## Sunday, June 26, 2005

### 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 * λ4(r-1) * μ24 + μ14 = β2.

(1) From the previous lemma, we know that:
ε * λ4r4 = 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 λ4r-2 divides either γ + β2 or γ - β2 since:

(a) From step #1 above, we know that: λr 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 4r2 = λ4r-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 = λ4r-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(η12) = 1.

(a) Assume that gcd(η12) = 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*λ4rα4 = λ4rη1η2 since:

ε*λ4rα4 =(γ - β2)(γ + β2) [From step #1 above]

λ4rη1η2 =(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 * λ4r-2 * μ24

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

(11) Then,
2 * β2 = ε2 * λ4r-2 * μ24 - ε1 * λ2 * μ14

(12) Which dividing both sides by i*λ2 gives us:
β2 = -i*ε24r-424 + i*ε114

[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*ε114.

(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*ε1λ4(r-1)μ24 + μ14 = (iβ)2.

(14) Combining Step (12) and Step(13) and setting ε = -i*ε1 gives us:
β2 = ε4(r-1)24 + μ14

QED