4. Modular
Fermat Fields
If the function xN is
one-to-one on the integers modulo M, then for any value of a and b, we
can solve Fermat's equation, modulo M:
aN+bN = cN mod M
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. It is therefore self‑inverse; ((x3)3) = x mod
5.
Now
consider the cube root of addition, mod 5;
x 3\/+ y =
((x3)+(y3))1/3 =
((x3)+(y3))3 mod 5
x 3\/+ y
= z if and only if x3+y3 = z3 mod 5
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
as well as +
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,...
Now consider cubing modulo 11:
x 0
1 2 3
4 5 6
7 8 9 10
x3 0
1 8 5
9 4 7
2 6 3 10
Here's the cube root of addition
mod 11:
3\/+ 0
1 2 3
4 5 6
7 8 9
10
0 0
1 2 3
4 5 6
7 8 9
10
1 1
7 4 8
10 3 2
9 6 5
0
2 2
4 3 7
8 1 5
10 9 0
6
3 3
8 7 10
9 4 1
6 0
2 5
4 4
10 8 9
6 7 3
0 5 1
2
5 5
3 1 4
7 2 0
8 10 6 9
6 6
2 5 1
3 0 9
4 7 10 8
7 7
9 10 6
0 8 4
5 2 3
1
8 8
6 9 0
5 10 7
2 1 4
3
9 9
5 0 2
1 6 10
3 4 8
7
10 10
0 6 5
2 9 8
1 3 7
4
This has the Fibonacci sequence
1, 1, 7, 9, 3, 2, 7, 10, 1, 0, 1, 1, ...
In the integers we solve Fermat's
equation for N=2 via the rational‑right‑triangle identity:
(m2 ‑ n2)2
+ (2mn)2 = (m2 + n2)2
This equation still applies in
cube-root addition mod 11, but with a change in the second term; for 1 3\/+ 1 =
7 mod 11. Therefore, in this
context, the correct equation is:
(m2 3\/‑ n2)2 3\/+ (7mn)2 =
(m2 3\/+ n2)2 mod 11
Cubing yields: (m2
3\/‑ n2)6 + (7mn)6 =
(m2 3\/+ n2)6 mod 11
We get these triples:
m n
(m2 3\/‑ n2)6 + (7mn)6 =
(m2 3\/+ n2)6 mod 11
2 1 2 6 + 3 6 = 10 6 mod 11
3 2 3 6 + 9 6 = 1 6 mod 11
3 1
7 6 + 10 6 = 5 6 mod 11
4 1 9 6 + 6 6 = 3 6 mod 11
4 3 1 6 + 7 6 = 6 6 mod 11
These in turn imply other sums:
1 6 + 7 6 = 5
6 mod 11
1 6 + 3 6 = 4
6 mod 11
3 6 + 2 6 = 1
6 mod 11
Note the colliding triplets
(1,7;6), (1,7;5). Mod 11, 6 = ‑ 5 and 62
= 52
This method obtains 2*3 power
Fermat equations, but not 4*3 power, nor 8*3, nor higher orders of evenness.
So modulo 5 and 11, we have
conjugate fields full of loopholes in 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.
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