dream_weaver

Legal Tender Status Another questioner asked: I thought that United States currency was legal tender for all debts. Some businesses or governmental agencies say that they will only accept checks, money orders or credit cards as payment, and others will only accept currency notes in denominations of $20 or smaller. Isn't this illegal? The answer found on the US Department of the Treasury is: The pertinent portion of law that applies to your question is the Coinage Act of 1965, specifically Section 31 U.S.C. 5103, entitled "Legal tender," which states: "United States coins and currency (including Federal reserve notes and circulating notes of Federal reserve banks and national banks) are legal tender for all debts, public charges, taxes, and dues." This statute means that all United States money as identified above are a valid and legal offer of payment for debts when tendered to a creditor. There is, however, no Federal statute mandating that a private business, a person or an organization must accept currency or coins as for payment for goods and/or services. Private businesses are free to develop their own policies on whether or not to accept cash unless there is a State law which says otherwise. For example, a bus line may prohibit payment of fares in pennies or dollar bills. In addition, movie theaters, convenience stores and gas stations may refuse to accept large denomination currency (usually notes above $20) as a matter of policy.

Short Circuit, Bicentennial Man. and one of my favorites, The Matrix are a few movies that deal with machines and aspects of human qualities. The argument from or appeal to ignorance does not warrant that it is possible for a machine to become self-aware. As to your ethical questions in your OP, the first two movies show a more benevolent approach while the latter malevolently pits man against the machines.

From the link: "The Soul of Atlas leads the reader through a powerful dialogue between two worlds. Both Christianity and Rand's system of thought have played key roles in shaping Western culture, particularly in America, and Henderson challenges readers to uncover the commonalities between them. What they will discover, as Henderson himself discovered, is that reason cannot rule out the need for faith. In fact, the human person needs both." Or is it that faith cannot rule out the need for reason. The human person who embraces faith, finds, in fact, that reason is still required. "The ideologies are miles apart on the surface. Mark's discussion brings out the point that political conclusions can be similar yet come from two differing theological perspectives." The article started,out earlier about weaving two philosophies together. exploring the compatability between faith and reason and transitions into positing as "two differing theological perspectives" Christianity and atheism, or Christianity and Objectivism? The introduction and some selected passages are available on Amazon.

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+ ≠ 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- ≠ 6(tetrahedral number)+a c3-b3 ≠ a3

We are saying the exact same things to one another. They are identical, at the same time, and in the same respect, and referring to the same referents too!

It's good to see you so wholeheatedly agree.

Yes, deduction can only go so far. It depends on the validity of your premises. Since you contend that you are unable to determine if your premises are valid or invalid, it would appear you are also unable to determine if anyone else premises are valid or invalid either. Then, like a standard of perfection based on an unattainable ideal, you reference a standard of induction, which is also based on an unattainable criteria. This looks strangely like a recipe for frustration and dissappointment. You should probably try to address sNerd's use of meritenacity. Rather than being frustrating and dissappointing, it is delicious and refreshing. It tastes like unobtainium and smells like time marching on.

Had not seen that variation. Out of curiosity, how do you arrive at your basic premises from whence to deduce from?

Right out of chapter 6 of ITOE.

Ah yes. The story that tries to focus in on a single aspect while ignoring the full context within which the aspect takes place. If the progression of time is ignored while focusing in on the distance, or the limited discrimination of perception is ignored while the focusing in on the precision made available by the language of mathematics, does the paradox that seems to arise reveal the futility of rationalization or of the failure to consider the full context?

Those who are not familiar with history may be bound to repeat it. Those who are familiar with it can help to ensure that.

This is reformatted from The Ayn Rand Society Concepts and Their Role in Knowledge: Reflections on Objectivist Epistemology Allan Gotthelf, editor James G. Lennox associate editor Gregory Salmieri, consulting editor Pittsburgh University Press, 2012 CONTENTS Preface PART ONE Ayn Rand's Theory of Concepts: Rethinking Abstraction and Essence Allan Gotthelf Conceptualization and Justification Gregory Salmieri Perceptual Awareness as Presentational Onkar Ghate Concepts, Context, and the Advance of Science James G. Lennox PART TWO Concepts and Kinds Rand on Concepts, Definitions, and the Advance of Science: Comments on Gotthelf and Lennox Paul E. Griffiths Natural Kinds and Rand's Theory of Concepts: Reflections on Griffiths Onkar Ghate Definitions Rand on Definitions--"One Size Fits All"? Jim Bogen Taking the Measure of a Definition: Response to Bogen Allan Gotthelf Concepts and Theory Change On Concepts that Change with the Advance of Science Richard Burian Conceptual Development versus Conceptual Change:Response to Burian James G. Lennox Perceptual Awareness In Defense of the Theory of Appearing: Comments on Ghate and Salmieri Pierre LeMorvan Forms of Awareness and "Three-Factor" Theories Gregory Salmieri Direct Realism and Salmieri's "Forms of Awareness" Bill Brewer Keeping up Appearances: Reflections on the Debate Over Perceptual Infallibilism Benjamin Bayer Uniform Abbreviations of Works References List of Contributors Index

Any of the precious metal sites with contributing commentators have their cut of bull and bear commentators. Many of the hide behind their charts and statistics as their divining rods, and separating the wheat from the chaff takes time and effort. The historians always bring more interesting observations to consider. Beck and Krugman are entertainers. They are paid to keep the audience tuned in to keep the commercial revenue streams flowing. The trend chasers or reactionary traders will weed themselves from any market they are ill-equipped to deal with. It seems that when the occasional ill wind blows, there will always be those who find an opportunity to fly their kites. Pulling on the string to keep the colorful piece of framed paper in the sky will always be more effective than pushing on the string to try the same.

Hopefully the shelves are still stocked. Sometimes these type of sales leave some shop-owners passing out rain-checks.

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+(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 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

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.

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

How Life May Have First Emerged On Earth: Foldable Proteins in a High-Salt Environment Another prevailing view holds that a high-temperature (thermophile) environment, such as deep-ocean thermal vents, may have been the breeding ground for the origin of life. "The halophile, or salt-loving, environment has typically been considered one that life adapted into, not started in," Blaber said. "Our study of the prebiotic amino acids and protein design and folding suggests the opposite."

Origin of Life: Power Behind Primordial Soup Discovered The scientists simulated the impact of such a meteorite with the hot, volcanically-active, early Earth by placing samples of the Sikhote-Alin meteorite, an iron meteorite which fell in Siberia in 1947, in acid taken from the Hveradalur geothermal area in Iceland. The rock was left to react with the acidic fluid in test tubes incubated by the surrounding hot spring for four days, followed by a further 30 days at room temperature. In their analysis of the resulting solution the scientists found the compound pyrophosphite, a molecular 'cousin' of pyrophosphate -- the part of ATP responsible for energy transfer. The scientists believe this compound could have acted as an earlier form of ATP in what they have dubbed 'chemical life'.

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?

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.

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.

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.

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.

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