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Let $a,b,c \in \mathbb Q_2^*$, where $\mathbb Q_2$ denotes the 2-adic field. How to compute the Hilbert symbol $(b^2-4ac,2a)$?

By definition, $(b^2-4ac,2a)=1$, if the equation $z^2-(b^2-4ac)x^2-2ay^2=0$ has a non-trivial solution and $(b^2-4ac,2a)=-1$ otherwise.

I know, that for $\alpha=2^mu$ and $\beta=2^n v$ (with $u,v\in U$) we have $(\alpha,\beta)=(-1)^{\varepsilon (u)\varepsilon (v)+m\omega (v)+n\omega (u)}$.

But I don't think, that it is helpful, because I need to write $b^2-4ac$ as $2^k w$ with $w\in U$.

Are there other ways to compute it?

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up vote 6 down vote accepted

On $\Bbb Q_p$ the Hilbert symbol $(a,b)$ depends only on the classes of $a$ and $b$ modulo $(\Bbb Q_p^\times)^2$. There are eight such classes when $p=2$. So, if nothing better, you can try to obtain the classes of $b^2-4ac$ and $2a$ modulo $(\Bbb Q_2^\times)^2$ depending on $a$, $b$ and $c$.

When $a$ and $b$ are in $\Bbb Q$, a way to compute $(a,b)_2$ is to compute $(a,b)_\infty$ and $(a,b)_p$ for odd $p$ (which is somewhat simpler) and use the fact that $$ \prod_{p\leq\infty}(a,b)_p=1. $$

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