Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 9G (Quadratic Mathematics)

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