Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 11I / Solution

Two integral binary quadratic forms are properly equivalent when one is obtained from the other by a change of variables in $S{L}_2(\mathbb{Z})$. For a negative discriminant $d\equiv 0,1(\mathrm{mod}4)$, the class number of binary quadratic forms $h(d)$ is the number of proper equivalence classes of primitive positive-definite forms of discriminant $d$.

If $f$ properly represents $m$, then $f(r,s)=m$ for coprime $r,s$. Extend $(r,s{)}^T$ to a matrix in $S{L}_2(\mathbb{Z})$; after the associated variable change, the coefficient of $x^2$ is $m$, so the transformed form isThe converse follows by evaluating this form at $(1,0)$.

Such a form has discriminant $d$ precisely whenso it exists exactly when $b^2\equiv d(\mathrm{mod}4m)$. Thus $m$ is properly represented by some form of discriminant $d$ exactly when $d$ is a square modulo $4m$.

Now let $d=1-4A$. If $n^2+n+A$ is composite for $0\le n\le A-2$, it has a prime divisor $p<A$, andConversely, if $d$ is a square modulo $4p$ for a prime $p<A$, choose an odd square root $2n+1$ with $0\le n<p$. Then $p\mid n^2+n+A$ and $n\le A-2$, so that value is composite.

Finally, reduction theory says that every nonprincipal positive-definite class of discriminant $1-4A$ has a reduced representative whose leading coefficient has a prime divisor $p<A$ for which $d$ is a square modulo $4p$. Conversely such a prime produces a represented form not equivalent to the principal form $x^2+xy+A{y}^2$. Combining this with the preceding equivalence gives

Solved by gpt-5.6-sol high.