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

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Using an Online Equation Solver, plugging in 3(C-A)(CA)+(C-A)3 returned (C-A)(C2+CA+A2)

Solving for A it returned the following three equations.

 

1.) A=-((31/2i+1)C)/2

2.) A=((31/2i-1)C)/2

3.) C=A

 

The earlier breakdown of An+Bn=Cn in post #69 yielded what was plugged into the online equation solver.

 

Borrowing from Wolfram's website: Fermat's last theorem is a theorem first proposed by Fermat in the form of a note scribbled in the margin of his copy of the ancient Greek text Arithmetica by Diophantus. The scribbled note was discovered posthumously, and the original is now lost. However, a copy was preserved in a book published by Fermat's son. In the note, Fermat claimed to have discovered a proof that the Diophantine equation xn+yn=zn(represented by An+Bn=Cn in post #69) has no integer solutions for n>2 and x,y,z≠0.

 

Would these three equations, more formally compiled, be sufficient demonstrate that fact? If so, it should be able to be done. If not, then the wrong tree is being barked up.

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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.

http://en.wikipedia.org/wiki/Theorem

 

Oops! The statement quoted above has been proven and therefor is a theorem.

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Yes, it was declared proven in 1993 with Andrew Wiles submission.

 

In post #7, you had stated: "I have not read the proof cited above, but I consider it possible that Fermat might have had, or someone else might discover, a much more elegant and insightful proof that is much shorter than 109 pages. I'm sure it wouldn't be the first time that a long proof could be replaced by a much shorter one, but I don't have any examples at hand to cite."

 

I'm wondering if the approach I'm considering might be a more elegant and insightful proof. Before reorganizing the material, if the formulas in post #76 would not clearly demonstrate this, why spend the cad time illustrating how the euclidean approach to the sum of the squares might be applied to cubes in a similar fashion to derive them?

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Just to be clear, this is what you need to prove:

 

If n is a natural number greater than 2, then there do not not exist natural numbers a, b, and c such each of a, b and c is greater than 0 and a^n+b^n=c^n.

Edited by GrandMinnow

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Just to be clear, this is what you need to prove:

 

If n is a natural number greater than 2, then there do not not exist natural numbers a, b, and c such each of a, b and c is greater than 0 and a^n+b^n=c^n./;

Indeed.  Some indication has been given that it may not work for some numbers / some ranges of numbers, however that is not sufficient.  At best it is an attempt to establish proof by enumeration.  We need to establish a more inductive proof here that does not basically enumerate.

Though the working we have seen does seem rather interesting.  I am sure this is how many theorems started, seemingly chaotic and impossible to understand.  Where really the originator just did not know exactly how to present the evidence in a convincing way.  Though such troubles can also indicate the originator does not fully understand the theory himself yet.

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Consider the nature or breakdown of the cubed number sequence.

 

13=1 is 0*6+1 more than 03 or (0*6)+1=1

23=8 is 1*6+1 more than 13 or (1*6)+1 = 7

33=27 is 2*6+1 more than 23 or (3*6)+1 = 19

43=64 is 3*6+1 more than 33 or (6*6)+1 = 37

53=125 is 4*6+1 more than 43 or (10*6)+1 = 61

63=216 is 5*6+1 more than 53 or (15*6)+1 = 91

73=343 is 6*6+1 more than 63 or (21*6)+1 = 127

83=512 is 7*6+1 more than 73 or (28*6)+1 = 169

93=729 is 8*6+1 more than 83 or (36*6)+1 = 217

103=1000 is 9*6+1 more than 93 or (45*6)+1 = 271

 

While the first multiplier is sequential, the second sequence is what is known as the Triangular Number Sequence.

 

The formula for the Triangular Number Sequence can be expressed as: n(n-1)/2

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, . . .

 

Consider also a formula for the sum of the squares expressed as: n(n-1)(2n-1)/6

0, 1, 5, 14, 30, 55, 91, 140, 204, 285

 

Adding these two formulas together and dividing by 2

(0+0)/2=0, (1+1)/2=1, (3+5)/2=4, (6+14)/2=10, (10+30)/2=20, (15+55)/2=35,

(21+91)/2=56, (28+140)/2=84, (36+204)/2=120, (45+285)/2=165, . . .

or

0, 1, 4, 10, 20, 35, 56, 84, 120, 165, . . .

