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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6 comments:
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
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
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
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
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.
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