怎样利用格理论，也就是 minkowski 基本定理来证明拉格朗日四平方和定理以及费马平方和定理?

好的，我们来详细阐述如何利用格理论（特别是 Minkowski 基本定理）来证明拉格朗日四平方和定理和费马平方和定理。

核心概念回顾

在深入证明之前，我们先回顾一下关键的数学概念：

1. 格 (Lattice): 一个格是一个 $n$ 维欧几里得空间 $mathbb{R}^n$ 中的一组离散的、周期性的点。更严谨地说，一个 $n$ 维格 $L$ 是由一组线性无关的 $n$ 个向量（基向量） ${mathbf{v}_1, mathbf{v}_2, ldots, mathbf{v}_n}$ 张成的所有整数线性组合的集合：
$L = {a_1 mathbf{v}_1 + a_2 mathbf{v}_2 + cdots + a_n mathbf{v}_n mid a_i in mathbb{Z}}$
我们通常将这些基向量写成一个矩阵 $V = [mathbf{v}_1, mathbf{v}_2, ldots, mathbf{v}_n]$ 的列。那么格中的点就是 $Vmathbf{a}$，其中 $mathbf{a} in mathbb{Z}^n$。

2. 基行列式 (Determinant of the Lattice) 或 体积 (Volume): 格的基行列式 $det(L)$ 是由格的基向量构成的矩阵的绝对值。它代表了由基向量张成的平行体的体积。如果 $V$ 是基矩阵，那么 $det(L) = |det(V)|$。这个体积是与格的基选择无关的。

3. Minkowski 基本定理 (Minkowski's First Theorem): 这是格理论中一个至关重要的定理。
定理陈述: 设 $L$ 是 $mathbb{R}^n$ 中的一个格，其基行列式为 $det(L)$。设 $K$ 是 $mathbb{R}^n$ 中的一个凸集，它关于原点对称（即如果 $mathbf{x} in K$，则 $mathbf{x} in K$）。如果 $K$ 的体积（Lebesgue measure）$ ext{vol}(K)$ 满足 $ ext{vol}(K) > 2^n det(L)$，那么 $K$ 中必然包含一个非零的格点。

直观理解: Minkowski 基本定理告诉我们，如果我们考虑一个足够大的（体积超过 $2^n det(L)$ 的）关于原点对称的凸集，它必然会“挤压”进一个非零的格点。想象一下，如果我们把这个凸集看作一个“球”或者“盒子”，当它的体积大到一定程度时，它就必然会包含一个除了原点之外的格点。

如何利用 Minkowski 基本定理证明数论问题？

证明方法的核心思想是将一个数论问题转化为在某个特定格中寻找非零格点的问题。通常，我们会构造一个维度与问题相关的格，并选择一个合适的凸集 $K$。如果这个凸集满足 Minkowski 基本定理的条件，那么我们就能证明存在一个非零格点，而这个格点恰好对应于数论问题的一个解。

一、 费马平方和定理 (Fermat's Theorem on Sums of Two Squares)

定理陈述: 一个大于 0 的整数 $n$ 可以表示为两个整数平方之和（即 $n = x^2 + y^2$）当且仅当 $n$ 的所有形如 $4k+3$ 的素因子在 $n$ 的素因数分解中出现的指数都是偶数。

我们主要关注如何证明“若 $n = x^2 + y^2$，则 $n$ 的形如 $4k+3$ 的素因子指数是偶数”的部分，或者更常见的证明方向是“若 $n$ 的形如 $4k+3$ 的素因子指数是偶数，则 $n$ 可以表示为两个平方和”。这里我们选择后者，因为它更直接地体现了格理论的应用。

证明思路:
我们需要证明：如果一个整数 $n$ 的素因数分解中，所有形如 $4k+3$ 的素因子的指数都为偶数，那么 $n$ 可以表示为两个整数的平方和。

1. 将问题转化为寻找格点:
我们考虑一个整数 $n$。我们需要找到整数 $x, y$ 使得 $n = x^2 + y^2$。
这可以改写成 $x^2 + y^2 n = 0$。

我们将在一个二维格中寻找满足这个条件的点。考虑复数域 $mathbb{C}$，它等价于 $mathbb{R}^2$。我们可以将整数对 $(x, y)$ 视为复数 $x+iy$。那么 $x^2 + y^2 = |x+iy|^2$。
我们要找的条件是 $|x+iy|^2 = n$。

然而，直接在 $mathbb{R}^2$ 的标准格 $mathbb{Z}^2$ 中寻找 $|x+iy|^2 = n$ 的点并不直接。Minkowski 基本定理通常处理的是“大于”或“小于”的条件，而不是精确等于的条件。

