"Fermat's Last Theorem" by Amir D. Aczel

[dream_weaver]

Nope.

 

I had a little pre-calc 30 years ago.

 

But sooner or later, (3n+1)=2^x, as does (5n+1) or (7n+1)

 

3n+1 does for 1, 5 and more,

5n+1 does for 3, 7 and more.

7n+1 does for 1, 5 and more.

 

I would tend to believe that is why it works.

[ruveyn1]

Nope.

 

I had a little pre-calc 30 years ago.

Do two things.

 

1  Learn what a proof is

2. Write you proof out carefully and have a professional mathematician vet your proof for errors.

 

If you can prove FLT  by elementary means and you are under 45 years of age,  I guarantee you will get a Fields Medal.

[dream_weaver]

Already past that point. I just enjoy tinkering with this stuff occasionally. The drawing I produced is more in line with my forte. I have to envision what it looks like in order to create it.

[dream_weaver]

The more I look at my final formula using the exponent, I do not think it accurate represents the number of floors. I think there are many more, and are influenced by the number being exponented in some way..

Up to the (X²+X²+X²) is the means of deriving that figure. Like the pythagorean triple cannot occur in a square, the cube exhibits similar properties concretized in this form.

[dream_weaver]

Another way of looking at the problem.

 

Take a cube of 10 X 10 X 10 boxes,

Take 8 X 8 X 8 boxes away,

Take 2 X 2 X 2 boxes away. 

What is left?

 

10³-8³-2³ cannot be 0

The remainder is not enough to make 8³ and 3³ or 9³ and 2³.

 

(A+B )³-A³-B³>0

(A+1)³+B³>(A+B )³

A³+(B+1)³>(A+B )³

[dream_weaver]

The solution to exponents greater than 3 appears it may be as simple.

 

2(Xⁿ)<(X+1)ⁿ

Aⁿ+Bⁿ<>Cⁿ

[dream_weaver]

Scratch that last one. I’ll settle with the simpler cube solution. That’s enough for me right now.

[dream_weaver]

Except that both fail.

[dream_weaver]

X⁽ⁿ⁾, X⁽ⁿ⁺¹⁾, X⁽ⁿ⁺²⁾ resolves into a repeating pattern of 3 basic prisms.

X is an integer used as a length.

The letter n is an integer.

When n is 1, n is solving for X³, n+1 is solving for X⁴, and n+2 is solving for X⁵.

When n is 2, n is solving for X⁶, n+1 is solving for X⁷, and n+2 is solving for X⁸.

When n is 3, n is solving for X⁹, n+1 is solving for X¹⁰, and n+2 is solving for X¹¹.

************************************************************************

The first prism is a cube that can be represented by a shape of

X⁽ⁿ⁾ x X⁽ⁿ⁾ x X⁽ⁿ⁾.

 

The second prism is a square of its base and rectangular in its extension.

The can be represented by as X⁽ⁿ⁾ x X⁽ⁿ⁾ x X⁽ⁿ⁺¹⁾.

 

The third prism also consists of a square base with a rectangular extension.

It will be represented as X⁽ⁿ⁺¹⁾ x X⁽ⁿ⁺¹⁾ x X⁽ⁿ⁾.

**********************************************************************

It will help to envision a building. There is a door located at one corner. If you walk diagonally across the floor to the corner opposite the door (located at the intersection of the two walls that do not share the door) is a stairwell.

The stairway connects the stairwell to the corner on the wall directly above the door.

************************************************************************

For the first and second prism, the distance from the door to the stairwell is represented by ((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²)^(1/2).

(2(X⁽ⁿ⁾)²)^(1/2).

2^(1/2)(X⁽ⁿ⁾).

 

For the third prism, the distance from the door to the stairwell is represented by the ((X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁺¹⁾)²)^(1/2).

(2(X⁽ⁿ⁺¹⁾)²)^(1/2).

2^(1/2)(X⁽ⁿ⁺¹⁾)

************************************************************************

The stairway for the first prism is the hypotenuse of the distance from the door to the stairwell and the distance from the door to the top of the prism.

