# Is -1 a sum of 2 squares in a certain field K?

Consider the field of fractions $K$ of the quotient algebra $\mathbb{R}[x,y,z,t]/(x^2+y^2+z^2+t^2+1)$, where $\mathbb{R}$ is the field of real numbers and $x,y,z,t$ are variables. Clearly $-1$ is a sum of 4 squares in $K$. How can one prove that $-1$ is not a sum of 2 squares in $K$?

Serre mentions without proof this (probably known or easy) fact in a letter to Eva Bayer of May 1, 2010, and I am stuck: I cannot prove it.

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I hope it's not impertinent to ask: how do you know about a letter Serre wrote four days ago? –  Graham Leuschke May 5 '10 at 19:48
@Graham: I asked a question to Eva Bayer about Galois cohomology of $G=PSU_n$ and maximal tori of $G$ over field extensions of $\mathbb{R}$, Eva forwarded my question to Serre, Serre answered it in a letter to Eva, and Eva forwarded me his letter. –  Mikhail Borovoi May 6 '10 at 6:17

Pfister, Albrecht, Zur Darstellung von $-1$ als Summe von Quadraten in einem Körper. (German) J. London Math. Soc. 40 1965 159--165.
In this same paper Pfister defines the "stufe" (which Lam has successfully campaigned to be called the "level") of a non-formally real field, namely the least positive integer $n$ such that $-1$ is a sum of $n$ squares. Among his other achievements, he proves that the level is always a power of $2$ (so that Kevin Buzzard's recollection is correct). It is also worth remarking that his work is an insightful and rapid response to previous work of J.W.S. Cassels.
The special case Kevin mentioned has a short proof which makes it easy to remember: if $-1=a^2+b^2+c^2$ then $-1-c^2=a^2+b^2$ implying $-1=(a^2+b^2)(1+c^2)/(1+c^2)^2$. Now use the fact that the product of the sum of two squares is also a sum of two squares. –  Keivan Karai May 6 '10 at 9:54