## Monday, May 30, 2005

### 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 * ... * 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.

QED