Exercise 3.7.1 (The representations of $a$ by $f$ are proper)

Proof. Let $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$ be a reduced positive definite form. Suppose that $a=f(u,v)$, where $u,v$ are integers, and put $\delta =u\wedge v$. Then $\delta \ne 0$, otherwise $u=v=0$, and then $a=f(u,v)=0$, but $f$ is a positive definite form, so $a>0$.

Since $a=a{u}^2+\mathit{buv}+c{v}^2$, we obtain ${\delta }^2\mid a$.

Put $u^{\prime }=u∕\delta ,v^{\prime }=v∕\delta$. Then $u^{\prime }\wedge v^{\prime }=1$, and

where $u^{\prime }\wedge v^{\prime }=1$. So the integer $a∕{\delta }^2$ is properly represented by $f$, and $a∕{\delta }^2\le a\le c$,

By Lemma 3.24, $a∕{\delta }^2=a$ or $a∕{\delta }^2=c$. But if $\delta \ne 1$, then $a∕{\delta }^2<a\le c$, so $a∕{\delta }^2\ne a,a∕{\delta }^2\ne c$. This contradiction shows that $\delta =1$.

If $f$ is a reduced positive definite form, all representations of $a$ by $f$ are proper. □