Fermat's Last Theorem: n = 3: More a2 + 3b2

Today's blog presents another lemma associated with the polynomial a² + 3b². This lemma is used in the proof for Fermat's Last Theorem: n=3 that was first presented by Leonhard Euler. It is believed that Pierre de Fermat knew this proof but it took the work of a genius like Euler to recreate it. You can see the full proof for Fermat's Last Theorem n=3 by starting here.

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

Then, there exists:

• 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

Here is the proof:

(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 * ... * p_n [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.

QED