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How to proof that a curve has no rational points

  1. Feb 5, 2012 #1
    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.
     
  2. jcsd
  3. Feb 5, 2012 #2

    AlephZero

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    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)
     
  4. Feb 5, 2012 #3
    Hi AlephZero, thanks for rewriting it.
    I'd like an opinion about the proof I did.

    First I proved, at least I think, that [itex]x^{2}[/itex][itex]\equiv[/itex]0(mod 4) or [itex]x^{2}[/itex][itex]\equiv[/itex]1(mod 4).

    Proof. Suppose x[itex]\in[/itex]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, [itex]x^{2}[/itex] = 4 [itex]a^{2}[/itex]. By definition of divisibility, 4 | [itex]x^{2}[/itex]. Thus [itex]x^{2}[/itex][itex]\equiv[/itex]0(mod 4).
    Case 2: Suppose x is odd. Then x = 2b + 1, for some integer b. Squaring both sides, [itex]x^{2}[/itex] = 4[itex]b^{2}[/itex] + 4b + 1 = 4([itex]b^{2}[/itex] + b) + 1, which means [itex]x^{2}[/itex][itex]\equiv[/itex]1(mod 4).
    So, [itex]x^{2}[/itex][itex]\equiv[/itex]0(mod 4) or [itex]x^{2}[/itex][itex]\equiv[/itex]1(mod 4) as we wanted to proof.

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

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

    As we already proved [itex]x^{2}[/itex][itex]\equiv[/itex]0(mod 4) or [itex]x^{2}[/itex][itex]\equiv[/itex]1(mod 4), it's easy to see that ([itex]x^{2}[/itex] + [itex]y^{2}[/itex])[itex]\equiv[/itex]0(mod 4) or ([itex]x^{2}[/itex] + [itex]y^{2}[/itex])[itex]\equiv[/itex]1(mod 4) or ([itex]x^{2}[/itex] + [itex]y^{2}[/itex])[itex]\equiv[/itex]2(mod 4).

    Let's analyze the 3[itex](qs)^{2}[/itex]. 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[itex](qs)^{2}[/itex]=3[itex](2a)^{2}[/itex]=12[itex]a^{2}[/itex]=4(3[itex]a^{2}[/itex]). So 3[itex](qs)^{2}[/itex] is divisible by 4, which means 3[itex](qs)^{2}[/itex][itex]\equiv[/itex]0(mod 4).
    Case 2: Suppose qs is odd. Then qs = 2b + 1, for some integer b. Then 3[itex](qs)^{2}[/itex]=3[itex](2b + 1)^{2}[/itex]= 12[itex]b^{2}[/itex]+12b+3 = 4(3[itex]b^{2}[/itex]+3b) + 3, which means 3[itex](qs)^{2}[/itex][itex]\equiv[/itex]3(mod 4).
    But we know that ([itex]x^{2}[/itex] + [itex]y^{2}[/itex])[itex]\equiv[/itex]0(mod 4) or ([itex]x^{2}[/itex] + [itex]y^{2}[/itex])[itex]\equiv[/itex]1(mod 4) or ([itex]x^{2}[/itex] + [itex]y^{2}[/itex])[itex]\equiv[/itex]2(mod 4), for any x,y [itex]\in[/itex] Z and that [itex](ps)^{2}[/itex] + [itex](rq)^{2}[/itex][itex]\equiv[/itex]3[itex](qs)^{2}[/itex](mod 4), so it must be the case that 3[itex](qs)^{2}[/itex] = 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 [itex]x^{2}[/itex] + [itex]y^{2}[/itex] - 3 = 0.

    Are these proofs right? What do you think?
    Thanks.
     
  5. Feb 5, 2012 #4
    You only proved [itex] 3(qs)^{2}[/itex] = 0 (mod 4) and not [itex] 3(qs)^{2}[/itex] = 0
     
  6. Feb 5, 2012 #5

    AlephZero

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    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...
     
  7. Feb 5, 2012 #6
    Thanks both!
    I did some mistakes.
    1. p,q,r,s [itex]\in[/itex]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[itex](qs)^{2}[/itex][itex]\equiv[/itex]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 [itex](ps)^{2}[/itex]+[itex](rq)^{2}[/itex][itex]\equiv[/itex]0(mod 4).
    Then [itex](ps)^{2}[/itex]+[itex](rq)^{2}[/itex] = 4f, for some natural f.
    Then [itex](ps)^{2}[/itex]+[itex](rq)^{2}[/itex] = 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!
     
  8. Feb 6, 2012 #7
    Actually I can prove this:
    by considering the remainder of a sum of squares (mod 3)
     
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