The Fermat Circle
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.
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).
--Inter Arma Enim Silent Leges--
“Science needs the light of free expression to flourish. It depends on the fearless questioning of authority, and the open exchange of ideas.” ― Neil deGrasse Tyson
"When photons interact with electrons, they are interacting with the charge around a "bare" mass, and thus the interaction is electromagnetic, hence light. This light slows the photon down." - BuleriaChk