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We consider the equation:

$Ax^2+Bxy+Cy^2+Dx+Ey-F=0$ with $A,B,C,D,E,F \in \mathbb{Q}$

If one has a rational point (a point whose coordinates are both rational) on the curve described by the equation, then one can find infinitely many rational points on it. (We ignore the case where it's a degenerate equation, such as a single point, or two lines and so on.)

My question is: Is it always possible to find such a point? And if so, is there some algorithm to find or construct it?

So far, all I came up with is pure brute force, i.e. checking for all rational $x$ if the associated $y$ (such that $(x,y)$ satisfies the equation) is rational too. That would work, and it would give me a guaranteed answer in a finite time (since $\mathbb{Q}$ is countable), but only if I could say with absolute certainty, that there was at least one rational point (and hence infinitely many) on the curve.

I read something about the Legendre theorem, but it was extremely confusing for me. I would appreciate it, if someone could help me with that issue.

EDIT: With "we ignore degenerate equations", I meant actual degenerate equations and the case where it's no point at all (when it has only complex solutions). So just the regular conics: circle, ellipsis, hyperbola, parabola.

K.A.
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  • The curve $x^2 + y^2 = -1$ does not have any rational points on it. – J. J. Dec 02 '15 at 12:09
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    That's what I meant with "we don't consider degenrate cases". So no single points, no two lines (parralel or intersecting), no single line and not $\mathbb{Q}^2$ (in case $A,...,F=0$)) – K.A. Dec 02 '15 at 12:11
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    This curve is a non-degenerate curve but only has complex points on it. (Or if you want to say that degeneracy includes no points, then what is the question?) – J. J. Dec 02 '15 at 12:14
  • Then I correct myself: I meant degenerate equations and the ones that give me either no solution in $\mathbb{R}^2$ or entire $\mathbb{R}^2$. – K.A. Dec 02 '15 at 12:16

1 Answers1

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$x^2+y^2=3$ has no rational points.

Gerry Myerson
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  • Hmm, okay. That would help me. Is there some proof of it? – K.A. Dec 02 '15 at 12:37
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    Assume that $x=p/q$ and $y=r/s$ is a solution with $\gcd(p,q)=\gcd(r,s)=1$. Then we have $p^2s^2 + r^2q^2 = 3q^2s^2$. Because squares are $0$ or $1$ modulo $3$, $ps$ and $rq$ must both be divisible by $3$. Now either $p$ and $r$ are divisible by $3$ or $q$ and $s$ are. If $p$ and $r$ are, then left hand side is divisible by $9$, so one of $q$ or $s$ are divisible by $3$, which is a contradiction. If $q$ and $s$ are, then we notice that the right hand side is divided by more threes than the left hand side, again a contradiction. – J. J. Dec 02 '15 at 13:15
  • Thank you a lot. That helped me so much! – K.A. Dec 02 '15 at 19:11