## Wednesday, November 27, 2013

### Exceptions to Fermat's Last Theorem

Exceptions to Fermat's Last Theorem

Recall Fermat's Last Theorem:
For no integer n > 2  is there an integer solution to   an + bn  =  cn
This theorem, though proven true by Wiles, has loopholes; negative exponents, fractional exponents, and mod N.

NEGATIVE EXPONENTS

We have these Diophantine formulas:
a1 + b1                                                  =          (a+b)1
[m2 ‑ n2]2 + [2mn]2                               =          [m2 + n2]2
[a(a+b)] ‑1 + [b(a+b)] ‑1                        =          [ab] ‑1
[m4 ‑ n4] ‑2  +  [2mn(m2 + n2)] ‑2            =          [2mn(m2 - n2)] ‑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 ax+bx-cx = 0 by Newton’s method.

MODULO N

If the function xn 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
x3  mod 5         0   1   3   2   4
It swaps 2 and 3, leaving the others unchanged. Therefore the function (x3) is its own inverse; (x3)3 = x mod 5. Now consider addition conjugated by cubing:
( x3 + y3 )3   =   x  "3|/+"   y    ;
This is the 'cube root of addition';
x 3|/+  y  =  z    iff   x3 + y3 =  z3 .

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, x3 is not 1 to 1.

This theorem applies:
x3 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
xp is one-to-one mod M           if and only if
M is a product of distinct primes, none of form 2pk+1.

xpq  = (xp)q  ;  therefore: If N is an odd composite number, then
xN 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.