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The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.

The three reduced primitive binary quadratic forms of discriminant $-23$ are $(1,1,6)$ and $(2,\pm 1,3)$, so the class number is three. The displayed cubic has discriminant $-23$ and is irreducible. Its splitting field has Galois group $S_3$ and is cyclic cubic over $\mathbb{Q}(\sqrt{-23})$. Modulo $23$ it factors as $(X-10{)}^2(X-3)$. Lift the simple factor; the remaining quadratic has discriminant valuation one, so its local inertia group has order two. Its intersection with $A_3$ is trivial. Thus the relative cyclic cubic extension is everywhere unramified and equals the Hilbert class field.

For a negative fundamental discriminant $D$, primitive positive definite integral binary quadratic forms of discriminant $D$, modulo proper equivalence, correspond bijectively to the ideal class group of $\mathbb{Q}(\sqrt{D})$. The displayed ideal gives one conventional choice of this correspondence. Reduced positive definite binary quadratic forms give unique class representatives with the usual boundary sign convention, making class-number computations finite.

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

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 $3x^2+2xy+3y^2=2(x^2+y^2)+(x+y{)}^2$.

The discriminant of a binary quadratic form is $d=b^2-4ac$; it is unchanged by a unimodular matrix change of integral variables and satisfies $d\equiv 0$ or $1(\mathrm{mod}4)$. For such a $d$, the discriminant criterion for prime representation by a binary quadratic form says that some integral binary quadratic form of discriminant of a binary quadratic form $d$ represents an odd prime $p$ exactly when $d$ is a square modulo $p$, including zero. Equivalently,These versions agree: a square root modulo $p$ can be chosen even when $d\equiv 0(\mathrm{mod}4)$ and odd when $d\equiv 1(\mathrm{mod}4)$, by adding $p$ if needed, and then the Chinese remainder theorem applies.

For completeness, if $f(x,y)=p$, then $(x,y)$ is primitive, since any common divisor would have its square divide $p$. Complete $(x,y)$ to a basis of ${\mathbb{Z}}^2$ with determinant one using Bézout's identity. The transformed binary quadratic form has leading coefficient $p$ and discriminant of a binary quadratic form $B^2-4pC=d$. Conversely, a suitable $B$ makes $[p,B,(B^2-d)/(4p)]$ an integral binary quadratic form representing $p$ at $(1,0)$.

For $x^2+3y^2$, the discriminant of a binary quadratic form is $-12$. The prime $3$ is represented. If $p\ne 3$ is represented, reducing modulo $3$ gives $p\equiv x^2\equiv 1(\mathrm{mod}3)$, which also excludes $p=2$. To prove sufficiency for $p\equiv 1(\mathrm{mod}3)$, quadratic reciprocity givesThe criterion constructs a positive definite binary quadratic form $[p,B,C]$ of discriminant of a binary quadratic form $-12$. It is primitive: a common divisor of its coefficients would divide $p$, and if that divisor were $p$, then $p^2$ would divide $12$, impossible for $p\ne 2,3$.

We must still establish that this form is equivalent to $x^2+3y^2$, rather than merely some form of the same discriminant of a binary quadratic form. In a positive definite integral binary quadratic form, replace $x$ by $x+ky$ to arrange $|b|\le a$. If $c<a$, apply $(x,y)\mapsto (-y,x)$ and repeat. Each such exchange strictly decreases the positive integer leading coefficient, so this reduction algorithm for a positive definite binary quadratic form terminates with $|b|\le a\le c$. Then $12=4ac-b^2\ge 3a^2$, so $a\le 2$. With $a=1$, parity forces $b=0$, giving $c=3$. With $a=2$, the only integral choices are $b=\pm 2,c=2$, and these are not primitive. Thus every primitive positive definite binary quadratic form of discriminant of a binary quadratic form $-12$ is equivalent to $[1,0,3]$, and preserves its represented integers. Therefore

Write a reduced positive definite binary quadratic form as $[a,b,c]=a{x}^2+bxy+c{y}^2$, with $|b|\le a\le c$ and $b\ge 0$ on the reduction boundaries. Since $4ac-b^2=20$, reduction gives $3a^2\le 20$, so $a=1$ or two. The middle coefficient is even. For $a=1$, only $b=0$ is possible, yielding $c=5$. For $a=2$, $b=0$ would give a noninteger $c$, while $b=\pm 2$ gives $c=3$; the boundary convention selects $b=2$. Thus the two reduced forms are

A positive odd integer $n$ is properly represented by some integer form of discriminant $d$ exactly when $b^2\equiv d(\mathrm{mod}4n)$ has a solution. For a specified proper equivalence class, the more precise criterion is that $[n,b,(b^2-d)/(4n)]$ belong to that class for some such $b$. Indeed, a primitive representing vector extends to an $S{L}_2(\mathbb{Z})$ basis change, making the leading coefficient $n$; conversely that form represents $n$ at the primitive vector $(1,0)$.

