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?