## Thursday, July 21, 2005

### Proof for n=3: Using Eisenstein Integers

If you are not familiar with Eisenstein Integers, please start here. The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

For those who would like to see this proof done in terms of rational integers only, check out this previous blog.

As in previous blogs, I will use Greek letters to represent quadratic integers and Latin letters to represent rational integers.

Theorem: The equation α3 + β3 = γ3 does not have any integer solutions where
α, β, γ, are Eisenstein integers and α * β * γ ≠ 0.

(1) Since Eisenstein Integers are Euclidean (proof), we know that they are characterized by a Division Algorithm (proof), Bezout's Identity (proof), and Unique Factorization (proof).

(2) We can assume that α, β and γ are coprime. [See here for proof]

(3) If we set ζ = -γ, then we have:
α3 + β3 + ζ3 = 0.

(4) Let:

J = (1 + i√3)/2
O = (1 - i√3)/2

(5) Then, J - O is an Eisenstein prime number (proof ) which divides α*β*ζ [See here for proof.]

(6) But if α * β * γ ≠ 0, then we have an infinite descent. [See here for proof.]

QED

## Tuesday, July 19, 2005

### Eisenstein Integers

Eistenstein integers are quadratic integers of the form Z[(-1 + √-3)/2]. They are named in honor of Ferdinand Eisenstein. For those, who need a review of quadratic integers, start here.

In today's blog, I will go over some basic properties of Eisenstein Integers. Later on, I will show how they can be used to prove that Fermat's Last Theorem has no solutions for n=3.

The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

For purposes of the proof, Dorrie proposed the idea of G-numbers. There are numbers that are based on two values: J,O where

J = (1 + i√3)/2
O = (1 - i√3)/2.

NOTE: In case this is not clear to anyone, i√3 = √-3.

Lemma 1: Every integer g is also a G-number.

g = gJ + gO = (g + gi√3)/2 + (g - gi√3)/2 = 2g/2 = g.

QED

Lemma 2: The following properties are true:

(a) J + O = 1

From the previous lemma.

(b) JO = 1

JO = (1 - [(-1)(3)])/4 = 4/4=1

(c) J2 + O = 0

J2 = (1 + i√3)2/4 = (1 + 2i√3 + (-1)(3)) /4 = (-2 + 2i√3)/4 =
(-1 + i√3) /2

(d) O2 + J = 0

O2 = (1 - i√3)2/4 = (1 - 2i√3 + (-1)(3))/4 = (-2 - 2i√3)/4 =
(-1 - i√3) /2

(e) J3 = -1

J2 = -O [From (c)]
So, J3 = -JO = -1 [From (b)]

(f) O3 = -1

O2 = -J [From (d)]
So, O3 = -OJ = -1 [From (b)]

QED

Lemma 3: The norm of a G-number aJ + bO is a2 + b2 - ab

(1) aJ + bO = (a + ai√3)/2 + (b - bi√3)/2 = (a + b)/2 + [(a - b)i√3]/2

(2) Norm(aJ + bO) = [(a + b)/2 + [(a - b)i√3]/2][(a + b)/2 - [(a - b)i√3]/2] =
(a + b)2/4 - (a - b)2(-1)(3)/4 =
[a2 + 2ab + b2]/4 - (-3)[a2 - 2ab + b2]/4 =
(a2 + 3a2 + b2 + 3b2 + 2ab - 6ab)/4
= a2 + b2 - ab.

QED

Corollary 3.1: The norm is always greater or equal to 0.

In other words, we can prove that a2 + b2 ≥ ab.

Case I: a=b

a2 + b2 = 2a2 which is ≥ ab = a2.

Case II: a is greater than b

In this case: a2 is greater than ab.

Case III: a is less than b

Same argument using b instead of a

QED

Lemma 4: The following numbers are units: J,-J, O, -O, 1,-1

(a) The norm of each of these values is 1. [See here for the explanation of why the norm of a unit is one]

J = 1(J) + 0(O) = 1 + 0 + 0 = 1
-J = (-1)J + 0(O) = 1 + 0 + 0 = 1
O = 0J + 1(O) = 0 + 1 + 0 = 1
-O = 0J + (-1)O = 0 + 1 + 0 = 1
1 = (1)J + (1)O = 1 + 1 - 1 = 1
-1 = (-1)J + (-1)O = 1 + 1 -1 = 1

(b) There are no other units

Norm(Z[(-1 + i√3)/2]) =[(a + bi√3)/2][(a - bi√3)/2] = (a2 - b2(-3) )/4= ±1

This means that: (since the norm in this case must be positive)

a2 + 3b2 = 4.

We know that absolute(b) has to be less than 2. Otherwise, 3b2 is greater than 4.