关键的转换：利用高维格和理想。
费马平方和定理的格理论证明通常涉及到 高斯整数环 (Gaussian Integers) $mathbb{Z}[i] = {a+bi mid a, b in mathbb{Z}}$。高斯整数环可以看作是一个二维格。
设 $n$ 是一个正整数，其素因数分解中所有形如 $4k+3$ 的素因子的指数都是偶数。我们可以将 $n$ 写成 $n = m^2 cdot q$，其中 $q$ 的所有素因子都是 $2$ 或形如 $4k+1$ 的素因子。

现在考虑素数 $p$ 的情况：
如果 $p = 2$，那么 $2 = 1^2 + 1^2$。
如果 $p equiv 1 pmod{4}$，根据费马小定理的延伸（或二次互反律），存在一个整数 $a$ 使得 $p mid (a^2+1)$，即 $a^2 equiv 1 pmod{p}$。这意味着在模 $p$ 下，$i^2 equiv 1$ 是可解的。因此，$p$ 在 $mathbb{Z}[i]$ 中可以分解：$p = (a+i)(ai)$。并且 $N(a+i) = a^2+1$ 并且 $N(ai) = a^2+1$。如果 $a+i$ 是一个素元，那么 $p$ 是不可约的。
如果 $p equiv 3 pmod{4}$，那么 $p$ 在 $mathbb{Z}[i]$ 中仍然是不可约的（并且是素元）。

费马平方和定理的证明可以转化为在 $mathbb{Z}[i]$ 的理想中寻找元素。
我们想要证明：如果 $n$ 的所有形如 $4k+3$ 的素因子指数都是偶数，则 $n$ 是一个模数（norm）为 $n$ 的高斯整数的模。换句话说，存在 $z = x+iy in mathbb{Z}[i]$ 使得 $N(z) = |z|^2 = x^2+y^2 = n$。

为了使用 Minkowski 基本定理，我们需要将其问题置于 $mathbb{R}^n$ 的格中。
设 $p$ 是一个素数。
如果 $p equiv 1 pmod{4}$，则 $p = pi ar{pi}$ 在 $mathbb{Z}[i]$ 中，其中 $pi, ar{pi}$ 是高斯素数。 $N(pi) = p$。
如果 $p equiv 3 pmod{4}$，则 $p$ 是 $mathbb{Z}[i]$ 中的素数。$N(p) = p^2$。
$2 = i(1+i)^2$ 是单位和高斯素数因子 $(1+i)$。$N(1+i)=2$.

如果 $n = p_1^{e_1} cdots p_k^{e_k}$ 且 $p_i equiv 3 pmod{4}$ 的 $e_i$ 都是偶数，那么 $n$ 可以写成 $n = m^2 d$，其中 $d$ 的素因子都是 $2$ 或形如 $4k+1$ 的素因子。
我们的目标是证明 $n = x^2 + y^2$。

更直接的格理论证明方式（聚焦于 Minkowski 基本定理的应用）：

考虑一个整数 $n$ 满足费马平方和定理的条件（即 $n$ 的形如 $4k+3$ 的素因子指数都是偶数）。
我们需要证明存在 $x, y in mathbb{Z}$ 使得 $n = x^2 + y^2$。

这等价于在复数域 $mathbb{C}$ 中寻找一个元素 $z = x+iy in mathbb{Z}[i]$ 使得 $N(z) = n$。

Let's try a slightly different approach using 2D lattice directly.

Suppose $n = x^2 + y^2$. This means the point $(x, y)$ is on the circle of radius $sqrt{n}$ centered at the origin. We are looking for a nonzero integer point on this circle.

Consider the Minkowski approach for proving sums of two squares. This usually involves constructing a lattice in $mathbb{R}^2$ or $mathbb{R}^4$. The standard proof of Fermat's twosquare theorem using Minkowski often relates to the properties of the ring of Gaussian integers $mathbb{Z}[i]$, which can be viewed as a lattice in $mathbb{R}^2$.

Let $n$ be a positive integer. We want to show that $n = x^2 + y^2$ for some $x, y in mathbb{Z}$ if and only if all prime factors of $n$ of the form $4k+3$ occur with an even exponent.

Let's focus on proving the forward direction: If all prime factors of $n$ of the form $4k+3$ occur with an even exponent, then $n = x^2 + y^2$.

This means that $n$ is a norm of an element in the Gaussian integers $mathbb{Z}[i]$.
A key idea is to use Minkowski's theorem to find an element in $mathbb{Z}[i]$ with a specific norm. However, Minkowski's theorem directly gives existence of a nonzero lattice point in a convex set of large volume. We need to relate the norm property to such a set.

The standard way to use Minkowski for $n=x^2+y^2$ involves constructing a specific lattice and a specific set.

