Fermat Versus Fermat
The Theorem versus
the Primes
A math essay by Nathaniel Hellerstein
City College of San Francisco
Consider Fermat’s Last Theorem:
AN
+ BN = CN
has no positive
integer solutions if N>2.
This theorem has finally been proven by Andrew Wiles.
But consider these equations:
13 + 23 = 1 +
8 =
9 = 4 (mod 5)
43
= 64 = 4
(mod 5)
So 13 + 23 = 43 (mod 5)
; a modular exception to Fermat’s
Last Theorem! Call such an exception a Fermat
Triple.
There
are many Fermat triples; for modulo 5, the cubing function is 1-to-1 and onto:
X mod 5 0 1 2 3 4
X3 mod 5 0 1 3 2 4
Therefore cubing is invertible, mod 5. In fact it is its
own inverse:
(X3)3 =
X mod 5
Let (x1/3) be the cube root of x, mod 5; and
define the ‘cube root of addition, mod 5’, by conjugating addition by the cube
root, as follows:
x +1/3 y = (x3 + y3)1/3
(mod 5)
x +1/3 y = z
if and only if x3+y3 = z3
.
The cube root of addition is a Fermat-triple addition; it
solves a3+b3 = c3
for any a and b. Thus mod 5 maximally violates Fermat’s Last
Theorem for cubes.
+1/3 0
1 2 3 4
0 0
1 2 3
4 A Fermat-triple
addition
1 1
3 4 2
0
2 2
4 1 0
3
3 3
2 0 4 1
4 4 0
3 1 2
The cube root of addition is commutative, associative, has identity
0, inverse –x, and multiplication distributes over it:
x*(y +1/3 z)
= x*(y3+z3)1/3
= (x3*(y3+z3))1/3
= ((xy)3+(xz)3)1/3 = (xy) +1/3 (xz)
So the cube root of
addition fits the field axioms. Mod 5 has not just an exception to Fermat’s
Last Theorem: it has an entire arithmetical system against it!
Better yet: modulo 5, any
odd power equals identity or cube:
x = x5 = x9 = …
x3 = x7
= x11 = …
So any odd power is invertible, mod 5; therefore there are Nth roots
of addition, mod 5, for any odd N:
x +1/N y =
(xN + yN)1/N (mod 5)
x +1/N y =
z if and only if xN+yN = zN
.
+1/N
also fits the field axioms including distribution. So mod 5 has a field of exceptions
to Fermat’s Last Theorem for any odd power.
There
are Fermat-triples for any odd power not just mod 5, but for any modulus in
which all odd powers are 1-to-1 and onto. Which moduli are these? The Fermat
primes; 2(2^n)+1.
Theorem: For
any Fermat prime F, the integers mod F have one-to-one odd-power functions.
Proof:
The integers mod F
(a.k.a. ZF) is a field since F is a prime; so its multiplicative
group (minus zero) is isomorphic to the cyclic addition group in ZF-1
= Z2^(2^n). Odd powers xN in ZF correspond to
odd multiples Nx in Z2^(2^n); if N is odd then it is relatively
prime to 2^(2^n), therefore it has a multiplicative inverse: an n such that nN
= 1 mod 2^(2^n). Na=Nb implies nNa=nNb implies a=b; so Nx is one-to-one. N(na)=a;
so Nx is onto. For odd N, the function Nx is one-to-one onto in Z2^(2^n);
therefore xN is one-to-one onto in ZF. QED.
The
integers modulo any Fermat prime have one-to-one onto odd power functions;
therefore they also have all odd roots of addition. Fermat primes maximally
violate Fermat’s last theorem; thus this article’s title!
The
five known Fermat primes are:
3, 5,
17, 257, 65537
By
Fermat’s Little Theorem, x3 = x mod 3; so xodd = x mod 3.
All odd powers are the identity mod 3, so any addition mod 3 is a Fermat triple
for all odd powers.
We have
seen odd roots of addition mod 5, above. They are either + or +1/3.
Each is a cube root of each other; (+1/3)1/3 = +, because
x9=x mod 5. So Z5’s additions have a cube-root symmetry.
