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