a lot, since you are assuming you have a Fermat triple. X, Y, and Z must be integers.

Quote:
Originally Posted by michaelcweir View Post
. This is purely a consideration of the number line and where this numbers are on the number line in relation to each other. That means that Z is larger than either X and Y. If Y is the larger of the two, than Z must be less than 2Y. ( I made a notation error and said c/d <2^1/2.The reasoning still holds. The range for Y is still is between 0.5 to 0.6 approximately.)
You're not addressing the key point. You claimed e/f is greater than 1, but I showed that it can be exactly 1. That's your second equation.

Quote:
Originally Posted by michaelcweir View Post
Equation 2 is derived from the fact that the squares of the numbers X,Y,Z have a a similar relationship to each other, the same as equation 1.
X^2 + Y^2 = e/f *Z^2 If X was as big as Y, then Y^2 +Y^2 = Z^2. 2Y^2 =Z^2 or

Z =2^1/2 Y
You claimed e/f > 1 but in fact e/f = 1.

Quote:
Originally Posted by michaelcweir View Post
As a side note, when n is 2, then e/f is 1 as you say. That's why Fermat specified that n be greater than 2.
e/f = 1 for n = 10 as well. Your false claim that e/f > 1 is the same as your first thread when you claimed alpha > 0. The exact same example of n = 10 shows that alpha = 0 whenever n is even.
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April 25th, 2019, 05:04 PM   #6
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Quote:
Originally Posted by Maschke View Post
I don't believe so, not to my satisfaction. I'm referring to the end of this thread:

Easy proof of Fermat's last Theorem

In that thread you start with an integer right triangle and then keep dropping a new triangle underneath it whose hypotenuse is the lower leg of the previous triangle.

After doing this a number of times, you triumphantly say, "And see, some triangle has an irrational side." If I'm understanding you correctly. And then you say that proves FLT.

There are two problems. First, every side of every one of your triangles is irrational, except in exceptional cases where the sides of the first triangle are all squares. Even so you will soon get to the case where all your sides of all your triangles are irrational.

So far from proving that some side will eventually be irrational, all the sides of all the triangles are irrational. But regardless, I don't see how this proves FLT. You haven't explained that. I do await that explanation.

It doesn't matter whether Y leg or Z leg is irrational or not. It only matters if the Y leg is getting shorter on the way down.Finally the Y leg is shorter than (Y^1/n) - this always happens when you have infinite descent.



Yes but you have the same problem for n = 10 and in fact for n being any even number. You will immediately have a Pythagorean triple in which e/f is exactly 1. You claimed that it must be greater than 1 but this is false as false can be.

For n=2, e/f is true. But for n > 2, Z^2 is always greater than X^n + Y^2 .
For example of 3,4,5 triangle.
n=1 , 3+4 =7 but Z =5
n=2, 9+16 =25 and Z=5
n=3 , 27 + 64= 91 but Z = 125

for n =2 e/f =1
for n=3 e/f = 1.37...

If you paste Z^n, (Y^n + X^n), Z^n-1, (Y^n-1 + X^n-1) on the number line,
if
Z^n = X^n +Y^n, then (Y^n-1 + X^n-1) is always less than Z^n-1 and e/f >1 and upper limit of e/f is determined by the largest number
(Y^n-1 + X^n-1) can achieve,
So for n=3, the maximum Z^2 can reach is Y^2 +Y^2. Z^2 = Y^2 +Y^2

or Z^2 = 2^1/2 or Z=2^1/2* Y

So the range of e/f for n=3 is 1 <e/f < 1.4



It's a lot to take in.


Integer, you mean.



It matters a lot, since you are assuming you have a Fermat triple. X, Y, and Z must be integers.



You're not addressing the key point. You claimed e/f is greater than 1, but I showed that it can be exactly 1. That's your second equation.



You claimed e/f > 1 but in fact e/f = 1.



e/f = 1 for n = 10 as well. Your false claim that e/f > 1 is the same as your first thread when you claimed alpha > 0. The exact same example of n = 10 shows that alpha = 0 whenever n is even.
Maybe it's time to be a little proactive.
Y = (e/f * h)/(c^2/d^2 * h^2 - 2 * c/d * h + 2) is the final result
To determine the range of Y we have to determine what values Y takes on when Y is close to Z, and when Y is close to X

When Y is close to Z

e/f close to 1
h is close to 1
c^2/d^2 close to 4
c/d is close to 2
Y = 1*1/4*1-2*2+2 = 1/2

Y = (e/f * h)/(c^2/d^2 * h^2 - 2 * c/d * h + 2)

Now when X is close to Y
e/f is close to 1.4
h is close to 1.4
c^2/d^2 close to 4
c/d close to 2
Y = 1.4 * 1.4 / 4*2 - 2*2*1.4 +2

or

Y = 2/ 8-5.6 + 2

or

Y = 2/4.4 = 0.45

So the range of Y is 0.45 < Y < 0.5

Now, since Y must be less than 1, FLT is proved.
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April 25th, 2019, 05:16 PM   #7
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Quote:
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Now, since Y must be less than 1, FLT is proved.
I have no idea what you're doing.

Is e/f still claimed to be greater than 1?

How does any of this prove FLT?

Where did you use the assumption that n > 2? It's the classic hallmark of a false proof of FLT that the claimed proof does not distinguish the case n = 2.

Quote:
Originally Posted by michaelcweir View Post
To determine the range of Y we have to determine what values Y takes on when Y is close to Z, and when Y is close to X
What are X, Y, and Z? I had assumed they must be integers but at some point you said they're rationals. You started in the middle of your argument here. I don't know what anything is supposed to be. Earlier you said e/f > 1 but now you've abandoned that idea without admitting you did or explaining that you didn't.
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Last edited by Maschke; April 25th, 2019 at 05:33 PM.
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April 26th, 2019, 12:05 PM   #8
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Here is the diagram that Fermat probably used to demonstrate the relationships between X<Y>Z<X^2 +Y^2, Z^2, Z^n, X^n +Y^n.

When n > 2, Z^2 < X^2 + Y^2 every time, it doesn't matter which n you talk about.
When n = 3 and you use the number 3,4,5 then e/f is greater than 1 and we can calculate the value of e/f exactly. If we use different numbers, then the value of e/f will change but always be in the range.

Last edited by skipjack; April 26th, 2019 at 12:52 PM.
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April 26th, 2019, 12:50 PM   #9
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The diagram would need to be uploaded to this site. Also, you seem to be ignoring most of the questions that you have been asked, and unable to express your reasoning clearly.
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April 26th, 2019, 03:49 PM   #10
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Quote:
Originally Posted by michaelcweir View Post
If we use different numbers, then the value of e/f will change but always be in the range.
No this is not true. If there were a Fermat triple for n = 10 you would have a right triangle with sides $x^5$, $y^5$, and $z^5$ and e/f would be exactly 1. You simply refuse to engage with this point but it breaks your proof.
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