 

13=(0*6)+1=1

23=(1*6)+2=8

33=(4*6)+3=27

43=(10*6)+4=64

53=(20*6)+5=125

63=(35*6)+6=216

73=(56*6)+7=343

83=(84*6)+8=512

93=(120*6)+9=729

103=(165*6)+10=1000

 

Looking at the relationship between these two formulas demonstrates the following:

 

13=(0*6)+1

or (0*6)+1=1

 

23=(0*6)+1+(1*6)+1

or (1*6)+2=8

 

33=(0*6)+1+(1*6)+1+(3*6)+1

or (4*6)+3=27

 

43=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1

or (10*6)+4=64

 

53=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1

or (20*6)+5=125

 

63=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1+(15*6)+1

or (35*6)+6=216

 

73=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1+(15*6)+1+(21*6)+1

or (56*6)+7=343

 

83=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1+(15*6)+1+(21*6)+1+(28*6)+1

or (84*6)+8=512

 

93=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1+(15*6)+1+(21*6)+1+(28*6)+1+(36*6)+1

or (120*6)+9=729

 

103=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1+(15*6)+1+(21*6)+1+(28*6)+1+(36*6)+1+(45*6)+1

or (165*6)+10=1000

 

Each sequential cube begins with begins with (0*6)+1+. . ., which n(n-1)/2+1 does not reintroduce additionally into the sequence while simultaneously satisfying the quantitative difference for the next cube in the sequence.

 

For sequential powers,

 

44=41*43=41*((0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1)

or 41*((10*6)+4)=256

 

45=42*43=42*((0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1)

or 42*((10*6)+4)=1024

 

46=43*43=43*((0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1)

or 43*((10*6)+4)=4096

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Each sequential cube begins with begins with (0*6)+1+. . ., which n(n-1)/2+1 does not reintroduce additionally into the sequence while simultaneously satisfying the quantitative difference for the next cube in the sequence.

 

should have read:

 

Each sequential cube begins with begins with (0*6)+1+. . ., which (n(n-1)/2)*6)+1 does not reintroduce additionally into the sequence while simultaneously satisfying the quantitative difference for the next cube in the sequence.

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Fermat’s Last Theorem

 

In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two.

 

This can be shown for n3 by demonstrating a few simple relationships that are inherent in every cubed positive integer along the cubed number sequence.

 

A. Consider the cubed sequence.

           

13, 23, 33, 43, 53, 63, 73, 83, 93, 103

 

This factors out to the cubed sequence:

 

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000

 

The difference between each two consecutive numbers can be expressed as the formula: (n(n-1)/2)*6+1 as a general principle.

 

1, 7, 19, 37, 61, 91, 127, 169, 217, 271

 

The only difference enumerated appearing in both the cube sequence and difference between them is 1.

 

B. The first portion expression (n(n-1)/2) yields the Triangular Number Sequence.

 

0, 1, 3, 6, 10, 15, 21, 28, 36, 45

 

(0*6+1)=1

(1*6+1)=7

(3*6+1)=19

(6*6+1)=37

(10*6+1)=61

(15*6+1)=91

(21*6+1)=127

(28*6+1)=169

(36*6+1)=217

(45*6+1)=271

 

Multiplying the Triangular Number Sequence by 6 and adding 1 to it, and summing this sequence in reverse order,.

 

(45*6+1)=271

(45*6+1)+(36*6+1)=81*6+2=488

(45*6+1)+(36*6+1)+(28*6+1)=109*6+3=657

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)=130*6+4=784

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)+(15*6+1)=145*6+5=875

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)+(15*6+1)+(10*6+1)=155*6+6=936

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)+(15*6+1)+(10*6+1)+(6*6+1)=161*6+7=973

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)+(15*6+1)+(10*6+1)+(6*6+1)+(3*6+1)=164*6+8=992

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)+(15*6+1)+(10*6+1)+(6*6+1)+(3*6+1)+(1*6+1)=165*6+9=999

 

Or putting it in another way, more exhaustively:

217+169 = 386

169+127 = 296

127+91 = 218

91+61 = 152

61+37 = 98

37+19 = 56

19+7 = 26

217+169+127 = 513

169+127+91 = 387

127+91+61 = 279

91+61+37 = 189

61+37+19 = 117

37+19+7 = 63

217+169+127+91 = 604

169+127+91+61 = 448

127+91+61+37 = 340

91+61+37+19 = 208

61+37+19+7 = 127

217+169+127+91+61 = 665

169+127+91+61+37 = 485

127+91+61+37+19 = 335

91+61+37+19+7 = 215

217+169+127+91+61+37 = 702

169+127+91+61+37+19 = 504

127+91+61+37+19+7 = 342

217+169+127+91+61+37+19 = 721

169+127+91+61+37+19+7 = 511

217+169+127+91+61+37+19+7 = 728

 

None of these result in a cube.

