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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 $x^2+(2n+1)xy+(n^2+n+17)y^2$.

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

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

Both forms have discriminant of a binary quadratic form $-60$. The reduction algorithm for a positive definite binary quadratic form gives $|b|\le a\le c$ and $a\le \sqrt{60/3}<5$ for a reduced positive definite representative. Enumerating $a=1,2,3,4$ givesThe second and fourth forms are imprimitive and represent only even integers. Every odd prime represented by a form of discriminant of a binary quadratic form $-60$ must therefore be represented by $f$ or $g$.

For a prime $p\nmid 30$, quadratic reciprocity givesEvery prime $p\equiv 1(\mathrm{mod}15)$ has this symbol equal to one and hence is represented by one of the two primitive forms, by part (b). But an odd prime represented by $g$ and different from three is $2(\mathrm{mod}3)$, so these primes must be represented by $f$. Similarly, primes $p\equiv 2(\mathrm{mod}15)$ have both symbols equal to $-1$, and are represented by $g$, since a prime represented by $f$ is $1(\mathrm{mod}3)$. Both residue classes are coprime to 15, so Dirichlet theorem on primes in arithmetic progressions proves each form represents infinitely many primes.

The same modulo-three distinction excludes every prime other than three from being represented by both. The form $f$ does not represent three; $g$ represents three and five but neither is represented by $f$, and two is represented by neither. Therefore there are no primes represented by both forms.

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