Exercise 3.4.10 (Representations of $0$ by a quadratic form)

Proof. Suppose that $f(x_0,y_0)=0$ , where $(x_0,y_0)\ne (0,0)$.

If $y_0\ne 0$, then the equality

shows that $d={\left(\frac{2a{x}_0+b{y}_0}{y_0}\right)}^2$ is the square of a rational $\delta =\frac{p}{q}$, where $p\wedge q=1,q>0$. Then $p^2=d{q}^2$, thus $q\mid p^2$ where $q\wedge p=1$, so $q\mid 1(q>0)$, therefore $q=1$ and $d=p^2$ is a perfect square.

Similarly, if $x_0\ne 0$, we prove that $d$ is a perfect square, using the equality

Conversely, suppose that $d$ is a perfect square. If $f=0$, then $f(1,1)=0$, where $(1,1)\ne 0$. Suppose now $f\ne 0$.

By Problem 9, there are integers $h_1,k_1,h_2,k_2$ such that

Note that $(h_1,k_1)\ne (0,0)$, otherwise $f=0$. Then $f(-k_1,h_1)=0$, and $(-k_1,h_1)\ne (0,0)$.

There exist integers $x_0,y_0$, not both $0$, such that $f(x_0,y_0)=0$, if and only if the discriminant $d$ of $f(x,y)$ is a perfect square □