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Applications to Minkowski's Theorem

Theorem 1   Any convex set (or body) in ${\mathbb{R}}^n$ that has central symmetry and volume greater than $2^n$ contains an integer lattice point other than $\overline{0}$.

If $A$ is an invertible matrix, then $Ax$ is a linear map. The lattice ${\mathbb{Z}}^n$ is mapped into a system of points in ${\mathbb{R}}^n$ which we call a lattice $L$, where $det(A)$ is equal to the determinant of the lattice $L$.

Theorem 2   Let $L$ be any lattice $L$ of determinant $\Delta\in{\mathbb{R}}^n$. Then any convex set symmetrical about the origin whose volume is greater than $2^n\Delta$ contains a point of $L$ other than $(0,0)$.

Example 3   Given any real number $\alpha$ and integer $t'>0$, there exists integers $p,q$ such that $\vert\frac{p}{q}-\alpha\vert \leq \frac{1}{\vert p\vert t'}$.

Take $M$-set parallelogram bounded by the four lines

$$y-\alpha x=k, y-\alpha x =-k, x=-t, x=t$$

where $t=\vert p\vert t'$.

This parallelogram has base $2y$, altitude $2k$, hence $Area=2t\cdot2k=4tk$. If we take $k=\frac{1}{t}$, then $area=4$. By Minkowski's fundamental theorem, there must be at least one lattice point $(p,q)$ other than $(0,0)$. Thus

$$\vert p\vert\leq t$$

or,

$$\alpha p-k \leq q \leq \alpha p+k.$$

Recall $k=\frac{1}{t}$, so

$$\alpha p-\frac{1}{t} \leq q \leq \alpha p+\frac{1}{t}\Rightarrow \vert q-\alpha p\vert \leq \frac{1}{t}=\frac{1}{\vert p\vert t'}.$$

QED

Example 4   Lagrange: Every positive integer $n$ can be expressed as a sum of four squares,

$$n=x_1^2+x_2^2+x_3^2+x_4^2$$

where $x_i$ are non-negative integers.

In view of the algebraic identity

$$(x_1^2 + x_2^2 + x_3^2 + x_4^2)(y_1^2 + y_2^2 + y_3^2 + y_4^2) = (x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4)^2$$

$$+ (x_1 y_2 - x_2 y_1 + x_3 y_4 - x_4 y_3)^2$$

$$+ (x_1 y_3 - x_2 y_4 - x_3 y_1 + x_4 y_2)^2$$

$$+ (x_1 y_4 + x_2 y_3 - x_3 y_2 - x_4 y_1)^2,$$

we see that $m,n$ are sums of four squares, thus $mn$ is the sum of four squares. Therefore, it is enough to prove the theorem for primes. Let

$$A= \left[ \begin{array}{cccc} p & 0 & r & s \\ 0 & p & s & -r \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$$

where $r,s$ are chosen so that $r^2 + s^2 + 1 \equiv 0 \,$   mod$\, p$. Let $\Lambda=A{\mathbb{Z}}^4$ and suppose that $x=At$ is a point of $\Lambda$. If $x\in\Lambda$, then

$$x_1^2 + x_2^2 + x_3^2 + x_4^2 = (pt_1 + rt_3 + st_4)^2 + (pt_2 - st_3 - rt_4)^2 + t_3^2 + t_4^2$$

$$\equiv (1+r^2+s^2)(t_3^2+t_4^2) \, \, p$$

$$\equiv 0 \, \, p$$

So we have $det(A)=p^2$. Now consider the ball

$$\zeta=\{\overline{x} \, \vert \, x_1^2+x_2^2+x_3^2+x_4^2<2p\}.$$

Thus, $\zeta$ is convex and symmetric about 0. A ball of radius $r\in{\mathbb{R}}^4$ has volume $\frac{1}{2}\pi^2 r^4$. If we take $r=\sqrt{2p}$, we see

$$V(\zeta) = \frac{1}{2}\pi^2 (2p)^2$$

$$= 2 \pi^2 p^2$$

$$> 2^4 p^2$$

$$\Rightarrow \exists \, \overline{x} \in \Lambda, \: \overline{x} \neq 0, \: \overline{x} \in \zeta$$

$$\Rightarrow 0<x_1^2+x_2^2+x_3^2+x_4^2<2p$$

$$\equiv 0 \, \, p,$$

so it has to be $p$!

QED

Theorem 5   There exists infinitely many positive integers that can be written as a sum of four positive perfect squares, but every integer $n>169$ is a sum of five positive perfect squares.