Fermat's Last Theorem: n = 3: Step 3

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-n-3-step-3.html

Fermat's Last Theorem: n = 3: Step 3

Today's blog continues the proof for Fermat's Last Theorem n=3 that I began in a previous blog.

With this step, we enter the heart of the proof for Fermat's Last Theorem n=3. This step is a bit complicated so I will only be able to provide an outline at this point. Later on, I will update this blog and insert all the details.

Lemma: Given the following conditions:

(a) x,y,z is a solution to x³ + y³ = z³
(b) two values of x,y,z are derived from p+q,p-q
(c) p,q coprime
(d) p,q opposite parity
(e) p,q positive
(f) 2p*(p² + 3q²) is a cube.
(g) gcd(2p,p² + 3q²) = 1

Then:
There exists a smaller solution A,B,C such that A³ = B³ + C³.

(1) 2p and p² + 3q² are cubes. [From Relatively Prime Divisors Lemma]

(2) First, we show that there exists two values a,b such that: [See here for the lemma.]
p = a³ - 9ab², q = 3a²b - 3b³, gcd(a,b)=1, a,b opposite parity

(3) This gives us that:
2p =2a³ - 18ab² =(2a)(a - 3b)(a + 3b)

(4) 2a, a - 3b,a + 3b are all coprime.

(a) First, 2a is coprime with a - 3b and a + 3b. Both a - 3b and a + 3b are odd since a,b have opposite parity. If a had a common factor with either, then this factor would also divide b which goes against our assumption.

(b) If any odd prime greater than 3 divides a - 3b and a + 3b, then it must divide a since 2a = a - 3b + a + 3b and it must divide b since 6b
= a + 3b - (a - 3b). But this is impossible.

(c) We already showed that 2 can't divide a - 3b, a + 3b. So all we need to prove is that 3 also can't divide both. If 3 divides both then it divides a since2a = a - 3b + a + 3b, then it also divides p since p = a³ - 9ab². But it can't divide p since gcd(2p,p² + 3q²)=1. So, 3 can't divide both.

(5) So, once again, all three are cubes. [By Relatively Prime Divisors Lemma] so that we have A,B,C such that:

2a = A³
a - 3b = B³
a + 3b = C³

(6) But this gives us another solution to Fermat's Last Theorem: n=3 since:

A³ = 2a = a - 3b + a + 3b = B³ + C³

(7) Now, this solution is necessarily smaller than x,y,z since:

(A³)(B³)(C³) = (2a)(a - 3b)(a + 3b) = 2p

And from our previous work, either:

x³ = (2p)*(p² + 3q²) or
z³ = (2p)*(p² + 3q²)

QED