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

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I think I have the right formulas now.

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((n-1)/3) and the longer one being 4X((n-1)/3). -->Sheet.6

X5 , X8, X11, . . . yeilds an isocoles triangle the sides being 4X((n-2)/3) with the shorter leg resolving to X((n-2)/3). -->Sheet.7

In these cases, X would be an integer, n would be the exponent >3

(a2-b2), (2ab), (a2+b2) is somehow contained in X in all three equations, much like it is here.

In the first equation, it would be getting X(n/3)and get the (a2+b2) deal out of it.

In the second equation, it would be getting X((n-1)/3) out of it.

And the third would come from X((n-2)/3).

The other sheets uploaded regarding this are in post 33 and post 45

Edited by dream_weaver

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Separate the cube as indicated and end up with

1 - b3, 3 - b2 x (a- , 3 - b x (a- 2, and 1 (a- 3

These don't appear to be able to be reassembled into two cubes of their own.

Those dang-nab emoticons.

Edited by dream_weaver
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The second group breaks down into the following volumes:

1 - b3

3 - b2 x (a- 3 - (b x (a- 2

1 - (a- 3

1 - b2 x 3a

2 - b x (a- x 3a

1 - (a- 2 x 3a

And the final one yeilds these sub-volumes.

1 - b3

3 - b2 x (a- 3 - b x (a- 2

1 - (a- 3

2 - b2 x 3a

4 - b x (a- x 3a

2 - (a- 2 x 3a

1 - 3a2 x b

1 - 3a2 x (a- Edited by dream_weaver
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Maybe Fermat was not trying to solve for An+Bn=Cn, but it came as a by-product of applying the Euclidean method to cube space.to derive the a, b, (a- relationship and noticed the resulting blocks could not be re-arranged into 2 separate cubes. Considering he is one of the fore-runners to Galileo, Leibniz and Newton, this comes to mind viewing how this can be disassembled so far.

Edited by dream_weaver
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Sorry, it should have read post 46.

I don't consider post 46 to be a summary of your ideas. There's no way to know, from that post alone, what you're talking about.

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This is becoming way more complex that I thought it was at first.

The block illustrations are like a three dimensional Euclidean puzzle. The quantity of blocks generated by X blocks to the nth power are assembled into the hexahedrens.

The a, b, a-b relationships are used to subdivide the hexahedrons into the list of blocks described in post 52 and post 53. Does that help any?

The size of the hexahedron would be explained by instead of triangles, a hexahedron X(n/3) x X(n/3) x X(n/3).,

one of X((n-1)/3) x X((n-1)/3) x 4X((n-1)/3)

and finally one of 4X((n-2)/3) x 4X((n-2)/3) x X((n-2)/3).

where X is the number of cubes began with.used in conjuntion with an nth power.

Edited by dream_weaver
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This is becoming way more complex that I thought it was at first.

The block illustrations are like a three dimensional Euclidean puzzle. The quantity of blocks generated by X blocks to the nth power are assembled into the hexahedrens.

The a, b, a-b relationships are used to subdivide the hexahedrons into the list of blocks described in post 52 and post 53. Does that help any?

No, it does not! I think you might get a few people interested in your ideas if you were to write a summary that explained things from scratch. If you don't do that I expect that you will continue to have a conversation all by yourself. If that's what you want, then go ahead. But if you want to interest anyone else in your ideas then please write a summary that starts from scratch and assumes the reader knows nothing about your subject. Got it?

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Thank-you.

I wish I could summarize it better.

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Ok. I've modified the last three sheets to discover that the block count in posts 52 and 53 were incorrect. Hopefully this will help tie some more of it together. It's hard to descibe 3 dimensional object in word, much alone discovering relational formulas between the various aspects.

Please keep in mind, I've never really attempted anything like this before.

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Oh, the block count was correct. I had not originated the diagram with that in mind at the time.

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Thank-you.

I wish I could summarize it better.

