Exercise 3.5.4 (New proof of Theorem 3.13)

Proof.

(a)

Suppose first that there is a form equivalent to $f$ in which the coefficient of $x^2$ is $n$, so that $f(x,y)=n{x}^2+\mathit{bxy}+c{y}^2$. Then $n=f(1,0)$, where $1\wedge 0=1$, so $f$ properly represents $n$.

Conversely let $f$ be the quadratic form $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$ which properly represents $n$. By definition there are integers $x_0,y_0$ such that $n=f(x_0,y_0)=a{x}_0^2+b{x}_0{y}_0+c{y}_0^2$ and $x_0\wedge y_0=1$. By Problem 3, there exist integers $u,v$ such that $A=\left(\begin{array}{cc}\hfill x_0\hfill & \hfill u\hfill \\ \hfill y_0\hfill & \hfill v\hfill \end{array}\right)\in \Gamma$ (since $A\in \Gamma ⟺A^t\in \Gamma$).

Put $g=f\cdot A$. Then $g\stackrel{+}{\sim }f$, and

$$\begin{array}{llll}\hfill g(x,y)& =f(x_{0}x+\mathit{uy},y_{0}x+\mathit{vy})& \hfill & \\ \hfill & =a{(x_{0}x+\mathit{uy})}^2+b(x_{0}x+\mathit{uy})(y_{0}x+\mathit{vy})+c{(y_{0}x+\mathit{vy})}^2& \hfill & \\ \hfill & =(a{x}_0^2+b{x}_0{y}_0+c{y}_0^2)x^2+\ast \mathit{xy}+\ast y^2& \hfill & \\ \hfill & =n{x}^2+\ast \mathit{xy}+\ast y^2,& \hfill & \end{array}$$

where $\text{∗}$ is in place of some coefficients whose value has no interest here. Thus $g=(n,\ast ,\ast )$ is (properly) equivalent to $f$: there is a form $g$ equivalent to $f$ in which the coefficient of $x^2$ is $n$.

A binary quadratic form $f$ properly represents an integer $n$ if and only if there is a form properly equivalent to $f$ in which the coefficient of $x^2$ is $n$.

(b)

We give a new proof of Theorem 3.13.

Theorem 3.13 (Louis Lagrange) Let $n$and $d$be given integers with $n\ne 0$. There exists a binary quadratic form of discriminant $d$that represents $n$properly if and only if the congruence $x^2\equiv d(\mathit{mod}4|n|)$has a solution.

The direct part is the same:

Suppose that $b$ is a solution of the congruence $x^2\equiv d(\mathit{mod}4|n|)$, with $b^2-d=4\mathit{nc}$, say. Then the form $f(x,y)=n{x}^2+\mathit{bxy}+c{y}^2$ has integral coefficients and discriminant $d$. Moreover, $f(1,0)=n$ is a proper representation of $n$.

Now we prove the converse, using part (a).

Suppose that there exists a binary quadratic form $f$ of discriminant $d$ that represents $n$ properly. By part (a), there is a quadratic form $g(x,y)=n{x}^2+\mathit{uxy}+v{y}^2$, properly equivalent to $f$, in which the coefficient of $x^2$ is $n$. Since $f$ and $g$ have same discriminant $d$, $d=u^2-4\mathit{nv}$, thus $d\equiv u^2(\mathit{mod}4|n|)$, and so the congruence $x^2\equiv d(\mathit{mod}4|n|)$ has a solution.