**3. Powers of Addition**

Define a

^{n}\/+ b =^{n}\/( a^{n}+ b^{n}) ; "Fermat addition"
a +

^{n}b = (^{n}\/(a) +^{n}\/(b) )^{n}; "Power addition"
a

^{n}\/* b =^{n}\/( a^{n}* b^{n}) = a * b
a *

^{n}b = (^{n}\/(a) *^{n}\/(b) )^{n }= a * b
Thus

^{n}\/+ and +^{n}are fields under * with identities 0 and 1.
a +

^{n}b = c if and only if^{n}\/a +^{n}\/b =^{n}\/c.
According to the Binomial
Theorem:

a +

^{2}b = a + 2 \/(ab) + b
a +

^{3}b = a + 3^{3}\/(a^{2}b) + 3^{3}\/(ab^{2}) + b
a +

^{4}b = a + 4^{4}\/(a^{3}b) + 6^{4}\/(a^{2}b^{2}) + 4^{4}\/(ab^{3}) + b
...

a +

^{N}b = S [ (_{N}C_{I})^{N}\/(a^{I}b^{N-I}) ]
Note: a

^{n}\/+ b = c if and only if a^{n}+ b^{n}= c^{n}.
Therefore we can state Fermat's
theorem in terms of

^{n}\/+ , Fermat addition.
Here are some "Fermat
near-misses":

95800

^{4}\/+ 217519^{4}\/+ 414560 = 422481
27

^{5}\/+ 84^{5}\/+ 110^{5}\/+ 133 = 144
In the third power we have the
identity:

(6b - 144 b

^{4})^{3}\/+ (144b^{4}) = ( 72b^{3}- 1 )^{3}\/+ 1
and these near-misses:

9

^{3}\/+ 10 = 12.002314...
720

^{3}\/+ 242 = 728.999999373...
729

^{3}\/+ 244 = 738.000000612...
The Fermat sum a

^{N}\/+ b = c is possible only if c is either less than both a and b, or more than both; for if N approaches positive infinity, then (a^{N}\/+ b) approaches their maximum from above; and if N approaches negative infinity, then (a^{N}\/+ b) approaches their minimum from below. If c>(a max b) or c<(a min b), then you can find a suitable n. (It will probably be transcendental.)
For instance:

a=1, b=2, c=5 ‑‑‑‑>
N=0.5638955243...

a=2, b=3, c=4 ‑‑‑‑>
N=1.507126592...

These N solve the equations:

1

^{N}+ 2^{N}- 5^{N}= 0
2

^{N}+ 3^{N}- 4^{N }= 0
The "Fermat
near-misses" cited above imply these near-integer powers:

a=9, b=10,
c=12 ‑‑‑‑> N = 3.0025515252785...

a=720, b=242, c=729 ‑‑‑‑> N = 2.999999373...

a=729, b=244, c=738 ‑‑‑‑> N = 3.000000048...

Exact Fermat sums are confined to
these Diophantine equations:

a

^{1}\/+ b = [a+b]
[m

^{2}‑ n^{2}]^{2}\/+ [2mn] = [m^{2}+ n^{2}]
[a(a+b)]

^{-1}\/+ [b(a+b)] = [ab]
[m

^{4}‑ n^{4}]^{-2}\/+ [2mn(m^{2}+ n^{2})] = [2mn(m^{2}‑ n^{2})]
For instance:

3

^{2}\/+ 4 = 5 ; 5^{2}\/+ 12 = 13 ; 15^{2}\/+ 8 = 17 ; 21^{2}\/+ 20 = 29 ; ...
3

^{-1}\/+ 6 = 2 ; 10^{-1}\/+ 15 = 6 ; 30^{-1}\/+ 70 = 10 ; 35^{-1}\/+ 14 = 10 ; ...
15

^{-2}\/+ 20 = 12 ; 65^{-2}\/+ 156 = 60 ; 255^{-2}\/+ 136 = 120 ; 609^{-2}\/+ 580 = 420 ; ...
Evidently the Fermat/Wiles
theorem should read "... for |n|>2..." Now the question is; what's
so special about ‑2, ‑1, 1, 2? Perhaps it is because we can give all four of
these roots of addition geometric interpretations.

The first root of addition is ordinary
linear addition; geometrically, this is adding lengths by joining parallel
lines. The second root of addition figures in the Pythagorean Theorem; for if
A, B, and C are the sides of a right-angle triangle, with C the hypotenuse,
then C
= A \/+ B . The negative root of addition is
reciprocal sum; geometrically constructible by the "Fence" figure of
the previous section. The negative-second root equals the reciprocal of
square-root of addition. Geometrically, this is "area over diagonal",
or "altitude to diagonal".

Fermat addition has applications
in dimensional analysis and fractal geometry.

Consider two circular oil slicks
that merge, combining areas. If we denote their radii by R

_{1}, R_{2}, and their surface areas by A_{1}, A_{2}, then:
radius of new slick = R

_{1}+^{1/2}R_{2}
Now consider two spherical water
droplets that merge, combining volumes. If we denote their radii by R

_{1}, R_{2}, their surface areas by A_{1}, A_{2}, and their volumes by V_{1}, V_{2}, then:
radius of new drop = R

_{1}+^{1/3}R_{2}
surface area of new drop =
A

_{1}+^{2/3}A_{2}
If a coastline has fractal
dimension d, then if a coast segment of length
L

_{1}(as the crow flies) joins to a segment of length L_{2}, then that forms a coast segment of length approximately L_{1}^{d}\/+ L_{2}.
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