Theorem: Lagrange's Four Square Theorem

Every positive integer can be expressed as a sum of four squares.

Proof:
=== 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:
{{begin-eqn}}
{{eqn | l = z_1
      | r = x_1 + y_1
      | c = 
}}
{{eqn | o = \equiv
      | r = \alpha x_3 + \beta x_4 + \alpha y_3 + \beta y_4
      | rr= \pmod p
      | c = 
}}
{{eqn | o = \equiv
      | r = \alpha \paren {x_3 + y_3} + \beta \paren {x_4 + y_4}
      | rr= \pmod p
      | c = 
}}
{{eqn | o = \equiv
      | r = \alpha z_3 + \beta z_4
      | rr= \pmod p
      | c = 
}}
{{eqn | l = z_2
      | r = x_2 + y_2
      | c = 
}}
{{eqn | o = \equiv
      | r = \beta x_3 - \alpha x_4 + \beta y_3 - \alpha y_4
      | rr= \pmod p
      | c = 
}}
{{eqn | o = \equiv
      | r = \beta \paren {x_3 + y_3} - \alpha \paren {x_4 + y_4}
      | rr= \pmod p
      | c = 
}}
{{eqn | o = \equiv
      | r = \beta z_3 - \alpha z_4
      | rr= \pmod p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \vec x + \vec y
      | r = \vec z
      | c = 
}}
{{eqn | o = \in
      | r = L
      | c = 
}}
{{end-eqn}}
So $L$ is closed under vector addition.
{{begin-eqn}}
{{eqn | l = -x_1
      | o = \equiv
      | r = -\paren {\alpha x_3 + \beta x_4}
      | rr= \pmod p
      | c = 
}}
{{eqn | o = \equiv
      | r = \alpha \paren {-x_3} + \beta \paren {-x_4}
      | rr= \pmod p
      | c = 
}}
{{eqn | l = -x_2
      | o = \equiv
      | r = -\paren {\beta x_3 - \alpha x_4}
      | rr= \pmod p
      | c = 
}}
{{eqn | o = \equiv
      | r = \beta \paren {-x_3} - \alpha \paren {-x_4}
      | rr= \pmod p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\vec x
      | o = \in
      | r = L
      | c = 
}}
{{eqn | ll= \leadsto
      | l = L
      | o = \le
      | r = \R^4
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \alpha c_1 + \beta c_2 + p c_3
      | r = 0
      | c = 
}}
{{eqn | l = \beta c_1 - \alpha c_2 + p c_4
      | r = 0
      | c = 
}}
{{eqn | l = c_1
      | r = 0
      | c = 
}}
{{eqn | l = c_2
      | r = 0
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \alpha c_1 + \beta c_2
      | r = 0
      | c = 
}}
{{eqn | l = \alpha c_1 - \beta c_2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c_3
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c_4
      | r = 0
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map {\dim_\R} {\span_\R L}
      | r = \dim_\Z L
      | c = 
}}
{{eqn | r = \card {\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} } }
      | c = 
}}
{{eqn | r = 4
      | c = 
}}
{{end-eqn}}
Thus:
:$\span_\R L = \R^4$ 
So $L$ is a point 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:
:$\map \det L = \# \paren {\Z^4 / L} < p^2$
{{explain|$\#$}}
Let $\norm {\, \cdot \,}$ denote the Euclidean metric.
{{explain|Work out exactly which version of the Definition:Euclidean Metric links is relevant here.}}
Consider $\map {B_{\sqrt {2 p} } } {\vec 0}$, the open ball of radius $\sqrt {2 p}$.
Then:
:$\forall \vec x, \vec y \in \map {B_{\sqrt {2 p} } } {\vec 0}, \forall t \in \closedint 0 1$:
{{begin-eqn}}
{{eqn | l = \norm {t \vec x + \paren {1 - t} \vec y}
      | o = \le
      | r = \norm {t \vec x} + \norm {\paren {1 - t} \vec y}
      | c = {{Defof|Metric}}
}}
{{eqn | r = \size t \norm {\vec x} + \size {1 - t} \norm {\vec y}
      | c = {{Defof|Metric}}
}}
{{eqn | o = <
      | r = \size t \sqrt {2 p} + \size {1 - t} \sqrt {2 p}
      | c = {{Defof|Open Ball}} $\map {B_{\sqrt{2 p} } } {\vec 0}$
}}
{{eqn | r = \paren {\size t + \size {1 - t} } \sqrt {2 p}
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = \paren {t + 1 - t} \sqrt{2 p}
      | c = since $t \in \closedint 0 1 \implies t \ge 0$
}}
{{eqn | r = \sqrt {2 p}
      | c = since $t + 1 - t = 1$ is the multiplicative identity of $\R$
}}
{{end-eqn}}
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 {{WP|Volume_of_an_n-ball|formula for the volume of an $n$-ball}} ({{WRT}} the standard measure on $\R^n$):
{{explain|Change the above to be a page on {{ProofWiki}} -- external links of material of a mathematical nature are disallowed by house rules.}}
:$\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } = \dfrac {\pi^2 \paren {\sqrt {2 p} }^4} 2 = 2 \pi^2 p^2$
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:
{{begin-eqn}}
{{eqn | l = a_1^2 + a_2^2 + a_3^2 + a_4^2
      | o = \equiv
      | r = \paren {\alpha a_3 + \beta a_4}^2 + \paren {\beta a_3 - \alpha a_4}^2 + a_3^2 + a_4^2
      | rr= \pmod p
      | c = Definition of $L$
}}
{{eqn | o = \equiv
      | r = \alpha^2 a_3^2 + 2 \alpha \beta a_3 a_4 + \beta^2 a_4^2 + \beta^2 a_3^2 - 2 \alpha \beta a_3 a_4 + \alpha^2 a_4^2 + a_3^2 + a_4^2
      | rr= \pmod p
      | c = Binomial Theorem
}}
{{eqn | o = \equiv
      | r = \paren {\alpha^2 + \beta^2 + 1} \paren {a_3 + a_4}
      | rr= \pmod p
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | o = \equiv
      | r = 0 \paren {a_3 + a_4}
      | rr= \pmod p
      | c = by $(1)$ from above
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod p
}}
{{end-eqn}}
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:
:$a_1^2 + a_2^2 + a_3^2 + a_4^2 = p$
{{qed|lemma}}
=== Proof for Composites ===
Suppose $x, y \in \Z$ are a sum of four squares with neither of $x$ or $y$ being primes.
Suppose either one is equal to $1$.
Then $x * 1 = x$ is a sum of four squares.
{{explain|$x * 1$? Is multiplication meant here?}}
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 {{iff}} $x$ 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 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 Fundamental Theorem of Arithmetic, every number can be written as a unique product of primes.  
Then 
{{finish}}