Proposition 7

— Let K be an ordered field in which every positive element is a square. Then every element of $K(i)$ is a square, and every polynomial of degree 2 over $K(i)$ has a root in $K(i)$.

Let us show first that the second assertion reduces to the first. One can put the second degree polynomial $aX^2 + bX + c$ ($a \neq 0$) in the following form, often called the canonical trinomial form:

If d is a square root of $(b^2 - 4ac)/4a^2$, then $d - (b/2a)$ is a root of the quadratic polynomial under consideration.

Now we show that every element $a + bi$ ($a \in K, b \in K$) is a square; we are looking for an element $x + yi$ such that

this translates into $x^2 - y^2 = a$ and $2xy = b$. From this we deduce that

Let c denote the positive square root of $a^2 + b^2$; then $c \geq |a|, c \geq |b|$ and $x^2 + y^2 = c$. Whence $x^2 = (c + a)/2$ and $y^2 = (c - a)/2$. Since $c \geq |a|$ these equations are soluble in K, and if $x_0$ and $y_0$ are two solutions then $x_0^2 - y_0^2 = a$ and $2x_0y_0 = \pm b$. We obtain the desired square root by taking $x = x_0$ and $y = b/2x_0$.