TITLE: Is it true in an arbitrary field that $-1$ is a sum of two squares iff it is a sum of three squares?
QUESTION [34 upvotes]: Here's a statement from Lam's First Course in Noncommutative Rings. (Paraphrased)

Let $k$ be a field. Then the following conditions are equivalent. $$(\forall a,b,c,d\in k)\;\;(a,b,c,d)\neq 0\implies a^2+b^2+c^2+d^2\neq 0;\tag 1$$ $$-1\text{ is not a sum of two squares.}\tag2$$

I can't prove this fact. If I replace $(2)$ with

$$-1\text{ is not a sum of three squares,}\tag3$$

then the equivalence between $(1)$ and $(3)$ is easy to prove. Indeed, suppose $(1)$ is false. Then we may assume without loss of generality that there are $a,b,c,d\in k$ with $a\neq 0$ such that $$a^2+b^2+c^2+d^2=0.$$ But then $$-1=\left(\frac ba\right)^2+\left(\frac ca\right)^2+\left(\frac da\right)^2,$$
and so $(3)$ is false. Conversely, if $(3)$ is false, then there exist $x,y,z\in k$ such that $$-1=x^2+y^2+z^2$$ and therefore $$0=x^2+y^2+z^2+1^2,$$ which means that $(1)$ is false.
I can also prove $(1)\implies (2).$ Suppose $(2)$ is false. Then $$-1=x^2+y^2,$$ whence $$0=x^2+y^2+1^2+0^2,$$ which implies that $(1)$ is false.
But I can't prove $(2)\implies (1)$ and I'm starting to suspect that it may actually be false. But I can't find a counterexample either. It would require finding a field in which $-1$ is a sum of three squares but not a sum of two squares. But in finite fields, every element is a sum of two squares, as proven here. In the subfields of $\mathbb R$ on the other hand, $-1$ is not a sum of three squares. And in $\mathbb C,$ it is a sum of two squares: $-1=0^2+i^2.$ In general, an example cannot be algebraically closed, which leaves me with the fields of rational functions as the only other fields I know. Also, the example cannot have characteristic $2$ as then $-1=1=1^2+0^2.$
Edit: The theorem Gerry Myerson cites in his answer is proven here. A Wikipedia article on the Stufe is here. Both links have been given by joriki in the comments.

REPLY [16 votes]: Hint $\ $ If $\rm\: -1 = a^2 + b^2 + c^2\:$ then either $\rm\:c^2 = -1\:$ or $\rm\:(a^2+b^2)/(c^2+1) = -1,\:$ which may be rewritten as a sum of two squares, since such sums $\ne 0$  form a group, using Brahmagupta for  multiplication, and inversion via $\rm\ 0\ne z = x^2\! + y^2\:$ $\Rightarrow$ $\rm\:1/z = (x/z)^2 + (y/z)^2.$ 
Remark $\ $ The proof generalizes.  Replacing the Brahmagupta composition identity by identities discovered by Pfister for quadratic forms in $\rm\:2^n\:$ variables, the proof shows that if $-1$ is a sum of $\rm\:m\:$ squares in a field, then the least value of $\rm\:m\:$ is necessarily a power of $2$, which is called the level (Stufe in German) of the field. Further, every power of $2$ is the level of some field.