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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

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 $Q(u,v)=A{u}^2+Buv+C{v}^2$, 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.

Necessity follows by reducing modulo three. For a prime $p\equiv 1(\mathrm{mod}3)$, quadratic reciprocity gives $(-3/p)=1$, so the discriminant criterion for prime representation by a binary quadratic form constructs a primitive positive definite form of discriminant of a binary quadratic form $-12$ representing $p$. Reduction gives $|b|\le a\le c$ and $a\le 2$. The only primitive reduced form is $[1,0,3]$: the alternatives $[2,\pm 2,2]$ are not primitive. Integral equivalence preserves represented integers, establishing sufficiency.

The reduced-form bound $3a^2\le 32$ leaves $a=1,2,3$. Checking $c=(b^2+32)/(4a)$ and the boundary sign convention gives precisely the displayed forms. The middle one is imprimitive and represents the prime two. Odd represented primes are $1$ or $3$ modulo eight by the discriminant criterion for prime representation by a binary quadratic form. The first form represents only $1$ modulo eight among odd integers, and the last only $3$, so a prime is represented by $3x^2+2xy+3y^2$ exactly when it is $3$ modulo eight.