The Hilbert symbol is a symmetric bilinear nondegenerate pairing

Step [L1] descends the symbol to square classes. Symmetry is immediate from the defining equation $a{x}^2+b{y}^2=1$. The explicit formulas of [L2] are multiplicative in each argument on the square-class group, so they give bilinearity at every rational place.

To prove nondegeneracy, fix a nonsquare class $c$. Over $\mathbb{R}$, [L2] shows that $(-1,c{)}_{\mathrm{\infty }}=-1$ when $c<0$. Over ${\mathbb{Q}}_p$ for odd $p$, write $c=p^{\gamma }w$: if $\gamma$ is odd, choose a nonsquare unit $u$ so that [L2] gives $(u,c{)}_p=-1$; if $\gamma$ is even, then $w$ is a nonsquare unit and [L2] gives $(p,c{)}_p=-1$. Over ${\mathbb{Q}}_2$, the classes of $-1,2,5$ generate the square-class group and [L2] shows that each nontrivial class is detected by one of them. Hence no nontrivial square class pairs trivially with every other one.