## Saturday, June 25, 2005

### Proof for n=4 using Gaussian Integers: Step 1

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 shows the details to Step 1 in Fermat's Last Theorem n = 4. As before, I use Greek letters to refer to Gaussian Integers and Latin letters to refer to rational integers. I introduced the Gaussian Prime λ in a previous blog.

Lemma: if Fermat's Last Theorem is true for n = 4, then there exist α, β, γ, ε, λ such that:
ε * λ4r * α4 + β4 = γ2 where r ≥ 2

(1) Assume that x4 + y4 = z2 where xyz ≠ 0.

(2) Setting α = x + 0*i, β = y + o*i, γ = z+0*i, we get:
α4 + β4 = γ2 where α, β, γ, are Gaussian Integers.

(3) Now, we can assume that α, β are coprime.

(a) Let δ = gcd(α,β). [See here for method of gcd for Gaussian Integers]
(b) Then, there exists α',β' such that α = δ * α', β = δ * β'.
(c) We know that gcd(α',β') = 1. [since we already divided out all common divisors]
(d) We also know that δ4 must divide γ2 since γ2 = δ4(α'4 + β'4)
(e) Which gives us that δ2 must divide γ [From here based on Euclid's Lemma for Gaussian Integers and Fundamental Theorem of Arithmetic for Gaussian Integers]
(f) Then, there exists γ' such that γ = γ' * δ2.
(g) So that we get α'4 + β'4 = γ'2
(h) This shows that we can always reduce α,β, to coprime values.

(4) We can further assume that α, β, γ are coprime.

(a) We know for example that α22, and γ are coprime. [See here ]
(b) Assume that there exists δ = gcd(α,γ) where δ is greater than 1.
(c) Then δ2 would necessarily divide α4 and γ2
(d) It would then also divide β2 which goes against (a) so we reject our assumption in (b).
(e) We can use the same argument to show that gcd(β,γ)=1.

(5) We know that λ divides either α or β [See here for proof.]

(6) Let us assume that it divides α. The same arguments will hold if it divides β

(7) Then, there exists α' such that α = α'*λr where r ≥ 1.

(8) Let ε be some unit such as 1, -1, i, or -i. We can now state that:
ε * α'4 * λ4*r + β4 = γ2.

(9) Now, since λ does not divide β, we know that β ≡ 1 (mod λ6) which also means that β ≡ 1 (mod λ4) since:

(a) β ≡ 1 (mod λ6) [See here for proof]
(b) β ≡ 1 (mod λ6) implies that λ6 divides β - 1. [Definition of ]
(c) Thus, there exists δ such that β - 1 = λ6 * δ = λ42 * δ).

(10) From (8) and (9), we conclude that:
γ2 ≡ 1 (mod λ4).

(11) Now, based on this we can conclude that γ ≡ 1 (mod λ2) or γ ≡ i (mod λ2) since:

(a) λ2 = -2i. [Since λ = 1-i]
(b) γ = a + bi which gives us 4 cases that we need to prove.
(c) Case 1: a is even, b is even. This is impossible since 2 would then divide γ and 2 = (-2i)*i.
(d) Case 2: a is odd, b is odd. This is also impossible since then γ ≡ λ (mod λ2) and we know that λ does not divide γ
(e) Case 3: a is odd, b is even. Since b is divisible by 2i and a is partially divisible by 2, this gives us: γ ≡ 1 (mod λ2)
(f) Case 4: a is even, b is odd. Since a is divisible by 2i and b is partially divisible by 2i, this gives us: γ ≡ i (mod λ2)

(12) But, γ ≡ i (mod λ2) is impossible since:

(a) λ2 would divide γ - i.
(b) Then, there exists δ such that: γ - i = λ2 * δ
(c) We can restate this as: γ = λ2 * δ + i.
(d) Squaring both sides gives us:
γ2 = λ42 + 2*λ2*δ*i - 1 = λ42 + (i*λ2)*λ2*δ*i - 1 =
λ42 - δ) - 1.
(e) But then γ2 + 1 is divisible by (λ4) which contradicts step #10 since γ2 - 1 is divisible by λ4.

(13) So, γ ≡ 1 (mod λ2) and there exists δ such that:
γ - 1 = λ2 * δ

(14) Adding 2 to both sides gives us:
γ + 1 = λ2 * δ + 2 = λ2(δ + i).

(15) Now we can conclude that λ divides δ(δ + i) since

(a) if λ divides δ, then this is true so we can assume that λ does not divide δ
(b) By the definition of Gaussian Integers, there exists a,b such that δ = a + bi.
(c) We know that a,b cannot both be even. If both are even, the δ is divisible by 2 which is divisible by λ2
(d) We know that a,b cannot both be odd. If both are odd, then we get a remainder of 1 + i which is equal to (λ)*i = (1 - i)*i = i + 1.
(d) If a is even, b is odd, the remainder is i, then λ divides i + i = 2i which is divisible by λ2.
(e) If a is odd, b is even, the remainder is 1. then λ divides 1 + i which is equal to λ * i.

(16) Multiplying Step#13 to both sides of Step#14 gives us:
γ2 - 1 = λ4(δ)(δ + i).

(17) From this, we can conclude that λ5 divides γ2 - 1. [From Step #15]

(18) And this gives us that λ5 divides ε * α'4 * λ4*r + (β4-1).

(19) λ5 divides β4-1 since:

(a) λ does not divide β [from step #5 and step #6 above]

(b) Then, β4 ≡ 1 (mod λ6) [See Lemma 5, here]

(c) So, λ6 divides β4 - 1

(d) Which means that λ5 must also divide β4 - 1

(20) So, λ5 divides ε * α'4 * λ4*r. [if any x divides A+B (step #18 above) and x divides B (step #19 above), then x divides A+B-B=A]

(21) But λ does not divide α' (step #7 above) and it does not divide ε (see step #8 above), so we are left with the conclusion that λ5 must divide λ4*r. [Note, ε is a unit and is only divisible by other units]

(22) And this shows that r ≥ 2.

QED

Scouse Rob said...

Step (18) is bit tricky.
No explaination given of why λ^5 divides(β^4-1).

Think I figured it:
λ doesn't divide β
So λ doesn't divide β^4
So β^4 ≡ 1 (mod λ^6)
So β^4 - 1 = λ^6*ζ (For some ζ)
So β^4 - 1 = λ^5*(λ*ζ)
So λ^5 divides(β^4-1)

Rob

Scouse Rob said...

Sorry, step (19)

Larry Freeman said...

Hi Rob,

Thanks for your comment. I've modified the blog to make the argument clearer and to provide the explanation that you are looking for.

Cheers,

-Larry

Scouse Rob said...

Enjoying this immensely.

In Step (7) you don't explicitly state that α' and λ are coprime, which is used in step (21).

Rob