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.
QED
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7 comments:
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
Hi Daan,
I looked it over and I believe that you are correct. I have modified the proof.
Thanks very much for noticing this!
Cheers,
-Larry
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,
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.
-Larry
Thanks a lot, it's very clear now.
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.
Hi Scouse Rob,
Thanks for noticing! I've fixed the typo.
Cheers,
-Larry
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