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Write an integral binary quadratic form asA positive definite form is a reduced positive definite binary quadratic form whenwith $b\ge 0$ on the boundary $|b|=a$ or $a=c$. The class number of a negative discriminant $h(d)$ is the number of proper equivalence classes of primitive positive definite integral forms of discriminant $d$.

Every such class has a reduced representative. For a reduced form,soThere are only finitely many possible integers $a$ and $b$, and thenis determined. Hence $h(d)<\infty$. There is at least one class: the principal form isandThus $1\le h(d)<\infty$, in agreement with the enumeration of reduced binary quadratic forms.

Now let $q\equiv 3(\mathrm{mod}8)$ be prime. The two formsare primitive, positive definite, reduced, and have discriminant $-8q$. They are not properly equivalent: $f_0$ represents one, while the least positive value of $f_1$ is two. Therefore

Assume henceforth that $h(-8q)=2$, so these are the only two classes. Iffor a prime $p>q$, then $x$ is odd andMoreover $p\equiv x^2(\mathrm{mod}q)$, so $p$ is a quadratic residue modulo $q$. This proves the necessity.

Conversely, suppose $p\equiv \pm 1(\mathrm{mod}8)$ and $p$ is a quadratic residue modulo $q$. By quadratic reciprocity, these two conditions implyIndeed, for $p\equiv 1(\mathrm{mod}8)$ both relevant signs are positive, while for $p\equiv 7(\mathrm{mod}8)$ the signs from $(-8/p)$ and reciprocity with $q\equiv 3(\mathrm{mod}4)$ cancel. Choose an even integer $b$ such thatThenis a primitive positive definite integral form of discriminant $-8q$ and represents $p$. It belongs to one of the two classes $f_0,f_1$.

The second form cannot represent such a prime. Ifthen $y$ is odd and, since $q\equiv 3(\mathrm{mod}8)$,This contradicts $p\equiv \pm 1(\mathrm{mod}8)$. Hence $[p,b,c]$ lies in the principal class $f_0$, and proper equivalence of binary quadratic forms preserves represented integers. ThereforeThis is the prime representation when h of minus eight q equals two.