Today's blog continues the proof that I started earlier. The full proof for Fermat's Last Theorem: n = 3 can be found in my previous blog.
Today, I will show the proof for this lemma:
Lemma: if p,q are coprime, p,q opposite parity, then gcd(2p,p2 + 3q2) = 1 or 3
(1) Assume that there is a prime f which divides both 2p and p2 + 3q2.
(2) We know that it can't be 2 since p2 + 3q2 is odd since p,q have opposite parity.
(3) Let's assume that f is greater than 3. So that there exist P,Q such that:
2p = fP, p2 + 3q2 = Qf
(4) Now, f isn't 2 so we know that 2 must divide P so there exists a value H that is half of P and:
p = fH
(5) So combining the two equations, we get:
3q2 = Qf - p2 = Qf - f2H2 = f(Q - fH2)
(6) f doesn't divide 3 since it is greater than 3. So by Euclid's Lemma, it must divide q.
(7) But this contradicts p,q being coprime since it also divides p so we reject our assumption.