Lagrange's Four Square Theorem/Proof 2

Proof for Odd Primes
Suppose $p$ is an odd prime.

Define:


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

Define:


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

Suppose for $\alpha, \alpha' \in S$:


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

Obviously:


 * $\paren {\alpha + \alpha'} \paren {\alpha - \alpha'} = \alpha^2 - \alpha'^2 \equiv 0 \pmod p$

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


 * $\alpha + \alpha' \not \equiv 0 \pmod p$

Therefore $\alpha - \alpha' \equiv 0 \pmod p$.

This shows that $\alpha = \alpha'$.

Thus we have $\size S = \size {\hointr 0 {\dfrac p 2} \cap \Z} = 1 + \dfrac {p - 1} 2 = \dfrac {p + 1} 2$.

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 by the same reasoning as above:


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

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:


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

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

So $L$ is closed under vector addition.

So $L$ has additive inverses.

By Two-Step Subgroup Test, $\struct {L, +}$ is a subgroup of $\struct {\R^4, +}$.

For each $\vec x = \tuple {x_1, x_2, x_3, x_4} \in L$, one can write:
 * $\vec x = x_3 \tuple {\alpha, \beta, 1, 0} + x_4 \tuple {\beta, -\alpha, 0, 1} + \floor {\dfrac {x_1} p} \tuple {p, 0, 0, 0} + \floor {\dfrac{x_2} p} \tuple {0, p, 0, 0}$

Thus:
 * $\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} }$

spans $L$.

Suppose we have for some $c_1, c_2, c_3, c_4 \in \Z$:
 * $\vec 0 = c_1 \tuple {\alpha, \beta, 1, 0} + c_2 \tuple {\beta, -\alpha, 0, 1} + c_3 \tuple {p, 0, 0, 0} + c_4 \tuple {0, p, 0, 0}$

Extracting the various coordinates:

and so:

So:
 * $\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} }$

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

Thus:

Thus:
 * $\span_\R L = \R^4$

So $L$ is a lattice.

Define the quotient map:
 * $\varphi: \N_p^2 \times \set {\tuple {0, 0} } \to \Z^4 / L$
 * $\tuple {x, y, 0, 0} \mapsto \sqbrk {\tuple {x, y, 0, 0} }$

Since quotient maps are surjective:


 * $\paren {\Z^4 / L} \le \size {\N_p^2 \times \set {\tuple {0, 0} } } = p^2$

So:


 * $\map \det L = \# \paren {\Z^4 / L} < p^2$

Let $\norm {\, \cdot \,}$ denote the Euclidean metric.

Consider $\map {B_{\sqrt {2 p} } } {\vec 0}$, the open ball of radius $\sqrt {2 p}$.

Then:
 * $\forall \vec x, \vec y \in \map {B_{\sqrt {2 p} } } {\vec 0}, \forall t \in \closedint 0 1$:

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

So $\map {B_{\sqrt{2 p} } } {\vec 0}$ is convex.

Let $\vec x \in \map {B_{\sqrt{2 p} } } {\vec 0}$.

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

That means:


 * $\norm {-\vec x} = \size {-1} \norm {\vec x} = \norm {\vec x} < \sqrt{2 p}$

So:
 * $-\vec x \in \map {B_{\sqrt{2 p} } } {\vec 0}$

Then $\map {B_{\sqrt{2 p} } } {\vec 0}$ is symmetric about the origin.

By the ( the standard measure on $\R^n$):


 * $\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } = \dfrac {\pi^2 \paren {\sqrt {2 p} }^4} 2 = 2 \pi^2 p^2$

Since $2\pi^2 > 1$:


 * $\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } > \map \det L$

By Minkowski's Theorem:


 * $L \cap \map {B_{\sqrt{2 p} } } {\vec 0} \ne \O$

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

then:


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

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 = \map N {\mu \lambda}$

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