Let $n$ be a positive integer such that all its prime factors of the form $4k+3$ appear with an even exponent. This implies that $n$ can be written as $n = s^2 m$, where $m$ is a product of primes of the form $4k+1$ and the prime $2$.
Crucially, for any prime $p equiv 1 pmod{4}$, there exists an integer $a_p$ such that $a_p^2 equiv 1 pmod{p}$. This means $p mid (a_p^2+1)$, so $p$ is the norm of $a_p + i$ in $mathbb{Z}[i]$.

Consider the case when $n=p$ is a prime, $p equiv 1 pmod{4}$. Then $p = a^2 + b^2$.
We know there exists $a$ such that $a^2 equiv 1 pmod{p}$.
Consider the lattice $L = {(x, y) in mathbb{Z}^2 mid x equiv ay pmod{p}}$.
This is not a standard basis lattice. We need to construct a lattice in $mathbb{R}^2$.

Let's use a standard construction of a lattice and a set to apply Minkowski's theorem.
Suppose $n$ is a positive integer, and all its prime factors of the form $4k+3$ have even exponents.
We want to show $n = x^2 + y^2$ for some $x, y in mathbb{Z}$.

This is equivalent to finding $x+iy in mathbb{Z}[i]$ such that $N(x+iy) = x^2+y^2 = n$.
The set of Gaussian integers $mathbb{Z}[i]$ can be viewed as a lattice in $mathbb{R}^2$ with basis vectors $(1, 0)$ and $(0, 1)$. The basis matrix is $V = egin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix}$. The determinant of this lattice is $|det(V)| = 1$.

We need to construct a suitable convex set $K$ to apply Minkowski's theorem.
Consider the set of complex numbers $z$ such that $N(z) le n$. This is a disk of radius $sqrt{n}$ centered at the origin. Its area is $pi (sqrt{n})^2 = pi n$.
Minkowski's theorem states that if $ ext{vol}(K) > 2^n det(L)$, then $K$ contains a nonzero lattice point. Here $n=2$.
So, if $ ext{vol}(K) > 2^2 cdot 1 = 4$, then the disk $D = {z in mathbb{R}^2 mid |z| le sqrt{n}}$ contains a nonzero lattice point in $mathbb{Z}^2$.
If $pi n > 4$, then the disk contains a nonzero integer point $(x, y)$, meaning $x^2+y^2 le n$.

This application of Minkowski's theorem tells us that for sufficiently large $n$, there's a point $(x,y)$ with $x^2+y^2 le n$. This is not exactly $x^2+y^2=n$.

The actual proof for Fermat's twosquare theorem using Minkowski often involves a slightly more sophisticated lattice or a specific condition that forces equality.

A common method involves constructing a lattice $L$ in $mathbb{R}^2$ with a specific determinant.
Let $n$ be a prime $p equiv 1 pmod{4}$. We know there exists $a$ such that $a^2 equiv 1 pmod{p}$.
Define a lattice $L$ in $mathbb{R}^2$ with basis vectors:
$mathbf{v}_1 = (p, 0)$
$mathbf{v}_2 = (a, 1)$

The determinant of this lattice is $|det(egin{pmatrix} p & a \ 0 & 1 end{pmatrix})| = |p cdot 1 a cdot 0| = p$.
This lattice consists of points $(px + ay, y)$ for $x, y in mathbb{Z}$.

Now, consider the convex set $K$:
$K = {(u, v) in mathbb{R}^2 mid u^2 + v^2 < 2p}$ (This is an open disk of radius $sqrt{2p}$)
The volume of $K$ is $ ext{vol}(K) = pi (sqrt{2p})^2 = 2pi p$.

According to Minkowski's theorem, if $ ext{vol}(K) > 2^2 det(L)$, then $K$ contains a nonzero lattice point.
Here, $2pi p > 4p$, which is true since $pi > 2$.
So, there exists a nonzero point $(u, v) in L$ such that $u^2 + v^2 < 2p$.

Since $(u, v) in L$, we can write $(u, v) = x(p, 0) + y(a, 1) = (px + ay, y)$ for some $x, y in mathbb{Z}$.
So, $u = px + ay$ and $v = y$.
The condition $u^2 + v^2 < 2p$ becomes $(px + ay)^2 + y^2 < 2p$.

Now consider $u pmod{p}$:
$u = px + ay equiv ay pmod{p}$.
Since $a^2 equiv 1 pmod{p}$, we have $a otequiv 0 pmod{p}$.
If $y equiv 0 pmod{p}$, then $u equiv 0 pmod{p}$.
If $y otequiv 0 pmod{p}$, then $u^2 equiv (ay)^2 equiv a^2 y^2 equiv y^2 pmod{p}$.
So, $u^2 + y^2 equiv 0 pmod{p}$.

We have a point $(u, v) in L$, $(u,v) eq (0,0)$, such that $u^2+v^2 < 2p$.
Let $y = v$. Then $u equiv av pmod{p}$.
So, $u^2 + v^2 equiv (av)^2 + v^2 equiv a^2 v^2 + v^2 equiv (1)v^2 + v^2 equiv 0 pmod{p}$.
This means $p mid (u^2 + v^2)$.

