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For an integral binary quadratic form $[a,b,c]=a{x}^2+bxy+c{y}^2$ with $a>0$ and $D=b^2-4ac<0$, the reduced positive definite binary quadratic form convention is $|b|\le a\le c$, with $b\ge 0$ on either boundary $|b|=a$ or $a=c$. Since $|D|=4ac-b^2\ge 3a^2$, the enumeration of reduced binary quadratic forms for $D=-23$ has $a\le \sqrt{23/3}<3$. The integer $b$ is odd. For $a=1$, only $b=1$ is retained and $c=6$; for $a=2$, both $b=\pm 1$ give $c=3$. Thus the complete list isA proper representation of a prime uses a primitive representing vector. Completing that vector to a matrix in ${\mathrm{SL}}_2(\mathbb{Z})$ transforms a representing form to $[p,b,c]$, so its discriminant of a binary quadratic form gives $b^2\equiv -23(\mathrm{mod}p)$. Conversely, choose an odd $b$ satisfying this congruence and put $c=(b^2+23)/(4p)$. This is integral, and $[p,b,c]$ is positive definite and represents $p$ at $(1,0)$. This proves the discriminant criterion for prime representation by a binary quadratic form here.

For $p\ne 23$, quadratic reciprocity gives $(-23/p)=(p/23)$. Including the ramified prime, the answer is thereforeFor example $[23,23,6]$ supplies a proper representation of $23$. Reduction transfers every such representation to one of the three displayed forms.

Use the reduction algorithm for a positive definite binary quadratic form. A determinant-one shear $(x,y)\mapsto (x+ky,y)$ puts the middle coefficient in $[-a,a]$. If the resulting last coefficient is smaller than $a$, the determinant-one substitution $(x,y)\mapsto (-y,x)$ decreases the positive integer leading coefficient. Repeating terminates with a reduced positive definite binary quadratic form satisfying $|b|\le a\le c$. These substitutions preserve proper equivalence of binary quadratic forms and the discriminant of a binary quadratic form.

The enumeration of reduced binary quadratic forms givesSince $b^2\equiv -163\equiv 1(\mathrm{mod}4)$, $b$ is odd. For $|b|=1,3,5,7$, the possible values of $(b^2+163)/4$ are respectively $41,43,47,53$. Each is a prime number greater than seven. But $ac=(b^2+163)/4$, so its positive divisor $a\le 7$ must be one. Then $|b|\le a$ forces $b=\pm 1$ and $c=41$. The two signs are properly equivalent: substituting $x\mapsto x-y$ in $x^2+xy+41y^2$ gives $x^2-xy+41y^2$. Thus every form in question is properly equivalent to

Write an integral binary quadratic form as $[a,b,c]=a{x}^2+bxy+c{y}^2$, with discriminant of a binary quadratic form $D=b^2-4ac$. It is positive definite when $a>0$ and $D<0$. Two forms are equivalent if one is obtained from the other by an integral linear change of variables whose matrix lies in ${\mathrm{SL}}_2(\mathbb{Z})$. A positive definite form is reduced whenwith $b\ge 0$ when $|b|=a$ or $a=c$.

To reduce a form, apply a unimodular shear $x\mapsto x+ny$ with $n$ chosen so that the new middle coefficient has $|b^{\prime }|\le a$. If the new final coefficient $c^{\prime }$ is smaller than $a$, exchange the variables, using a determinant-one rotation, so that the positive leading coefficient decreases from $a$ to $c^{\prime }$. Repeating must stop because a positive integer cannot decrease indefinitely. At termination $|b|\le a\le c$, and elementary sign changes impose the boundary convention. This is the reduction algorithm for a positive definite binary quadratic form, and proves that every positive definite form is equivalent to a reduced one.

The displayed forms have discriminant $1-36=-35$, so the intended discriminant is $-35$. For any reduced form of this discriminant, enumeration of reduced binary quadratic forms givesThe boundary convention and direct checking of $a=1,2,3$ leave only

Suppose first that a prime $p>7$ is represented by $f_1$. The identityshows modulo both $5$ and $7$ that $4p$ is a nonzero square. HenceSimilarly,Modulo $5$, this says $2p$ is a square, while modulo $7$ it says $5p$ is a square. Since $(2/5)=(5/7)=-1$, the multiplicativity of the Legendre symbol gives

Conversely, suppose the two displayed Legendre symbols have the same sign. By quadratic reciprocity,Choose an odd integer $B$ satisfying $B^2\equiv -35(\mathrm{mod}p)$. Since $B$ is odd, $B^2+35$ is also divisible by $4$, sois a positive definite integral form of discriminant $-35$ that represents $p$. Reducing $F$ yields either $f_1$ or $f_2$, and equivalence preserves represented integers. The necessary symbol calculation above determines which one: the positive pair selects $f_1$, and the negative pair selects $f_2$. Therefore

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.