$polyhedron$

$LattE$

background

Introduction to Lattice Points

Introduction

Suppose we are given a Polygon P inscribed on a grid with squares of unit length in $\mathrm{I\!R\!}^2$ . Now, define a lattice point to be an integer coordinate point lying on such a grid. Refer to figures 1 and 2 for examples of the above.

**Figure:** (1,1) denotes a lattice point
$\scalebox{1}{\includegraphics{point.eps}}$

**Figure:** a polygon P inscribed on a grid in $\mathrm{I\!R\!}^2$
$\scalebox{1}{\includegraphics{plys.eps}}$

A natural question arises, how many lattice points lie inside and on the boundary of a given polygon P? We can explore this question by first experimenting with circles. Indeed according to, Karl Friedrich Gauss, given a circle of radius r

s.t. r $\ge$ 1

the number of lattice points that lie on the boundary and inside the circle is related to the area of the circle. i.e.,

$\left\vert G - \pi r^2 \right\vert$ $\le 2$ $\pi r$ .

the accompanying error function is as follows:

$\begin{displaymath} E(r) \le Cr^\theta ; \frac{1}{2} < \theta \le \frac{46}{73} \end{displaymath}$

Interestingly enough the exact value of $\theta$ is not known. If we let G(r) denote the number of lattice points that lie inside and on the boundary of a given circle, Gauss calculated that: G(1) = 5 G(2) = 13 G(3) = 29 G(4) = 49 G(5) = 81 G(6) = 113 G(7) = 149 . . . G(30) = 2821 . . . G(300) = 282,697, Gauss cared because the function G(r) can be used to approximate $\pi$ . i.e.,

$\begin{displaymath} \lim_{r\rightarrow\infty} G(r)/r^2 = \pi \end{displaymath}$

Figure 3 offers an illustration of circle of radius 3 with exactly 29 lattice points inside and on the boundary of the circle.

**Figure:** circle of radius 3 with 29 lattice points
$\scalebox{1}{\includegraphics{lat.eps}}$

Other theorems that help us count the number of lattice points given by circles are the following:

theorem 1 (Steinhaus) For every positive integer n, there exists a circle of area n which contains exactly n lattice points in its interior.

However this just tells us that such circles exist, but the following theorem describes how such circles can be found explicitly.

theorem 2 (Schinzel's) For every positive integer n, there is a circle in the plane having exactly n lattice points on its circumference. i.e.,

$\begin{displaymath} (x - \frac{1}{2})^2 + y^2 = \frac{5^(k-1)}{4} ; n = 2k \end{displaymath}$

$\begin{displaymath} (x - \frac{1}{3})^2 + y^2 = \frac{5^(2k)}{9} ; n = 2k + 1 \end{displaymath}$

Consider the circle described by figure 4, in this case n is even, k = 1 radius is .5, and G(r) = 2.

**Figure:** a case where n is even, k = 1, radius is .5
$\scalebox{1}{\includegraphics{cir1.eps}}$

We have counted the number of lattice points that lie inside and on the boundary of a given circle. Suppose now we wanted to count the number of lattice points of other curvy regions such as hyperbolas. For this, consider, the hyperbola $y = \frac{n}{x}$ as illustrated in figure 5. Let k(n) denote the number of lattice points given by the region k(n). Likewise, let k(n+1) be the number of lattice points given by the region k(n+1). Moreover, let L be the strip as denoted in the figure. However, the boundary of L does not include the boundary of k(n) nor of k(n+1).

**Figure:** regions k(n) and k(n+1)
$\scalebox{1}{\includegraphics{arc.eps}}$

Given all of this, we arrive at the following theorem:

$\left\vert k(n+1) - k(n) \right\vert = 2n+1$ iff n is prime

Before proving observe that: 1) n is not prime $\Leftrightarrow$ $\exists$ (a,b) s.t. (a,b) = n $\Leftrightarrow$ $\exists$ a lattice point (a,b) on the lower boundary of k(n) s.t. (a,b) $\ne$ (n,1) or (1,n). 2) L has no lattice points $\rightarrow$ Proof: suppose n is not prime, then by observation one we have that $\exists$ (a,b) s.t. ab = n, s.t. (a,b) $\exists$ on the lower boundary of k(n) and (a,b) $\ne$ (n,1) or (1,n). Now suppose L has a lattice point (c,d) then

$\begin{displaymath} \frac{n+1}{n} > d > \frac{n}{c} \end{displaymath}$

However, this implies that n < cd < n+1 which is a contradiction since no integer can exist between consecutive integers. Thus we have that given the premise n must be prime. $\leftarrow$ Now suppose n is prime then by observation one we have that n is prime $\Leftrightarrow$ (n,1) and (1,n) are lattice points. Also we have that region R does not have any lattice points. So if we count we have n+1 lattice points on the upper boundary of region k(n+1) + n lattice points that exist on the right hand side boundary of region k(n+1) + Q - [2 + Q] where Q denotes the intersection between regions k(n) and k(n+1). After simplification we have that:

$\left\vert k(n) - k(n+1) \right\vert$ = 2n-1. qed.

• Mathematics • The University of California, Davis •
• One Shields Avenue • Davis, CA 95616 •