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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.

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This is an example of a post which is well written. I have been discussing this with the user on chat. I did not have this clear frame in my mind--that I read this post makes me organise what I know. +1 for the question... –  user21436 Apr 19 '12 at 12:12
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2 Answers

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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.

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Thanks a lot! I was having a bit of trouble lessening the generality of Pfister's argument given in a comment by joriki. Now everything is clear! –  user23211 Apr 19 '12 at 14:12
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The "stufe" of a field is the smallest $m$ such that $-1$ can be expressed as a sum of $m$ squares. It is a theorem of Pfister that the stufe (if it exists) is always a power of 2.

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Thank you! So if I understand this correctly, Lam's statement is correct. Is is difficult to prove? It would be strange for Lam to make such a statement without proof if it requires advanced field theory... –  user23211 Apr 19 '12 at 12:42
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There's a proof here. There's also a German Wikipedia article on the "Stufe", but it presupposes some of the facts about Pfister forms that the other text proves. –  joriki Apr 19 '12 at 12:52
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@ymar: Yes, sorry, I just realized that that paper contains only a part of this slide show that has the whole proof (part III beginning on p. 81). I was trying to avoid linking to a slide show :-) –  joriki Apr 19 '12 at 13:14
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I've created an English Wikipedia article on Albrecht Pfister (translated from the German article). –  joriki Apr 19 '12 at 13:46
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@ymar: Here's the new English Wikipedia article on Stufe. –  joriki Apr 19 '12 at 15:22
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