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For a negative discriminant of a binary quadratic form $D\equiv 0$ or $1(\mathrm{mod}4)$, the class number of a negative discriminant $h(D)$ counts proper equivalence classes of primitive binary quadratic forms that are positive definite and have discriminant $D$.

Uniqueness of the reduced positive definite binary quadratic form in each class makes this the number of reduced primitive forms. Enumeration of reduced binary quadratic forms proves finiteness and gives an explicit procedure. For example, $h(-4)=1$, represented by $[1,0,1]$, while $h(-39)=4$, with the four forms listed under reduced positive definite binary quadratic forms of discriminant minus thirty-nine.

Imprimitive forms do not enter $h(D)$. At $D=-32$, the reduced form $[2,0,4]$ has content two, so the three reduced forms listed under reduced forms of discriminant minus thirty-two contribute only two primitive classes. If $D$ is a fundamental discriminant, this form class number agrees with the class number of the imaginary quadratic field of discriminant $D$. For nonfundamental $D$, the associated order in a number field need not be the full ring of integers, so that distinction must be retained.

Use the standard reduced-form convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. To reduce a positive definite integral binary quadratic form, choose a primitive integer vector at which the positive integer value is least. Complete that vector to a unimodular basis; it makes the leading coefficient $a$ that least value. A shear of the second basis vector by an integer multiple of the first replaces $b$ by $b+2ka$, allowing $|b|\le a$. The new $c$ is the value at another primitive vector, hence $c\ge a$. If $b=-a$, a shear makes $b=a$ without changing $a,c$; if $a=c$ and $b<0$, the determinant-one swap $(x,y)\mapsto (-y,x)$ reverses its sign. Thus every form is properly equivalent to a reduced positive definite binary quadratic form. This argument also gives the reduction algorithm for a positive definite binary quadratic form by successively decreasing its positive leading coefficient.

For a negative integral discriminant $d\equiv 0$ or $1(\mathrm{mod}4)$, a prime $p$ is represented by some integral form of discriminant $d$ exactly whenIndeed a representing vector is primitive, since a common divisor of its coordinates would have its square divide $p$. A unimodular change takes it to the first basis vector, giving a form $[p,b,c]$ and $d=b^2-4pc$. Conversely a solution $b$ gives the positive definite form $[p,b,(b^2-d)/(4p)]$, which represents $p$ at $(1,0)$. For odd $p$ this is the discriminant criterion for prime representation by a binary quadratic form: $d$ must be a square modulo $p$, with parity chosen to satisfy the modulo-four condition.

For $d=-32$ and odd $p$, the Legendre symbol is $(-32/p)=(-2/p)=(-1/p)(2/p)$. The supplementary laws make it positive exactly for $p\equiv 1,3(\mathrm{mod}8)$. The prime $2$ also works directly, since $[2,0,4]$ has discriminant $-32$ and represents $2$. ThereforeThe class $2(\mathrm{mod}8)$ here consists only of the prime $2$.

For a reduced form, $32=4ac-b^2\ge 3a^2$, so $a\le 3$. The integer checks yield the reduced forms of discriminant minus thirty-two:If one omits the boundary sign convention, $[3,-2,3]$ also appears, but is properly equivalent to $[3,2,3]$. The middle form is imprimitive and represents no odd integer. An odd integer represented by $x^2+8y^2$ has $x$ odd and is $1(\mathrm{mod}8)$. For $3x^2+2xy+3y^2$ to be odd, one of $x,y$ must be odd and the other even; substitution modulo eight gives $3(\mathrm{mod}8)$ in either case. If both are odd or both even, the value is even. Thus a prime represented by the last form is necessarily $3(\mathrm{mod}8)$.

Conversely, a prime $p\equiv 3(\mathrm{mod}8)$ is represented by some discriminant-$-32$ form, which reduces to one of the three displayed forms without changing its represented integers. The first and second cannot represent it, so the third must. ConsequentlyThe prime $2$ is not an exception for this particular form: its minimum positive value is $3$, as also follows from $3x^2+2xy+3y^2=2(x^2+y^2)+(x+y{)}^2$.