I don't understand the discussion in exercise 2.8 of Cassels and Frohlich (page 352) beginning with "more generally". Why should it matter whether the formula for $c$ has a power of $-1$ in it if this would be fixed by any automorphism anyway?
My attempt at carrying out the computation they suggest is as follows:
Write $a=a_0 \pi^{v(a)}, b=b_0 \pi^{v(b)}$. Then $(a,b)_v=(\pi^{v(a)}, \pi^{v(b)})(a_0, \pi^{v(b)})(\pi^{v(a)},b_0)(a_0,b_0)$ by bilinearity of the norm residue symbol. Since $|a_0|_v=|b_0|_v=1$, we get that the last term is just $1$. By exercise 2.6 and the first part of 2.8, we get that $(a_0, \pi^{v(b)})=(\frac{a_0}{v})^{v(b)}$ and $(\pi^{v(a)}, b_0)=(\frac{b_0}{v})^{-v(a)}$ and by bilinearity, $(\pi^{v(a)}, \pi^{v(b)})=( \pi, \pi)^{v(a)v(b)}$. By symmetry, this is $1^{v(a)v(b)}$ or $(-1)^{v(a)v(b)}$ but it is not clear to me if either is forced.
Edit: The book is located here http://math.arizona.edu/~cais/scans/Cassels-Frohlich-Algebraic_Number_Theory.pdf