Lagrange's Four Square Theorem/Proof 2

Proof for Primes
Suppose $p$ is a prime.

Define:


 * $S := \set {\alpha^2 \pmod p: \alpha \in \hointr 0 {\dfrac p 2} \cap \Z}$

Define:


 * $S' := \set {-1 - \beta^2 \pmod p: \beta \in \hointr 0 {\dfrac p 2} \cap \Z}$

Choose $\alpha, \alpha' \in S$:


 * $\alpha^2 \equiv \alpha'^2 \pmod p$

Obviously:


 * $\left({\alpha + \alpha'}\right) \left({\alpha - \alpha'}\right) = \alpha^2 - \alpha'^2 \equiv 0 \pmod p$

Since $0 \le \alpha$, $\alpha' < \dfrac p 2$:


 * $\alpha + \alpha' \not \equiv 0 \pmod p \implies \alpha - \alpha' \equiv 0 \pmod p$ we have $\left\vert{S}\right\vert = \left\vert{\hointr 0 {\dfrac p 2} \cap \Z}\right\vert = 1 + \dfrac {p - 1} 2 = \dfrac {p + 1} 2$

Choose $\beta, \beta' \in S'$:


 * $-1 - \beta^2 \equiv -1 - \beta'^2 \pmod p$

By simple algebraic manipulation:


 * $-1 - \beta^2 \equiv -1 - \beta'^2 \pmod p \iff \beta^2 \equiv \beta'^2 \pmod p$

Then:


 * $\card {S'} = \card S = \dfrac {p + 1} 2$

By the Pigeonhole Principle:


 * $S \cap S' \ne \O$

Thus $\exists \alpha, \beta \in \Z$:


 * $(1): \quad \alpha^2 + \beta^2 + 1 \equiv 0 \pmod p$

Define:


 * $L = \set {\vec x = \left({x_1, x_2, x_3, x_4}\right) \in \Z^4: x_1 \equiv \alpha x_3 + \beta x_4 \pmod p, x_2 \equiv \beta x_3 - \alpha x_4 \pmod p}$

If $\vec x$, $\vec y \in L$ and $\vec z = \left({z_1, z_2, z_3, z_4}\right)$, then:

So $L$ is closed under vector addition.

So $L$ has additive inverses.

Thus $L$ is a subgroup of $\R^4$ (associativity is trivial).

Since:
 * $\vec x = x_3 \left({\alpha, \beta, 1, 0}\right) + x_4 \left({\beta, -\alpha, 0, 1}\right) + \left\lfloor{\dfrac {x_1} p }\right\rfloor \left({p, 0, 0, 0}\right) + \left\lfloor{\dfrac{x_2} p}\right\rfloor \left({0, p, 0, 0}\right)$

Thus:
 * $\set {\left({\alpha, \beta, 1, 0}\right), \left({\beta, -\alpha, 0, 1}\right), \left({p, 0, 0, 0}\right), \left({0, p, 0, 0}\right)}$

spans $L$.

We have:
 * $c_1 \left({\alpha, \beta, 1, 0}\right) + c_2 \left({\beta, -\alpha, 0, 1}\right) + c_3 \left({p, 0, 0, 0}\right) + c_4 \left({0, p, 0, 0}\right)$

Extracting the various coordinates:

and so:

So:
 * $\set {\left({\alpha, \beta, 1, 0}\right), \left({\beta, -\alpha, 0, 1}\right), \left({p, 0, 0, 0}\right), \left({0, p, 0, 0}\right)}$

is linearly independent and thus a basis for $L$.

Thus:
 * $\displaystyle \dim_\R \left({\operatorname{span}_\R L}\right) = \dim_\Z L = \left\vert{\left\{ {\left({\alpha, \beta, 1, 0}\right), \left({\beta, -\alpha, 0, 1}\right), \left({p, 0, 0, 0}\right), \left({0, p, 0, 0}\right)}\right\} }\right\vert = 4$

Thus:
 * $\operatorname{span}_\R L = \R^4$

So $L$ is a lattice.

