Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? This isn't a homework question but one I found online.

Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? 

We just started doing field theory in my class and I want extra practice, but I have no idea how to even start this problem. It's no, right? I'm not sure why.

On the other hand, $-1$ is the sum of two squares here. What Lam calls Siegel's Theorem is that, in an algebraic number field, the stufe is one of $1,2,4,\infty.$ Either $-1$ is not the sum of any number of squares in the field, or it is itself a square, or it is the sum of two or four squares. 
I will see if I can find an explicit expression; others would be quicker than I, though.
There we go, looked at Berrick's expressions:
$$ \left( \frac{1}{2} \sqrt{-2} \right)^2 +  \left( \frac{1}{2} \sqrt{-2} \right)^2 =  \frac{-1}{2} +  \frac{-1}{2} = -1  $$
or
$$ 1^2 +  \left(  \sqrt{-2} \right)^2 =  1-2 = -1  $$
http://en.wikipedia.org/wiki/Stufe_%28algebra%29#Examples
Extra: found details in a book I don't have by Rajwade; the stufe of a quadratic field, $\mathbb Q(\sqrt {-D})$ with $D>0$ squarefree, is $1$ if $D=1,$ is $2$ if $D \neq 8n+7,$ is $4$ if $D = 8n+7.$ Theorem 3.2 on pages 31-32 in his 1993 book. 
$$ \left( \frac{1}{2}  \right)^2 +  \left( \frac{1}{2}  \right)^2 +  \left( \frac{1}{2}  \right)^2 +  \left( \frac{1}{2} \sqrt{-7} \right)^2 =  \frac{1}{4} +  \frac{1}{4}  +  \frac{1}{4}  +  \frac{-7}{4}  = -1  $$
or
$$ 2^2 + 1^2 + 1^2 +  \left(  \sqrt{-7} \right)^2 =  4+1+1-7 = -1  $$