 

Only when the entire reverse sequence is taken into consideration is the result a cubed positive integer among the enumerated sequences.

 

(45*6+1)+(36*6+1)+(28*6+1)+(21*6+1)+(15*6+1)+(10*6+1)+(6*6+1)+(3*6+1)+(1*6+1)+(0*6+1)=165*6+10 = 1000 = 103

or

271+217+169+127+91+61+37+19+7+1 = 1000 or 103

 

It is only summing the entire sequence in order that yields the sequential cube sequence.

 

(0*6+1) =

   (0+1) = 1 or 13

 

(0*6+1)+(1*6+1) =

   (0+1)+    (6+1) = 8 or 23

 

(0*6+1)+(1*6+1)+(3*6+1) =

   (0+1)+    (6+1)+  (18+1) = 27 or 33

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1) =

   (0+1)+    (6+1)+  (18+1)+  (36+1) = 64 or 43

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+(10*6+1) =

   (0+1)+    (6+1)+  (18+1)+  (36+1)+    (60+1) = 125 or 53

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+(10*6+1)+(15*6+1) =

   (0+1)+    (6+1)+  (18+1)+  (36+1)+    (60+1)+    (90+1) = 216 or 63

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+(10*6+1)+(15*6+1)+(21*6+1) =

   (0+1)+    (6+1)+  (18+1)+  (36+1)+    (60+1)+    (90+1)+  (126+1) = 343 or 73

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+(10*6+1)+(15*6+1)+(21*6+1)+(28*6+1) =

   (0+1)+    (6+1)+  (18+1)+  (36+1)+    (60+1)+    (90+1)+  (126+1)+  (168+1) = 512 or 83

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+ . . . +((n(n-1)/2)*6+1)=n3

 

C. Adding two cubed numbers together demonstrates in this way why a3+b3 cannot = c3

           

Assuming b is greater than a,

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+ . . . +(a(a-1)/2)*6+1 = a3

 

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+ . . . +(a(a-1)/2)*6+1+ . . . +(b(b-1)/2)*6+1 = b3

 

Add these two sequences together

 

2((0*6+1))+2((1*6+1))+2((3*6+1))+2((6*6+1))+ . . .+2((a(a-1)/2)*6+1)+. . . +(b(b-1)/2)*6+1 

 

This simply cannot not be a cubed number. It violates the identified form.

 

D. For exponents greater than 3, the same reasoning applies.

           

Consider 44:

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)

(0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)

 

Summed up is:

4(0*6+1)+4(1*6+1)+4(3*6+1)+4(6*6+1)

 

Consider 34:

(0*6+1)+(1*6+1)+(3*6+1)

(0*6+1)+(1*6+1)+(3*6+1)

(0*6+1)+(1*6+1)+(3*6+1)

 

Summed up is:

3(0*6+1)+3(1*6+1)+3(3*6+1)

 

To add 44+34:

4(0*6+1)+4(1*6+1)+4(3*6+1)+4(6*6+1)

3(0*6+1)+3(1*6+1)+3(3*6+1)

sums up to

7(0*6+1)+7(1*6+1)+7(3*6+1)+4(6*6+1)

 

But, 74 would be:

7(0*6+1)+7(1*6+1)+7(3*6+1)+7(6*6+1)+. . .+7((7(7-1)/2)6+1)

 

For a4 and b4 (again, b>a):

a(0*6+1)+a(1*6+1)+a(3*6+1)+a(6*6+1)+. . .+a((a(a-1)/2)6+1)

b(0*6+1)+b(1*6+1)+b(3*6+1)+b(6*6+1)+. . .+b((a(a-1)/2)6+1)+. . .+b((b(b-1)/2)6+1)

 

adds up to:

(a+B)(0*6+1)+(a+B)(1*6+1)+(a+B)(3*6+1)+(a+B)(6*6+1)+. . .+(a+B)((a(a-1)/2)6+1)+...+b((b(b-1)/2)6+1)

or if a=b it would add up to:

2a(0*6+1)+2a(1*6+1)+2a(3*6+1)+2a(6*6+1)+. . .+2a((a(a-1)/2)6+1)

both of which violate the identified form.

