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For negative discriminant $-67$, a reduced positive definite binary quadratic form satisfies $1\le a\le \sqrt{67/3}<5$ and $b$ odd with $|b|\le a$. Checking $4a\mid b^2+67$ leaves only $(a,b,c)=(1,1,17)$ under the boundary convention. The reduction algorithm for a positive definite binary quadratic form therefore puts every form in this one proper equivalence of binary quadratic forms class. Its unimodular shears give infinitely many distinct forms $x^2+(2n+1)xy+(n^2+n+17)y^2$.

Write an integral binary quadratic form as $[a,b,c]$. For proper equivalence of binary quadratic forms, the allowed changes of variables have determinant one. Here is the needed reduction rather than an appeal to class-number tables. A shear $(x,y)\mapsto (x+ny,y)$ changes $b$ to $b+2an$ without changing $a$ or the discriminant of a binary quadratic form. Choose $n$ so $-a<b\le a$. If the resulting $c<a$, the determinant-one change $(x,y)\mapsto (-y,x)$ replaces $[a,b,c]$ by $[c,-b,a]$, strictly decreasing the positive integer leading coefficient. Repeating must terminate with $|b|\le a\le c$. If $a=c$ and $b<0$, the same rotation makes $b$ positive; the choice $-a<b\le a$ already fixes the other boundary sign. This is a reduced positive definite binary quadratic form.

Since $4ac-b^2=67$, the reduction inequalities give $67\ge 3a^2$, so $a\in \{1,2,3,4\}$. The integer $b$ is odd. For $a=1$, only $b=1$ survives the boundary convention and $c=17$. For $a=2$, $b=\pm 1$ gives $c=68/8$, not an integer. For $a=3$, $b=\pm 1,\pm 3$ gives $68/12$ or $76/12$; for $a=4$, $b=\pm 1,\pm 3$ gives $68/16$ or $76/16$. None is integral. Therefore every form is properly equivalent to $x^2+xy+17y^2$, establishing class number one for discriminant minus sixty-seven.

For infinitude, its determinant-one shears produceEach has discriminant $-67$ and is positive definite because it is a change of variables of the displayed positive form. Their middle coefficients are distinct, so they are infinitely many distinct elements of the set.

Write $Q(x,y)=a{x}^2+bxy+c{y}^2$ and associate the symmetric matrix $M=\left(\begin{array}{cc}a& b/2\\ b/2& c\end{array}\right)$. Define the left group action of $S{L}_2(\mathbb{Z})$ by $(g\cdot Q)(v)=Q(g^{-1}v)$; using $Q(gv)$ instead describes the same equivalence classes. The transformed matrix is $g^{-T}M{g}^{-1}$, so its determinant is unchanged. Hence the discriminant of a binary quadratic formis invariant under proper equivalence of binary quadratic forms.

A proper representation here means $Q(x,y)=n$ with $\gcd (x,y)=1$. Such a vector extends to the first column of an integer determinant-one matrix by Bezout identity. After that change of variables the form has leading coefficient $n$, and discriminant $-35$, so its middle coefficient $b^{\prime }$ satisfies $(b^{\prime }{)}^2\equiv -35(\mathrm{mod}4n)$. Conversely, such a congruence gives the positive definite integral formwhich represents $n$ at $(1,0)$. It is primitive: a common divisor of all three coefficients would have its square dividing the squarefree discriminant 35.

For clarity, enumerate all the proper equivalence classes at this discriminant. Substitutions $(x,y)\mapsto (x+my,y)$ first arrange $|b|\le a$; if $c<a$, the determinant-one substitution $(x,y)\mapsto (-y,x)$ makes the leading coefficient smaller. Repeating terminates because this coefficient is a positive integer. Thus every class has a reduced representative with $|b|\le a\le c$. From $4ac-b^2=35$ it follows that $3a^2\le 35$, giving $a\le 3$. Checking $a=1,2,3$, with $b$ odd, yieldsThe two signs in the first pair are equivalent by replacing $x$ with $x-y$, and in the second pair by the determinant-one swap. Thus the two forms in the question exhaust the classes.

The primitive representation criterion at discriminant minus thirty-five now follows. For odd $n$ prime to 35, the congruence modulo $4n$ is solvable exactly when $-35$ is a square modulo every prime divisor of $n$. Necessity is immediate. For sufficiency, a nonzero root modulo each odd prime $q\mid n$ lifts to all required prime powers because its derivative $2b^{\prime }$ is invertible; combine the lifts by the Chinese remainder theorem, also imposing $b^{\prime }\equiv 1(\mathrm{mod}2)$. This gives the modulus-four condition since $-35\equiv 1(\mathrm{mod}4)$. We concludeBy quadratic reciprocity, the symbol equals $(q/5)(q/7)$. In particular each prime divisor must be a quadratic residue at both 5 and 7, or a nonresidue at both. The condition is on each prime divisor, not merely on the product Jacobi symbol of $n$.

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

A reduced positive definite binary quadratic form $a{x}^2+bxy+c{y}^2$ satisfies $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. These inequalities select representatives under proper equivalence of binary quadratic forms; the changes of variables have integer entries and determinant $1$.

For the first binary quadratic form, put $x=u-v$, $y=v$. For the second, put $x=-u-3v$, $y=u+2v$. Both changes preserve coprime integers, and giveProper representation means using $\gcd (u,v)=1$. For $f$, $|v|\ge 2$ gives a value greater than $5$, while $v=0$ forces $u=\pm 1$. For $|v|=1$, the smallest values are $4$ and $5$. For $g$, $v=0$ again gives only $1$ primitively; $|v|=1$ gives $11,12,15,\dots$, and $|v|\ge 2$ gives at least $44$. Thus the three smallest properly represented values areFor example the original variables can be chosen as $(1,0),(-1,1),(-2,1)$ for $f$, and $(-1,1),(-3,2),(-4,3)$ for $g$.

Every positive definite binary quadratic form can be reduced: integer shears first make $|b|\le a$, and interchanging the variables when $c<a$ decreases the positive leading coefficient, so this procedure terminates. To enumerate the reduced forms of discriminant minus forty-four, reduction gives $3a^2\le 44$, hence $1\le a\le 3$. Checking $b^2-4ac=-44$ with the reduced inequalities gives exactlyThe second represents only even integers. The last two have the same represented values, by $u\mapsto -u$. The first and third are the reduced versions of $g$ and $f$. Therefore every represented odd integer occurs in at least one of $f,g$, whether or not its original representation was primitive.

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

To prove reduction, choose in a proper equivalence class a form whose positive leading coefficient $a$ is minimal. Replacing $x$ by $x+ny$ changes $b$ by $2an$, so choose $n$ to arrange $|b|\le a$. If the resulting $c<a$, the determinant-one substitution $(x,y)\mapsto (-y,x)$ would put $c$ in the leading position, contradicting minimality. Thus $a\le c$. Sign changes on the boundary give the stated convention. Hence every positive definite form is properly equivalent to a reduced one.

A unimodular integral substitution is a bijection of ${\mathbb{Z}}^2$, so properly equivalent forms represent exactly the same integers.

Both given forms have discriminant $-23$. Reduction givesThe two reduced forms represent the same integers becausean integral change of variables of determinant $-1$. They are not properly equivalent: the uniqueness theorem for reduced positive definite forms says that each proper class has one reduced representative, and $[2,1,3]\ne [2,-1,3]$. Therefore the original forms represent the same integers but are not equivalent in the required proper sense.

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.