Now for
mod 17. Consider these tables:
X 0 1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16
X3 0 1 8 10
13 6 12
3 2 15 14 5 11
4 7 9 16
X5 0 1 15
5 4 14 7 11
9 8 6 10 3 13
12 2 16
X7 0 1 9 11
13 10 14 12 15 2 5 3 7
4 6 8 16
From
these we can derive Z17
root-addition tables.
+1/3 0 1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16
0 0 1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16
1 1 8 15 12 10 14
4 13 7 16 9 5 6 11
2 3 0
2 2 15 16 1 13 10
7 12 3 5 11 4 8
6 9 0 14
3 3 12 1
7 6 16 11 4
6 2 14 9 13 10
0 8 15
4 4 10 13 6 15
8 2 16 9 12
3 1 14 0 7
11 6
5 5 14 10 16 8
6 1 15 2 13 7
12 0
3 4 9 11
6 6 4 7
11 2
1 14 9 10 3 15
0 5 16 8 13 12
7 7 13 12 4 16 15
9 5 11 1
0 8 10 14 3
6 8
8 8 7
3 6 9 2 10
11 13 0 16 2
4 5 15 12 1
9 9 16 5 2 12
13 3
1 0 4
6 7 15 8 11 14 10
10 10 9 11 14 3 7
15 0 16
6 12 8 2 1
13 5
4
11 11 5 4
9 1 12 0
8 2 7
8 3 16 15 6 10 13
12 12 6 8 13 14
0 5 10 4 15 2
16 11 9
1 7 3
13 13 11 6 10 0 3 16
14 5
8 1 15 9 2
12 4
7
14 14 2 9 0
7 4 8 3 15
11 13 6
1 12 10 16 5
15 15 3 0 8
11 9 13
6 12 14 5 19 7 4
16 1
2
16 16 0 14 15 6 11 12
8 1 10 4 13
3 7 5
2 9
+1/5 0 1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16
0 0 1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16
1 1 15 16 10 3
2 9 14 11 8
6 7 4 5 13
12 0
2 2 16 13 12 15 13 3
8 6 10 4 9 1 7
11 0
5
3 3 10 12 11 8 15 14 16
5 13 7 2 9 1
0 6 4
4 4 3 15
8 9 1
7 2 13 14 11 5
6 0 16 10 12
5 5 2 13 15
1 7 4 9
10 3 12
6 0 11 8 16 13
6 6 9 3
14 7
4 5 1 16 2
13 0 11 12 15 8 10
7 7 14 8 16
2 9 1 3 12
15 0
4 5 6 10 13 11
8 8 11 6 5 13
10 16 12 1 0 2 15
14 3
4 7 9
9 9 8 10 13 14
3 2 15 0 16
5 1 7 4 12
11 6
10 10 6 4 7 11
12 13 0
2 5 14 16 8 15
1 9 3
11 11 7 9
2 5 6
0 4 15 1 16 12 13 10
3 14 8
12 12 4 1
9 6 0 11 5
14 7
8 13 10 16 2 3 15
13 13 5 7
1 0 11 12 6
3 4 15 10 16 8
9 2 14
14 14 13 11 0 16 8 15 10
4 12 1 3 2 9
6 5 7
15 15 12 0 6 10 16
8 13 7 11 9 14 3 2
5 4 1
16 16 0 5 4 12
13 10 11 9 6
3 8 15 14 7
1 2
+1/7 0 1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16
0 0 1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16
1 1 9 5
7 6 3
8 4 16 11 14 13 15 10 12 2 0
2 2 5 1 11
10 9 14 13 12 3
6 7 16 4
8 0 15
3 3 7 11 10 12 13 15 14 2 4
16 6
1 8 0
9 5
4 4 6 10 12
2 14 5 15 3
8 1 16 11 0 9
13 7
5 5 3 9 13
14 11 12 10 15 7 8
4 0 6 16
1 2
6 6 8 14 15
5 12 3 2 7
16 9
0 13 1 11 10 4
7 7 4 13 14 15 10
2 12 5 6
0 8 9 16 1
11 3
8 8 16 12 2 3
15 7
5 4 0 11 1
10 9 13 14 6
9 9 11 3
4 8 7 16
6 0 13 12 10 2 14 15
5 1
10 10 4 6 16
1 8 9 0 11
12 5 15
7 2 3 4 13
11 11 13 7 6 16 4
0 8 1 10 15 14
5 12 2 3 9
12 12 15 16 1 11 0 13 9
10 2
7 5 6
3 4 8 14
13 13 10 4 8
0 6 1 16 9
14 2 12
3 15 5 7 11
14 14 12 8 0 9 16
11 1 13 15 3
2 4 5
7 6 10
15 15 2 0 9
13 1 10 11 14 5
4 3 8
7 6 16 12
16 16 0 15 5
7 2 4
3 6 1 13 9
14 11 10 12 8
I leave
computing tables for +1/9, +1/11, +1/13, and +1/15,
mod 17, as an exercise for the ambitious student.
