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 * λ^4r * 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 α²,β², 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'*λ^r where r ≥ 1.

(8) Let ε be some unit such as 1, -1, i, or -i. We can now state that:
ε * α'⁴ * λ^4*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 = λ⁴(λ² * 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 = λ⁴*δ² + 2*λ²*δ*i - 1 = λ⁴*δ² + (i*λ²)*λ²*δ*i - 1 =
λ⁴(δ² - 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 * λ^4*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 * λ^4*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 λ^4*r. [Note, ε is a unit and is only divisible by other units]

(22) And this shows that r ≥ 2.

QED