Saturday, May 28, 2005

Fermat's Last Theorem: n= 3: Key Lemma

One of the surprising discoveries in the Fermat's Last Theorem n=3 is how much more complicated it is than n=4. In this blog, I will present a seemingly obscure lemma that is very important in Euler's proof for n=3. The full proof for n=3 can be found at this link.

In my opinion, this obscure lemma is the most beautiful part of the proof. It is surprisingly elegant.

Here is the lemma:

Lemma: Given that there exist p,q with the following properties:
(a) gcd(p,q)=1 (gcd = greatest common denominator)
(b) p,q have opposite parities (one is odd, one is even)
(c) p2 + 3q2 is a cube

Then there exists a,b such that:
(a) p = a3 - 9ab2
(b) q = 3a2b - 3b3
(c) gcd(a,b)=1

(d) a,b have opposite parities


(1) Let u3 = p2 + 3q2. [Since we know that p2 + 3q2 is a cube.]

(2) We know that u is odd since p,q have opposite parities [that is, one is odd, one is even].

(3) We know then that u must be of the form a2 + 3b2. [Since any odd factor would also have to have the same form, see here for the lemma regarding odd factors of p2 + 3q2]

(4) Now, (a2 + 3b2)3 = (a2 + 3b2)[(a2 - 3b2)2 + 3(2ab)2]
since (a2 + 3b2)2 = a4 + 6a2b2 + 9b4 =
= a4 + 12a2b2 - 6a2b2 + 9b4 =
(a2 - 3b2)2 + 3(2ab)2

(5) And, (a2 + 3b2)[(a2 - 3b2)2 + 3(2ab)2] =
[ a(a2 - 3b2) - 3b(2ab)]2 + 3[a(2ab)+b(a2-3b2)]2
[See here for the proof.]

(6) And: [ a(a2 - 3b2) - 3b(2ab)]2 + 3[a(2ab)+b(a2-3b2)]2 =
= [a3 - 3ab2 - 6ab2]2 + 3(2a2b + a2b - 3b3)2 =
=[a3 -9ab2]2 + 3(3a2b - 3b3)2

(7) Which combined with step (1) gives us:
p2 + 3q2 = [a3 -9ab2]2 + 3(3a2b - 3b3)2

(8) Which means that we could define a,b such that:
p = a3 -9ab2.
q =
3a2b - 3b3.
[since otherwise, any common factor would divide p and q].

(9) We also know that a,b have opposite parities since:

(a) If a,b are both odd, then, p is even since p = odd - odd and q is even since q = odd - odd which is impossible since p,q have opposite parities.

(b) If a,b are both even, then p is even since p = even - even and q is even since q = even - even which is impossible.



Matthew Thistle said...

Dear Larry

thanks for your efforts but even as i layman did a year of mathmatics at Uni I can't follow! wish I could...! ha

God Bless
Mahdi bin Doud Thistle

Larry Freeman said...

Hi Matthew,

I have filled in more details in order to make this lemma clearer.

I hope that this helps.



Xavier Gonze said...

Dear Larry,

In going from step 7 to step 8, you make an unjustified
hypothesis. You have not proven that if
a^2 + 3 b^2 = c^2 + 3 d^2
then one must have
a^2 = c^2
b^2 = d^2
separately ...

Larry Freeman said...

Hi Xavier,

The step is valid. Here's why:

I am substituting based on form.

In other words, in any math equation, if I show:

a^5 + b^3 = (c + d + e)^5 + (f + g)^3,

then it follows that:

a = c + d + e
b = f + g


Karl said...

Hey Larry

About the following claim:

if a^2 + 3 b^2 = c^2 + 3 d^2
then one must have
a^2 = c^2 and b^2 = d^2

But, for example,
28= 1^2 + 3*(3^2)
28= 4^2 + 3*(2^2)

Larry Freeman said...

Hi Karl,

Thanks for your comments.

You are right. I should correct that to say that one solution is to set a,b = to those values.

This is one solution to the equation but there may be *other* solutions as you point out.

Since this is really an existence proof, the proof only requires us to show that this is one possible solution.

I have changed the proof to reflect this point.


Anonymous said...

