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 λ 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 = λ⁶*δ₁ + 1.
β⁴ = λ⁶*δ₂ + 1.

(5) Then:
γ² = λ⁶*δ₁ + 1 + λ⁶*δ₂ + 1 = λ⁶(δ₁ + δ₂) + 2.

(6) Since 2 = i*λ², we get: [From a lemma, found here]
γ² = λ⁶(δ₁ + δ₂) + i*λ² =λ²[λ⁴(δ₁ + 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 = λ⁴(λ²δ₁ + 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*ζ
(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 = λ⁶*δ₁ + 1 + λ⁶*δ₂ + 1 = λ⁶(δ₁ + δ₂) + 2.

(16) Letting μ = δ₁ + δ₂ and letting γ = λ * η, we get:
(λ*η)² = λ²*η² = 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 = λ⁸*μ² + 2*λ⁴*μ*i -1 =
λ⁸*μ² + ( i*λ²)*λ⁴*μ*i -1 =
λ⁶[λ²*μ² -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 = λ⁶[ν₁ + ν₂]=
i*(λ²)*η = λ⁶[ν₁ + ν₂]
(e) Dividing both sides by λ² gives us:
i*η = λ⁴[ν₁ + 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