dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 (edited) I think I have the right formulas now. X^{3} , X^{6}, X^{9}, . . . yeilds an equalateral triangle the sides being X^{(n/3)}. -->Sheet.5 X^{4} , X^{7}, X^{10}, . . . yeilds three sides, two of which are X^{((n-1)/3)} and the longer one being 4X^{((n-1)/3)}. -->Sheet.6 X^{5} , X^{8}, X^{11}, . . . 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 (a^{2}-b^{2}), (2ab), (a^{2}+b^{2}) 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 (a^{2}+b^{2}) 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 March 16, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 (edited) Separate the cube as indicated and end up with 1 - b^{3}, 3 - b^{2 }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 March 16, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 (edited) The second group breaks down into the following volumes: 1 - b^{3} 3 - b^{2 }x (a- 3 - (b x (a-^{2} 1 - (a-^{3} 1 - b^{2 }x 3a 2 - b x (a- x 3a 1 - (a-^{2 }x 3a And the final one yeilds these sub-volumes. 1 - b^{3} 3 - b^{2 }x (a- 3 - b x (a-^{2} 1 - (a-^{3} 2 - b^{2 }x 3a 4 - b x (a- x 3a 2 - (a-^{2 }x 3a 1 - 3a^{2 }x b 1 - 3a^{2 }x (a- Edited March 16, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 (edited) Maybe Fermat was not trying to solve for A^{n}+B^{n}=C^{n}, 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 March 16, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

John Link Posted March 16, 2013 Report Share Posted March 16, 2013 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. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 (edited) 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 n^{th} 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 n^{th} power. Edited March 16, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

John Link Posted March 16, 2013 Report Share Posted March 16, 2013 (edited) 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 n^{th} 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? Edited March 16, 2013 by John Link Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 Thank-you. I wish I could summarize it better. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 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. FermatsLC-Final_Sheet_5.pdfFermatsLC-Final_Sheet_6.pdfFermatsLC-Final_Sheet_7.pdf Please keep in mind, I've never really attempted anything like this before. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 16, 2013 Author Report Share Posted March 16, 2013 Oh, the block count was correct. I had not originated the diagram with that in mind at the time. Quote Link to comment Share on other sites More sharing options...

John Link Posted March 17, 2013 Report Share Posted March 17, 2013 (edited) 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. Edited March 17, 2013 by John Link Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 17, 2013 Author Report Share Posted March 17, 2013 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. Quote Link to comment Share on other sites More sharing options...

John Link Posted March 17, 2013 Report Share Posted March 17, 2013 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. That makes sense. I look forward to reading your summary when you're ready to write it. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 17, 2013 Author Report Share Posted March 17, 2013 Thanks. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 17, 2013 Author Report Share Posted March 17, 2013 (edited) 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. C^{n} produces the three distinct hexahedron patterns illustrated, again for the example shown using 4^{n}. 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 A^{2},(A^{2}-A) that show up in the 5^{th} and 6^{th} 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. FermatsLC-Final_Sheet_5.pdfFermatsLC-Final_Sheet_6.pdfFermatsLC-Final_Sheet_7.pdf edits: Emoticons and math do not get along. Also, on sheet 7, the (A^{2}- should have been (A^{2}-A). Edited March 17, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 17, 2013 Author Report Share Posted March 17, 2013 (edited) Case 1.) A^{3, 6, 9 . . .} (A-^{3}=A^{3}+3(AB^{2}-A^{2}-B^{3} B^{3}=A^{3}+3(AB^{2}-A^{2B)}-(A-^{3} Case 2.) A^{4, 7, 10 . . .} (A-^{4}=(A-^{4}+2(2AB^{2}-A^{2} B^{4}=B^{4}+2(2AB^{2}-A^{2} Case 3.) A^{5, 8, 11 . . .} (A-^{5}=(A-^{5}+2(A^{5}-2A^{4}+A^{3}-A^{4}B+2A^{3}B-AB) B^{5}=B^{5}+2(A^{5}-2A^{4}+A^{3}-A^{4}B+2A^{3}B-AB) I think that's QED. Edited March 17, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

