How to proof that a curve has no rational points

[AndreAo]

Hello, I'm trying to do exercise number 20 from chapter 6 of this http://www.people.vcu.edu/~rhammack/BookOfProof/index.html, it asks to show that the curve x² + y² - 3 = 0 has no rational points. In the answer it has this tip: first show that a² + b² = 3c has no solutions, other than the trivial. To do this, investigate the remainders of a sum of squares (mod 4). After you’ve done this, prove that the only solution is indeed the trivial solution...
I'm in trouble with this part, how can I use the information from the tip?

Thanks.

 

[AlephZero]

Writing the hint a different way:

If there is a rational point x = p/q, y= r/s, then (ps)² + (qr)² = 3(qs)²

Any square has remainder 0 or 1 mod 4 (consider (2k)² and (2k+1)²)

 

[AndreAo]

Hi AlephZero, thanks for rewriting it.
I'd like an opinion about the proof I did.

First I proved, at least I think, that x^{2}\equiv0(mod 4) or x^{2}\equiv1(mod 4).

Proof. Suppose x\inZ. Then either x is even or x is odd. We consider theses cases separately.
Case 1: Suppose x is even. Then x = 2a, for some integer a. Squaring both sides, x^{2} = 4 a^{2}. By definition of divisibility, 4 | x^{2}. Thus x^{2}\equiv0(mod 4).
Case 2: Suppose x is odd. Then x = 2b + 1, for some integer b. Squaring both sides, x^{2} = 4b^{2} + 4b + 1 = 4(b^{2} + b) + 1, which means x^{2}\equiv1(mod 4).
So, x^{2}\equiv0(mod 4) or x^{2}\equiv1(mod 4) as we wanted to proof.

The problem: Show that the curve x^{2} + y^{2} - 3 = 0 has no rational points.

Proof. Suppose for the sake of contradiction that there exists an rational point (x_{0}, y_{0})\inQ^{2}. Then we can write x_{0} = \frac{p}{q} and y_{0}=\frac{r}{s}, with p,q,r,s\inQ and q, s\neq 0.
Replacing (x_{0}, y_{0}) in the equation, (ps)^{2} + (rq)^{2} = 3(qs)^{2}.

As we already proved x^{2}\equiv0(mod 4) or x^{2}\equiv1(mod 4), it's easy to see that (x^{2} + y^{2})\equiv0(mod 4) or (x^{2} + y^{2})\equiv1(mod 4) or (x^{2} + y^{2})\equiv2(mod 4).

Let's analyze the 3(qs)^{2}. We have two cases, qs is even or qs is odd.
Case 1: Suppose qs is even. Then qs = 2a, for some integer a. Then 3(qs)^{2}=3(2a)^{2}=12a^{2}=4(3a^{2}). So 3(qs)^{2} is divisible by 4, which means 3(qs)^{2}\equiv0(mod 4).
Case 2: Suppose qs is odd. Then qs = 2b + 1, for some integer b. Then 3(qs)^{2}=3(2b + 1)^{2}= 12b^{2}+12b+3 = 4(3b^{2}+3b) + 3, which means 3(qs)^{2}\equiv3(mod 4).
But we know that (x^{2} + y^{2})\equiv0(mod 4) or (x^{2} + y^{2})\equiv1(mod 4) or (x^{2} + y^{2})\equiv2(mod 4), for any x,y \in Z and that (ps)^{2} + (rq)^{2}\equiv3(qs)^{2}(mod 4), so it must be the case that 3(qs)^{2} = 0, but this would only be true, if q or s equal zero, as we already said that q,s must be different from zero, we have a contradiction.
Therefore, it's not the case that exists a rational point in the curve x^{2} + y^{2} - 3 = 0.

Are these proofs right? What do you think?
Thanks.

 

[willem2]

But we know that (x^{2} + y^{2})\equiv0(mod 4) or (x^{2} + y^{2})\equiv1(mod 4) or (x^{2} + y^{2})\equiv2(mod 4), for any x,y \in Z and that (ps)^{2} + (rq)^{2}\equiv3(qs)^{2}(mod 4), so it must be the case that 3(qs)^{2} = 0, but this would only be true, if q or s equal zero, as we already said that q,s must be different from zero, we have a contradiction.

You only proved 3(qs)^{2} = 0 (mod 4) and not 3(qs)^{2} = 0

 

[AlephZero]

You only proved 3(qs)^{2} = 0 (mod 4) and not 3(qs)^{2} = 0

If might help to be a bit more precise about what you assume for a solution.

For example you can put both fractions over the same denominator and write
x = p/q, y = r/q
You can also cancel out any common factors between p q and r, so at least one of the three numbers must be odd...

 

[AndreAo]

If might help to be a bit more precise about what you assume for a solution.

For example you can put both fractions over the same denominator and write
x = p/q, y = r/q
You can also cancel out any common factors between p q and r, so at least one of the three numbers must be odd...

Thanks both!
I did some mistakes.
1. p,q,r,s \inZ
AlephZero, as you suggest, I think I should have said that p,q have no common factors, and r,s have no common factors, so they're reduced fractions.

Until the point willem2 said, I think it's ok.
Then 3(qs)^{2}\equiv0(mod 4).
Thus, q or s have to be even. Suppose that q is even. Then p must be odd, as we said the fraction is already reduced. We know that (ps)^{2}+(rq)^{2}\equiv0(mod 4).
Then (ps)^{2}+(rq)^{2} = 4f, for some natural f.
Then (ps)^{2}+(rq)^{2} = 2(2f), which implies that both square numbers have to be even. We have rq even, as we suppose q to be even, so we must have s even, as p can't be. Until now we found that, p is odd, q is even, s is even, so r is odd.
I can't figure out any way to show a contradiction from this. Any ideas?

Thanks!

 

[willem2]

Actually I can prove this:

first show that a^2 + b^2 = 3c has no solutions, other than the trivial.

by considering the remainder of a sum of squares (mod 3)