= Solution
For a point with $u\ne0$, consider the <valuation> $v_\ell$ at any rational <prime number>[prime]. If $v_\ell(u)<0$, the term $u^3$ has strictly lower <valuation> than $a_1u^2$ and $b_1u$, since the <coefficients>[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 $u^2+a_1u+b_1$ is a <unit in a ring>[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 number>[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 integer>[squarefree] representative also divides $b_1$; the identity has representative one. Hence
$$
\boxed{\operatorname{im}q\subseteq\{[r]:r\text{ is a signed squarefree divisor of }b_1\}.}
$$
The brackets stress that these <integers> represent <square classes>, which is the target of $q$.
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