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- 12-22-2016, 12:47 PM #1

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## Fermat's Theroem and the Relativistic Unit Circle

The Fermat Circle

See also

The Relativistic Unit Circle

The Binomial Expansion yields the result:

where rem(a,b,n)> 0.

This by itself proves Fermat's Theorem for a,b, and c integers and n >2

(for n = 2, the expression can be that of a circle if a,b,c form a Pythagorean triple; otherwise the Binomial Theorem applies.

If Fermat's expression , it would mean that rem(a,b,n)= 0.

Since Rem(a,b,n) consists of additive terms which are all composed of multiplicative products of a and b

(e.g. , it can only vanish if a=0 or b=0, so that

or , respectively

Since the Binomial Theorem is valid for all positive numbers (e.g. fractions, transcendentals) and even complex numbers, the question of whether a and b are integers is moot - it is the remainder term that must vanish.

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For the relativistic circle in two dimensions (a,b) positive numbers are generated in the positive quadrant of the circle by the prescription when , and vice versa, so that n = or at or at v=c, respectively.

(for v=0, , , but at , , , so for each integer generated represented by a, and .

(for v = c, , , and b=b for each integer b.

The product can bet thought of as the area of either a concentric circle or the diagonal of an inscribed polygon, where the area of its equivalent square is

For , the product cannot be an integer, since , corresponding to the result of the Binomial Expansion of , since

For the corresponding expression is:

which can only be valid for and (i.e.,

(For QFT, it is instructive to think of or = h, where h is Planck's constant).Last edited by BuleriaChk; 12-26-2016 at 04:12 PM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

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- 03-04-2017, 09:24 PM #2
## Re: Fermat's Theroem and the Relativistic Unit Circle

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