Sunday, June 19, 2005

Gaussian Integers: properties of 1 - i

Today's blog continues a proof that was first presented in a previous blog. If you are new to unique factorization, start here. If you are new to Gaussian Integers, start here. To begin this proof, start here.

Today's result is based on work presented by Paulo Ribenboim's Fermat's Last Theorem for Amateurs.

Today's blog focuses on properties of the Gaussian prime λ which we will need to prove Fermat's Last Theorem for n = 4.

Definion 1: λ is a Gaussian Integer that is equal to 1 - i.

Lemma 1: λ is a prime.

Since Norm(λ) = (1 - i)(1 + i) = 1 + 1 = 2. [See here for further details on why this proves that λ is a prime.]

QED

Lemma 2: λ divides 2

i * λ2 = (i)(1 - i)2 = (i)(1 -2i - 1) = 2.

QED

Corollary 2.1: λ6 divides 8

From Lemma 2 above,
(i*λ2)3 =(-i)*λ6= 8

QED

Lemma 3: λ*i = (1 + i)

λ * i = i(1 - i) = i + 1

QED

Definition 2: is relation such that α ≡ β (mod γ) means that α - β is divisible by γ.

We describe this relationship by saying that α modulus γ is equal to β

This definition is true for Gaussian Integers or standard integers. For example, we know that 6 ≡ 2 (mod 4) since 4 divides 6 - 2. We then can say that 6 modulus 4 is equal to 2.

Lemma 4: Let α be any Gaussian Integer. α modulus 2 equals 0, 1, i, or λ

(1) For any α, there exists a,b such that α = a + bi. [Definition of Gaussian Integer]

Case I: a is even, b is even

In this case, α ≡ 0 (mod 2) since a is even, b is even implies there exists A, B such that a = 2A, b = 2B and α = 2A + 2Bi = 2(A+Bi)

Case II: a is odd, b is even

In this case, α ≡ 1 (mod 2) since there exists A, B such that a = 2A+1, b = 2B and α - 1 = 2A + 1 - 1 + 2Bi = 2(A + Bi)

Case III: a is even, b is odd

In this case, α ≡ i (mod 2) since there exists A, B such that a = 2A, b = 2B + 1 and α - i = 2A + (2B+1)i - i = 2A + 2Bi = 2(A+Bi)

Case IV: a is odd, b is odd

In this case, α ≡ λ (mod 2) since there exists A,B such that a = 2A + 1, b = 2B + 1, and α - λ = 2A + 1 + (2B + 1)i - (1 - i) = 2A + 1 + 2Bi + i + i - 1 = 2A + 2B + 2i = 2(A + B + i)

QED

Lemma 5: if α is a Gaussian Integer that is not divisible by λ, then α4 ≡ 1 (mod λ6).

(1) Since α is not divisible by λ, it cannot be divisible by 2. [See Lemma 2 above]

(2) We also know that modulo 2, it cannot be λ. If α modulo 2 is λ then it would imply that α is divisible by λ which it is not.

(3) So, we are left with α ≡ 1 or i (mod 2) from Lemma 4.

(4) α4 ≡ 1 (mod 8) since:

Case I: α ≡ 1 (mod 2)

(a) (α + 1) ≡ (α - 1) ≡ 0 (mod 2)

(b) 2 - 1) ≡ (α + 1)(α - 1) ≡ 0 (mod 4)

(c) 2 + 1) ≡ 0 (mod 2)

(d) 4 - 1) ≡ (α2 + 1)(α2 - 1) ≡ 0 (mod 8)

Case II: α ≡ i (mod 2)

(a) (α + i) ≡ (α - i) ≡ 0 (mod 2)

(b) 2 + 1) ≡ (α + i)(α - i) ≡ 0 (mod 4)

(c) 2 - 1) ≡ 0 (mod 2)

(d) 4 - 1) ≡ (α2 + 1)(α2 - 1) ≡ 0 (mod 8)

(5) And this proves that α4 ≡ 1 (mod λ6) using Corollary 2.1 above.

QED

7 comments:

Jose Brox said...

Why don't just say that:

(1+i)^6 = -8i

Since -i is a unit, asking if a^4-1 is divisible by -8i is the same as asking if a^4-1 is divisible by 8.

Now, a^4-1 = (a^2+1)(a^2-1) =
= (a+1)(a-1)(a+i)(a-i)

a) If a==1(mod 2), then
(a+1)==(a-1)==0 (mod 2) and a^2==1 (mod 2), so a^2+1==0 (mod 2) and therefore, a^4-1==0 (mod 8)

b) If a==i(mod 2), then
(a+i)==(a-i)==0 (mod 2) and a^2==-1==1 (mod 2), so a^2-1==0 (mod 2) and therefore, a^4-1==0 (mod 8)

I find this way to be easier to understand in a single look.

Regards. Jose Brox

Jose Brox said...

Sorry, I meant (1-i)^6 = 8i.

The rest is ok, I think.

Jose

Larry Freeman said...

Hi Jose,

Thank you very much for your comments! I agree with your thinking and made the changes that you suggested.

Cheers,

-Larry

Scouse Rob said...

In Lemma 2 shouldn't it be:

(i)(1 -2i - 1) = 2.

Instead of:
(i)(1 -2i + 1) = 2.

Rob

Scouse Rob said...

Definition 2: ≡ is relation such that α ≡ β (mod γ) means that α - β is divisible by α.

Shouldn't that be divisible by γ?

Scouse Rob said...

In Lemma 4 (Case IV) shouldn't α - λ be:

2A + 1 + (2B + 1)i - (1 - i)

Instead of:
2A + 1 + (2B + 1)i - (i - 1)

Larry Freeman said...

Hi Rob,

You are correct on all accounts. I've updated the blog to correct the typos.

I really appreciate all your comments! :-)

Cheers,

-Larry