Actually you've supplied no summary at all! A summary would require you to start from scratch explaining your ideas. It seems you might be on to something interesting, but I have no desire to wade through your meanderings which I suspect are full of dead ends. And your posts 59 and 60 are just more of your conversation with yourself.

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Sorry John. I'm getting glimpses of what I am onto, and am trying to combine them into a combination of explaination, supplimented by picture where I'm having difficulty explaining.

What I'm seeing here is like a 3-d application of Euclid's Pythagorean proof. Right now that's the best I can do.

I think why this thread looks like a conversation with myself, is that I'm still trying to flesh out the connections in my mind.

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Sorry John. I'm getting glimpses of what I am onto, and am trying to combine them into a combination of explaination, supplimented by picture where I'm having difficulty explaining.

What I'm seeing here is like a 3-d application of Euclid's Pythagorean proof. Right now that's the best I can do.

I think why this thread looks like a conversation with myself, is that I'm still trying to flesh out the connections in my mind.

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Ok. these should provide some clarifying corrections to the last three sheets, and this explaination may help to clarify the approach to the problem.

Remember, the Euclidians used the method found here to use the square to solve for the area of a right triangle, which essentially decombiles C into the A and B counterparts.

If I am correct, Fermat is using the repeating shape he observed in post 33 which also appears using 4 as the number of blocks to apply the process.

Cn produces the three distinct hexahedron patterns illustrated, again for the example shown using 4n. The three patterns then can be broken down into 8 smaller pieces in terms of A, (A- , B, and in the case of the intermediate patterns between the block shaped on, adding the addition working the addition of the A2,(A2-A) that show up in the 5th and 6th powers.

To confirm his conclusion, the list of the 8 sub-volumes identified in this last update are what need to be taken through the Euclidian exercise referenced to ascertain it.

edits:

Emoticons and math do not get along. Also, on sheet 7, the (A2- should have been (A2-A).

Edited by dream_weaver
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Case 1.) A3, 6, 9 . . .

(A- 3=A3+3(AB2-A2 -B3

B3=A3+3(AB2-A2B)-(A- 3

Case 2.) A4, 7, 10 . . .

(A- 4=(A- 4+2(2AB2-A2 B4=B4+2(2AB2-A2 Case 3.) A5, 8, 11 . . .

(A- 5=(A- 5+2(A5-2A4+A3-A4B+2A3B-AB)

B5=B5+2(A5-2A4+A3-A4B+2A3B-AB)

I think that's QED.

Edited by dream_weaver
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Case 1.) A3, 6, 9 . . .

(A- 3=A3+3(AB2-A2 -B3

B3=A3+3(AB2-A2B)-(A- 3

Case 2.) A4, 7, 10 . . .

(A- 4=(A- 4+2(2AB2-A2 B4=B4+2(2AB2-A2 Case 3.) A5, 8, 11 . . .

(A- 5=(A- 5+2(A5-2A4+A3-A4B+2A3B-AB)

B5=B5+2(A5-2A4+A3-A4B+2A3B-AB)

I think that's QED.

If it is, then please provide a clear statement of what you intend to prove and then carefully go through the steps of your proof.

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It looks like the A2-A relationship needs to be broke down into the proper (A- , B setup. It may be a couple of more days to put it in order.

Edited by dream_weaver
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While this may not be in "Offical Proof" organization, the results do work.

Here goes:

Case 1

Beginning with a cube having 3 sides of C, Remove a cube having sides of A

What remains can be broken into 4 subgroups, 3 units having a sides of C, A, & (C-A), and one unit having 3 sides of (C-A)

3(C-A)(CA)+(C-A)3

Do the same bye removing a cube having 3 sides of B.

3(C- (CB)+(C- 3

Case 2

Next take a hexahedron having 2 sides of C and one side of C2. Remove a hexahedron having 2 sides of A and one side of A2. Using the example of the cube from before, multiply the remains by A and add (C-A)C3 to it.