Since $(u, v)$ is a nonzero point in $L$, $u^2 + v^2$ is a positive integer.
We have $u^2 + v^2 = kp$ for some integer $k$.
Also, $u^2 + v^2 < 2p$.
So, $kp < 2p$, which implies $k < 2$.
Since $u^2+v^2 > 0$, we must have $k=1$.
Therefore, $u^2 + v^2 = p$.

This shows that any prime $p equiv 1 pmod{4}$ can be written as a sum of two squares.
The proof for a general integer $n$ uses the multiplicative property of sums of two squares: $(a^2+b^2)(c^2+d^2) = (acbd)^2 + (ad+bc)^2$.
If $n = p_1 cdots p_k$, where each $p_i equiv 1 pmod{4}$ or $p_i=2$, then $n$ is a sum of two squares.
If $n = m^2 cdot q$ where $q$ has only prime factors $p equiv 1 pmod{4}$ or $p=2$, and $m^2$ is the part with prime factors $p equiv 3 pmod{4}$ raised to even powers, then $n$ is a sum of two squares.

In summary for Fermat's TwoSquare Theorem: The Minkowski method proves that if $p$ is a prime $equiv 1 pmod{4}$, then $p=x^2+y^2$. This is achieved by constructing a lattice whose determinant is $p$, and a disk of radius $sqrt{2p}$. By Minkowski's theorem, this disk contains a nonzero lattice point $(u,v)$. The structure of the lattice forces $u^2+v^2$ to be a multiple of $p$, and the size of the disk forces $u^2+v^2$ to be exactly $p$.

二、 拉格朗日四平方和定理 (Lagrange's FourSquare Theorem)

定理陈述: 任何非负整数都可以表示为四个整数的平方之和。即，对于任何非负整数 $n$，都存在整数 $x_1, x_2, x_3, x_4$ 使得 $n = x_1^2 + x_2^2 + x_3^2 + x_4^2$。

证明思路:
要证明这个定理，我们可以将问题转化为在 $mathbb{R}^4$ 中的一个特定格里寻找一个非零格点，该格点满足一定的条件。

1. 高维格与四元数 (Quaternions):
这个证明通常利用四元数 (Quaternions) 的概念，特别是 整数四元数环 (Hurwitz Integers)。
一个四元数可以写成 $q = a + bi + cj + dk$，其中 $a, b, c, d$ 是实数。
一个整数四元数是形如 $a + bi + cj + dk$ 的四元数，其中 $a, b, c, d$ 要么都是整数，要么都是半整数（即 $k + 1/2$ 的形式）。
令 $mathcal{H}$ 表示四元数。

我们考虑一个特殊的格，它与整数四元数有关。
整数四元数构成一个四维格。一个常用的基是：
$1, i, j, k$ （标准基，对应格点 $(a, b, c, d)$）
或者更一般地，包含形如 $frac{1+i+j+k}{2}$ 这样的元素。

更直接的方法是使用四维欧几里得空间 $mathbb{R}^4$ 和一个特殊的格。

设 $n$ 是一个非负整数。我们需要找到 $x_1, x_2, x_3, x_4 in mathbb{Z}$ 使得 $n = x_1^2 + x_2^2 + x_3^2 + x_4^2$。

考虑一个四维格 $L$。我们需要选择一个基，使得其格点对应于和为平方和的整数。
我们可以在 $mathbb{R}^4$ 中构建一个特殊的格 $L$，其基向量是：
$mathbf{v}_1 = (n, 0, 0, 0)$
$mathbf{v}_2 = (0, n, 0, 0)$
$mathbf{v}_3 = (0, 0, n, 0)$
$mathbf{v}_4 = (0, 0, 0, n)$
这是 $mathbb{Z}^4$ 格的一个缩放版本。其行列式是 $n^4$。Minkowski 定理在这种情况下需要一个非常大的凸集。这也不是最有效的方式。

使用 Hurwitz Integers 的格构造:
Hurwitz Integers 构成一个四维格，其基可以是：
$mathbf{e}_1 = (1, 0, 0, 0)$
$mathbf{e}_2 = (0, 1, 0, 0)$
$mathbf{e}_3 = (0, 0, 1, 0)$
$mathbf{e}_4 = (0, 0, 0, 1)$
以及 $mathbf{b} = (frac{1}{2}, frac{1}{2}, frac{1}{2}, frac{1}{2})$。
这些元素可以看作是 $mathbb{R}^4$ 中的向量。

一个更常用的证明方法是利用一个特定的四维格和线性条件。

考虑整数 $n$。我们需要找到 $x_1, x_2, x_3, x_4 in mathbb{Z}$ 使得 $n = x_1^2 + x_2^2 + x_3^2 + x_4^2$。
这等价于在 $mathbb{R}^4$ 中寻找一个非零向量 $(x_1, x_2, x_3, x_4)$，其范数为 $sqrt{n}$。

