Wednesday, June 01, 2005

Fermat's Last Theorem: n = 3: Division of a2 + 3b2 by 2

In today's blog, I will go over the case of where a2 + 3b2 is an even number. This result is part of the proof of Fermat's Last Theorem for n=3. If you would like to see the full proof, start here.

Lemma: if 2 divides a polynomial of form a2 + 3b2, then 4 also divides the polynomial and this division by 4 results in another polynomial of the same form.

In other words, we will prove two propositions:

  • if 2 divides a2 + 3b2, then 4 also divides it.
  • if 4 divides a2 + 3b2, then there exists c,d such that a2 + 3b2 = 4*(c2 + 3d2)

(1) We know that a,b have the same parity (they are both odd or they are both even). Otherwise, a2 + 3b2 would not be divisible by 2

(2) If they are both even, then it is easy to show that 4 divides a2 + 3b2 and that the result is of the same form:


(a) There exists c,d such that a = 2c and b = 2d
(b) a2 + 3b2 = (2c)2 + 3(2d)2 = 4(c2 + 3d2)


(3) So, we can assume that they are both odd.

(4) Now, there exists m,n such that a = 4m ± 1, b = 4n ± 1 [See here for proof]

(5) From this, we know that 4 divides either a + b or a - b

(6) I will now show that in both cases, the lemma is proven.

Case I: 4 divides a + b

(a) First, 4(a2 + 3b2) = (12 + 3*12)(a2 + 3b2) = (a - 3b)2 + 3(a + b)2 [See here for proof]

(b) Since a - 3b = (a + b) - 4b, we know that 4 divides a - 3b

(c) So, this gives us that 42 divides (a - 3b)2 + 3(a + b)2 which also means that:
4 divides a2 + 3b2 [Since 42 divides (a - 3b)2 + 3(a+b)2 implies 42 divides 4(a2 + 3b2)]

(d) Since 4 divides a - 3b and a + b, we know that there exists u,v such that: u = (a - 3b)/4 and v = (a + b)/4

(e) 4(u2 + 3v2) = 4{[(1/4)(a - 3b)]2 + 3[(1/4)(a + b)]2} =
= (1/4)(a2 + 9b2 + 3a2 + 3b2) =
= (1/4)(4a2 + 12b2) = a2 + 3b2


Case II: 4 divides a - b

(a) The argument here is the same as for case I except that we set 4(a2 + 3b2) = (12 + 3*(-1)2)(a2 + 3b2) = (a + 3b)2 + 3(a - b)2 [See here for proof.]

(b) Then after proving that 42 divides this result, we let u = (1/4)(a + 3b) and let v = (1/4)(a - b).

(c) Then 4(u2 + 3v2) = a2 + 3b2

QED

4 comments:

Daan Hoek said...

Dear Larry,
By showing that:

"(b) a^2 + 3b^2 = (2c)^2 + 3(2d)^2 = 4(c^2 + 3d^2)", you've already proven the lemma. You needn't go further than that. Or am I wrong?

Daan

Daan Hoek said...

I'm sorry, I was mistaken... Silly me...
Sorry!

Daan

Scouse Rob said...

In case I (d)

Shouldn't it be:
4[(1/4)(a - 3b)^2 + 3((1/4)(a + b))^2]

Instead of:
4[((1/4)a - 3b)^2 + 3((1/4)(a + b))^2]

Rob

Larry Freeman said...

Hi Rob,

Thanks very much for the comment.

You are correct that there is an error in Case I.

I found it in Case I(e) and changed it so that it is correct.

Cheers,

-Larry