Notice:
 * $\exists \varphi: \N_p^2 \times \set {\left({0, 0}\right)} \twoheadrightarrow \Z^4 / L, \left({x, y, 0, 0}\right) \mapsto \left[{\left({x, y, 0, 0}\right)}\right]$

Then:


 * $\left({\Z^4 / L}\right) \le \left\vert{\N_p^2 \times \left\{ {\left({0, 0}\right)}\right\} }\right\vert = p^2$

So:


 * $\det \left({L}\right) = \# \left({\Z^4 / L}\right) < p^2$

Let $\Vert \cdot \Vert$ be the Euclidean metric.

Consider $B_{\sqrt {2 p} } \left({\vec 0}\right)$, the open ball of radius $\sqrt {2 p}$.

Then:
 * $\forall \vec x, \vec y \in B_{\sqrt{2p}} \left({\vec 0}\right), \forall t \in \left[{0 \,.\,.\, 1}\right]$:

Thus the line between $\vec x$ and $\vec y$ is contained in $B_{\sqrt{2 p} } \left({\vec 0}\right)$.

So $ B_{\sqrt{2 p} } \left({\vec 0}\right)$ is convex.

Let $\vec x \in B_{\sqrt{2 p} } \left({\vec 0}\right)$.

Then $\vec x < \sqrt {2 p}$.

That means:


 * $\left\Vert{ -\vec x}\right\Vert = \left\vert{-1}\right\vert \left\Vert{\vec x}\right\Vert = \left\Vert{\vec x}\right\Vert < \sqrt{2 p}$

So:
 * $-\vec x \in B_{\sqrt {2 p} } \left({\vec 0}\right)$

Then $B_{\sqrt{2 p} } \left({\vec 0}\right)$ is symmetric about the origin.

By the | formula for the volume of an n-ball (with respect to the standard measure on $\R^n$):


 * $\operatorname{Vol} \left({B_{\sqrt{2 p} } \left({\vec 0}\right) }\right) = \dfrac {\pi^2 \left({\sqrt{2p} }\right)^4} 2 = 2 \pi^2 p^2$

Since $2\pi^2 > 1$:


 * $\operatorname{Vol} \left({B_{\sqrt{2 p} } \left({\vec 0}\right) }\right) > \det \left({L}\right)$

By Minkowski's Theorem:


 * $L \cap B_{\sqrt{2 p} } \left({\vec 0}\right) \ne \varnothing$

Thus, if:
 * $\vec a = \left({a_1, a_2, a_3, a_4}\right) \in L \cap B_{\sqrt{2 p} } \left({\vec 0}\right)$

then:


 * $0 < a_1^2 + a_2^2 + a_3^2 + a_4^2 = \left\Vert{\vec a}\right\Vert^2 < 2 p$

However:

So $\exists k \in \Z$:


 * $(2): \quad 0 < a_1^2 + a_2^2 + a_3^2 + a_4^2 = k p \le 2 p$

Dividing $(2)$ by $p$:


 * $0 < k < 2 \implies k = 1$

Thus:


 * $a_1^2 + a_2^2 + a_3^2 + a_4^2 = p$

Proof for Composites
Suppose $x$, $y \in \Z$ are a sum of four squares with neither of $x$, $y$ being primes.

Suppose one of them is equal to $1$.

Then $x * 1 = x$ is a sum of four squares.

Suppose neither of them is equal to $1$.

Let $\mathbb H$ denote the set of Hurwitz quaternions.

Let $N: \mathbb H \to \R, a + b i + c j + d k \mapsto a^2 + b^2 + c^2 + d^2$ be the standard norm on $\mathbb H$.

Notice:


 * $x$ is a sum of four squares $x$ is a norm of a Hurwitz quaternion

Then:


 * $\exists \mu, \lambda \in \mathbb H: x = \map N \mu, y = \map N \lambda$

From Norm is a Homomorphism:


 * $ x y = \map N \mu \, \map N \lambda = N \left({\mu \lambda}\right)$

Since $\mathbb H$ is a ring, we have:
 * $\mu \lambda \in \mathbb H$

and thus $x y$ is a sum of four squares.

From the Unique Factorization Theorem, every number can be written as a unique product of primes.

Then