Saw this question in NZM and have a lot of difficulties trying to start.

Consider the binary quadratic form $f(x,y) = ax^2 + bxy + cy^2$ with $a > 0$ and $d = b^2 − 4ac < 0$. Show that there exist integers $x$ and $y$, not both zero, such that $\lvert ax^2 + bxy + cy^2\rvert \leq \frac{2}{\pi{}}\sqrt{−d}$.

I tried to use Minkowski's convex body theorem, but to no avail. What could be the possible convex symmetric subset of $\mathbb{R}^2$ to choose in this case such that the volume is greater than 4?


1 Answer 1


This is 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.

First note that the absolute value signs are superfluous because the assumptions that $d < 0$ and $a > 0$ imply that $f(x, y) \ge 0$. (See Definition 3.5 and Theorem 3.11. Those without the book can refer to the Wikipedia article on binary quadratic forms.)

The assumption $d < 0$ also implies that $$ax^2 + bxy + cy^2 \leq \frac{2}{\pi}\sqrt{−d}$$ represents an ellipse and its interior. That answers your question except for the following subtlety:

The area of our ellipse is equal to 4, not greater than 4 as you asked for in order to use Minkowski's convex body theorem to complete the problem. Instead, use a modified version of the theorem, also given in Corollary 6.22 and Problem 9, to complete the problem. The modification allows the area to be greater than or equal to 4 if the set of points is closed, which is true for an ellipse and its interior.


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