there's a circle with area at least n with n+1 lattice points

Question

Prove that we can perform a translation on a circle of area at least $n$, for $n$ being a positive integer, such that there are at least $n+1$ points enclosed or in the boundary of the circle.

Answer 1

For any $n \in \mathbb{Z}_{+}$ and $R \in \mathbb{R}_{+}$ such that $\pi R^2 \ge n$. Define functions $\phi, \Phi : \mathbb{R}^2 \to \mathbb{R}$ as: $$\phi(p) = \begin{cases} 1 & |p| \le R\\0, & \text{otherwise}\end{cases} \quad\text{ and }\quad \Phi(p) = \sum_{c\in\mathbb{L}}\phi(p - c) $$ where $\mathbb{L} = \mathbb{Z}^2 \subset \mathbb{R}^2$ is the square lattice at hand.

It is clear for any $q \in \mathbb{R}^2$, there are only finitely many $c\in\mathbb{L}$ which are within a distance $R$ to $q$. This means in the definition of $\Phi(q)$, the number of $c$ which makes $\phi(p - c) \ne 0$ are always finite. As a result, $\Phi(q)$ is well defined. It is clear

• $\Phi(q)$ is the number of lattice points within a distance $R$ from $q$.
• $\Phi(q)$ is invariant under any translation in $\mathbb{L}$. i.e $$\Phi(q) = \Phi(q + t),\quad \forall t \in \mathbb{L}$$
• $\Phi(q)$ is Lebesgue integrable over any finite rectangles.

Consider the integral of $\Phi(q)$ over a square $[-M,M]^2$ for some big integer $M$:

$$I_{M} \stackrel{def}{=} \int_{[-M,M]^2}\Phi(q) dq = \sum_{c\in\mathbb{L}} I(c)\quad\text{ where }\quad I(c) \stackrel{def}{=} \int_{[-M,M]^2} \phi(q-c) dq$$

It is clear for any $c$, $0 \le I(c) \le \pi R^2$. Furthermore,

$$I(c) = \begin{cases} \pi R^2, & c \in [-M+R,M-R]^2\\ 0, & c \notin [-M-R,M+R]^2 \end{cases}$$

This allows us to bound $I_M$ as

$$(2M+1-2R)^2 \le \frac{1}{\pi R^2}I_M \le (2M+1+R)^2 \quad\implies\quad \lim_{M\to\infty} \frac{1}{4M^2} I_M = \pi R^2 $$ Since $\Phi(q)$ is periodic, this implies

$$\int_{q \in [0,1]^2} \Phi(q) dq = \pi R^2 \ge n\tag{*1}$$

If $\pi R^2 > n$, then there is a $q \in [0,1]^2$ with $\Phi(q) > n$ and we are done.

This leaves us the limiting case where $\pi R^2 = n$.

For any $p \in \mathbb{R}^2$ and $r > 0$, let $B(p,r)$ be the open disk $\big\{\; q \in \mathbb{R}^2 : |q-p| < r \;\big\}$.
Notice for any $c \in \mathbb{L}$, $p \in \mathbb{R}^2$, we can find a $r > 0$ such that

• For every $s \in ( 0, r )$, $\phi(p - c) \ge \phi(q - c)$ for all $q \in B(p,s)$.
• If $p \in \partial B(c,R)$, then $\phi(p - c) > \phi(q - c)$ for $q$ over a subset of $B(p,r)$ with non-zero measure.

This implies if we pick any $p \in B(0,R)$, then either $\Phi(p) > n$ or there exists a set $U$ of non-zero measure near $p$ and $\Phi(q) < n$ for $q \in U$. To satisfy $(*1)$, this forces $\Phi(q) > n$ over other set with non-zero measure. In both cases, there is a $p$ such that $\Phi(p) > n$.

Random Mumbling

If one look back at above proof, the "circle" part isn't that essential. We can replace the "circle" by any bounded measurable set $D$ and $\pi R^2$ by its measure $\mu(D)$. The same argument continue to work when $\mu(D) > n$. For the limiting case when $\mu(D) = n$, if $D$ is closed, a trivial modification is enough to fix the argument.

In short, in addition to "circle", the statement in question is valid over a wide range of geometric shapes (compact if $\mu(D) = n$, bounded measurable if $\mu(D) > n$). In particular, it works for ellipses and polygons.

Notice for any lattice in the plane, we can transform it to a square lattice by affine transformation. Under such a transform, circle/ellipse get mapped to circle/ellipse and polygon to polygon. This means in general we can drop the requirement that the lattice is a square one. The statement in question continue to work for "circle/ellipse/polygon" against any lattice whose fundamental domain has area $1$.