Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 11H (Number Theory)

Write a reduced positive definite binary quadratic form as $[a,b,c]=a{x}^2+bxy+c{y}^2$, with $|b|\le a\le c$ and $b\ge 0$ on the reduction boundaries. Since $4ac-b^2=20$, reduction gives $3a^2\le 20$, so $a=1$ or two. The middle coefficient is even. For $a=1$, only $b=0$ is possible, yielding $c=5$. For $a=2$, $b=0$ would give a noninteger $c$, while $b=\pm 2$ gives $c=3$; the boundary convention selects $b=2$. Thus the two reduced forms are

A positive odd integer $n$ is properly represented by some integer form of discriminant $d$ exactly when $b^2\equiv d(\mathrm{mod}4n)$ has a solution. For a specified proper equivalence class, the more precise criterion is that $[n,b,(b^2-d)/(4n)]$ belong to that class for some such $b$. Indeed, a primitive representing vector extends to an $S{L}_2(\mathbb{Z})$ basis change, making the leading coefficient $n$; conversely that form represents $n$ at the primitive vector $(1,0)$.

For primes $p\nmid 10$, the existence criterion is $(-20/p)=1$, because the congruence modulo four is compatible with an even middle coefficient. Quadratic reciprocity gives $(-20/p)=(-1/p)(p/5)$, so the split possibilities are $p\equiv 1,3,7,9(\mathrm{mod}20)$. Any proper representing form reduces to one of the two listed forms. Odd values of $x^2+5y^2$ are one modulo four; odd values of $2x^2+2xy+3y^2$ require odd $y$ and are three modulo four. These distinguish the two classes. ConsequentlyA representation of a prime is automatically proper: a common divisor of $x,y$ would have its square divide $p$.