构造合适的格和凸集：
设 $n$ 是一个非负整数。我们考虑一个格 $L$。

关键思想：利用 Hurwitz integers 的性质来构造一个格和凸集。
Hurwitz integers 环 $mathcal{H}$ 构成 $mathbb{R}^4$ 中的一个格，其体积（格行列式）为 $1$（如果选择合适的单位）。
更准确地说， Hurwitz integers 的单位格的体积是 $( ext{vol}( ext{group of units})/2)^2 = (24/2)^2 = 144$? 不对，这是关于单值群的。

Hurwitz integers 构成一个格，其 basis vectors can be expressed as:
$(1, 0, 0, 0)$
$(0, 1, 0, 0)$
$(0, 0, 1, 0)$
$(0, 0, 0, 1)$
and $(1/2, 1/2, 1/2, 1/2)$.
This set forms a lattice.
However, the standard Minkowski theorem proof for Lagrange's theorem is more direct.

Let's consider the specific lattice and set construction for Lagrange's theorem.

We want to show that any integer $n ge 0$ can be written as $x_1^2 + x_2^2 + x_3^2 + x_4^2$.
This is equivalent to finding an integer (Gaussian) quaternion $q = x_1 + x_2i + x_3j + x_4k$ such that $N(q) = x_1^2 + x_2^2 + x_3^2 + x_4^2 = n$.

We can use the Hurwitz integers which form a lattice in $mathbb{R}^4$.
The set of Hurwitz integers is $mathcal{O}_{mathcal{H}} = {a + bi + cj + dk mid a,b,c,d in mathbb{Z} ext{ or } a,b,c,d in mathbb{Z} + 1/2}$.
This set forms a lattice in $mathbb{R}^4$.
A possible basis for this lattice is:
$mathbf{v}_1 = (1, 0, 0, 0)$
$mathbf{v}_2 = (0, 1, 0, 0)$
$mathbf{v}_3 = (0, 0, 1, 0)$
$mathbf{v}_4 = (1/2, 1/2, 1/2, 1/2)$

The determinant of this lattice can be calculated. The matrix is:
$V = egin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 1/2 & 1/2 & 1/2 & 1/2 end{pmatrix}$
The determinant is $|det(V)| = 1 cdot 1 cdot 1 cdot (1/2) = 1/2$.

Now we need to choose a suitable convex set $K$.
We want to find a nonzero Hurwitz integer $q$ such that $N(q) = n$.
Consider the set $K = {q in mathbb{R}^4 mid N(q) le n}$. This is a 4dimensional ball of radius $sqrt{n}$.
The volume of this ball is $ ext{vol}(K) = frac{pi^2}{2} (sqrt{n})^4 = frac{pi^2 n^2}{2}$.

Minkowski's theorem states that if $ ext{vol}(K) > 2^4 det(L)$, then $K$ contains a nonzero lattice point.
Here, $2^4 det(L) = 16 cdot (1/2) = 8$.
So, if $frac{pi^2 n^2}{2} > 8$, i.e., $pi^2 n^2 > 16$, or $n^2 > 16/pi^2 approx 1.6$, then the ball $K$ contains a nonzero Hurwitz integer $q_0$.
This means $N(q_0) le n$.

This tells us that for $n$ large enough, there exists a Hurwitz integer whose norm is less than or equal to $n$. This is not yet $n$.

The key is to use a slightly modified set or condition to ensure the norm is exactly $n$.

This proof often uses a specific construction for a given $n$.
Let $n$ be a nonnegative integer.
Consider the lattice $L_n$ in $mathbb{R}^4$ formed by points $(x_1, x_2, x_3, x_4)$ satisfying:
$x_1 equiv x_2 equiv x_3 equiv x_4 pmod{m}$ for some integer $m$. This is getting complicated.

Let's use a simpler argument structure that leads to the foursquare theorem.

The standard approach for the foursquare theorem using Minkowski's theorem often involves the Lipschitz quaternions, which form a subset of Hurwitz integers and are easier to work with for lattice construction.
Lipschitz quaternions: $a + bi + cj + dk$ where $a, b, c, d in mathbb{Z}$.
These form a lattice $L_{Lipschitz}$ in $mathbb{R}^4$ with basis vectors $(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)$.
The determinant of this lattice is $1$.

Let $n$ be a positive integer. We want to find $x_1, x_2, x_3, x_4 in mathbb{Z}$ such that $n = x_1^2 + x_2^2 + x_3^2 + x_4^2$.
This is equivalent to finding a Lipschitz quaternion $q = x_1 + x_2i + x_3j + x_4k$ such that $N(q) = n$.

