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}$$\equiv$0(mod 4) or $x^{2}$$\equiv$1(mod 4).

Proof. Suppose x$\in$Z. 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}$$\equiv$0(mod 4).
Case 2: Suppose x is odd. Then x = 2b + 1, for some integer b. Squaring both sides, $x^{2}$ = 4$b^{2}$ + 4b + 1 = 4($b^{2}$ + b) + 1, which means $x^{2}$$\equiv$1(mod 4).
So, $x^{2}$$\equiv$0(mod 4) or $x^{2}$$\equiv$1(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 ($x_{0}$, $y_{0}$)$\in$$Q^{2}$. Then we can write $x_{0}$ = $\frac{p}{q}$ and $y_{0}$=$\frac{r}{s}$, with p,q,r,s$\in$Q 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}$$\equiv$0(mod 4) or $x^{2}$$\equiv$1(mod 4), it's easy to see that ($x^{2}$ + $y^{2}$)$\equiv$0(mod 4) or ($x^{2}$ + $y^{2}$)$\equiv$1(mod 4) or ($x^{2}$ + $y^{2}$)$\equiv$2(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$(2a)^{2}$=12$a^{2}$=4(3$a^{2}$). So 3$(qs)^{2}$ is divisible by 4, which means 3$(qs)^{2}$$\equiv$0(mod 4).
Case 2: Suppose qs is odd. Then qs = 2b + 1, for some integer b. Then 3$(qs)^{2}$=3$(2b + 1)^{2}$= 12$b^{2}$+12b+3 = 4(3$b^{2}$+3b) + 3, which means 3$(qs)^{2}$$\equiv$3(mod 4).
But we know that ($x^{2}$ + $y^{2}$)$\equiv$0(mod 4) or ($x^{2}$ + $y^{2}$)$\equiv$1(mod 4) or ($x^{2}$ + $y^{2}$)$\equiv$2(mod 4), for any x,y $\in$ Z and that $(ps)^{2}$ + $(rq)^{2}$$\equiv$3$(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 ($x^{2}$ + $y^{2}$)$\equiv$0(mod 4) or ($x^{2}$ + $y^{2}$)$\equiv$1(mod 4) or ($x^{2}$ + $y^{2}$)$\equiv$2(mod 4), for any x,y $\in$ Z and that $(ps)^{2}$ + $(rq)^{2}$$\equiv$3$(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 $\in$Z
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}$$\equiv$0(mod 4).
Thus, q or s have to be even. Suppose that q is even. Then p must be odd, as we said the fraction is already reduced. We know that $(ps)^{2}$+$(rq)^{2}$$\equiv$0(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)