=== 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} {\operatorname{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: :$\operatorname{span}_\R L = \R^4$  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: :$\map \det L = \# \paren {\Z^4 / L} < p^2$ {{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 [http://en.wikipedia.org/wiki/Volume_of_an_n-ball| formula for the volume of an n-ball] (with respect to 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.}} {{Disambiguate|Definition:Measure}} :$\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } = \dfrac {\pi^2 \paren {\sqrt {2 p} }^4} 2 = 2 \pi^2 p^2$ 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 $x$, $y$ being primes. Suppose one of them 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 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  {{finish}}