 

To solve for a5+b5 (again, b>a):

a2(0*6+1)+a2(1*6+1)+a2(3*6+1)+a2(6*6+1)+. . .+a2((a(a-1)/2)6+1)

b2(0*6+1)+b2(1*6+1)+b2(3*6+1)+b2(6*6+1)+. . .+b2((a(a-1)/2)6+1)+. . .+b2((b(b-1)/2)6+1)

 

sums up to

(a2+b2)(0*6+1)+(a2+b2)(1*6+1)+(a2+b2)(3*6+1)+(a2+b2)(6*6+1)+. . .+(a2+b2)((a(a-1)/2)6+1)+...+b2((b(b-1)/2)6+1)

or if a=b it would add up to:

2a2(0*6+1)+2a2(1*6+1)+2a2(3*6+1)+2a2(6*6+1)+. . .+2a2((a(a-1)/2)6+1)

both of which violate the identified form.

 

QED

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Some further elaborations on Fermat’s Last Theorem

 

I. The counting numbers, or integers are simply:

 

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 . . .

 

II. These can be added together into what is known as the triangular number sequence:

 

0, 0+1=1, 1+2=3, 3+3=6, 6+4=10 . . .

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55 . . .

 

III. The triangular number sequence can be added together into what is known as the tetrahedral number sequence:

 

0, 0+1=1, 1+3=4, 4+6=10, 10+10=20 . . .

0, 1, 4, 10, 20, 35, 56, 84, 120, 165, 220 . . .

 

IV. The formula for the triangular number sequence used for this exploration is: n(n-1)/2

See Induction of Triangular Numbers by Dr. Math.

 

V. The formula for the tetrahedral number sequence will be: n(n-1)(n+1)/6

See Induction of Tetrahedral numbers by Dr. Math.

 

VI. Multiplying the nth triangular number by 6 and adding 1 to it provides the difference between n3 and (n-1)3.

 

For example, n=7 where 7(7-1)/2=7(6)/2=21, and 6(21)+1=127 with the difference between 73-(7-1)3=343-(6)3=343-216=127

 

VII. Multiplying the nth tetrahedral number by 6 and adding n to it is n3.

 

For example, n=7 where 7(7-1)(7+1)/6=7(6)(8)/6=56, and 6(56)+7=343

 

VIII. Since the tetrahedral number sequence is derived from the triangular number sequence, any nth tetrahedral number is the sum of the entire triangular number sequence up to n.

 

For example, n=7 in the tetrahedral sequence is 56, and using the formula n(n-1)/2 starting with 1, working sequential up the counting numbers to 7 adds up as follows: 0+1+3+6+10+15+21=56

 

IX. Extending the usage of the difference between the cubed numbers from section VI, multiplying by 6 and adding 1, and applying it to how the tetrahedral numbers are generated from section III, works out as follows for n=7:

 

6(0)+1+6(1)+1+6(3)+1+6(6)+1+6(10)+1+6(15)+1+6(21)+1 which can be rearranged as 6(0)+6(1)+6(3)+6(6)+6(10)+6(15)+6(21)+1+1+1+1+1+1+1 which can be combined as 6(0+1+3+6+10+15+21)+7 which contains the same sequence shown back in VIII, which is also the same as 6(56)+7 as shown in section VII.

(This is as close to an induction of the cubed number sequence as I’ve discovered how to generate.)

 

X. This clearly illustrates the relationship of the cubed number sequence to the tetrahedral number sequence, and how the difference between them relates to the triangular number sequence, which in turn relates back to the integers.

 

XI. Adding two tetrahedral numbers together does not produce a tetrahedral number.

A tetrahedral number is the sum of the triangular number sequence.

 

Triangular number sequence a and b

0+1+3+6+10+15+21+…=tetrahedral number a

0+1+3+6=tetrahedral number b

Added together yield

0+2+6+12+10+15+21+… is not a tetrahedral number

 

XII. Subtracting a smaller tetrahedral number from a larger does not produce a tetrahedral number.

 

0+1+3+6+10+15+21+…=tetrahedral number c

0+1+3+6=tetrahedral number b or a

Subtracted from one another yield

0+0+0+0+10+15+21+… is not a tetrahedral number

 

XIII. Considering the observations in sections XI & XII,

 

6(sum of two tetrahedral numbers)+(a+B) ≠ 6(tetrahedral number)+c

a3+b3 ≠ c3

 

6(difference between two tetrahedral numbers)+(c-a) ≠ 6(tetrahedral number)+b

c3-a3 ≠ b3

 

6(difference between two tetrahedral numbers)+(c-B) ≠ 6(tetrahedral number)+a

c3-b3 ≠ a3

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