Let $F$ be an unramified extension of $Q_2$. How can I compute the Hilbert symbol (a,b) for $a,b \in F^*$. Here, (a,b) is 1 if $ax^2+by^2=z^2$ has a nontrivial solution, and -1 otherwise.

In the case of $F=Q_2$, I can do this by finding representatives of $Q_2^*/(Q_2^*)^2$, and calculating the Hilbert symbol for each possible pair. However, this becomes a bit unmanageable for larger fields $F$. Is there a better way of approaching this?

Sur les lois de réciprocité explicites. I.J. Reine Angew. Math.329(1981), 177–203. $\endgroup$ – Chandan Singh Dalawat Nov 17 '11 at 16:52