**Exceptions to Fermat's Last Theorem**

Recall Fermat's Last Theorem:

For no integer n > 2 is there an integer solution to a

^{n}+ b^{n}= c^{n}
This theorem, though proven true by
Wiles, has loopholes; negative exponents, fractional exponents, and mod N.

*NEGATIVE EXPONENTS*
We have these Diophantine formulas:

a

^{1}+ b^{1}= (a+b)^{1}
[m

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

^{ ‑1}+ [b(a+b)]^{ ‑1}= [ab]^{ ‑1}
[m

^{4}‑ n^{4}]^{ ‑2}+ [2mn(m^{2}+ n^{2})]^{ ‑2}= [2mn(m^{2}- n^{2})]^{ ‑2}
The last two equations involve
'reciprocal addition':

x "1/+" y = 1
/ ( (1/x) + (1/y) ) = xy /
( x+ y )

This 'conjugate of addition' is
involved in Kirkhoff's Law and flow‑rate algebra.

Evidently the Fermat/Wiles theorem
should read "... for |n|>2..." Now the question is; what's so
special about ‑2, ‑1, 1, 2?

*FRACTIONAL EXPONENTS*
For practically any a,b,c you can
find a suitable n, provided that it may be fractional ‑ and probably
transcendental. For instance:

a=1, b=2, c=5 ‑‑‑‑>
n=0.5638955243...

a=2, b=3, c=4 ‑‑‑‑>
n=1.507126592...

You can find these n’s by solving a

^{x}+b^{x}-c^{x}= 0 by Newton’s method.

*MODULO N*
If the function x

^{n}is a one-to-one function on the integers modulo k, then we can find c in Fermat's equation for*any*value of a and b.
For instance, consider the cube mod
5:

x mod 5 0 1
2 3 4

x

^{3}mod 5 0 1 3 2 4
It swaps 2 and 3, leaving the
others unchanged. Therefore the function (x

^{3}) is its own inverse; (x^{3})^{3}= x mod 5. Now consider addition conjugated by cubing:
( x

^{3}+ y^{3})^{3}= x "3|/+" y ;
This is the 'cube root of
addition';

x 3|/+
y = z
iff x

^{3}+ y^{3}= z^{3}.
3|/+ 0 1
2 3 4

0 0
1 2 3
4

1 1
3 4 2
0 this is a field under *
mod 5;

2 2
4 1 0
3 * distributes over this

3 3
2 0 4 1

4 4
0 3 1 2

In this 'modular Fermat field', the
Fibonacci sequence goes:

1,1,3,2,0,2,2,1,4,0,4,4,2,3,0,3,3,4,1,0,1,1,...

So mod 5, we have a whole conjugate
field full of exceptions to Fermat's theorem! This trick doesn't work for every modulus; in some moduli, x

^{3}is not 1 to 1.
This theorem applies:

x

^{3}is one-to-one mod M (and M has a cube root of addition) if and only if
M is a product of distinct primes, none of form 6k+1.

These moduli include 2, 3, 5, 6, 10, 11, 15, 17, 22, 23, 29,
30, 33, 34, 41, 43, 46, 47, 53, 55, 57, 58, 66, 69, 71, 82, 83, 86, 87, 89,
94... A motley crew!

This generalizes: If p
is an odd prime, then

x

^{p}is one-to-one mod M if and only if
M is a product of distinct primes, none of form 2pk+1.

x

^{pq}= (x^{p})^{q}; therefore: If N is an odd composite number, then
x

^{N}is one-to-one mod M if and only if
M is a product of distinct primes, none of form 2pk+1, for
any prime factor p of N.

Any modulus equal to a Fermat prime (M = 2

^{(2^n)}+1) will pass all these tests: therefore Fermat-prime moduli have one-to-one odd-power functions; and therefore all Fermat-prime moduli have all odd roots of addition. (So do moduli equalling distinct products of Fermat primes.)
There are infinitely many Fermat
primes if and only if there are infinitely many moduli which have every odd
root of addition.

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