Proof for n=3: Eistenstein Integers: Step Two

If you are not familiar with Eisenstein Integers, please start here. The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma: There is no Eisenstein integer whose norm is 2.

(1) The norm of a G-number aJ + bO is a² + b² - ab.

(2) We note first that both abs(a) and abs(b) need to be greater than 1.

Case I: 0,0 --> 0² + o² - (0)(0) = 0

Case II: 1,0 --> 1² + o² - (1)(0) = 1

Case III: 0,1 --> 0² + 1² - (0)(1) = 1

(3) There is no solution where the min(a,b) ≥ 2.

(a) In all cases, a² + b² - ab ≥ min(a,b)²

Case I: a=b: a² + b² - ab = a² + a² - a² = a²

Case II: b greater than a: this implies that b² is greater than ab and therefore: a² + (b² - ab) is at least equal or greater than a²

Case III: The same argument applies only swap a for b.

(b) So, in this case, if min(a,b) ≥ 2 → Norm ≥ 4.

(4) But there is also no solution for max(a,b) ≥ 2.

Assume b greater than a: this implies that 2b + 1 is greater than a which implies that 2b + 1 - a is greater than 0. Adding the equation to both sides, this gives us that a² + (b+1)² - a(b+1) is greater than a² + b² - ab.

So all we need to do is to show that the equation doesn't hold for b=2, a=0 and b=2,a=1. We know that any other value of b will be greater than our result here and a must be less than 2.

Case I: b=2,a=0 (0)² + (2)² - (0)(2) = 4.

Case II: b=2,a=1 (1)² + (2)² - (1)(2) = 3.

QED

Lemma: If J - O divides αβγ where αβγ ≠ 0, then we have reached a position of infinite descent.

(1) Let us suppose that J-O divides α. [We can apply the same argument where it divides β or γ since we are assuming a form where α³ + β³ + γ³ = 0]

(2) We can assume that J-O does not divide β nor γ since all three are relatively prime to each other.

(3) For purposes of clarity, let's let ζ = J - O.

(4) So, α³ ≡ 0 (mod ζ³). [J-O divides ζ implies (J-O)³ divides ζ³; Start here if you unfamiliar with the '≡' symbol.]

(5) We also know that β³ + γ³ ≡ 0 (mod ζ³) [See Definition 1, here if this point is not clear]

(6) Since J-O doesn't divide β³ or γ³, we can conclude that:

β³ ≡ e (mod 9), γ³ ≡ f (mod 9) where e=±1 and f= ±1. [See Lemma 8, here for proof]

(7) Now since ζ³ divides 9 (see Corollary 9.1 here), we also have:

β³ ≡ e (mod ζ³), γ³ ≡ f (mod ζ³) [From step #6 above]

(8) Now e + f ≡ 0 (mod ζ³). [From Step #5 above since f + e ≡ β³ + γ³ ≡ -α³ ≡ 0 (mod ζ³)]

(9) So, f = -e . [From step #8 above and step #6]

(10) But then e + f = 0 and e + f ≡ 0 (mod 9); and therefore β³ + γ³ ≡ 0 (mod 9). [Since from step #6, e + f ≡ β³ + γ³ (mod 9)

(11) Which gives us that α ≡ 0 (mod 9) [since -α = β³ + γ3] and since 9 = ζ⁴:
α³ ≡ 0 (mod ζ⁴). [See here for proof of 9 = (J-O)⁴]

(12) And since J-O divides α, we know that:
α ≡ 0 (mod ζ²).

