- A lattice $L(B) \equiv \{ Bx : x \in \mathbb Z^n \}$ for some basis $B \in \mathbb R^n$. The lattice $L$ is spanned by integer linearcombinations of rows of $B$.
- A body $S \subseteq R^n$ which
**need not be convex!**, which has volume greater than $\det(B)$. Recall that for a lattice $L(B)$,the volume of a fundamental unit / fundamental parallelopiped is $det(B)$.

- Chop up sections of $S$ across all translates of the fundamental parallelopipedthat have non-empty intersections with $S$ back to the origin. This makesall of them overlap with the fundamental parallelopiped with the origin.
- Since $S$ has volume great that $\det(B)$, but the fundamental paralellopipedonly has volume $\det(B)$, points from two different parallelograms
**must**overlap. - "Undo" the translation to find two points which are of the form $x_1 = l_1 + \delta$,$x_2 = l_2 + \delta$. they must have the same $\delta$ since they overlappedwhen they were laid on the fundamental paralellopiped. Also notice that $l_1 \neq l_2$since they came from two different parallograms on the plane!
- Notice that $x_1 - x_2 = l_1 - l_2\in L \neq 0$, since we already arguedthat $l_1 \neq l_2$. This gives us what we want.

$T \equiv S/2 = \{ (x_1/2, x_2, \dots, x_n/2) : (x_1, x_2, \dots, x_n) \in S \}$

We now see that $Vol(T) > det(B)$ to invoke Blichfeldt's theorem.
Formally:
$Vol(T) = 1/2^n Vol(S) > 1/2^n (2^n det(B)) = det(B)$

We can apply Blichfeldt's theorem to get our hands on two points $x_1, x_2 \in T$
such that $x_1 - x_2 \in L$.
$\begin{aligned}
&x_1 \in T \Rightarrow 2x_1 \in S ~(S = 2T) \\
&x_2 \in T \Rightarrow 2x_2 \in S ~(S = 2T) \\
&2x_2 \in S \Rightarrow -2x_2 \in S~\text{($S$ is symmetric about origin)} \\
&\frac{1}{2}(2x_1) + \frac{1}{2} (-2x_2) \in S~\text{($S$ is convex)}\\
&x_1 - x_2 \in S~\text{(Simplification)}\\
&\text{nonzero lattice point}~\in S \\
\end{aligned}$