For primes $p\nmid 10$, the existence criterion is $(-20/p)=1$, because the congruence modulo four is compatible with an even middle coefficient. Quadratic reciprocity gives $(-20/p)=(-1/p)(p/5)$, so the split possibilities are $p\equiv 1,3,7,9(\mathrm{mod}20)$. Any proper representing form reduces to one of the two listed forms. Odd values of $x^2+5y^2$ are one modulo four; odd values of $2x^2+2xy+3y^2$ require odd $y$ and are three modulo four. These distinguish the two classes. ConsequentlyA representation of a prime is automatically proper: a common divisor of $x,y$ would have its square divide $p$.

Use the usual positive-definite convention for binary quadratic forms with negative discriminant. Reduction by integral changes of variables of determinant one produces a form $a{x}^2+bxy+c{y}^2$ with $|b|\le a\le c$, and with $b\ge 0$ on the boundary. For completeness, replacing $x$ by $x+ny$ makes $|b|\le a$; if then $c<a$, interchange the variables, reversing one sign to keep determinant one. The new positive leading coefficient is smaller. Repeating terminates and gives a reduced form.

For discriminant $-7$, $4ac-b^2=7$ and the reduction inequalities give $3a^2\le 7$. Thus $a=1$, $b$ is odd and $|b|\le 1$, and $c=2$. The boundary convention chooses $b=1$. Hence there is exactly one positive-definite equivalence class, represented by $x^2+xy+2y^2$.

For an odd prime $p\ne 7$ represented by this form, the identity $4p=(2x+y{)}^2+7y^2$ shows that $-7$ is a square modulo $p$. Here $p\nmid y$, because otherwise $p\mid x$ and $p^2$ would divide the represented number. Conversely, if $-7$ is a square modulo $p$, choose odd $b$ with $b^2\equiv -7(\mathrm{mod}p)$. Then $(p,b,(b^2+7)/(4p))$ is an integral positive-definite form of discriminant $-7$. By the preceding reduction it is equivalent to the given form, so its value $p$ at $(1,0)$ is represented by that form. Quadratic reciprocity gives $(-7/p)=(p/7)$, whose value is one precisely for $p\equiv 1,2,4(\mathrm{mod}7)$. The exceptional primes are represented: $2=Q(0,1)$ and $7=Q(1,-2)$. ThereforeWithout the positive-definite convention, the first printed assertion is false: $-x^2-xy-2y^2$ also has discriminant $-7$ but cannot be equivalent to a positive-definite form, because an invertible change of variables preserves its sign.

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.

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.

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.

Let $K=\mathbb{Q}(\sqrt{-31})$. Its ring of integers of a quadratic field is $\mathbb{Z}[(1+\sqrt{-31})/2]$, and its fundamental field discriminant is $-31$.

First compute the class number. The ideal-form correspondence for imaginary quadratic fields identifies the ideal classes with proper equivalence classes of primitive positive definite binary quadratic forms of discriminant $-31$. Each has a unique reduced positive definite binary quadratic form representative under the convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\ge 3a^2$ we get $1\le a\le \sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactlyFor $a=3$, neither odd residue choice $b=\pm 1,\pm 3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, soThis computes the ideal class group of Q of square root minus thirty-one.

Now take a root $\theta$ of $F(T)=T^3+T-1$, and let $H$ be its splitting field over $\mathbb{Q}$. There is no rational root: the only candidates are $\pm 1$, and neither vanishes. Thus the cubic is irreducible. Its polynomial discriminant isSince this is nonsquare, the Galois group of an irreducible cubic is $S_3$. Its quadratic subfield is $K$: the product of pairwise root differences squares to $-31$ and changes sign under odd permutations. Thus $H/K$ is cyclic of degree three, and $H=K(\theta )$. Indeed the cubic field and $K$ have coprime degrees, so their compositum already has the full splitting-field degree six. The squarefree polynomial discriminant also shows, by the discriminant-index formula for an integral lattice, that $\mathbb{Z}[\theta ]$ is the full integer ring of the cubic field.

It remains to prove that $H/K$ is unramified everywhere. At any rational prime $\ell \ne 31$, the reduction of $F$ is separable. Over a finite residue extension in which it splits, all residue roots are simple and lift by Hensel lemma to the corresponding unramified extension of ${\mathbb{Q}}_{\ell }$. Therefore the local splitting field is unramified over ${\mathbb{Q}}_{\ell }$, and its extension over a completion of $K$ is unramified as well.