(13) Let us now define 3 values: η, μ, χ where:
η = βJ + γO
μ = βO + γJ
χ = β + γ

(14) Now, it follows that: β³ + γ³ = ημχ since:

(a) (βJ + γO)(βO + γJ)(β + γ) =
(β²OJ + βγJ² + βγO² + γ²OJ)(β + γ) = [β² + γ² + βγ(-O) + βγ(-J)](β + γ) =
(β² + γ² - βγ)(β + γ)

(b) (β² + γ² - βγ)(β + γ) =
β³ + β²γ + βγ² + γ³ - β²γ - βγ² = β³ + γ³

(15) Now, it also follows that each value η, μ, χ is divisible by J-O since:

(a) (J-O)³ divides ημχ. [Since J-O divides α and α³ = ημχ]

(b) J - O divides μ - η since μ - η = ( βO + γJ) - (βJ + γO)=(J-O)(γ - β)

(c) Also note that μ + η = βO + γJ + βJ + γO = β + γ = χ

(d) From (a), we know that J-O divides μ, η, or χ. [See Euclid's Generalized Lemma for details since J-O is a prime.]

(e) This is enough to show that J-O divides all three since:

Case I: J-O divides μ

(i) Step 15b gives us that J - O divides μ - η

(ii) Since J-O divides μ, it must also divide η.

(iii) Then J-O divides χ, since χ = μ + η from step #c above.

Case II: J-O divides η,

(i) Step 15b gives us that J - O divides μ - η

(ii) Since J-O divides η, it must also divide μ.

(iii) Then J-O divides χ, since χ = μ + η from step #c above.

Case III: J-O divides χ

(i) Then J-O divides μ + η since χ = μ + η from step #c above.

(ii) But J-O divides μ - η from step 15b above.

(iii) This means that J-O divides both 2μ = (μ + η) + (μ - η) and 2η = (μ + η) - (μ - η).

(iv) Now, J-O is prime (see Lemma 5, here), we can apply Euclid's Lemma for Gaussian Integers (see Corollary 3.3, here) to establish that for each J-O divides 2 or J-O divides both μ and η.

(v) J-O does not divide 2.

In order for J-O to divide 2, then N(J-O) must divide N(2) [This is the same as the argument for norms for Gaussian Integers, see Corollary 7.1, here ]. N(J-O) = 3 [See Lemma 5, here] and N(2) = a² + b² - ab = 2² + 0 - 0 = 4 [See Lemma 3, here] And clearly, 3 does not divide 4.

(16) So, there exists: μ', η', χ' such that:
μ = μ'(J-O)
η = η'(J-O)
χ = χ'(J-O)

(17) And we can further show that μ', η', χ' are all coprime since:

(a) If any two have a common divisor, then μ', η' would have a common divisor.

Case I: μ', χ' have common divisor δ: μ' = μ''δ, χ' = χ''δ. (η')(J-O) + (μ')(J-O) =(χ')(J-O) so η' + μ' = χ' so that η' = χ' - μ' = δ(χ'' - μ'').

Case II: η', χ' have common divisor. You can apply the same argument as above.

(b) Assume η', μ' have a common divisor δ which is not a unit.

(c) Then δ divides β - γ since:
μ' - η' = (β - γ) [From 13b above]

(d) And δ divides β + γ since:
μ + η = β + γ [From 13c above and the fact that δ divides μ + η]

(e) So, δ divides both 2β and 2γ since:
2β = (β + γ) + (β - γ)
2γ = (β + γ) - (β - γ)

(f) Now, since β, γ are relatively prime and 2 is an Eisenstein prime (see below), we can conclude that δ = 2.

We know that any number that is only divisible by itself and 1 is a prime. We also know that if a number is divisible by another number, then its norm is likewise divisible by the norm of the other number.

The norm(2) = 2J+2O = 2² + 2² - (2)(2) = 4.

4 is divisible by 1,2,4. So, to prove that 2 is a prime, we only need to prove that there is no Eisenstein integer whose norm is 2.

We already proved this in the lemma above.

(g) But the we have a contradiction since η is not divisible by 2. Here is the details:

Since δ = 2, both β + γ and β - γ are divisible by 2.

