Proof for n=4 using Gaussian Integers: Step 2

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 2 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 Gaussian integers: μ₁, μ₂, β, ε, λ such that:

ε * λ^4(r-1) * μ₂⁴ + μ₁⁴ = β².

(1) From the previous lemma, we know that:
ε * λ^4r*α⁴ = γ² - β⁴ = (γ - β²)(γ + β²)

(2) We know that λ² divides both (γ - β²) and (γ + β²) since:

(a) Assume λ² does not divide γ - β²

(b) Then, it must divide γ + β² [From Euclid's Generalized Lemma for Gaussian Integers and (step #1 above)]

(c) But then it necessarily divides γ - β² since
(γ + β²) + (γ - β²) = 2*γ = (λ²)(i * γ) [Since 2= i*λ²]

(d) Having found a contradiction we reject our assumption in step #2a above. We can make the same exact argument if we assume that λ² does not divide γ + β²

(e) So, we are left concluding that λ² necessarily divides both γ - β² and γ + β².

(3) We can also conclude that gcd(γ - β², γ + β²) cannot be greater than 2.

(a) Assume that gcd(γ - β², γ + β²) is greater than 2. Then gcd = λ³ or there exists a prime μ that is greater than 2. [Since we know that λ divides 2, see Lemma 2, here and further, we know that there is no other prime less than 2; see Lemma 4, here]

(b) First, let's assume that there exists a prime μ that is greater than 2.

(c)Then there exists values δ₁, δ₂ such that:
γ - β² = μ * δ₁ and γ + β² = μ * δ₂.

(d) And this means that: μ divides γ since:
2 * γ = (γ - β²) + (γ + β²) = μ(δ₁ + δ₂).

(e) And this also means that μ divides β² since:
2 * β² = (γ + β²) - (γ - β²) = μ(δ₂ - δ₁).

(f) But this is a contradiction since from the lemma cited in step #1 above (see step #4, here), we know that gcd(β²,γ) = 1.

(g) So we reject our assumption in step #3b above and assume that gcd = λ³.

(h) But, then λ³ divides 2 * γ = (γ - β²) + (γ + β²) and λ³ divides 2 * β².

(i) Since λ² is an associate of 2 (see Lemma 2, here), this gives us that λ divides both γ and β².

(j) But this is impossible since we know that β and γ are relatively prime from the lemma cited in step #1 above (see step #4, here)

(k) So we can also reject the assumption in step #3g above and for that matter, the assumption in step #3a above.

(l) So, we can conclude that gcd(γ - β², γ + β²) cannot be greater than 2.

(4) So, we can conclude that λ^4r-2 divides either γ + β² or γ - β² since:

(a) From step #1 above, we know that: λ^r divides (γ - β²)(γ + β²).

(b) From step #2 above, we know that λ² divides both (γ - β²) and (γ + β²).

(c) But, from step #3 above, we know that (λ^4r/λ² = λ^4r-2) can only divide (γ + β²) or (γ - β²) but not both.

(5) So, there exist two values: let's call them η₁ and η₂ such that:
γ ± β² = λ² * η₁
γ ± β² = λ^4r-2 * η₂

NOTE: Please read ± as + or -. So that one of these values is γ + β² and another of these values is γ - β².

(6) We also can conclude that gcd(η₁,η₂) = 1.

(a) Assume that gcd(η₁,η₂) = δ which is not a unit.

(b) So that δ ≥ λ [Since there is no other prime less than λ, see Lemma 4, here]

(c) But then δ*λ² would be greater than 2 which contradicts Step (3) above.

(7) Thus, ε*λ^4rα⁴ = λ^4rη₁η₂ since:

ε*λ^4rα⁴ =(γ - β²)(γ + β²) [From step #1 above]

λ^4rη₁η₂ =(γ - β²)(γ + β²) [From step #5 above]

(8) By the lemma of relatively prime divisors of n-powers, we can conclude that there exists μ₁ and μ₂ such that:

η₁ = ε₁ * μ₁⁴
η₂ = ε₂ * μ₂⁴

NOTE: ε₁ and ε₂ are units.

(9) Now, we know that:
2*β² = γ + β² - (γ - β²)

(10) From step(5) above and step(8) above, let's assume that:
γ - β² = ε₁ * λ² * μ₁⁴
γ + β² = ε₂ * λ^4r-2 * μ₂⁴

NOTE: We can make a similiar argument if the other case is true.

(11) Then,
2 * β² = ε₂ * λ^4r-2 * μ₂⁴ - ε₁ * λ² * μ₁⁴

(12) Which dividing both sides by i*λ² gives us:
β² = -i*ε₂*λ^4r-4*μ₂⁴ + i*ε₁*μ₁⁴

[Note: since 2=i*λ², see Lemma 2, here and since 1/i=-i since -i*i=-(-1)=1]

(13) Now, we can assume that i*ε₁ = 1 since:

(a) r ≥ 2, so we know that λ⁴ divides β² - i*ε₁*μ₁⁴.

(b) But, λ does not divide β

[if it did, from step #2 above, this would mean λ also divides γ, but this is impossible since they are relatively prime from the lemma cited in step #1 above (see step #4, here)]

(c) So, λ cannot divide μ₁

[if λ divided μ₁, then it would also divide η₁ from step #8 above and this is impossible since gcd(η₁,λ)=1 from step #5 above.]

(d) So μ₁ ≡ 1 (mod λ⁶) [See Lemma 5, here]

(e) So that, λ⁴ divides β² - i*ε₁.

(f) But, λ⁶ divides β⁴ - 1 = (β² - 1)(β² + 1)

(g) So, β² ≡ 1 or -1 (mod λ⁴)

(h) This shows that i*ε₁ = ± 1.

(i) If μ₁ = -1, by multiplication with -1, we obtain the relation:
-i*ε₁λ^4(r-1)μ₂⁴ + μ₁⁴ = (iβ)².

(14) Combining Step (12) and Step(13) and setting ε = -i*ε₁ gives us:
β² = ε*λ^4(r-1)*μ₂⁴ + μ₁⁴

QED