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
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 since 2a = 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)