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.