Fermat's Last Theorem: n = 3: Step 2

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-n-3-step-2.html

Fermat's Last Theorem: n = 3: Step 2

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 Pand:
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.

QED