What of
even powers?
X 0 1
2 3 4
5 6 7 8 9 10 11 12 13 14 15 16
X2 0 1
4 9 16 8 2 15
13 13 15 2 8 16
9 4 1
X4 0 1 16 13
1 13 4 4 16 16
4 4 13 1 13 16 1
X8 0 1 1
16 1 16 16 16 1 1 16
16 16 1 16 1 1
X16 0 1
1 1 1
1 1 1
1 1 1 1
1 1 1
1 1
There are many sums among
the squares table:
1+1=2; that is, (1 or 16)2
+ (1 or 16)2 = (6 or 11)2
1+8=9; that is, (1 or 16)2
+ (5 or 12)2 = (3 or 14)2
1+15=16; that is, (1 or
16)2 + (7 or 10)2
= (4 or 13)2
1+16=0; that is, (1 or
16)2 + (4 or 13)2
= 02
2+2=4; that is, (6 or 11)2
+ (6 or 11)2 = (2 or
15)2
2+13=15; that is, (6 or
11)2 + (8 or 9)2
= (7 or 10)2
2+15=0; that is, (6 or 11)2
+ (7 or 10)2 = 02
2+16=1; that is, (6 or 11)2
+ (4 or 13)2 = (1 or 16)2
4+4=8; that is, (2 or 15)2
+ (2 or 15)2 = (5 or 12)2
4+9=13; that is, (2 or 15)2
+ (3 or 14)2 = (8 or 9)2
4+13=0; that is, (2 or 15)2
+ (8 or 9)2 = 02
4+15=2; that is, (2 or 15)2
+ (7 or 10)2 = (6 or 11)2
8+8=16; that is, (5 or 12)2
+ (5 or 12)2 = (4 or 13)2
9+16=8; that is, (3 or
14)2 + (4 or 13)2
= (5 or 12)2
9+8=0; that is, (3 or 14)2
+ (5 or 12)2 = 02
9+9=1; that is, (3 or 14)2
+ (3 or 14)2 = (1 or 16)2
13+13=9; that is, (8 or 9)2
+ (8 or 9)2 = (3 or 14)2
15+15=13; that is, (7 or
10)2 + (7 or 10)2
= (8 or 9)2
16+16=15; that is, (4 or
13)2 + (4 or 13)2
= (7 or 10)2
Some of these (maybe all)
are the classic Pythagorean triples:
(m2 – n2)2
+ (2mn)2 = (m2 + n2)2
m=2, n=1 yields the triple 32+42=52
;
m=3, n=2 yields the triple 52+122=132
;
m=3, n=1 yields the triple 82+62=102
;
m=4, n=1 yields the triple 152+82=02
;
m=4, n=2 yields the triple 122+162=32
;
m=5, n=4 yields the triple 92+62=72
;
and so on.
