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Like everyone, I have sensed that Fermat's Last Theorem is true. It was in a rather geometrical way--when I saw the picture of the shells or nexus numbers that add up to form cubes:
The proof was somehow already there. Shell/nexus numbers, that "grow" an xn value, are a grasp-able concept. And it does lead to a proof-by-contradiction for Fermat's famous statement, which is roughly:
Figure 1. For an + bn = cn, n must be 1 or 2 to have integer solutions; In fact, if n > 2 there cannot be integer solutions for a, b, and c.
The formulas for shells for each power level are visibly related to Pascal's Triangle as is established within the binomial theorem. The Powers' Pattern also shows a basis of the formulas by a very interesting depiction of accumulations.
Figure 2. The first several shell/nexus number formulas.
1 = "Shell" difference formula S1 for x1 2k – 1 = "Shell" difference formula S2 for x2 3k2 – 3k + 1 = "Shell" difference formula S3 for x3 4k3 – 6k2 + 4k – 1 = "Shell" difference formula S4 for x4 5k4 – 10k3 + 10k2 – 5k + 1 = "Shell" difference formula S5 for x5 6k5 – 15k4 + 20k3 – 15k2 + 6k – 1 = "Shell" difference formula S6 for x6 etc.
The concept of nexus number is that "shells" consecutively surround an original 1, and add up to create an xn value. This includes the values of an, bn and cn of Fermat's Last Theorem. If cn will serve as the total value of two terms that are power-of-n values, then the larger of the two terms further completes the smaller value's initial run of shells and the smaller value should complete the run of shells after the larger term's shells, totaling up to the cn value. Due to the commutative law of addition, either of the two terms must be able to occur firstly or secondly.
So, assuming integers, 0 < a < b < c and n > 0 and assuming an + bn = cn the commutative law of addition, as well as the shell/nexus-number-nature of any integer raised to a power, are the basis of Figure 3.
Figure 3. Diagrammed shells accumulate to a cn total. Either an or bn may occur firstly in shell/nexus number sets, due to the commutative law of addition. S-sub-n represents the specific formula that generates shell/nexus number values for the power of n. The large sigma symbols indicate the procedure: apply integer number or variables, from the value located at the symbol's bottom through the variable located at the top, to the S-sub-n formula; sum-total values to establish a bracket's value. Shells are 'counted' by variables located on the top brackets. a + x + u = c. a + x = b. The equation at the bottom depicts known and/or assumed values of bracketed shell sets. "x-shells-value" is equivalent to [bn value less an value]. Carets "^" are placed in terms to clearly reinforce that term's variable as raised to the power of n.
x-shells-value = bn - an
Figure 4a. So 2an + (bn - an) is to cn (an equivalency) Figure 4b. As an + (bn - an) is to bn (an equivalency)
Figure 5a. THUS cn / (2an + (bn - an)) equals bn / (an + (bn - an)) Figure 5b. OR cn (an + (bn - an)) = bn (2an + (bn - an)) Figure 5c. OR ancn + bncn - ancn = 2anbn + (bn)2 - anbn Figure 5d. OR bncn = anbn + (bn)2
Figure 6a. For n = 1 bc = ab + b2 OR c = a + b (No Contradiction) Figure 6b. For n = 2 b2c2 = a2b2 + (b2)2 OR c2 = a2 + b2 (No Contradiction)
under construction Figure 6c. For n = 3 b3c3 = a3b3 + (b3)2 OR c3 = a3 + b3 Figure 6d. For n > 2 bncn = anbn + (bn)2 OR cn = an + bn
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain. --Pierre de Fermat Cecilia December 15, 2005
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