The actual proof for Lagrange's theorem via Minkowski's theorem is a bit more intricate and relies on constructing a specific lattice for a given $n$.

Let's try to explain the core idea using Minkowski's theorem for $n = x_1^2 + x_2^2 + x_3^2 + x_4^2$.

The general strategy is to show that for any $n$, we can find a nonzero lattice point within a specific ball of radius $sqrt{2n}$ centered at the origin. The structure of the lattice is crucial.

Consider the set of all vectors $(x_1, x_2, x_3, x_4) in mathbb{Z}^4$ such that $x_1 + x_2 + x_3 + x_4$ is an even number. This set itself is not a single lattice, but it has structure.

A more elegant approach uses the properties of quaternions and their norms.
For any integer $n ge 0$, we can find a Hurwitz quaternion $q$ such that $N(q) = n$.
This is proven by induction. Base case $n=0$ is trivial.
Assume all integers $k < n$ can be written as sum of four squares.
Consider $n$. We can find $q_0$ such that $N(q_0) le n$. Let $m = n N(q_0)$. $m ge 0$.
If $m=0$, we are done.
If $m>0$, we need to show that $m$ can be represented. This requires a way to "reduce" the problem or construct a specific quaternion.

Let's focus on the Minkowski theorem part for the foursquare theorem.
The proof of Lagrange's foursquare theorem using Minkowski's theorem is typically done by constructing a lattice in $mathbb{R}^4$ tailored for a specific integer $n$.

Consider a positive integer $n$. We want to show $n = x_1^2 + x_2^2 + x_3^2 + x_4^2$.
This is equivalent to finding a vector $(x_1, x_2, x_3, x_4) in mathbb{Z}^4$ such that the squared Euclidean norm is $n$.

The proof goes by first proving it for primes $p$.
If $p$ is a prime. Can we find $x_1, x_2, x_3, x_4 in mathbb{Z}$ such that $p = x_1^2 + x_2^2 + x_3^2 + x_4^2$?
By properties of finite fields, for any prime $p$, the equation $x_1^2 + x_2^2 + x_3^2 + x_4^2 equiv 0 pmod{p}$ always has a nontrivial solution (not all $x_i=0$).
Let $(x_1, x_2, x_3, x_4)$ be such a nontrivial solution.
Let $k = x_1^2 + x_2^2 + x_3^2 + x_4^2$. Then $k equiv 0 pmod{p}$, so $k = mp$ for some integer $m$.
Since the solution is nontrivial, $k > 0$. So $m > 0$.

Now, we need to construct a lattice $L$ and a set $K$ such that we can guarantee $k=p$.
Consider the lattice $L$ in $mathbb{R}^4$ formed by all vectors $(x_1, x_2, x_3, x_4)$ such that $x_1 equiv x_2 equiv x_3 equiv x_4 pmod{p}$. This is not a single lattice, but a set of parallel lattices.

The actual proof uses the existence of a nonzero solution to $x_1^2 + x_2^2 + x_3^2 + x_4^2 equiv 0 pmod{p}$ to define a lattice.

Let $p$ be a prime. There exists a nonzero integer solution $(x_1, x_2, x_3, x_4)$ to $x_1^2 + x_2^2 + x_3^2 + x_4^2 equiv 0 pmod{p}$.
Let $k = x_1^2 + x_2^2 + x_3^2 + x_4^2$. Then $k$ is a positive multiple of $p$.
Consider the lattice $L$ formed by all integer linear combinations of the vectors:
$mathbf{v}_1 = (p, 0, 0, 0)$
$mathbf{v}_2 = (0, p, 0, 0)$
$mathbf{v}_3 = (0, 0, p, 0)$
$mathbf{v}_4 = (x_1, x_2, x_3, x_4)$

The determinant of this lattice is $|det(egin{pmatrix} p & 0 & 0 & x_1 \ 0 & p & 0 & x_2 \ 0 & 0 & p & x_3 \ 0 & 0 & 0 & x_4 end{pmatrix})| = |p^3 cdot x_4| = p^3 x_4$.
Wait, the last vector must be $(x_1, x_2, x_3, x_4)$ where the sum of squares is $kp$.
Let the vectors be:
$mathbf{v}_1 = (p, 0, 0, 0)$
$mathbf{v}_2 = (0, p, 0, 0)$
$mathbf{v}_3 = (0, 0, p, 0)$
$mathbf{v}_4 = (x_1, x_2, x_3, x_4)$ where $x_1^2+x_2^2+x_3^2+x_4^2 = kp$ for some $k$.
Let's choose the solution $(x_1, x_2, x_3, x_4)$ such that $0 < k < p$. Such a solution exists.

The determinant of $L$ is $|det(egin{pmatrix} p & 0 & 0 & x_1 \ 0 & p & 0 & x_2 \ 0 & 0 & p & x_3 \ 0 & 0 & 0 & x_4 end{pmatrix})| = p^3 x_4$. This choice of basis isn't quite right.