This means that there are four possible states where λ, θ are half the value - a unit.:

Case I: β has a form 2λ + unit, γ has a form 2θ + unit.
Case II: β has a form 2λ + unit, γ has a form 2θ - unit.
Case III: β has a form 2λ - unit, γ has a form 2θ + unit.
Case IV: β has a form 2λ - unit, γ has a form 2θ - unit.

Let's look at η = βJ + γO. So:

CaseI: μ = (2λ + unit)J + (2θ + unit)O = 2λJ + (unit)*J + 2θO + (unit)*O = 2(λJ + θO) + unit.

Case II: gets us: 2λJ + unitJ + 2θO = unitO = 2(λJ + θO) + unit(J-O)

Case III: gets us: 2λJ - unitJ + 2θO = unitO = 2(λJ + θO) - unit(J-O).

Case IV. gets us: 2(λJ + θO) - unit.

(h) So we reject our assumption (b).

(18) Since J-O divides α, we know that there exists a value ω such that:

α = (J-O)*ω

Which also gives us that:

(α)³ = (J-O)³(ω)³ = -ημχ

And dividing both sides by (J-O)³ gives us:

(ω)³ = -η'μ'χ'

(19) Since η', μ', and χ' are coprime, we can know that they are equal to a cube multiplied to a unit. [See here for the proof on this.]

So we could suppose that there exist: ι, ν, ξ which are Eisenstein Integers and ο, ρ, τ which will represent units. With this, we can assume:

μ' = ο * ι³
η' = ρ * ν³
-χ' = τ * ξ³

(20) Putting it all together, gives:

ω³ = ορτ*ι³ν³ξ³ where ορτ = a unit.

(21) Since μ' + η' = χ', we also have:

ο * ι³ + ρ * ν³ + τ * ξ³ = 0.

(22) Since μ', η', and χ' are coprime so: ι, ν, ξ must be coprime.

(23) Now, we also know that there exists κ such that:
ω = κ * ινξ.

So that κ³ = ορτ = a G-unit. Since J-O is not a unit, we know it doesn't divide κ and we have κ ≡ E (mod 9) which means that ορτ ≡ E (mod 9).

So that ορτ = E and E² = 1.

(24) Since we know that from step (10) that α ≡ 0 (mod ζ²), we know that:

ω ≡ 0 (mod J-O).

So we know that J-O divides ι, ν, or ξ. Let us suppose that it divides ι so that we have:

ι ≡ 0 (mod J-O)
ν not ≡ 0 (mod J-O)
ξ not ≡ 0 (mod J-O)

So that we have:

ν³ ≡ e (mod 9) where e² = 1
ξ³ ≡ f (mod 9) where f² = 1.

And then since ρ * ν³ + τ * ξ³ ≡ 0 (mod [J-O]³), and since (J-O)³ equals 9, we know that:

ρ *ν³ + τ * ξ³ ≡ 0 (mod 9) and applying above, we he have: e*ρ + f*τ ≡ 0 (mod 9) and since e*ρ + f*τ is less than 9, we know that e*ρ + f*τ = 0.

If we set F = -e/f then we have:
τ = Fρ

Combining this with E = ορτ, we get:
E = ορ²F.

And we know that F² = (-e/f)² = e²/f² = 1.

Multiplying Fρ² to both sides of ο * ι³ + ρ * ν³ + τ * ξ³ = 0, we get:

E* ι³ + Fρ³* ν³ + (F²)τ³ * ξ³ = E* ι³ + Fρ³* ν³ +τ³ * ξ³ = 0

(25) Now, if we set α' = Fρν, β' = τξ, γ' = Eι, then we get:

α'³ + β'³ + γ³ = (F²)Fρ³* ν3 + τ³ * ξ³ + (E²)Eι^3.

(26) But, we know that α'*β'*γ' is divisible fewer times by J-O than α*β*γ is since:

α³ = (J-O)³(μ'η'χ') and
μ'η'χ' = ε(α'³*β'³*γ'³) where ε is a unit.

(27) We can then apply the same reasoning to these values which prove infinite descent.

QED