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

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

2(Xn)<(X+1)n

An+Bn<>Cn

Edited by dream_weaver

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

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Except that both fail.

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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 (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))+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 (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))+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 (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))+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.

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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 (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+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 (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+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 (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+4 (multiply by 4)

(X(2n))≠4(2(1/2)(X(n+1)))/m+4

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&nbsp;

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

&nbsp;

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

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

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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.

Edited by dream_weaver

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Edited by dream_weaver

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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

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!

Edited by dream_weaver

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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"

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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.

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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.

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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 :

Edited by Math Bot

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.<delete>

Edited by dream_weaver

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Posts 29, 30, 33 and 35 turn out to be looking at a cube, not the triangle. While an interesting exercise, it turns out that in trying to see what made it tick, I lost track of what the "it" was.

Edited by dream_weaver

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Posts 29, 30, 33 and 35 turn out to be looking at a cube, not the triangle. While an interesting exercise, it turns out that in trying to see what made it tick, I lost track of what the "it" was.

Might you summarize anything useful you've discovered in your posts above?

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Yes, it might help if you were able to collect it into a unified, step-by-step proof.  Being scattered throughout so many posts makes it very hard to follow you know...

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<delete>

Edited by dream_weaver

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Ok. Went looking at some proofs on Pythagorean theorem last night. I ran across this site which inspired this illustration.

This should show the pattern I am seeing. Putting it into the language of mathematica is not exactly my strong suit. It uses 43, 44, 45 developed along the line illustrated in the pdf's contained in post 33.

Edited by dream_weaver

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I'm not sure what value this has, but it appears these three can be restated something like:

X3 , X6, X9, . . . yeilds an equalateral triangle the sides being X(n/3). -->Sheet.5

X4 , X7, X10, . . . yeilds three sides, two of which are X and the longer one being nX. -->Sheet.6

X5 , X8, X11, . . . yeilds an isocoles triangle the sides being (n-1)X with the shorter leg resolving to X. -->Sheet.7

I'm finding juggling the exponents a bit confusing to manage.

Edited by dream_weaver

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dream_weaver, are you going to write a summary of your ideas so that others might understand your thoughts, or are you simply having a conversation with yourself?

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I think I just posted a summary in post 46.

Edited by dream_weaver

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I think I just posted a summary in post 47.

I'm the one who wrote post 47, and I don't see any summary in your recent posts.

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Sorry, it should have read post 46.

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