A(3(C-A)(CA)+(C-A)3)+C3(C-A)

Do the same by removing a hexahedron having 2 sides of B and one side of B2.

B(3(C- (CB)+(C- 3)+C3(C- Case 3

Finally, start with a hexahedron having 2 sides of C2 and one of C. Remove a hexahedron having 2 sides of A2 and one side of A. Using the example of the cube from before, multiply the remains by A2 and add 2C3A(C-A) +C3(C-A)2 to it.

A2(3(C-A)(CA)+(C-A)3)+2C3A(C-A)+C3(C-A)2

Do the same by removing a hexahedron having 2 sides of B2 and one of B.

B2(3(C- (CB)+(C- 3)+2C3B(C- +C3(C- 2

Working in the exponent factor led to:

1A.) Cn-An=3(C(n/3)-A(n/3))(C(n/3)A(n/3))+(C(n/3)-A(n/3))3

1B.) Cn-Bn=3(C(n/3)-B(n/3))(C(n/3)B(n/3))+(C(n/3)-B(n/3))3

2A.) C(n+1)-A(n+1)=A(3(C(n/3)-A(n/3))(C(n/3)A(n/3))+(C(n/3)-A(n/3))3)+C3n/3(C-A)

2B.) C(n+1)-B(n+1)=B(3(C(n/3)-B(n/3))(C(n/3)B(n/3))+(C(n/3)-B(n/3))3)+C3n/3(C- 3A.) C(n+2)-A(n+2)=A2(3(C(n/3)-A(n/3))(C(n/3)A(n/3))+(C(n/3)-A(n/3))3)+2C3n/3A(C-A)+C3n/3(C-A)2

3B.) C(n+2)-B(n+2)=B2(3(C(n/3)-B(n/3))(C(n/3)B(n/3))+(C(n/3)-B(n/3))3)+2C3n/3B(C- +C3n/3(C- 2

Case 1 for n=3, 6, 9, . . .

An+Bn≠Cn because it fell short by 3(C(n/3)-A(n/3))(C(n/3)A(n/3))+(C(n/3)-A(n/3))3-Bn or conversely 3(C(n/3)-B(n/3))(C(n/3)B(n/3))+(C(n/3)-B(n/3))3-An

Case 2 for n=4, 7, 10, . . .

An+Bn≠Cn because it fell short by A(3(C(n/3)-A(n/3))(C(n/3)A(n/3))+(C(n/3)-A(n/3))3)+C3n/3(C-A)-Bn or conversely B(3(C(n/3)-B(n/3))(C(n/3)B(n/3))+(C(n/3)-B(n/3))3)+C3n/3(C- -An

Case 3 for n=5, 8, 11 . . .

An+Bn≠Cn because it fell short by

A2(3(C(n/3)-A(n/3))(C(n/3)A(n/3))+(C(n/3)-A(n/3))3)+2C3n/3A(C-A)+C3n/3(C-A)2-Bn

or conversely

B2(3(C(n/3)-B(n/3))(C(n/3)B(n/3))+(C(n/3)-B(n/3))3)+2C3n/3B(C- +C3n/3(C- 2-An

Edited by dream_weaver
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Looking this over, it bears noting that testing for n=3, 6, 9 etc, simultaneously solves the other two cases. C3 becomes C4 or C5 by the inclusion of C or C2 in the equation.

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I'm still waiting for a statement of a theorem, including all the necessary definitions and examples, and then a proof.

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As previously mentioned, I've never dealt with math proofs with the exception of coming up with formulas that do what they are intended to do. Fermat's margin stated

"On the other hand, it is impossible to separate a cube into two cubes, or a biquadrate into two biquadrats, or generally any power except a square into two powers with the same exponent. I have discovered a truly marvelous proof of this, which, however, the margin is not large enough to contain."

Is this considered a theorem?

