dream_weaver Posted March 27, 2013 Author Report Share Posted March 27, 2013 Using an Online Equation Solver, plugging in 3(C-A)(CA)+(C-A)^{3} returned (C-A)(C^{2}+CA+A^{2}) Solving for A it returned the following three equations. 1.) A=-((3^{1/2}i+1)C)/2 2.) A=((3^{1/2}i-1)C)/2 3.) C=A The earlier breakdown of A^{n}+B^{n}=C^{n }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 x^{n}+y^{n}=z^{n}(represented by A^{n}+B^{n}=C^{n} 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. Quote Link to comment Share on other sites More sharing options...

John Link Posted March 27, 2013 Report Share Posted March 27, 2013 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. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 27, 2013 Author Report Share Posted March 27, 2013 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? Quote Link to comment Share on other sites More sharing options...

GrandMinnow Posted March 28, 2013 Report Share Posted March 28, 2013 (edited) 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 March 28, 2013 by GrandMinnow Quote Link to comment Share on other sites More sharing options...

Guest Math Bot Posted March 28, 2013 Report Share Posted March 28, 2013 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. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted April 6, 2013 Author Report Share Posted April 6, 2013 Consider the nature or breakdown of the cubed number sequence. 1^{3}=1 is 0*6+1 more than 0^{3} or (0*6)+1=1 2^{3}=8 is 1*6+1 more than 1^{3} or (1*6)+1 = 7 3^{3}=27 is 2*6+1 more than 2^{3} or (3*6)+1 = 19 4^{3}=64 is 3*6+1 more than 3^{3} or (6*6)+1 = 37 5^{3}=125 is 4*6+1 more than 4^{3} or (10*6)+1 = 61 6^{3}=216 is 5*6+1 more than 5^{3} or (15*6)+1 = 91 7^{3}=343 is 6*6+1 more than 6^{3} or (21*6)+1 = 127 8^{3}=512 is 7*6+1 more than 7^{3} or (28*6)+1 = 169 9^{3}=729 is 8*6+1 more than 8^{3} or (36*6)+1 = 217 10^{3}=1000 is 9*6+1 more than 9^{3} 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, . . . 1^{3}=(0*6)+1=1 2^{3}=(1*6)+2=8 3^{3}=(4*6)+3=27 4^{3}=(10*6)+4=64 5^{3}=(20*6)+5=125 6^{3}=(35*6)+6=216 7^{3}=(56*6)+7=343 8^{3}=(84*6)+8=512 9^{3}=(120*6)+9=729 10^{3}=(165*6)+10=1000 Looking at the relationship between these two formulas demonstrates the following: 1^{3}=(0*6)+1 or (0*6)+1=1 2^{3}=(0*6)+1+(1*6)+1 or (1*6)+2=8 3^{3}=(0*6)+1+(1*6)+1+(3*6)+1 or (4*6)+3=27 4^{3}=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1 or (10*6)+4=64 5^{3}=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1 or (20*6)+5=125 6^{3}=(0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1+(10*6)+1+(15*6)+1 or (35*6)+6=216 7^{3}=(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 8^{3}=(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 9^{3}=(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 10^{3}=(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, 4^{4}=4^{1}*4^{3}=4^{1}*((0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1) or 4^{1}*((10*6)+4)=256 4^{5}=4^{2}*4^{3}=4^{2}*((0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1) or 4^{2}*((10*6)+4)=1024 4^{6}=4^{3}*4^{3}=4^{3}*((0*6)+1+(1*6)+1+(3*6)+1+(6*6)+1) or 4^{3}*((10*6)+4)=4096 Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted April 7, 2013 Author Report Share Posted April 7, 2013 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. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted April 11, 2013 Author Report Share Posted April 11, 2013 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 a^{n} + b^{n} = c^{n} for any integer value of n greater than two. This can be shown for n^{3} by demonstrating a few simple relationships that are inherent in every cubed positive integer along the cubed number sequence. A. Consider the cubed sequence. 1^{3}, 2^{3}, 3^{3}, 4^{3}, 5^{3}, 6^{3}, 7^{3}, 8^{3}, 9^{3}, 10^{3} 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 = 10^{3} or 271+217+169+127+91+61+37+19+7+1 = 1000 or 10^{3} It is only summing the entire sequence in order that yields the sequential cube sequence. (0*6+1) = (0+1) = 1 or 1^{3} (0*6+1)+(1*6+1) = (0+1)+ (6+1) = 8 or 2^{3} (0*6+1)+(1*6+1)+(3*6+1) = (0+1)+ (6+1)+ (18+1) = 27 or 3^{3} (0*6+1)+(1*6+1)+(3*6+1)+(6*6+1) = (0+1)+ (6+1)+ (18+1)+ (36+1) = 64 or 4^{3} (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 5^{3} (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 6^{3} (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 7^{3} (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 8^{3} (0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+ . . . +((n(n-1)/2)*6+1)=n^{3} C. Adding two cubed numbers together demonstrates in this way why a^{3}+b^{3 }cannot = c^{3} Assuming b is greater than a, (0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+ . . . +(a(a-1)/2)*6+1 = a^{3} (0*6+1)+(1*6+1)+(3*6+1)+(6*6+1)+ . . . +(a(a-1)/2)*6+1+ . . . +(b(b-1)/2)*6+1 = b^{3} 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 4^{4}: (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 3^{4}: (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 4^{4}+3^{4}: 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, 7^{4} 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 a^{4} and b^{4 }(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+(0*6+1)+(a+(1*6+1)+(a+(3*6+1)+(a+(6*6+1)+. . .+(a+((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 a^{5}+b^{5} (again, b>a): a^{2}(0*6+1)+a^{2}(1*6+1)+a^{2}(3*6+1)+a^{2}(6*6+1)+. . .+a^{2}((a(a-1)/2)6+1) b^{2}(0*6+1)+b^{2}(1*6+1)+b^{2}(3*6+1)+b^{2}(6*6+1)+. . .+b^{2}((a(a-1)/2)6+1)+. . .+b^{2}((b(b-1)/2)6+1) sums up to (a^{2}+b^{2})(0*6+1)+(a^{2}+b^{2})(1*6+1)+(a^{2}+b^{2})(3*6+1)+(a^{2}+b^{2})(6*6+1)+. . .+(a^{2}+b^{2})((a(a-1)/2)6+1)+...+b^{2}((b(b-1)/2)6+1) or if a=b it would add up to: 2a^{2}(0*6+1)+2a^{2}(1*6+1)+2a^{2}(3*6+1)+2a^{2}(6*6+1)+. . .+2a^{2}((a(a-1)/2)6+1) both of which violate the identified form. QED Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted April 29, 2013 Author Report Share Posted April 29, 2013 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 n^{th} triangular number by 6 and adding 1 to it provides the difference between n^{3} 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 7^{3}-(7-1)^{3}=343-(6)^{3}=343-216=127 VII. Multiplying the n^{th} tetrahedral number by 6 and adding n to it is n^{3}. 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 n^{th} 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+ ≠ 6(tetrahedral number)+c a^{3}+b^{3} ≠ c^{3} 6(difference between two tetrahedral numbers)+(c-a) ≠ 6(tetrahedral number)+b c^{3}-a^{3} ≠ b^{3} 6(difference between two tetrahedral numbers)+(c- ≠ 6(tetrahedral number)+a c^{3}-b^{3} ≠ a^{3} Quote Link to comment Share on other sites More sharing options...

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