Discriminant criterion for prime representation by a binary quadratic form ($p\text{ represented}⟺d\text{ is a square modulo }p$, $p\text{ represented by some form of discriminant }D⟺D\text{ is a square mod }p$)

Let $p$ be an odd prime and let $D$ be an integer congruent to zero or one modulo four. Some integral binary quadratic form of discriminant of a binary quadratic form $D$ represents $p$ if and only if $D$ is a square modulo $p$.

For necessity, a representing vector $(x,y)$ must have $\gcd (x,y)=1$, since a common divisor greater than one would make its square divide $p$. Bézout's identity completes this vector to the first column of a determinant-one integer matrix. Under that proper equivalence, the leading polynomial coefficient becomes $p$, so the transformed form is $[p,b,c]$. Its discriminant gives $D=b^2-4pc\equiv b^2(\mathrm{mod}p)$.

Conversely, choose a root $b$ modulo $p$ and adjust its parity so $b^2\equiv D(\mathrm{mod}4)$. Adding $p$ changes parity because $p$ is odd. Thenhas integer polynomial coefficients, discriminant $D$ and value $p$ at $(1,0)$. If $D<0$, it is positive definite. If $p\nmid D$, it is also a primitive binary quadratic form: a common divisor of its coefficients would have to be $p$, contradicting $p\nmid D$.

The conclusion concerns some form, not every form of that discriminant. For $D=-20,p=3$, the construction gives $[3,2,2]$, which represents three. The same-discriminant form $[1,0,5]$ does not represent three. Deciding the particular form requires the proper equivalence classes, not only the local square test. When $p\mid D$, imprimitive forms can occur, so primitiveness must be checked separately.