ZF supports higher
Pythagorean triples; for if N is odd then:
(m2 +1/N (-n2))2N + (21/N
mn)2N = (m2 +1/N n2)2N
Proof:
Left side = (m2 +1/N
(-n2))2N + (21/N mn)2N
= (m2N + (-n2N))2 + (2 mNnN)2
= (m2N - n2N)2 + (2 mNnN)2 if
N is odd
= m4N – 2m2Nn2N+
n4N + 4 m2Nn2N
= m4N + 2m2Nn2N+
n4N
= (m2N + n2N)2
= (m2 +1/N n2)2N
= right
side
21/3
= 8 and 21/5 = 15, mod 17; therefore:
(m2
+1/3 (-n2))6
+ (8 mn)6 = (m2
+1/3 n2)6
(m2
+1/5 (-n2))10
+ (15 mn)10 = (m2 +1/5 n2)10
so m=2, n=1
yields 66
+ 166 = 106
and 1210 + 1310 =
310
m=3,
n=1 yields 106
+ 76 = 166
and 610 + 1110 =
810
m=3,
n=2 yields 86 + 146
=
126
and 410 + 510 =
1410
You can
derive higher Pythagorean triples by applying odd roots to Pythagorean triples.
For instance, 22+72=62, and 2=83=155,
and 7= 143=65, and 6=53=105;
therefore 86+146=56, and 1510+610=1010
.
There
are sums amongst the 4th and 8th powers:
1+16=0; that is,
(1 or 4 or 13 or 15)4
+ (2 or 8 or 9 or 15)4 = 04
13+4=0; that is,
(3 or 5 or 12 or 14)4
+ (6 or 7 or 10 or 11)4
= 04
1+16=0; that is,
(1,2,4,8,9,13,15,16)8+(3,5,6,7,10,11,12,14)8 = 08
But all 16th
powers are 1 or 0, so the only possible sums amongst the 16th powers
are trivial: 0+B=B.
The higher binary powers involve
2kth roots of -1:
If r(2^k)
= -1 mod F, then
(a)(2^k)
+ (ra)(2^k) = 0(2^k)
And more generally
(a*(rodd))(2^k)
+ (a*(reven))(2^k)
= 0(2^k)
So since 24 =
-1 mod 17, and 38 = -1 mod 17:
(a*(2odd))4 + (a*(2even))4 = 04
(a*(3odd))8
+ (a*(3even))8
= 08
These ‘hypercomplex’ triples
include
14+24 = 24+44
= 34+64 =44+84 = 54+104
=64+124 =74+144 = 04
18+38 = 28+68
= 38+98 =48+128 = 58+158
=68+18 = 08
Since
(-1)odd = -1, hypercomplex triples for power 2k are also
triples for power odd*2k. For instance, 4th roots of -1
are also 12th roots; therefore
112+212 = 212+412 = 312+612
=412+812 = 512+1012 =612+1212
=712+1412 = 012
Conjecture:
modulo a Fermat prime F, the Fermat triples are:
For
any power N: the trivial case
0N + BN = BN
For
odd power N: odd roots of addition:
AN +
BN =
( A +1/N B )N
For
power 2: the Pythagorean triples:
(m2
- n2)2 + (2 mn)2 = (m2
+ n2)2
For
power Odd*2: the higher Pythagorean triples:
(m2
+1/N (-n2))2N
+ (21/N mn)2N = (m2
+1/N n2)2N
For
power Odd*2k: hypercomplex triples:
If r(2^k)
= -1 mod F, then
(a)(2^k)
+ (ra)(2^k) = 0(2^k)
And
more generally
(a*(rodd))(Odd*2^k)
+ (a*(reven))(Odd*2^k)
= 0(Odd*2^k)
This seems to apply to
mod 3, 5 and 17, but I haven’t checked mod 257, let alone mod 65537. I leave
investigating Fermat triples in those moduli to the ambitious student. For instance, do they
have fourth powers that add to a fourth power that isn’t zero? Or 8th
powers and up?
I end with a Speculation: that there are only
finitely many Fermat primes. My guess is that if there were infinitely many
Fermat prime moduli, then they would tend towards something resembling the
rational numbers; but all odd roots of addition are defined in the Fermat prime
moduli, though not in the rationals.
