=== $(2)$ iff $(3)$ === Suppose there exist $\set {x_i: i \in I}$ such that :$\displaystyle -1 = \sum_{i \mathop \in I} x_i^2$ Then :$\displaystyle 0 = 1^2 + \sum_{i \mathop \in I} x_i^2$ a non-empty sum of non-zero squared of $k$. Conversely, suppose that there is a set $\set {x_i: i \in I} \ne \O$ with $x_i \ne 0$ for all $i \in I$ such that: :$\displaystyle 0 = \sum_{i \mathop \in I} x_i^2$ Then for any $j \in I$: :$\displaystyle -x_j^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } x_i^2$ Dividing through by $x_j^2$ we find that: :$\displaystyle -1 = -\paren {\frac {x_j} {x_j} }^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } \paren {\frac {x_i} {x_j} }^2$ === $(1)$ implies $(2)$ === By Properties of Ordered Field, in an ordered field we have, $-1 < 0$, and squares are non-negative. Therefore for any subset $\set {x_i : i \mathop \in I} \subseteq k$, :$\displaystyle -1 < 0 \le \sum_{i \mathop \in I} x_i ^2$ === $(2)$ implies $(1)$ === Suppose that $-1$ is not a sum of squares in $k$.  Let $S$ be the set of non-empty sums of squares of non-zero elements of $k$. Then by supposition and  $(2) \iff (1)$: :$0, 1 \notin S$ {{explain|but $(2) \iff (1)$ has not yet been established.}} Trivially, $S$ is closed under addition. Also, for any subsets $\set {x_i: i \in I}, \set {y_j: j \in J} \subseteq k$: :$\displaystyle \paren {\sum_{i \mathop \in I} x_i^2} \cdot \paren {\sum_{j \mathop \in J} y_j^2} = \sum_{i, j} \paren {x_i y_j}^2 \in S$ so $S$ is closed under multiplication.  It follows that $S$ is a multiplicative subgroup of the set difference $k \setminus \set 0$. Now let $\Gamma$ be the collection of all subsets $M$ of $k$ such that: :$S \subseteq M$ :$M$ is closed under addition :$M$ is a multiplicative subgroup of $k \setminus \set 0$ Then every chain $\mathscr C = \set {M_i : i \in \N}$ has an upper bound: :$\displaystyle \bigcup_{i \mathop \in \N} M_i \in \Gamma$ So by Zorn's Lemma there is a maximal element $M$ with these properties. Clearly $0 \notin M$. {{explain|why "clearly"?}} Let: :$\paren {-M} := \set {x \in k : -x \in M}$ Suppose $x, -x \in M$. Then: :$x - x = 0 \in M$ and this contradicts $0 \notin M$. Hence $M$, $\set 0$ and $-M$ are pairwise disjoint. At this point the reader should think of $M$ as a partition of $k$ into positive elements, $\set 0$ and negative elements. The following two claims justify this statement. ==== Lemma 1 ==== $k = M \cup \set 0 \cup \paren {-M}$ ==== Proof of Lemma 1 ==== Let $a \in k$, $a \ne 0$, $-a \notin M$, and: :$M' = \set {x + a y: x, y \in M \cup \set 0, \paren {x = 0 \lor y = 0} }$ We have: :$S \subseteq M \subseteq M'$ and it is trivial to check that $M'$ is closed under addition. {{explain|Trivial though it be, it needs to be demonstrated.}} Let $x + a y$, $z + a w \in M'$. Then: :$\paren {x + a y} \paren {z + a w} = \paren {x z + a^2 y w} + a \paren {y z + x w}$ Since: :$a^2 \in S \subseteq M'$ and: :$M$ is closed under multiplication and addition it follows that: :$M'$ is also closed under multiplication. Since $a$ and at least one of $x, y \ne 0$, it follows that: :$0 \notin M'$ Since $1 \in S \subseteq M'$: :$1 \in M'$ Let $t = x + a y \in M'$. Then: {{begin-eqn}} {{eqn | l = t \paren {\frac x {t^2} + a \frac y {t^2} }       | r = \frac x t + a \frac y t }} {{eqn | r = \frac {x + a y} {x + a y} }} {{eqn | r = 1 }} {{end-eqn}} Therefore every $t \in M'$ has a multiplicative inverse in $M'$ So $M'$ is a multiplicative subgroup of $k \setminus \set 0$. Therefore $M'$ satisfies all the conditions subject to which we chose $M$. This and the fact that $M \subseteq M'$ imply that: :$M = M'$ Therefore: :$a = 1 + 1 a \in M$ Let $-a \in M$. Then: :$a \in -M$ so every $a \in k$ lies in exactly one of $M$, $\set 0$ and $-M$. Thus $k = M \cup \set 0 \cup \paren {-M}$. {{qed|lemma}} Now define a binary relation on $k$ by $x < y \iff y - x \in M$. Equivalently: :$x \le y \iff \paren {x < y} \lor \paren {x = y}$ ==== Lemma 2 ==== $\struct {k, \le}$ is an ordered field. ==== Proof of Lemma 2 ==== The proof is an elementary check of the relevant axioms. First we check that $\le$ is a total order on $k$. Reflexivity: Trivial from the definition. Antisymmetry: Suppose that $x \le y$ and $y \le x$. We cannot have $x - y \in M$ and $y - x \in M$, because since $M$ is closed under addition, this would imply $0 \in M$. Therefore, $x - y = y - x = 0$ and $y = x$. Transitivity: Suppose $x \le y$ and $y \le z$. Then $y - x \in M$ and $z - y \in M$. $M$ is closed under addition, so $z - x = \paren {y - x} + \paren {z - y} \in M$. Therefore $x \le z$. Comparability: The comparability of each pair of elements of $\struct {k, \le}$ is immediate from the above mentioned fact that $M$, $\set 0$ and $\paren {-M}$ partition $k$: Suppose $x \nleq y$. Then: :$y - x \notin M$ and: :$y - x \ne 0$ Then: :$y - x \in \paren {-M}$ so: :$x - y \in M$ and: :$y \le x$ Thus $x$ and $y$ are comparable. It remains to show that the total order $\le$ is compatible with the field structure. Let $x, y, z \in k$ such that $x < y$. Then: :$\paren {y + z} - \paren {x + z} = y - x \in M$ So: :$x + z \le y + z$. Now suppose further that $z > 0$. We have that: :$z \in M$ and $y - x \in M$ by hypothesis and: :$M$ is closed under multiplication. Therefore: :$y z - x z = z \paren {y - x} \in M$ This establishes that $k$ is an ordered field. {{qed|lemma}} {{qed}}