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Suppose that I have a quadratic polynomial $p(x,y)$ over the reals with two variables.

How many rational solutions can $p(x,y)$ have?

Can this be any natural number, or if it is large enough, then it will be infinite?

I know that deciding whether a solution exists or not can be difficult, I'm only interested in the cardinality.

domotorp
  • 909

1 Answers1

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Given any real quadratic polynomial $$p(x,y) = ax^2 + bxy + cy^2 + dx + ey + f \in \mathbb{R}[x,y]$$ where $a,b,c$ not all zero. Let

  • $\mathcal{C} = \{ (x,y) \in \mathbb{R}^2 : p(x,y) = 0 \}$ be its zero set,
  • $\mathcal{C}_r = \mathcal{C} \cap \mathbb{Q}^2$ and $N = |\mathcal{C}_r |$ be its number of rational solutions.

In general, $N$ can be $0,1,2,3,4$ or $\infty$.

In the special case that $p(x,y) \in \mathbb{Q}[x,y]$ and represents a non-degenerate conic, then $N = 0$ or $\infty$.

Geometrically, $\mathcal{C}$ has several possible forms:

  1. $\mathcal{C} = \emptyset \implies N = 0.\quad$ (eg. $x^2+y^2+1 = 0$)

    Since $p(x,y)$ has no real solutions, it has no rational solutions.

  2. $\mathcal{C}$ is a single point $\implies N = 0$ or $1$.

    • $N = 0\quad$ eg. $(x-\pi)^2 + y^2 = 0$.
    • $N = 1\quad$ eg. $x^2 + y^2 = 0$.

  3. $\mathcal{C}$ is a single line $\implies N = 0, 1$ or $\infty$

    • $N = 0\quad$ eg. $(y-\pi)^2 = 0$.

    • $N = 1\quad$ eg. $(y - \pi x)^2 = 0$.

    • $N > 1 \implies N = \infty.$

      Let's say we have found two rational solutions $u,v$ for $p(x,y) = 0$. Since $\lambda u + (1-\lambda) v \in \mathcal{C}_r$ for all $\lambda \in \mathbb{Q}$, $p(x,y)$ has infinitely many rational solutions.

  4. $\mathcal{C}$ is a pair of lines $\implies N = 0, 1, 2$ or $\infty$.

    • $N = 0\quad$ eg. $(y-\pi)(y - \sqrt{2}) = 0$

    • $N = 1\quad$ eg. $(y - \pi x)(y - \pi x - \sqrt{2}) = 0$.

    • $N = 2\quad$ eg. $(y - \pi x)(y - \pi x - 1) = 0$.

    • $N > 2\implies N = \infty$.

      If $p(x,y)$ has more than two rational solutions, at least two of them lie on same line. We can repeat argument of the single line case to deduce $p(x,y)$ has infinitely many rational solutions.

  5. $\mathcal{C}$ is non-degenerate (ie. is a parabola/ellipse/hyperbola) $\implies N = 0, 1, 2, 3, 4$ or $\infty$.

    • $N = 0,\quad$ eg. $x^2+y^2 - 3 = 0$.

    • $N = 1,\quad$ eg. $x^2+y^2 - \pi x = 0$.

    • $N = 2,\quad$ eg. $(x^2+y^2-1) + \pi x^2 = 0$.

    • $N = 3,\quad$ eg. $(x^2 +y^2 - 2x) + \pi(x^2-y^2) = 0$.

    • $N = 4,\quad$ eg. $(x^2+y^2-2) + \pi(x^2-y^2) = 0$.

    • When $N > 4$, let $u_i = (x_i,y_i), i = 1,\ldots, 5$ be five rational solutions of $p(x,y)$. Since the conic is non-degenerate, $\mathcal{C}$ is the zero set of another quadratic polynomial with rational coefficients: $$q(x,y) \stackrel{def}{=}\left|\begin{matrix} x^2 & xy & y^2 & x & y & 1\\ x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1\\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1\\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1\\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1\\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1\end{matrix}\right| \in \mathbb{Q}(x,y)$$ and there is a $\lambda \ne 0$ such that $p(x,y) = \lambda q(x,y)$.

      Consider the collection of lines passing through $u_1$ with rational slope. With at most one exception (the line coincides with the tangent line of $\mathcal{C}$ at $u_1$), such a line will intersect $\mathcal{C}$ at another point $u' =(x',y')$. Since $p(x',y') = 0 \implies q(x',y') = 0$ and coefficients of $q(x,y)$ are rational, one can use this to deduce $u' \in \mathcal{C}_r$. Since there are infinitely many such lines, $q(x,y) = 0$ and hence $p(x,y) = 0$ has infinitely many rational solutions.

      In the special case $p(x,y) \in \mathbb{Q}[x,y]$ and represents a non-degenerate conic, the same argument tell us when $p(x,y) = 0$ has a rational solution, it will have infinitely many rational solutions.

achille hui
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