Wednesday, May 25, 2005

Fermat's Last Theorem: n = 3: Step 2

Today's blog continues the proof that I started earlier. The full proof for Fermat's Last Theorem: n = 3 can be found in my previous blog.

Today, I will show the proof for this lemma:

Lemma: if p,q are coprime, p,q opposite parity, then gcd(2p,p2 + 3q2) = 1 or 3

(1) Assume that there is a prime f which divides both 2p and p2 + 3q2.

(2) We know that it can't be 2 since p2 + 3q2 is odd since p,q have opposite parity.

(3) Let's assume that f is greater than 3. So that there exist P,Q such that:
2p = fP, p2 + 3q2 = Qf

(4) Now, f isn't 2 so we know that 2 must divide P so there exists a value H that is half of P and:
p = fH

(5) So combining the two equations, we get:
3q2 = Qf - p2 = Qf - f2H2 = f(Q - fH2)

(6) f doesn't divide 3 since it is greater than 3. So by Euclid's Lemma, it must divide q.

(7) But this contradicts p,q being coprime since it also divides p so we reject our assumption.



Daan Hoek said...

Dear sir,
In proving Fermat the case n = 3,
Step 2, you state that: "combining the two equations, we get:3q^2 = f^2H^2." I don't think that's correct: It would mean that 3q^2 equals p^2 and 3q^2-p^2 equals zero. I'm writing a paper on this subject and I replaced the discussed passage by the following: "(5) So combining the two equations, we get: 3q^2 = f(Q-fH^2)
I admit it looks a little less elegant, but I think it sticks... Would you please correct me if I'm wrong about any of the things I wrote? Thank you very much,
Daan, Holland

Larry Freeman said...

Hi Daan,

I looked it over and I believe that you are correct. I have modified the proof.

Thanks very much for noticing this!



Le Barbu Rasé said...

Dear Larry Freeman,

I don't understand why gcd(2p,p^2+3q^2) could not be any power of 3, rather than 3 or 1.

I understand well that 3 is the only prime which can divides gcd(2p,p^2+3q^2), but I can't go any further :'(

It doesn't seem to affect the rest of the proof, but I would be very happy if you could explain this.

Thanks for your blog,

Larry Freeman said...

The reason why it can't be any power of 3 is because p,q are coprime.

It can be 3 since if 3 divides p (so that p=3p' but doesn't divide q, 3 is common factor of 2(3p'),(3p')^2 + 3q^2.

On the other hand, in order for 3^n to be a common factor, 3 would have to divide both p and q which is impossible since they are coprime.


Le Barbu Rasé said...

Thanks a lot, it's very clear now.

Scouse Rob said...

I think that there is a typo in (6), an extra 'and'.

I think the sentence(s) should be:

f doesn't divide 3, since it is greater than 3.
So, by Euclid's Lemma, f must divide q.

Larry Freeman said...

Hi Scouse Rob,

Thanks for noticing! I've fixed the typo.



Louis De Man said...

Dear sir,
I have a problem at (6). It seems to me that f must divide q^2 and not q. And that's a problem. Ex: 9divides36 but 9 does not divide 6. It is maybe very stupid but that disturbs me.
Ps: sorry for my approximative english

Larry Freeman said...

If f divides q^2, then it must divide q. It comes from Euclid's Lemma. If f is prime and f divides ab, then f divides a or f divides b. For q^2, this means f divides q or f divides q. In other words, if a prime f divides q^2, then a prime f divides q.