This means that we only need to consider b=-1,0,1.

If b = 0, a = ±2.
If b = 1 or b = -1 a = ±1

So, the only units are:

a=2,b=0: (2+0i√3)/2 = 1
a=-2,b=0 (-2+0i√3)/2=-1
a=1,b=1 (1 + i√3)/2 = J
a=-1,b=1 (-1 + i√3)/2 = -O
a=1, b=-1 (1 - i√3)/2 = O
a=-1, b=-1 (-1 - i√3)/2 = -J

QED

Lemma 5: J-O is a prime.

(1) Norm(J-O) = (1)2 + (-1)2 - (1)(-1) = 3.
(2) Only primes have norms that are rational integer primes. [The argument is the same as the one for Gaussian Integers, see here]

QED

Lemma 6: If aJ+bO is divisible by J-O, then a + b is divisible by 3.

(1) Since aJ + bO is divisible by J-O, there exists a value mJ + nO such that:

aJ + bO = (mJ + nO)(J - O) = mJ2 - mJO + nJO - nO2

(2) From the properties of G-Numbers (see above), we know that:

J2 = -O
O2 = -J
OJ = 1
-m = -mJ -mO
n = nJ + nO

(3) Applying these properties to (1), gives us:

aJ + bO = m(-O) -mJ - mO +nJ + nO -n(-J) = (2n-m)J + (n - 2m)O

(4) So that:
a = 2n-m
b = n-2m

(5) a + b = 2n -m + n - 2m = 3(n - m)

QED

Lemma 7: If 3 divides a + b, then J - O divides aJ + bO.

(1) So there exists g such that a + b = 3g
(2) We also support that there exists m,n such that n - m = g and 2n -m = a.
(3) This gives us that b = 3g -a = 3n - 3m - 2n + m = n - 2m.
(4) aJ + bO = (2n - m)J + (n - 2m)O
(5) Applying the reverse logic from above, we get:
(2n - m)J + (n - 2m)O = m(-O) -mJ - mO +nJ + nO -n(-J) =
=
mJ2 - mJO + nJO - nO2 = (mJ + nO)(J - O)

QED

Lemma 8: If J - O doesn't divide aJ + bO, then (aJ + bO)3 ≡ ±1 (mod 9)

(1) Since J - O doesn't divide aJ + bO, we know that 3 doesn't divide a + b since:

(a) Assume that J-O doesn't divide aJ + bO
(b) Assume that 3 divides a + b
(c) Then by the Lemma above J-O divides aJ + bO
(d) But this is not true by the assumption in (a) so we have a contradiction and we reject the assumption in (b) and conclude that 3 doesn't divide a+b.

(2) This means that we have the following possible cases:

(a) a = 3g, b = 3k ±1
(b) a = 3g ±1, b = 3k
(c) a = 3g ±1, b = 3k ±1

(3) Let λ = gJ + kO. Then in each of the three cases we have:

(a) (3g)J + (3k ±1)O = 3λ ±O
(b) (3g ±1)J + (3k)O = 3λ ±J
(c) (3g ±1)J + (3k ±1)O = 3λ ±1

(4) We can rewrite this as 3λ + ε where ε = ±O, ±J, or ±1.

(5) We can cube this result to get:
(3λ + ε)3 = (9λ2 + 6λε + ε2)(3λ + ε) =
=27λ3 + 18λ2ε + 3λε2 + 9λ2ε + 6λε2 + ε3 =
= 9 (3λ3 + 3λ2ε + λε2) + ε3

(6) Additionally, we note that ε3 = ±1 since:

(a) O3 = -1
(b) (-O)3 = 1
(c) J3 = -1
(d) (-J)3 = 1
(e) 13 = 1
(f) (-1)3 = -1

(7) Hence, we have shown that:
(aJ + bO)3 ≡ ±1 (mod 9).

QED

Lemma 9: J - O divides 3.

(1) 3 = 3J + 3O
(2) Since 3 + 3 is divisible by 3, then 3J + 3O is also divisible by J - O. [See above]

QED

Corollary 9.1: (J-O)3 divides 9.

(1) (J - O)3 = (J2 - 2JO + O2)(J - O) =
J3 - 2J2O + JO2 - J2O +2JO2 - O3 =
-1 -3J2O + 3JO2 + 1 =
3(JO2 - J2O)

(2) Since O2 = -J and J2 = -O, we get:
3(O2 - J2) = 3(O - J)(O + J) = 3(-1)(1)(J - O)

QED

Lemma 10: (J - O)2 = -3

(1) (J - O)2 = J2 - 2JO + O2

(2) Since OJ = 1, -2JO = -2.

(3) Since O2 = -J and J2 = -O, we get J2 + O2 = -O + -J = -1(O + J) = -1.

QED