((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²+ (X⁽ⁿ⁾)²)^(1/2)

or (3(X⁽ⁿ⁾)²)^(1/2)

or 3^(1/2)(X⁽ⁿ⁾).

 

The stairway for the second prism is:

((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)^(1/2)

or (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)^(1/2).

 

The stairway for the third prism is:

((X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)^(1/2)

or (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)^(1/2).

 

In summary:

For the first prism:

(X⁽ⁿ⁾)²+((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²)²=((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²)²

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁾), 3^(1/2)(X⁽ⁿ⁾) (Two legs, followed by the hypotenuse.)

 

For the second prism:

((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²)²+(X⁽ⁿ⁺¹⁾)²=((X⁽ⁿ⁾)²+(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)²

2^(1/2)(X⁽ⁿ⁾), (X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)

For the third prism:

((X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁺¹⁾)²)²+(X⁽ⁿ⁾)²=(2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)²

2^(1/2)(X⁽ⁿ⁺¹⁾), (X⁽ⁿ⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)

Fermat concluded that these were not Pythagorean triples. I agree, and here is why.

******************************************************************

Pythagorean triples have their own repeating pattern.

 

If P is odd,

P, P²/2-1/2, P²/2+1/2

If P is even,

P, (P/2)²-1, (P/2)²+1

******************************************************************

Starting with the cube

 

If (X⁽ⁿ⁾) is odd,

 

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁾), 3^(1/2)(X⁽ⁿ⁾)

P, P²/2-1/2, P²/2+1/2

If P is (X⁽ⁿ⁾), is P²/2-1/2=2^(1/2)(X⁽ⁿ⁾)?

 

((X⁽ⁿ⁾)²)/2-1/2=2^(1/2)(X⁽ⁿ⁾)

(X⁽²ⁿ⁾)/2-1/2=2^(1/2)(X⁽ⁿ⁾) (factor the ²)

(X⁽²ⁿ⁾)/2=2^(1/2)(X⁽ⁿ⁾)+1/2 (add 1/2)

(X⁽²ⁿ⁾)=2*2^(1/2)(X⁽ⁿ⁾)+1 (multiply by 2)

 

(X⁽²ⁿ⁾)≠2*2^(1/2)(X⁽ⁿ⁾)+1

 

If (X⁽ⁿ⁾) is even,

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁾), 3^(1/2)(X⁽ⁿ⁾)

P, (P/2)²-1, (P/2)²+1

 

If P is (X⁽ⁿ⁾), is (P/2)²-1=2^(1/2)(X⁽ⁿ⁾)?

 

((X⁽ⁿ⁾)/2)²-1=2^(1/2)(X⁽ⁿ⁾)

(X⁽²ⁿ⁾)/4-1=2^(1/2)(X⁽ⁿ⁾) (factor the ²)

 (X⁽²ⁿ⁾)/4=2^(1/2)(X⁽ⁿ⁾)+1 (add 1)

(X⁽²ⁿ⁾)=4*2^(1/2)(X⁽ⁿ⁾)+4 (multiply by 4)

 

(X⁽²ⁿ⁾)≠4*2^(1/2)(X⁽ⁿ⁾)+4

******************************************************************

The second prism.

 

If (X⁽ⁿ⁺¹⁾) is odd,

 

2^(1/2)(X⁽ⁿ⁾), (X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²) (Two legs, followed by the hypotenuse.)

(X⁽ⁿ⁺¹⁾), 2^(1/2)(X⁽ⁿ⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)

P, P²/2-1/2, P²/2+1/2

If P is (X⁽ⁿ⁺¹⁾), is P²/2-1/2=2^(1/2)(X⁽ⁿ⁾)?

