Tuesday, May 31, 2005

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

In today's blog, I will show that if you divide a polyonomial of form a2 + 3b2 with a prime of this same form, then you end up a polynomial that still has the same form.

This result is needed for the proof of Fermat's Last Theorem for n=3 that was first presented by Leonhard Euler. For those interested in seeing the entire proof, please start here.

Lemma: if a prime of form p2 + 3q2 divides a2 + 3b2, then there exists c,d such that:
a2 + 3b2 = (p2 + 3q2)(c2 + 3d2)


(1) There exists f such that a2 + 3b2 = (p2 + 3q2)f.

(2) (pb - aq)(pb + aq) = p2b2 - a2q2 + (3q2b2 - 3q2b2) =
= p2b2 + 3q2b2 - 3q2b2 - a2q2 =
= b2 ( p2 + 3q2 ) - q2 ( a2 + 3b2 )


(3) Now the prime p2 + 3q2 divides either pb - aq or pb + aq since:
(pb - aq)(pb+aq) = (p2 + 3q2)(b2 - q2f) [From Euclid's Lemma and steps (1), (2)]

(4) So that there exists a value F such that:
(p2 + 3q2)F equals either pb + aq or pb - aq

(5) Now, from multiplication of polynomials with this form, we know that:
(p2 + 3(±q)2)(a2 + 3b2) = (pa ± 3qb)2 + 3(pb ± aq)2

(6) So, p2 + 3q2 divides pa ± 3qb since:
(pa ± 3qb)2 = (p2 + 3q2)(a2 + 3b2) - 3(pb ± aq)2 =
(p2 + 3q2)[(a2 + 3b2) - 3(F)2]


(7) So, there exists c,d such that:
pa ± 3qb = c(p2 + 3q2)
pb ± aq = d(p2 + 3q2)


(8) Which means that:
(pa ± 3qb)2 + 3(pb ± aq)2 =
(c * (p2 + 3q2))2 + 3[d * (p2 + 3q2)]2


(9) Putting it altogether means that:
(a2 + 3b2) divided by (p2 + 3q2) =
(pa ± 3qb)2 + 3(pb ± aq)2
divided by (p2 + 3q2)2 [From step (5)]
= [c * (p2 + 3q2)]2 + 3[d * (p2 + 3q2)]2 divided by (p2 + 3q2)2 =
c2 + 3d2


QED

2 comments:

Nikolaos Chavaranis said...

I don't see why (pb - aq) and (pb + aq) should be relatively prime numbers for Euclid's Lemma to apply in step (3).

Larry Freeman said...

If the two values are relatively prime, then they do not share any primes. If a prime divides their product, then by Euclid's Lemma (which applies to products of relatively prime numbers), then it can only divide one number or the other but not both.