# Whether an area contains finitely many lattice points

Based on Finding the area between two regions in the plane I know that the area between $$y>\sqrt{2}x - \frac{1}{4x}$$ and $$y< \sqrt{2}x + \frac{1}{4x}$$ is infinite. However, I actually have a different question to ask. Are there only finitely many lattice points contained in the area? How could I show this? Looking at the graph presented in the link, it's clear that the infinitely many lattice points could only come from the area around the asymptote going off to the top right, so I would need to show that this section only contains finitely many lattice points.

## 2 Answers

There are no such lattice points at all.

The question is whether there are infinitely many pairs of positive integers $$(x,y)$$ such that $$\sqrt{2} x - \frac{1}{4x} < y < \sqrt{2} x + \frac{1}{4x}$$ or equivalently $$\left| \sqrt{2} - \frac{y}{x}\right| < \frac{1}{4 x^2}$$ This is a question about Diophantine approximation. The best Diophantine approximations of an irrational number $$\alpha$$ (and any where $$|\alpha - x/y| < 1/(2 y^2)$$) are convergents of the continued fraction. The basic inequality is

$$\frac{1}{x_n^2 (a_{n+1}+2)} < \left| \alpha - \frac{y_n}{x_n} \right| < \frac{1}{x_n^2 a_{n+1}}$$ where $$y_n/x_n$$ are the convergents and $$a_n$$ the elements of the continued fraction of $$\alpha$$.

Since $$\sqrt{2}$$ is a quadratic irrational, its continued fraction is eventually periodic: in fact $$\sqrt{2} = [1; 2,2,2,\ldots]$$

Since the elements of the continued fraction are at most $$2$$, we get $$\left| \sqrt{2} - \frac{y_n}{x_n}\right| > \frac{1}{4 x_n^2}$$

On the other hand, if you replaced $$4$$ by $$2$$ you would get infinitely many lattice points.

• I'm not really familiar with diophantine approximation or continued fractions. I am trying to do this problem in the context of the geometry of numbers. Given this, I don't really understand your solution. – Wesley Mar 27 at 23:33
• +1 Nice answer. Do you have any recommended references for your statements about the relationship between best Diophantine approximation and convergents? – Yly Mar 27 at 23:33
• A. Ya. Khinchin, Continued Fractions. – Robert Israel Mar 27 at 23:47

There is an elementary argument (not using continued fractions) that shows $$|\sqrt 2- \frac{y}{x}| \geq \frac{1}{3x^2}$$ for all $$x, y \in \mathbb{Z}, x \neq 0$$

We start from $$|2x^2-y^2| \geq 1$$ under the above assumptions ($$x \neq 0$$), so $$|\sqrt 2- \frac{y}{x}||\sqrt 2+ \frac{y}{x}| \geq \frac{1}{x^2}$$.

But if $$|x|=1, |\sqrt 2\pm y| \geq .4 \geq \frac{1}{3}$$, while if $$|x| \geq 2, \frac{1}{10} \geq \frac{1}{3x^2}$$, so the inequality is automatically satisfied unless $$|\sqrt 2- \frac{y}{x}| \leq \frac{1}{10}$$, which means $$1.3 \leq \frac{y}{x} \leq 1.52, |\sqrt 2+ \frac{y}{x}| \leq 3$$, so $$|\sqrt 2- \frac{y}{x}| \geq \frac{1}{|\sqrt 2+ \frac{y}{x}|x^2} \geq \frac{1}{3x^2}$$ and we are done.