(X⁽ⁿ⁺¹⁾)²/2-1/2=2^(1/2)(X⁽ⁿ⁾)

(X⁽²ⁿ⁺²⁾)/2-1/2=2^(1/2)(X⁽ⁿ⁾) (factor the ²)

(X⁽²ⁿ⁺²⁾)/2=2^(1/2)(X⁽ⁿ⁾)+1/2 (add 1/2)

(X⁽²ⁿ⁺²⁾)=2*2^(1/2)(X⁽ⁿ⁾)+1 (multiply by 2)

 

(X⁽²ⁿ⁺²⁾)≠2*2^(1/2)(X⁽ⁿ⁾)+1

 

If (X⁽ⁿ⁺¹⁾) is even,

(X⁽ⁿ⁺¹⁾), 2^(1/2)(X⁽ⁿ⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)

P, (P/2)²-1, (P/2)²+1

 

If P is (X⁽ⁿ⁺¹⁾), is (P/2)²-1=2^(1/2)(X⁽ⁿ⁾)?

 

((X⁽ⁿ⁺¹⁾)/2)²-1=2^(1/2)(X⁽ⁿ⁾)

(X⁽²ⁿ⁺²⁾)/4-1=2^(1/2)(X⁽ⁿ⁾) (factor the ²)

(X⁽²ⁿ⁺²⁾)/4=2^(1/2)(X⁽ⁿ⁾)+1 (add 1)

(X⁽²ⁿ⁺²⁾)=4*2^(1/2)(X⁽ⁿ⁾)+4 (multiply by 4)

 

(X⁽²ⁿ⁺²⁾)≠4*2^(1/2)(X⁽ⁿ⁾)+4

******************************************************************

The third prism.

 

If (X⁽ⁿ⁾) is odd,

2^(1/2)(X⁽ⁿ⁺¹⁾), (X⁽ⁿ⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²) (Two legs, followed by the hypotenuse.)

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)

P, P²/2-1/2, P²/2+1/2

If P is (X⁽ⁿ⁾), is P²/2-1/2=2^(1/2)(X⁽ⁿ⁺¹⁾)?

(X⁽ⁿ⁾)²/2-1/2=2^(1/2)(X⁽ⁿ⁺¹⁾)

(X⁽²ⁿ⁾)/2-1/2=2^(1/2)(X⁽ⁿ⁺¹⁾) (factor the ²)

(X⁽²ⁿ⁾)/2=2^(1/2)(X⁽ⁿ⁺¹⁾)+1/2 (add 1/2)

(X⁽²ⁿ⁾)=2*2^(1/2)(X⁽ⁿ⁺¹⁾)+1 (multiply by 2)

(X⁽²ⁿ⁾)≠2*2^(1/2)(X⁽ⁿ⁺¹⁾)+1

 

If (X⁽ⁿ⁾) is even,

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)

P, (P/2)²-1, (P/2)²+1

 

If P is (X⁽ⁿ⁾), is (P/2)²-1=2^(1/2)(X⁽ⁿ⁺¹⁾)?

 

((X⁽ⁿ⁾)/2)²-1=2^(1/2)(X⁽ⁿ⁺¹⁾)

((X⁽²ⁿ⁾)/4)-1=2^(1/2)(X⁽ⁿ⁺¹⁾) (factor the ²)

(X⁽²ⁿ⁾)/4=2^(1/2)(X⁽ⁿ⁺¹⁾)+1 (add 1)

(X⁽²ⁿ⁾)=4*2^(1/2)(X⁽ⁿ⁺¹⁾)+4 (multiply by 4)

 

(X⁽²ⁿ⁾)≠4*2^(1/2)(X⁽ⁿ⁺¹⁾)+4

*****************************************************************

I would also have to agree with Fermat, it would not fit easily into a margin.

 

Note that a stack of blocks 5 wide x 4 wide by 3 tall can be rearranged into three separate groups consisting of 25 blocks, 16 blocks, and 9 blocks respectively.

[dream_weaver]

To capture other triples, any triple can be increased in magnitude by multiplying the result times an integer, in this case, m.

