Lemma: if a2 + 3b2 has an odd factor that is not of this form, then the quotient has an odd factor which is not of this form.
In other words, given the following:
- There exists f which is a factor of a value a2 + 3b2
- f does not have the form p2 + 3q2
- A value F such that fF = a2 + 3b2
- A value f' such that f' divides F and such that it does not have the form p2 + 3q2
(1) So there exists f,g such that fg = a2 + 3b2, f is odd and f is not in the form p2 + 3q2
(2) Let's assume that all of the odd factors of g have the form p2 + 3q2
(3) Now g = a series of primes p1 * p2 * ... * pn [By the Fundamental Theoreom of Arithmetic]
(4) In this series, we can replace all instances of 2 with instances of 4.
(a) We know that if 2 divides a2 + 3b2, then 4 divides it. [See here for proof]
(b) We know that f is odd so each instance of 4 necessarily divides g
(5) Now we can divide all instances of 4 from g and from a2 + 3b2 and still have a result in the form of p2 + 3q2. [See here for proof]
(6) Likewise, we can divide all odd primes from g since we are assuming that all odd factors take the form p2 + 3q2. [The proof is found here.]
(7) But then this leaves f = p2 + 3q2 which is a contradiction.
(8) Therefore, we reject our assumption.