John Link Posted March 18, 2013 Report Share Posted March 18, 2013 Case 1.) A^{3, 6, 9 . . .} (A- ^{3}=A^{3}+3(AB^{2}-A^{2}-B^{3} B^{3}=A^{3}+3(AB^{2}-A^{2B)}-(A- ^{3} Case 2.) A^{4, 7, 10 . . .} (A- ^{4}=(A- ^{4}+2(2AB^{2}-A^{2} B^{4}=B^{4}+2(2AB^{2}-A^{2} Case 3.) A^{5, 8, 11 . . .} (A- ^{5}=(A- ^{5}+2(A^{5}-2A^{4}+A^{3}-A^{4}B+2A^{3}B-AB) B^{5}=B^{5}+2(A^{5}-2A^{4}+A^{3}-A^{4}B+2A^{3}B-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. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 18, 2013 Author Report Share Posted March 18, 2013 (edited) It looks like the A^{2}-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 March 18, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 24, 2013 Author Report Share Posted March 24, 2013 (edited) 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 C^{2}. Remove a hexahedron having 2 sides of A and one side of A^{2}. Using the example of the cube from before, multiply the remains by A and add (C-A)C^{3} to it. A(3(C-A)(CA)+(C-A)^{3})+C^{3}(C-A) Do the same by removing a hexahedron having 2 sides of B and one side of B^{2}. B(3(C-(CB)+(C-^{3})+C^{3}(C- Case 3 Finally, start with a hexahedron having 2 sides of C^{2} and one of C. Remove a hexahedron having 2 sides of A^{2} and one side of A. Using the example of the cube from before, multiply the remains by A^{2} and add 2C^{3}A(C-A) +C^{3}(C-A)^{2} to it. A^{2}(3(C-A)(CA)+(C-A)^{3})+2C^{3}A(C-A)+C^{3}(C-A)^{2} Do the same by removing a hexahedron having 2 sides of B^{2} and one of B. B^{2}(3(C-(CB)+(C-^{3})+2C^{3}B(C-+C^{3}(C-^{2} Working in the exponent factor led to: 1A.) C^{n}-A^{n}=3(C^{(n/3)}-A^{(n/3)})(C^{(n/3)}A^{(n/3)})+(C^{(n/3)}-A^{(n/3)})^{3} 1B.) C^{n}-B^{n}=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})+C^{3n/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})+C^{3n/3}(C- 3A.) C^{(n+2)}-A^{(n+2)}=A^{2}(3(C^{(n/3)}-A^{(n/3)})(C^{(n/3)}A^{(n/3)})+(C^{(n/3)}-A^{(n/3)})^{3})+2C^{3n/3}A(C-A)+C^{3n/3}(C-A)^{2} 3B.) C^{(n+2)}-B^{(n+2)}=B^{2}(3(C^{(n/3)}-B^{(n/3)})(C^{(n/3)}B^{(n/3)})+(C^{(n/3)}-B^{(n/3)})^{3})+2C^{3n/3}B(C-+C^{3n/3}(C-^{2} Case 1 for n=3, 6, 9, . . . A^{n}+B^{n}≠C^{n} 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}-B^{n} or conversely 3(C^{(n/3)}-B^{(n/3)})(C^{(n/3)}B^{(n/3)})+(C^{(n/3)}-B^{(n/3)})^{3}-A^{n} Case 2 for n=4, 7, 10, . . . A^{n}+B^{n}≠C^{n} 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})+C^{3n/3}(C-A)-B^{n} or conversely B(3(C^{(n/3)}-B^{(n/3)})(C^{(n/3)}B^{(n/3)})+(C^{(n/3)}-B^{(n/3)})^{3})+C^{3n/3}(C--A^{n} Case 3 for n=5, 8, 11 . . . A^{n}+B^{n}≠C^{n} because it fell short by A^{2}(3(C^{(n/3)}-A^{(n/3)})(C^{(n/3)}A^{(n/3)})+(C^{(n/3)}-A^{(n/3)})^{3})+2C^{3n/3}A(C-A)+C^{3n/3}(C-A)^{2}-B^{n} or conversely B^{2}(3(C^{(n/3)}-B^{(n/3)})(C^{(n/3)}B^{(n/3)})+(C^{(n/3)}-B^{(n/3)})^{3})+2C^{3n/3}B(C-+C^{3n/3}(C-^{2}-A^{n} Edited March 24, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 25, 2013 Author Report Share Posted March 25, 2013 Looking this over, it bears noting that testing for n=3, 6, 9 etc, simultaneously solves the other two cases. C^{3} becomes C^{4} or C^{5} by the inclusion of C or C^{2} in the equation. Quote Link to comment Share on other sites More sharing options...

John Link Posted March 25, 2013 Report Share Posted March 25, 2013 I'm still waiting for a statement of a theorem, including all the necessary definitions and examples, and then a proof. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 25, 2013 Author Report Share Posted March 25, 2013 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 6^{th}, 9^{th}, 12^{th} etc, the number of blocks can be arranged into a cube shape. This is the value taken as C. For illustration purposes, use 7^{3} or 343 blocks arrange into 7 blocks wide by 7 blocks deep by 7 blocks tall. Remove 6^{3} 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 6^{3} 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 5^{3} 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 5^{3} blocks can be represented as B 3(C-(CB)+(C-^{3} or 3(7-5)(7*5)+(7-6)^{3}=218 Does A^{3}+B^{3}=C^{3}? Does 6^{3}+5^{3}=7^{3}? Does 216+125=343? No. 216+125=341, precisely 2 less than 343. 3(C-A)(CA)+(C-A)^{3}-B^{3} = 3(7-6)(7*6)+(7-6)^{3}-5^{3} = 3(1*42)+1^{3}-125 = 126+1-125=2 conversely 3(C-B(CB)+(C-^{3}-A^{3} = 3(7-5)(7*5)+(7-5)^{3}-6^{3} = 3(2*35)+2^{3}-216 = 210+8-216=2 Starting with 2 arrangement of blocks A & B or 6^{3} & 5^{3}, these two stacks of blocks would be 2 short of being rearranged toward a 7^{3} stack. Starting with a 7^{3} arrangement, a rearrangement of 6^{3} & 5^{3} 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 A^{n}+B^{n}≠C^{n} reminds me a bit of trying to prove that something does not exist, rather than via the identification of what does. Quote Link to comment Share on other sites More sharing options...

Guest Math Bot Posted March 26, 2013 Report Share Posted March 26, 2013 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. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 26, 2013 Author Report Share Posted March 26, 2013 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 (3C^{2}A-3CA^{2})+(C^{3}-3C^{2}A+3CA^{2}-A^{3}) or (3C^{2}B-3CB^{2})+(C^{3}-3C^{2}B+3CB^{2}-B^{3}). In both cases, the 3C^{2}A-3CA^{2} and 3C^{2}B-3CB^{2} 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. Quote Link to comment Share on other sites More sharing options...

John Link Posted March 26, 2013 Report Share Posted March 26, 2013 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. http://en.wikipedia.org/wiki/Theorem Quote Link to comment Share on other sites More sharing options...

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