******************************************************************

Starting with the cube

 

If (X⁽ⁿ⁾) is odd,

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁾), 3^(1/2)(X⁽ⁿ⁾)

m(P), m(P²/2-1/2), m(P²/2+1/2)

If P is (X⁽ⁿ⁾), is m(P²/2-1/2)=2^(1/2)(X⁽ⁿ⁾)?

 

m(((X⁽ⁿ⁾)²)/2-1/2)=2^(1/2)(X⁽ⁿ⁾)

((X⁽ⁿ⁾)²)/2-1/2=(2^(1/2)(X⁽ⁿ⁾))/m (divide by m)

(X⁽²ⁿ⁾)/2-1/2=(2^(1/2)(X⁽ⁿ⁾))/m (factor the ²)

(X⁽²ⁿ⁾)/2=(2^(1/2)(X⁽ⁿ⁾))/m+1/2 (add 1/2)

(X⁽²ⁿ⁾)=2(2^(1/2)(X⁽ⁿ⁾))/m+1 (multiply by 2)

 

(X⁽²ⁿ⁾)≠2(2^(1/2)(X⁽ⁿ⁾))/m+1

 

If (X⁽ⁿ⁾) is even,

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁾), 3^(1/2)(X⁽ⁿ⁾)

mP, m((P/2)²-1), m((P/2)²+1)

 

If P is (X⁽ⁿ⁾), is m((P/2)²-1)=2^(1/2)(X⁽ⁿ⁾)?

m(((X⁽ⁿ⁾)/2)²-1)=2^(1/2)(X⁽ⁿ⁾)

((X⁽ⁿ⁾)/2)²-1=(2^(1/2)(X⁽ⁿ⁾))/m (divide by m)

(X⁽²ⁿ⁾)/4-1=(2^(1/2)(X⁽ⁿ⁾))/m (factor the ²)

 (X⁽²ⁿ⁾)/4=(2^(1/2)(X⁽ⁿ⁾))/m+1 (add 1)

(X⁽²ⁿ⁾)=4(2^(1/2)(X⁽ⁿ⁾))/m+4 (multiply by 4)

 

(X⁽²ⁿ⁾)≠4(2^(1/2)(X⁽ⁿ⁾))/m+4

******************************************************************

The second prism.

 

If (X⁽ⁿ⁺¹⁾) is odd,

2^(1/2)(X⁽ⁿ⁾), (X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²) (Two legs, followed by the hypotenuse.)

(X⁽ⁿ⁺¹⁾), 2^(1/2)(X⁽ⁿ⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)

mP, m(P²/2-1/2), m(P²/2+1/2)

 

If P is (X⁽ⁿ⁺¹⁾), is m(P²/2-1/2)=2^(1/2)(X⁽ⁿ⁾)?

 

m((X⁽ⁿ⁺¹⁾)²/2-1/2)=2^(1/2)(X⁽ⁿ⁾)

(X⁽ⁿ⁺¹⁾)²/2-1/2=(2^(1/2)(X⁽ⁿ⁾))/m (divide by m)

(X⁽²ⁿ⁺²⁾)/2-1/2=(2^(1/2)(X⁽ⁿ⁾))/m (factor the ²)

(X⁽²ⁿ⁺²⁾)/2=(2^(1/2)(X⁽ⁿ⁾))/m+1/2 (add 1/2)

(X⁽²ⁿ⁺²⁾)=2(2^(1/2)(X⁽ⁿ⁾))/m+1 (multiply by 2)

 

(X⁽²ⁿ⁺²⁾)≠2(2^(1/2)(X⁽ⁿ⁾))/m+1

 

If (X⁽ⁿ⁺¹⁾) is even,

(X⁽ⁿ⁺¹⁾), 2^(1/2)(X⁽ⁿ⁾), (2(X⁽ⁿ⁾)²+(X⁽ⁿ⁺¹⁾)²)

mP, m((P/2)²-1), m((P/2)²+1)

 

If P is (X⁽ⁿ⁺¹⁾), is m((P/2)²-1)=2^(1/2)(X⁽ⁿ⁾)?

