Minkowski: Representing Integers as Sums of 4 Squares

Here, we proved Minkowski's theorem. Here's the statement again:

Minkowski's Convex Body Theorem In an $n$-dimensional lattice spanned by the $n$ vectors

$$\begin{gather} \nu_1=(k_{(1,1)},k_{(1,2)},\cdots,k_{(1,n)})\\ \nu_2=(k_{(2,1)},k_{(2,2)},\cdots,k_{(2,n)})\\ \vdots\\ \nu_n=(k_{(n,1)},k_{(n,2)},\cdots,k_{(n,n)})\\ \end{gather},$$ let $\Delta=\left\lvert\begin{matrix} k_{(1,1)}&k_{(1,2)}&\cdots&k_{(1,n)}\\ k_{(2,1)}&k_{(2,2)}&\cdots&k_{(2,n)}\\ \vdots&\vdots&\ddots&\vdots\\ k_{(n,1)}&k_{(n,2)}&\cdots&k_{(n,n)}\\ \end{matrix}\right\rvert$.

Then, if we have a convex region $R$ symmetric about the origin with volume greater than $2^n \Delta$, then there exists a non-zero point of the lattice in $R$.

It turns out, that we can show that any integer is ths sum of four squares using this!

First, consider the following lemma:

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Lemma For an odd prime $p$, there exists $a,b$ such that $a^2\equiv -1-b^2\pmod{p}$.

For the sake of contradiction, suppose that $1-b^2$ is never a quadratic residue modulo $p$. Note that there are $\frac{p-1}{2}$ non-zero square modulo $p$, so there are $\frac{p+1}{2}$ squares modulo $p$. Let them be $\left\{s_1,s_2,\cdots,s_{\frac{p+1}{2}}\right\}$. Let's subtract them all from 1. We know that since all $s_i$ are distinct, all $-1-s_i$ are distinct, and consequently there are $\frac{p+1}{2}$ distinct values for $1-b^2$. However, there are only $\frac{p-1}{2}$ quadratic non-residues, so the values $s_i$ cannot be all distinct, which is a contradiction. Thus, there exists $a,b$ such that $a^2\equiv 1-b^2\pmod{p}$.

Note: This holds for all $n-b^2$, not just $-1$

Theorem Every positive integer is the sum of four squares.

Proof From Euler's four-square identity, if we show that every prime is the sum of four squares, then we can 'construct' all the other integers as the sum of four squares.

By our lemma, there exists $x,y$ such that $x^2+y^2\equiv -1\pmod{p}$. Thus, consider the lattice spanned by $(p,0,0,0)$, $(0,p,0,0)$, $(x,y,1,0)$, and $(y,-x,0,1)$. Note that every point on the lattice is a multiple of $p$ away from the origin: $$a(p,0,x,y)+b(0,p,y,-x)+c(0,0,1,0)+d(0,0,0,1)=(ap+cx+dy,bp+cy-dx,c,d)$$ $$\begin{align} |(ap+cx+dy,bp+cy-dx,c,d)|&\equiv (ap+cx+dy)^2+(bp+cy-dx)^2+(c)^2+(d)^2\\ &\equiv c(x^2+y^2+1)+d(x^2+y^2+1)\\ &\equiv 0\pmod{p} \end{align}$$ Now, consider the set of points $(w,x,y,z)$ such that $0\leq w^2+x^2+y^2+z^2<2p$. This is a convex region symmetric about the origin. Additionally, we have that $$\Delta=\left\lvert\begin{matrix}p&0&0&0\\ 0&p&0&0\\ x&y&1&0\\ y&-x&0&1\end{matrix}\right\rvert=p^2$$ Since the volume of a 4-dimensional sphere with radius $r$ is $\frac{1}{2}\pi^2r^4$, we also know that this is a 4-dimensional sphere with radius $\sqrt{2p}$ and volume $2\pi^2p^2$. Since $\pi^2>8$, we have $$2\pi^2p^2>16p^2,$$ so by Minkowski's theorem, there exists $(w,x,y,z)\in\mathbb{Z}^4$ with $p\mid w^2+x^2+y^2+z^2$ and $0<w^2+x^2+y^2+z^2<2p$. Thus, $w^2+x^2+y^2+z^2=p$, which is what we wanted to show. $\hspace{1cm}\Box$