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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.

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.