At the only possible ramified prime, direct reduction givesThe simple root lifts to $r\in {\mathbb{Q}}_{31}$. Write $F(T)=(T-r)Q(T)$, with $Q$ monic quadratic over ${\mathbb{Q}}_{31}$. The discriminant product formula givesHere $Q(r)=F^{\prime }(r)$ is a unit, so the quadratic discriminant has square class $-31$. Hence the entire local splitting field isIt is quadratic over ${\mathbb{Q}}_{31}$, but its extension over the completion of $K$ is trivial. Thus the prime over $31$ splits completely in $H/K$; in particular it is not ramified there. This is the local splitting-field argument behind an unramified cyclic cubic extension from a prime-discriminant cubic. There are no real places of the imaginary quadratic field $K$, so no infinite-place ramification remains to check.

The Hilbert class field theorem identifies the maximal everywhere unramified abelian extension with an extension of degree $h_K$. Our cyclic unramified extension already has that degree, so it must be the whole Hilbert class field. ThereforeEquivalently, it is the splitting field of $T^3+T-1$, of degree six over $\mathbb{Q}$ and degree three over $K$. Replacing $\theta$ by ${\theta }^{-1}$ gives the alternative defining cubic $T^3-T^2-1$. This is the Hilbert class field of Q of square root minus thirty-one.

Write an integral binary quadratic form as $Q(u,v)=A{u}^2+Buv+C{v}^2$, with discriminant of a binary quadratic form $d=B^2-4AC$. If $Q(u,v)=p$ is prime, then $\gcd (u,v)=1$: any common divisor would have its square dividing $p$. Complete $(u,v)$ to a matrix in ${\mathrm{SL}}_2(\mathbb{Z})$ using Bezout identity. After the associated integral change of variables, the properly equivalent form has leading coefficient $p$, say $[p,b,c]$. Its discriminant of a binary quadratic form gives $d=b^2-4pc$, hence $b^2\equiv d(\mathrm{mod}p)$.

Conversely suppose $b_0^2\equiv d(\mathrm{mod}p)$. Since $d$ is an integral-form discriminant of a binary quadratic form, $d\equiv 0$ or $1(\mathrm{mod}4)$. Choose $b\equiv b_0(\mathrm{mod}p)$ with even integer parity for $d\equiv 0(\mathrm{mod}4)$ and odd integer parity for $d\equiv 1(\mathrm{mod}4)$. This is possible because $p$ is odd. Then $b^2-d$ is divisible by $4p$, andis an integral form of discriminant of a binary quadratic form $d$ representing $p=Q(1,0)$. This proves the discriminant criterion for prime representation by a binary quadratic form, including the case $p\mid d$; no assumption that $d$ is a fundamental discriminant was used.

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

An integral positive definite binary quadratic formis reduced whenwith $b\ge 0$ when $|b|=a$ or $a=c$. Its discriminant is $D=b^2-4ac$.

For $D=-20$, reduction gives $3a^2\le 20$, so $a=1$ or $2$. Since $b^2\equiv D(\mathrm{mod}4)$, $b$ is even. If $a=1$, then $|b|\le 1$ forces $b=0$, and $c=5$. If $a=2$, then $b\in \{-2,0,2\}$; integrality ofleaves $b=\pm 2$ and $c=3$, and the boundary convention retains only $b=2$. Hence the complete list is

Now suppose the prime $p\ne 5$ is represented by $x^2+5y^2$. If $p=2$, the equation is impossible modulo four, so $p$ is odd. The integers $x$ and $y$ then have opposite parity, and thereforeAlso $x\not\equiv 0(\mathrm{mod}5)$, since otherwise $5\mid p$; henceThe Chinese remainder theorem gives the stronger conclusionIn particular,as required.

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.

Let $I$ be an integral ideal of an imaginary quadratic number field and $(e_1,e_2)$ an integral basis of $I$. Its normalized norm is a positive definite binary quadratic form. An unramified rational prime $p$ is represented by this form exactly when there is an integral prime ideal $P$ of norm $p$ with $[P]=[I{]}^{-1}$. Indeed, a representation gives an element $\alpha \in I$ with $N\alpha =pN(I)$, so $P=(\alpha )I^{-1}$ is integral of norm $p$. Conversely, $IP=(\alpha )$ with $\alpha \in I$ produces a representation by expressing $\alpha$ in the basis. Artin reciprocity then detects this ideal class by Frobenius in the Hilbert class field.

For odd $n$ prime to 35, proper representation by a positive definite binary quadratic form of discriminant $-35$ is equivalent to $-35$ being a square modulo every prime divisor of $n$. A primitive representing vector extends to a determinant-one integer matrix, producing a form with leading coefficient $n$ and middle coefficient $b$ satisfying $b^2\equiv -35(\mathrm{mod}4n)$. Conversely this congruence constructs such a form. Its local roots lift to prime powers and combine by the Chinese remainder theorem. Every class reduces to one of $x^2+xy+9y^2$ and $3x^2+xy+3y^2$. The condition is local at each prime divisor; a product Jacobi symbol equal to one alone is insufficient.