Hi Larry,

In the step (6) you say that

But I think that [a(a^2)-3b(2ab)]^2=[a^3-6ab^2]

So p=a^3-6ab^2 that changes all your proof

Larry Freeman said...

Hi Alvaro,

Thanks for noticing that. The mistake was in step #5. I corrected it.

Please let me know if you see any more typos.



Scouse Rob said...

I have no idea why but in step 4 the cubed in:

(a^2 + 3b^2)^3 = (a^2 + 3b^2)[(a^2 - 3b^2)2 + 3(2ab)^2]

shows as a backwards 2.


Larry Freeman said...

Hi Rob,

In Firefox, it looks fine to me. Which browser are you using?



Scouse Rob said...

I'm using Internet Explorer 7
(IE 7.0.5730.11)

xhantt said...

Hi Larry!,

I'm not convinced yet with the final step.

We obtain such a and b such that u = a^2 + 3b^2, but is not true that they will be such that p = a^3 - 9ab^2.

Even it is not clear that we can choose another pair a', b' such that p = a'^3 - 9a'b'^2.


Larry Freeman said...

Hi xhanntt,

Let's assume that we have the following equation:

(p^2 + 3q^2) = [a^3 -9ab^2]^2 + 3(3a^2b - 3b^3)^2

From this equation, it follows that there are numerous solutions for p,q in terms of a,b that will solve this equation.

It is very easy to find one such solution (that's all we need).

Let p=(a^3-9ab^2) and q=(3a^2b - 3b^3)

Then, it follows that:

p^2 + 3q^2 = (a^3-9ab^2)^2 + 3(3a^2b - 3b^3)^2

I am not arguing that this is the ONLY solution, I am arguing that it is ONE of the possible solutions.

That's all I need for the proof.

xhantt said...

Let say we have this equation
p+3q = a^2+3b^2,
Then I'd agree if p, q are variables then p=a^2, q=b^2 is a solution.

But the converse is not true, if p, q are given values then it not necessarily true that p=a^2, q=b^2, for any pair a, b, even if such pair exists.

Unknown said...

I think that this proof is wrong in this step

if a solution to x^3+y^3=z^3 exist
you have proven that exist p and q, such that

2p( p^2+3q^2)=z^3

it means that p and q are given.

So you have proven that

and not imply that p=g(a,b)

but you use that fact in the next step.


Alejandro Jimenez

Larry Freeman said...

Hi Alex,

Thanks for your comment. I don't fully understand your reasoning but let me think about it and I'll try to get back to you with a more detailed response.


Anders H said...

Dear Larry and others,

Sorry about this very late comment, but we are forced to have a Google account before we can post, you know...

I fully agree with Xavier, Karl, Ismael and Alex. Step (8) in the proof appears flawed. Perhaps the following helps to clarify the problem.

Up to step (7) you have shown

(p^2 + 3q^2) = [a^3 -9ab^2]^2 + 3(3a^2b - 3b^3)^2
= (p')^2 + 3(q')^2


p' = a^3 -9ab^2
q' = 3a^2b - 3b^3

However, there is no convincing proof that p=p' and q=q' for at least one choice of a and b that is *consistent* with requirements.

It is important that p and q are not free variables (they are functions of the assumed solution x,y,z to FLT3), nor are a and b.
a and b must satisfy
u = a^2 + 3 b^2
where u is also not free.
There may be multiple solutions here (Karl posted one example) but the possible choice of a and b is still severely restricted.

Therefore, it is not sufficient to show "one possible solution" to the equation since it involves an incorrect assumption that variables p,q,a,b are free to choose.

In case we are right in our complaints, one wonders: Did Euler himself make this mistake too?


Anders H said...

Addition to my previous post:

If we can show that the equation
u^3 = p^2 + 3q^2
must have a unique solution in p and q, the we are home safe.
Because that would indeed force p=p' and q=q'.

Unknown said...

Hi Larry,

I am writing my dissertation on Fermat's Last Theorem. I understand majority of your steps expect step 4. (a^2 + 3b^2)^2 = a^4 + 6a^2b^2 + 9b^4 =
= a^4 + 12a^2b^2 - 6a^2b^2 + 9b^4 =
(a^2 - 3b^2)^2 + 3(2ab)^2

Where and how did you get 12a^2b^2. I just don't see it and it is bothering me.

Thank you,