The illustration's used in the earlier pdf's, demonstrated in post #33 could be achieved using ordinary children's blocks. Starting with any quantity of blocks, cubing, or raising to the 6th, 9th, 12th etc,  the number of blocks can be arranged into a cube shape. This is the value taken as C.

For illustration purposes, use 73 or 343 blocks arrange into 7 blocks wide by 7 blocks deep by 7 blocks tall.

Remove 63 blocks or 216. Leave a row on the floor of 7 by 7 by 1 deep, and two adjacent walls 1 wide by 7 long by 7 tall, each sharing a common edge with the each of the other two.

The three walls can be broken into 4 separate groups, 3 groups of 7 by 6 by 1 with 1 block left over, a total of 127 blocks. The 63 blocks can be represented as A

3(C-A)(CA)+(C-A)3 or 3(7-6)(7*6)+(7-6)3=127

Repeat the procedure using 53 blocks or 125. Leaving on the floor 7 by 7 by 2 deep with two adjacent walls 2 wide by 7 long by 7 tall, each sharing the common edge with each of the other two.

The three walls can be broken into 4 separate groups, 3 groups of 7 by 5 by 2 with 8 blocks left over, a total of 218 blocks. The 53 blocks can be represented as B

3(C- (CB)+(C- 3 or 3(7-5)(7*5)+(7-6)3=218

Does A3+B3=C3? Does 63+53=73? Does 216+125=343? No. 216+125=341, precisely 2 less than 343.

3(C-A)(CA)+(C-A)3-B3 = 3(7-6)(7*6)+(7-6)3-53 = 3(1*42)+13-125 = 126+1-125=2

conversely

3(C-B(CB)+(C- 3-A3 = 3(7-5)(7*5)+(7-5)3-63 = 3(2*35)+23-216 = 210+8-216=2

Starting with 2 arrangement of blocks A & B or 63 & 53, these two stacks of blocks would be 2 short of being rearranged toward a 73 stack.

Starting with a 73 arrangement, a rearrangement of 63 & 53 could be created, with 2 extra blocks remaining.

That being said, as seductive a problem it is, the equations discovered describe a valid relationship. It has been a refresher course into this venue of life, and provided many hours of seeking out this particular aspect to it.

As to proof, the premise that has been kept in focus or in the forefront of the mind here is that it should consist of identifying what something is. The idea of proving that An+Bn≠Cn reminds me a bit of trying to prove that something does not exist, rather than via the identification of what does.

##### Share on other sites All you have to do is show that A^n + B^n = C^n is an invalid relationship for certain values.  You just have to prove that for values outside of a range, A^n + B^n = m, and that m =/= C^n. It does not require you to prove something does not exist, it merely requires you to prove for a range of values of n, an alleged relationship cannot hold because that would be a contradiction with the actual values of m.

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It looks like if 3(C-A)(CA)+(C-A)3 or 3(C- (CB)+(C- 3 could be shown not to be a cube root, since all 3 cases share that aspect in common, all three should fail for the same reason.

Factoring them yeild

(3C2A-3CA2)+(C3-3C2A+3CA2-A3) or (3C2B-3CB2)+(C3-3C2B+3CB2-B3). In both cases, the 3C2A-3CA2 and 3C2B-3CB2 get canceled out by the two middle terms resulting from the factoring of (C-A)3 or (C- 3. If anything could demonstrate an invalid relationship, it ought to be this.

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As previously mentioned, I've never dealt with math proofs with the exception of coming up with formulas that do what they are intended to do. Fermat's margin stated

"On the other hand, it is impossible to separate a cube into two cubes, or a biquadrate into two biquadrats, or generally any power except a square into two powers with the same exponent. I have discovered a truly marvelous proof of this, which, however, the margin is not large enough to contain."

Is this considered a theorem?

The assertion that "it is impossible to separate a cube into two cubes, or a biquadrate into two biquadrats, or generally any power except a square into two powers with the same exponent." would be a theorem once it has been proven.

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