 Solution

For a point with $u\ne 0$, consider the valuation $v_{\ell }$ at any rational prime. If $v_{\ell }(u)<0$, the term $u^3$ has strictly lower valuation than $a_1{u}^2$ and $b_{1}u$, since the coefficients are integral. Thus $2v_{\ell }(v)=3v_{\ell }(u)$, forcing $v_{\ell }(u)$ even. If $v_{\ell }(u)>0$ and $\ell \nmid b_1$, the factor $u^2+a_{1}u+b_1$ is a unit, so $2v_{\ell }(v)=v_{\ell }(u)$, again even. At valuation zero there is no odd exponent to remove.

Therefore every odd valuation in the square class of $u$ lies at a prime dividing $b_1$, and all negative valuations are even. Removing rational squares gives a signed squarefree integer $r$ with $r\mid b_1$. For $(0,0)$ the assigned class is that of $b_1$, whose squarefree representative also divides $b_1$; the identity has representative one. HenceThe brackets stress that these integers represent square classes, which is the target of $q$.