I am trying to solve this problem :
Let $K$ be a field not containing $\sqrt{-1}$, and $L$ be a Galois extension of $K$ containing $\sqrt{-1}$ and having Galois group $\mathbb Z/4\mathbb Z$. Show that there exists $x,y \in K$ such that $x^2+y^2=-1$.
The way to approach the solution is as $\sqrt{-1}\in L$ we can consider $K(\sqrt{-1}):K$, as the elements over there are of the form $x+y\sqrt{-1}$, we can try to look for an element whose norm is $-1$ in $K(\sqrt{-1}):K$. But what could be that element?
Let $F = K(i)$ be the quadratic subextension, and let $\sigma$ denote an extension of the nontrivial automorphism (complex conjugation) to $L/K$. Write $L = F(\sqrt{\alpha})$ for some $\alpha \in F$. Since $L/K$ is normal we must have $\sqrt{\alpha}^\sigma = \beta \sqrt{\alpha}$ for some $\beta \in F$. Applying $\sigma$ to this equation we get $\sqrt{\alpha}^{\sigma^2} = \beta^{1+\sigma} \sqrt{\alpha}$. Since $L/K$ is cyclic of order $4$, we must have $\sqrt{\alpha}^{\sigma^2} = - \sqrt{\alpha}$ because $\sigma^2$ must be a nontrivial automorphism, hence $\beta^{1+\sigma} = -1$. Write $\beta = x + yi$; then $-1 = \beta^{1+\sigma} = x^2 + y^2$.