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Two integral binary quadratic forms are properly equivalent when one is obtained from the other by a change of variables in $S{L}_2(\mathbb{Z})$. For a negative discriminant $d\equiv 0,1(\mathrm{mod}4)$, the class number of binary quadratic forms $h(d)$ is the number of proper equivalence classes of primitive positive-definite forms of discriminant $d$.

If $f$ properly represents $m$, then $f(r,s)=m$ for coprime $r,s$. Extend $(r,s{)}^T$ to a matrix in $S{L}_2(\mathbb{Z})$; after the associated variable change, the coefficient of $x^2$ is $m$, so the transformed form isThe converse follows by evaluating this form at $(1,0)$.

Such a form has discriminant $d$ precisely whenso it exists exactly when $b^2\equiv d(\mathrm{mod}4m)$. Thus $m$ is properly represented by some form of discriminant $d$ exactly when $d$ is a square modulo $4m$.

Now let $d=1-4A$. If $n^2+n+A$ is composite for $0\le n\le A-2$, it has a prime divisor $p<A$, andConversely, if $d$ is a square modulo $4p$ for a prime $p<A$, choose an odd square root $2n+1$ with $0\le n<p$. Then $p\mid n^2+n+A$ and $n\le A-2$, so that value is composite.

Finally, reduction theory says that every nonprincipal positive-definite class of discriminant $1-4A$ has a reduced representative whose leading coefficient has a prime divisor $p<A$ for which $d$ is a square modulo $4p$. Conversely such a prime produces a represented form not equivalent to the principal form $x^2+xy+A{y}^2$. Combining this with the preceding equivalence gives

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.

A reduced positive definite binary quadratic form $[a,b,c]$ satisfieswith $b\ge 0$ when either $|b|=a$ or $a=c$. For $d\equiv 0$ or $3(\mathrm{mod}4)$, the class number of a negative discriminant $h(-d)$ is the number of proper equivalence classes of primitive positive definite integral forms of discriminant $-d$, equivalently the number of their reduced representatives.

WritePartitioning the $k$ complete prime-power factors between $r$ and $s$ gives $2^k$ ordered factorizations $d=rs$ with $\gcd (r,s)=1$. After identifying $(r,s)$ with $(s,r)$, there are $2^{k-1}$ choices with $r\le s$. Each gives the primitive reduced formwhose discriminant is $-4rs=-4d$. The uniqueness of reduced representatives makes these classes distinct, proving the coprime-factorization lower bound for a quadratic-form class number

The inequality can be strict. For $d=5$, there is one distinct prime factor, while the reduced primitive forms of discriminant $-20$ areThus $h(-20)=2>1=2^{1-1}$.