Let $a,b,c$ be real numbers with $a > 0$. Suppose that $d = b^2 - 4ac > 0$. Is there a constant $e$ such that there exist integers $x,y$ not both $0$ and that $\left|ax^2 + bxy + cy^2\right|\le e\sqrt d$ ?

That question is a modified form of Problem 12 in Section 6.4, "The Geometry of Numbers," in I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991.

The solution to Problem 12 was relatively easy because the region is that of an ellipse, which is convex and has an easy-to-calculate area, so all we had to do was use a modified version of Minkowski's convex body theorem.

The region of the question posed above, however, is not convex and has a weird shape; part of its boundary is hyperbolic.

  • $\begingroup$ My comment is inspired by the following : math.stackexchange.com/questions/3675922/… This problem is equivalent to the following : Consider a region $R$ which is enclosed by two hyperbolas s.t. they share same asymptotes $l_i$ (Here $l_i$ is in $R$) Here $R$ contains a rectangle $G$ whose center is origin s.t. four vertices in hyperbolas are mid points of sides in $G$. Then is there a constant i.e. area of $G$ s.t. $R$ contains $(x,y)$, not origin, where $x,\ y$ are integers. $\endgroup$
    – HK Lee
    Commented Jun 12, 2020 at 10:48

1 Answer 1


$\begin{array}\\ f(x, y) &=ax^2+bxy+cy^2\\ &=ay^2((x/y)^2+(b/a)(x/y)+(c/a))\\ &=ay^2((x/y)-r)((x/y)-s)\\ \end{array} $

where $r$ and $s$ are the roots of $x^2+(b/a)x+(c/a) =0$.

These roots are

$\begin{array}\\ r, s &=\dfrac{-b/a\pm\sqrt{b^2/a^2-4c/a}}{2}\\ &=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-b\pm\sqrt{d}}{2a}\\ &= u \pm v \qquad\text{where } u = -\dfrac{b}{2a}, v = \dfrac{\sqrt{d}}{2a}\\ \end{array} $

Since $d > 0$, $r$ and $s$ are real.

Therefore $f(x, y) =ax^2+bxy+cy^2 =ay^2((x/y)-r)((x/y)-s) =a(x-ry)(x-sy) $. So we want $|a(x-ry)(x-sy)| \le e\sqrt{d} $.

Choosing $r = u+v $ and $s = u-v $,

$\begin{array}\\ a(x-ry)(x-sy) &=a(x-(u+v)y)(x-(u-v)y)\\ &=a(x-uy-vy)(x-uy+vy)\\ &=a((x-uy)^2-(vy)^2)\\ \end{array} $

so we want $e\sqrt{d} \ge a|(x-uy)^2-(vy)^2| $.

For any integer $y$, we can choose an integer $x$ such that $|x-uy| \lt 1$.

If $|y| = 1$ then $|(x-uy)^2-(vy)^2| \lt 1+v^2 $ so $e\sqrt{d} \ge a(1+v^2) $ will certainly do.

This is $e\sqrt{d} \ge a(1+\frac{d}{4a^2}) $ or $e \ge \dfrac{a}{\sqrt{d}}(1+\dfrac{d}{4a^2}) = \dfrac{a}{\sqrt{d}}+\dfrac{\sqrt{d}}{4a} $.

  • $\begingroup$ I had in mind that $e$ does not depend on $a,b,c$. For example, in the original Problem 12 where $d < 0$, $e = 2/\pi$. (If we are allowed to let $e$ depend on $a,b,c$, we can simply choose an $x,y$ and solve for $e$. For instance, choose $x = 1 = y$. Then any $e$ greater than $|a + b + c|/\sqrt d$ would do.) $\endgroup$
    – user0
    Commented Jun 12, 2020 at 17:31

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