 

m(((X⁽ⁿ⁺¹⁾)/2)²-1)=2^(1/2)(X⁽ⁿ⁾)

((X⁽ⁿ⁺¹⁾)/2)²-1=2^(1/2)(X⁽ⁿ⁾)/m (divide by m)

(X⁽²ⁿ⁺²⁾)/4-1=(2^(1/2)(X⁽ⁿ⁾))/m (factor the ²)

(X⁽²ⁿ⁺²⁾)/4=(2^(1/2)(X⁽ⁿ⁾))/m+1 (add 1)

(X⁽²ⁿ⁺²⁾)=4(2^(1/2)(X⁽ⁿ⁾))/m+4 (multiply by 4)

 

(X⁽²ⁿ⁺²⁾)≠4(2^(1/2)(X⁽ⁿ⁾))/m+4

******************************************************************

The third prism.

 

If (X⁽ⁿ⁾) is odd,

2^(1/2)(X⁽ⁿ⁺¹⁾), (X⁽ⁿ⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²) (Two legs, followed by the hypotenuse.)

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)

mP, m(P²/2-1/2), m(P²/2+1/2)

 

If P is (X⁽ⁿ⁾), is m(P²/2-1/2)=2^(1/2)(X⁽ⁿ⁺¹⁾)?

 

m((X⁽ⁿ⁾)²/2-1/2)=2^(1/2)(X⁽ⁿ⁺¹⁾)

(X⁽ⁿ⁾)²/2-1/2=(2^(1/2)(X⁽ⁿ⁺¹⁾))/m (divide by m)

(X⁽²ⁿ⁾)/2-1/2=(2^(1/2)(X⁽ⁿ⁺¹⁾))/m (factor the ²)

(X⁽²ⁿ⁾)/2=(2^(1/2)(X⁽ⁿ⁺¹⁾))/m+1/2 (add 1/2)

(X⁽²ⁿ⁾)=2(2^(1/2)(X⁽ⁿ⁺¹⁾))/m+1 (multiply by 2)

 

(X⁽²ⁿ⁾)≠2(2^(1/2)(X⁽ⁿ⁺¹⁾))/m+1

 

If (X⁽ⁿ⁾) is even,

(X⁽ⁿ⁾), 2^(1/2)(X⁽ⁿ⁺¹⁾), (2(X⁽ⁿ⁺¹⁾)²+(X⁽ⁿ⁾)²)

mP, m((P/2)²-1), m((P/2)²+1)

 

If P is (X⁽ⁿ⁾), is m((P/2)²-1)=2^(1/2)(X⁽ⁿ⁺¹⁾)?

 

m(((X⁽ⁿ⁾)/2)²-1)=2^(1/2)(X⁽ⁿ⁺¹⁾)

((X⁽ⁿ⁾)/2)²-1=(2^(1/2)(X⁽ⁿ⁺¹⁾))/m (divide by m)

((X⁽²ⁿ⁾)/4)-1=(2^(1/2)(X⁽ⁿ⁺¹⁾))/m (factor the ²)

(X⁽²ⁿ⁾)/4=(2^(1/2)(X⁽ⁿ⁺¹⁾))/m+1 (add 1)

(X⁽²ⁿ⁾)=4(2^(1/2)(X⁽ⁿ⁺¹⁾))/m+4 (multiply by 4)

 

(X⁽²ⁿ⁾)≠4(2^(1/2)(X⁽ⁿ⁺¹⁾))/m+4

[dream_weaver]

 

I really doubt it.  Some of the greatest mathematicians of all time have had a go at it,  and Wiles was the one who finally did it after 300+ years.  I will believe there is a short elementary proof when I see it with my own eyes.l

 

I’m not positive that I got all of the Pythagorean triples.

Could you identify or provide one that the formula might have missed?

[dream_weaver]

A Pythagorean triple has just been suggeted which indeed my formula does not take into consideration. 20, 21, 29. If there is one, there are more.