Let's use the set $K$ and $det(L)$ directly.
Let $p$ be a prime. We want to find $x_1, x_2, x_3, x_4 in mathbb{Z}$ s.t. $x_1^2 + x_2^2 + x_3^2 + x_4^2 = p$.

Consider the set of points in $mathbb{R}^4$:
$K = {(y_1, y_2, y_3, y_4) mid y_1^2 + y_2^2 + y_3^2 + y_4^2 < 2p}$.
This is an open ball of radius $sqrt{2p}$ in $mathbb{R}^4$.
Its volume is $ ext{vol}(K) = frac{pi^2}{2}(sqrt{2p})^4 = frac{pi^2}{2} (4p^2) = 2pi^2 p^2$.

Now, we need to construct a lattice $L$ such that its determinant $det(L)$ is related to $p$.
We know there exists a nonzero solution $(x_1, x_2, x_3, x_4)$ to $x_1^2 + x_2^2 + x_3^2 + x_4^2 equiv 0 pmod{p}$, and we can choose this solution such that $0 < x_1^2 + x_2^2 + x_3^2 + x_4^2 < p^2$.
Let $k = x_1^2 + x_2^2 + x_3^2 + x_4^2$. Then $k$ is a positive multiple of $p$. We can choose the solution such that $k=p$. This is a subtle point and requires specific algebraic number theory arguments for existence.

If we have a solution $(x_1, x_2, x_3, x_4)$ where $x_1^2 + x_2^2 + x_3^2 + x_4^2 = kp$ and $0 < k < p$.
Consider the lattice $L$ generated by:
$mathbf{v}_1 = (p, 0, 0, 0)$
$mathbf{v}_2 = (0, p, 0, 0)$
$mathbf{v}_3 = (0, 0, p, 0)$
$mathbf{v}_4 = (x_1, x_2, x_3, x_4)$

The determinant of $L$ is $|det(egin{pmatrix} p & 0 & 0 & x_1 \ 0 & p & 0 & x_2 \ 0 & 0 & p & x_3 \ 0 & 0 & 0 & x_4 end{pmatrix})| = |p^3 x_4|$. This is incorrect because the basis vectors are columns.
Let's consider the matrix whose rows are the basis vectors.
$M = egin{pmatrix} p & 0 & 0 & 0 \ 0 & p & 0 & 0 \ 0 & 0 & p & 0 \ x_1 & x_2 & x_3 & x_4 end{pmatrix}$.
The determinant is $|det(M)| = |p^3 x_4|$. This is still not right. The determinant of the lattice is $|det(V)|$ where $V$ is the matrix of basis vectors as columns.

Let's rethink the lattice definition. A lattice $L$ in $mathbb{R}^n$ is spanned by $n$ linearly independent vectors ${mathbf{v}_1, ldots, mathbf{v}_n}$.
Let $p$ be a prime.
Consider the lattice $L$ in $mathbb{R}^4$ generated by:
$mathbf{u}_1 = (p, 0, 0, 0)$
$mathbf{u}_2 = (0, p, 0, 0)$
$mathbf{u}_3 = (0, 0, p, 0)$
$mathbf{u}_4 = (x_1, x_2, x_3, x_4)$ where $x_1^2+x_2^2+x_3^2+x_4^2 = kp$ for some $k$ with $0 < k < p$.

The determinant of the lattice formed by these vectors is $|det(egin{pmatrix} p & 0 & 0 & x_1 \ 0 & p & 0 & x_2 \ 0 & 0 & p & x_3 \ 0 & 0 & 0 & x_4 end{pmatrix})| = |p^3 x_4|$. This determinant depends on $x_4$.
A better choice of basis vectors that gives a fixed determinant is needed.

Let's use the property of Hurwitz integers more directly.
Let $n$ be any positive integer.
Consider a special lattice $L$ in $mathbb{R}^4$ with determinant 1:
$mathbf{v}_1 = (1, 0, 0, 0)$
$mathbf{v}_2 = (0, 1, 0, 0)$
$mathbf{v}_3 = (0, 0, 1, 0)$
$mathbf{v}_4 = (1/2, 1/2, 1/2, 1/2)$

The set of all integer linear combinations of these vectors forms a lattice $L$.
The volume of the fundamental parallelepiped (the basic cell) is $det(L) = 1/2$.

Now, consider the ball $K$ centered at the origin with radius $sqrt{2n}$:
$K = {(y_1, y_2, y_3, y_4) in mathbb{R}^4 mid y_1^2 + y_2^2 + y_3^2 + y_4^2 < 2n }$.
The volume of $K$ is $ ext{vol}(K) = frac{pi^2}{2} (2n)^2 = 2pi^2 n^2$.

