Class number problem

For a squarefree integer $d \neq 1$ put $K=\mathbb{Q}(\sqrt{d})$, let $\mathcal{O}_K$ be the ring of integers of $K$, and let

$$h(K)=\#\mathrm{Cl}(\mathcal{O}_K)$$

be the order of the ideal class group of $\mathcal{O}_K$, i.e. the class number of $K$. The discriminant of $K$ is $D=d$ if $d\equiv 1 \pmod 4$ and $D=4d$ otherwise; the integers $D$ arising in this way are called fundamental discriminants, and for such $D$ write $h(D)=h(\mathbb{Q}(\sqrt{D}))$. More generally, for an arbitrary integer $D<0$ with $D\equiv 0,1\pmod 4$, let $h(D)$ denote the number of $\mathrm{SL}_2(\mathbb{Z})$-equivalence classes of primitive positive definite integral binary quadratic forms $ax^2+bxy+cy^2$ with $b^2-4ac=D$.

The following hold.

1. For every integer $n\ge 1$ the set

$$S^-(n)=\{\, D<0 \;:\; D \text{ a fundamental discriminant},\ h(D)=n \,\}$$

is finite, and it is effectively determinable: there is an algorithm which, given $n$, outputs a finite list of discriminants together with a proof that this list is exactly $S^-(n)$. Equivalently, $h(D)\to\infty$ as $D\to-\infty$ through negative fundamental discriminants, with an effective bound $B(n)$ such that $h(D)=n$ forces $|D|\le B(n)$.

2. For every integer $n\ge 1$ the set of all integers $D<0$ with $D\equiv 0,1\pmod 4$ (not necessarily fundamental) and $h(D)=n$ is finite and likewise effectively determinable; in particular the set of even such $D$ with $h(D)=n$ can be listed completely.

3. The set

$$\{\, D>0 \;:\; D \text{ a fundamental discriminant},\ h(D)=1 \,\}$$

is infinite; that is, there are infinitely many real quadratic fields $\mathbb{Q}(\sqrt{d})$, $d>1$ squarefree, with class number $1$.