[dream_weaver]

One last go. FermatsLC_Sheet_1.pdf (60.6 KB)FermatsLC_Sheet_2.pdf (156 KB)FermatsLC_Sheet_3.pdf (176 KB)

[Plasmatic]

Your PDF made it much more concrete for me.

[dream_weaver]

Another aspect to take into consideration.

 

While there are many configurations that can be factored from 60,
60 is the only quantity which satisfies the Pythagorean triple equation.

FermatsLC_Sheet_4.pdf (174 KB)

 

It is my contention that this should provide a perceptually graspable solution to Fermat’s Last Conjecture/Theorem.

 

The exercise certainly punctuated the objectivity of number elucidated by Dr. Corvini, for me, It comes from reality. That’s all there is!

[dream_weaver]

A further clarification generated from some recent feedback -

 

The hexahedrons are arranged the way they are presented, was to show the 3 repeating patterns that emerged from that approach. It is from those patterns, three hexahedrons Xⁿ x Xⁿ x Xⁿ, X⁽ⁿ⁺¹⁾ x Xⁿ x Xⁿ and X⁽ⁿ⁺¹⁾ x X⁽ⁿ⁼¹⁾ x X⁽ⁿ⁾ is taken.

This pattern repeats within the X^(n>3).

The “Pythagorean hexahedron” block representation showed the classic 3 x 4 x 5 relationship as the height width and depth.

Just as only one triangle of 3 specifically different length sides can be constructed, the Pythagorean triange bears its relationship of the specificity of their sides relationship to each other.

The repeating pattern in the X^(n>3) shows a pattern that any two sides are like a triangle whose legs are equal, the third being equal or 1 exponent more or less.

On page 4, the 3 x 4 x 5 hexahedrons is analysed. Several different hexahedrons can be made of the 60 blocks, but only one that satisfies the Pythagorean side of the equation.

I think the reason is relational between the 3 legs of the Pythagorean triangle and the height width and depth of the “Pythagorean hexahedron”

[guest_math_bot]

You know, Bertrand Russel once spent 300 pages proving 1 + 1 = 2 …for some reason this reminded me of that.

Of course, a person with a brain, unlike a well-educated moron such as Bertand, would just hold use their fingers to prove that and skip the abstract algebra.

[dream_weaver]

Thanks. I have no intention of going 300 pages on this. What I see is clear. Trying to articulate it clearer is more difficult.

The Pythagorean hexahedron  resolves into a Polyhedgron triangle of height wiidth and depth that satisfies (a²-b²), 2ab, (a²+b²), while the Xⁿ hexhedron height, width, depth resolves into two essential triangles, an equilateral triangle where the sides are all Xⁿ or X⁽ⁿ⁺¹⁾ or an isoceles triange where the 2 respectives sides are either Xⁿ  and X⁽ⁿ⁺¹⁾.

 

The 4 patge abstract algebra does not concretise this as well as the 4 page pictorial illustration.

[guest_math_bot]

Thanks. I have no intention of going 300 pages on this. What I see is clear. Trying to articulate it clearer is more difficult.

The Pythagorean hexahedron  resolves into a Polyhedgron triangle of height wiidth and depth that satisfies (a²-b²), 2ab, (a²+b²), while the Xⁿ hexhedron height, width, depth resolves into two essential triangles, an equilateral triangle where the sides are all Xⁿ or X⁽ⁿ⁺¹⁾ or an isoceles triange where the 2 respectives sides are either Xⁿ  and X⁽ⁿ⁺¹⁾.

 

The 4 patge abstract algebra does not concretise this as well as the 4 page pictorial illustration.

Not accusing you of anything, just to be clear.  Just saying, the thread reminds me of that.  Probably the 100 or so proof mentioned before, which I am sure is overly complicated in a similar manner.  Though I doubt the theorem is important enough to justify all that.

I am not sure how well your ideas work here as I am not quite convinced it qualifies as proof as such.  Though it is certainly interesting.

I am not sure if anyone mentioned this, but you might find it interesting :

http://www.fermatproof.com/

 

[dream_weaver]

.<delete>