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

(1) The norm of a G-number aJ + bO is a

^{2}+ b

^{2}- ab.

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

Case I: 0,0 --> 0

^{2}+ o

^{2}- (0)(0) = 0

Case II: 1,0 --> 1

^{2}+ o

^{2}- (1)(0) = 1

Case III: 0,1 --> 0

^{2}+ 1

^{2}- (0)(1) = 1

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

(a) In all cases, a

^{2}+ b

^{2}- ab ≥ min(a,b)

^{2}

Case I: a=b: a

^{2}+ b

^{2}- ab = a

^{2}+ a

^{2}- a

^{2}= a

^{2}

Case II: b greater than a: this implies that b

^{2}is greater than ab and therefore: a

^{2}+ (b

^{2}- ab) is at least equal or greater than a

^{2}

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 a

^{2}+ (b+1)

^{2}- a(b+1) is greater than a

^{2}+ b

^{2}- 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 α

^{3}+ β

^{3}+ γ

^{3}= 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, α

^{3}≡ 0 (mod ζ

^{3}). [J-O divides ζ implies (J-O)

^{3}divides ζ

^{3}; Start here if you unfamiliar with the '≡' symbol.]

(5) We also know that β

^{3}+ γ

^{3}≡ 0 (mod ζ

^{3}) [See Definition 1, here if this point is not clear]

(6) Since J-O doesn't divide β

^{3}or γ

^{3}, we can conclude that:

β

^{3}≡ e (mod 9), γ

^{3}≡ f (mod 9) where e=±1 and f= ±1. [See Lemma 8, here for proof]

(7) Now since ζ

^{3 }divides 9 (see Corollary 9.1 here), we also have:

β

^{3}≡ e (mod ζ

^{}

^{3}), γ

^{3}≡ f (mod ζ

^{3}) [From step #6 above]

(8) Now e + f ≡ 0 (mod ζ

^{3}). [From Step #5 above since f + e ≡ β

^{3}+ γ

^{3}≡ -α

^{3}≡ 0 (mod ζ

^{3})]

(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 β

^{3}+ γ

^{3}≡ 0 (mod 9). [Since from step #6, e + f ≡ β

^{3}+ γ

^{3 }(mod 9)

(11) Which gives us that α ≡ 0 (mod 9) [since -α = β

^{3}+ γ3] and since 9 = ζ

^{4}:

α

^{3}≡ 0 (mod ζ

^{4}). [See here for proof of 9 = (J-O)

^{4}]

(12) And since J-O divides α, we know that:

α ≡ 0 (mod ζ

^{2}).

(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: β

^{3}+ γ

^{3}= ημχ since:

(a) (βJ + γO)(βO + γJ)(β + γ) =

(β

^{2}OJ + βγJ

^{2}+ βγO

^{2}+ γ

^{2}OJ)(β + γ) = [β

^{2}+ γ

^{2}+ βγ(-O) + βγ(-J)](β + γ) =

(β

^{2}+ γ

^{2}- βγ)(β + γ)

(b) (β

^{2}+ γ

^{2}- βγ)(β + γ) =

β

^{3}+ β

^{2}γ + βγ

^{2}+ γ

^{3}- β

^{2}γ - βγ

^{2}= β

^{3}+ γ

^{3}

(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 α

^{3}= 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) = a

^{2}+ b

^{2}- ab = 2

^{2}+ 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 = 2

^{2}+ 2

^{2}- (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 * ι

^{3}

η' = ρ * ν

^{3}

-χ' = τ * ξ

^{3}

(20) Putting it all together, gives:

ω

^{3}= ορτ*ι

^{3}ν

^{3}ξ

^{3 }where ορτ = a unit.

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

ο * ι

^{3 }+ ρ * ν

^{3 }+ τ * ξ

^{3}= 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 κ

^{3}= 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 E

^{2}= 1.

(24) Since we know that from step (10) that α ≡ 0 (mod ζ

^{2}), 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:

ν

^{3}≡ e (mod 9) where e

^{2}= 1

ξ

^{3}≡ f (mod 9) where f

^{2}= 1.

And then since ρ * ν

^{3 }+ τ * ξ

^{3}≡ 0 (mod [J-O]

^{3}), and since (J-O)

^{3}equals 9, we know that:

ρ *ν

^{3 }+ τ * ξ

^{3}≡ 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 = ορ

^{2}F.

And we know that F

^{2}= (-e/f)

^{2}= e

^{2}/f

^{2}= 1.

Multiplying Fρ

^{2}to both sides of ο * ι

^{3 }+ ρ * ν

^{3 }+ τ * ξ

^{3}= 0, we get:

E* ι

^{3}+ Fρ

^{3}* ν

^{3 }+ (F

^{2})τ

^{3}* ξ

^{3}= E* ι

^{3}+ Fρ

^{3}* ν

^{3 }+τ

^{3}* ξ

^{3}= 0

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

α'

^{3}+ β'

^{3}+ γ

^{3}= (F

^{2})Fρ

^{3}* ν3 + τ

^{3}* ξ

^{3 }+ (E

^{2})Eι

^{3.}(26) But, we know that α'*β'*γ' is divisible fewer times by J-O than α*β*γ is since:

α

^{3}= (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

^{}