By Minkowski's theorem, if $ ext{vol}(K) > 2^4 det(L)$, then $K$ contains a nonzero lattice point.
$2^4 det(L) = 16 cdot (1/2) = 8$.
So, if $2pi^2 n^2 > 8$, i.e., $pi^2 n^2 > 4$, which is true for $n ge 1$, the ball $K$ contains a nonzero lattice point $mathbf{p} in L$.
So, there exists $mathbf{p} = a_1mathbf{v}_1 + a_2mathbf{v}_2 + a_3mathbf{v}_3 + a_4mathbf{v}_4 in L$ such that $mathbf{p} eq mathbf{0}$ and $y_1^2 + y_2^2 + y_3^2 + y_4^2 < 2n$, where $mathbf{p} = (y_1, y_2, y_3, y_4)$.

A Hurwitz integer is of the form $a + bi + cj + dk$ where $a,b,c,d$ are halfintegers (or integers).
The lattice $L$ spanned by the given vectors corresponds to the Hurwitz integers.
The elements of $L$ are Hurwitz integers.
So, there exists a nonzero Hurwitz integer $q_0 = y_1 + y_2i + y_3j + y_4k$ such that $N(q_0) < 2n$.

This is not enough. We need to reach exactly $n$.

The proof of Lagrange's theorem using Minkowski typically focuses on constructing a lattice such that if it contains a point with norm $m$, then we can "reduce" $m$ to a smaller value, eventually reaching a norm of $n$.

The actual proof for Lagrange's theorem using Minkowski's theorem involves a more specific construction of the lattice for a given $n$, or a different variant of the theorem. The general idea is to show that for any $n$, you can find a point in a specific lattice within a ball of radius $sqrt{2n}$, and this point corresponds to a sum of four squares.

A more common proof of the foursquare theorem uses quaternions and the division algorithm, not directly Minkowski's theorem in the standard form. However, the underlying principle of finding points in a lattice within certain geometric regions is what Minkowski's theorem captures.

Summary for Lagrange's FourSquare Theorem:
The proof sketch involves:
1. Considering $mathbb{R}^4$ as the space of quaternions.
2. Identifying Hurwitz integers as forming a lattice in $mathbb{R}^4$ with a small determinant (e.g., $1/2$).
3. Choosing a ball $K$ of radius $sqrt{2n}$.
4. Applying Minkowski's theorem: If $ ext{vol}(K) > 2^4 det(L)$, $K$ contains a nonzero lattice point. This gives a Hurwitz integer $q_0$ with $N(q_0) < 2n$.
5. The subtle part is to show that if there's a Hurwitz integer with norm $m < 2n$, we can construct another one with a smaller norm, eventually hitting $n$. This part often involves the division algorithm for Hurwitz integers, which is related to the geometry of the lattice.

总结

格理论，特别是 Minkowski 基本定理，为证明一些著名的数论定理提供了强大的工具。其核心思想是将数论问题转化为在特定格中寻找满足特定几何条件的点的问题。

费马平方和定理: 利用 $mathbb{R}^2$ 中的一个格（与高斯整数环相关）和一个半径为 $sqrt{2p}$ 的圆盘，Minkowski 定理可以证明一个素数 $p equiv 1 pmod{4}$ 可以表示为两个平方和。
拉格朗日四平方和定理: 在 $mathbb{R}^4$ 中考虑一个与整数四元数（或 Hurwitz 整数）相关的格。通过选择一个合适的球体并应用 Minkowski 定理，可以证明存在一个格点（对应一个 Hurwitz 整数）其范数（平方和）小于 $2n$。进一步的论证（通常涉及除法算法）表明可以从这个点出发找到一个范数为 $n$ 的点。

需要注意的是，直接应用 Minkowski 基本定理来精确地获得等于 $n$ 的平方和可能需要一些技巧来构造恰当的格和凸集，或者结合其他代数工具（如除法算法）来完成证明的最后一步。但格理论提供了一个强大的框架来理解这些问题的几何结构。

哪个 Minkowski 定理？

这个定理在不同学科里有不同的表示方式，比如格子理论和泛函分析里都有该定理。

Lattice 理论表示：

设 是一个 维格， 是一个对称凸面集合，其体积满足：

那么 中包含一个非零的格向量。

很多书上给出的证明都是通过压缩映射原理的平移搬迁给出的，但是不太具有代表性。我是这么考虑的，考虑 整数格，那么原点附近的什么样的中心对称图形才能一定包含格中非零点呢？很显然，如果是边长为 的正方形，面积为基本平行体的体积之四倍，它肯定是满足的，而这这个图形绕中心旋转也仍然能覆盖一个非零格子点。

如果你要把这个正方形揪成长条，那么不可避免地会接触到离原点更远的格子点，总得来说就是无处可逃，满地都是钉子，当你的脚足够大时，怎么踩都会被扎到。（好恐怖）