## Wednesday, June 22, 2005

### i - 1 and Fermat's Last Theorem n = 4

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 applies the Gaussian prime to 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 α4 + β4 = γ2 where gcd(α,β,γ)=1, then we can assume that λ divides α * β

(1) Assume λ does not divide α * β

(2) Then, from a previous lemma, we know that α4 ≡ 1 (mod λ6), β4 ≡ 1 (mod λ6)

(3) Then, there exists δ1, δ2 where: [based on the definition of found here]
α4 - 1 = λ6 * δ1.
β4 - 1 = λ6 * δ2

(4) So that:
α4 = λ61 + 1.
β4 = λ62 + 1.

(5) Then:
γ2 = λ61 + 1 + λ62 + 1 = λ61 + δ2) + 2.

(6) Since 2 = i*λ2, we get: [From a lemma, found here]
γ2 = λ61 + δ2) + i*λ2241 + δ2) + i]

(7) Which shows that:
λ2 divides γ2

(8) And from this, that:
λ divides γ [Since we have a Gaussian Integer proofs for gcd, unique factorization, and Euclid's Lemma, we can use the proof here. ]

(10) This means that λ4 does not divide γ2.

(a) Assume that λ4 does divide γ2.
(b) Then λ4 would divide 2 since γ2 = λ42δ1 + λ2δ2) + 2. [Since d divides a + b and d divides a, implies d divides b]
(c) Then there exists ζ such that 2 = λ4 * ζ
(d) Since 2 = i *λ2, we get:
i *λ2= λ4
(e) Dividing both sides from λ2 gives:
i = λ2*ζ = (1-i)2*ζ = -2i * ζ
(f) Which is impossible since ζ is a Gaussian Integer and not a fraction.

(11) This also gives us that λ2 does not divide γ [if it did, then λ4 would divide γ2 which it cannot.]

(12) Now, since λ divides γ, there must exist a value η such that γ = λ * η

(13) We know that λ cannot divide η. [If it did, then λ2 would divide γ]

(14) Now, we have our contradiction. Here are the details

(15) In step(5), we showed that:
γ2 = λ61 + 1 + λ62 + 1 = λ61 + δ2) + 2.

(16) Letting μ = δ1 + δ2 and letting γ = λ * η, we get:
(λ*η)2 = λ22 = λ6 * μ + i*(λ2)

(17) Dividing both sides by λ2, we get:
η2 = λ4 * μ + i.

(18) Squaring both sides gives us:
η4 = (λ4 * μ + i)2 = λ82 + 2*λ4*μ*i -1 =
λ82 + ( i*λ2)*λ4*μ*i -1 =
λ622 -μ] - 1

(19) So, that we get:
η4 ≡ -1 (mod λ6)

(20) But, since λ does not divide η, from a previous lemma, we get:
η4 ≡ 1 (mod λ 6)

(21) It is impossible for both of these modulo λ6 values to be true.

(a) Assume η4 ≡ -1 (mod λ6) and η4 ≡ 1 (mod λ6).
(b) λ6 divides both η4 - 1 and η4 + 1.
(c) Then there exists ν1 and ν2 such that;
η - 1 = λ6 * ν1
η + 1 = λ6 * ν2
(d) Adding the two equations together gives us:
2*η = λ61 + ν2]=
i*(λ2)*η = λ61 + ν2]
(e) Dividing both sides by λ2 gives us:
i*η = λ41 + ν2]
(f) But this then implies that λ divides η which goes against assumption. [Note, it doesn't divide i because it is not a unit. It therefore divides η by Euclid's Lemma for Gaussian Integers]

QED

Scouse Rob said...

I get:
α^4 + β^4 = γ2

α^4 + β^4 = γ^2

In the Lemma description/definition.

And y2 instead of y^2 in (5), (10b) and (15) as well.

In (18) shouldn't it be:
η^4 = (λ^4 * μ + i)^2

η^4 = (λ^4 * μ + i)

In (19) δ1 and δ2 have been used elsewhere (3) for a different purpose, which is a bit confusing.

Rob

Larry Freeman said...

Hi Rob,