Exercise 3.4.5 (Finite number of representations by a positive definite quadratic form)

Proof.

(a)

If $A=B=0$, then $F=0$, and $\underset{0\le \varphi \le 2\pi }{\max <!--\; FUNCTION\; APPLICATION\; -->}F(\varphi )=\underset{0\le \varphi \le 2\pi }{\min <!--\; FUNCTION\; APPLICATION\; -->}F(\varphi )=0=\sqrt{A^2+B^2}$.

Suppose now that $(A,B)\ne (0,0)$, so that $\sqrt{A^2+B^2}\ne 0$.

Put $\alpha =\frac{A}{\sqrt{A^2+B^2}},\beta =\frac{B}{\sqrt{A^2+B^2}}$. Since ${\alpha }^2+{\beta }^2=1$, there is some real ${\varphi }_0$ such that $\alpha =\cos <!--\; FUNCTION\; APPLICATION\; -->{\varphi }_0,\beta =\sin <!--\; FUNCTION\; APPLICATION\; -->{\varphi }_0$. Then

$$\begin{array}{llll}\hfill F(\varphi )& =A\cos <!--\; FUNCTION\; APPLICATION\; -->\varphi +B\sin <!--\; FUNCTION\; APPLICATION\; -->\varphi & \hfill & \\ \hfill & =\sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\cos <!--\; FUNCTION\; APPLICATION\; -->\varphi +\frac{B}{\sqrt{A^2+B^2}}\sin <!--\; FUNCTION\; APPLICATION\; -->\varphi \right)& \hfill & \\ \hfill & =\sqrt{A^2+B^2}\left(\cos <!--\; FUNCTION\; APPLICATION\; -->{\varphi }_0\cos <!--\; FUNCTION\; APPLICATION\; -->\varphi +\sin <!--\; FUNCTION\; APPLICATION\; -->{\varphi }_0\sin <!--\; FUNCTION\; APPLICATION\; -->\varphi \right)& \hfill & \\ \hfill & =\sqrt{A^2+B^2}\cos <!--\; FUNCTION\; APPLICATION\; -->(\varphi -{\varphi }_0).& \hfill & \end{array}$$

Since $|\cos <!--\; FUNCTION\; APPLICATION\; -->(\varphi -{\varphi }_0)|\le 1$ for all real $\varphi$, $|F(\varphi )|\le \sqrt{A^2+B^2}$, thus

$$-\sqrt{A^2+B^2}\le F(\varphi )\le \sqrt{A^2+B^2}.$$

Moreover, $F({\varphi }_0)=\sqrt{A^2+B^2},F({\varphi }_0+\pi )=-\sqrt{A^2+B^2}$, hence

$$\begin{array}{llll}\hfill \underset{0\le \varphi \le 2\pi }{\max <!--\; FUNCTION\; APPLICATION\; -->}F(\varphi )& =+\sqrt{A^2+B^2},& \hfill & \\ \hfill \underset{0\le \varphi \le 2\pi }{\min <!--\; FUNCTION\; APPLICATION\; -->}F(\varphi )& =-\sqrt{A^2+B^2}.& \hfill & \end{array}$$

(b)

Here $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$. For any ordered pair $(x,y)\in {ℝ}^2$ (even if $x=y=0$), there exist real numbers $r,𝜃$ such that $x=r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,y=r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃$. Then, using the formulas $${\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃=\frac{1+\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃}{2},{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃=\frac{1-\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃}{2},\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃=\frac{1}{2}\sin <!--\; FUNCTION\; APPLICATION\; -->2𝜃,$$

we obtain

$$\begin{array}{llll}\hfill f(x,y)& =f(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)& \hfill & \\ \hfill & =r^2(a{\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃+b\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃+c{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃)& \hfill & \\ \hfill & =\frac{r^2}{2}\left(a(1+\cos <!--\; FUNCTION\; APPLICATION\; -->2𝜃)+b\sin <!--\; FUNCTION\; APPLICATION\; -->2𝜃+c(1-\cos <!--\; FUNCTION\; APPLICATION\; -->2𝜃)\right)& \hfill & \\ \hfill & =\frac{r^2}{2}(a+c+(a-c)\cos <!--\; FUNCTION\; APPLICATION\; -->2𝜃+b\sin <!--\; FUNCTION\; APPLICATION\; -->2𝜃).& \hfill & \end{array}$$

Using part (a),

$$\begin{array}{llll}\hfill \underset{0\le \varphi \le 2\pi }{\max <!--\; FUNCTION\; APPLICATION\; -->}f(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)& =\frac{r^2}{2}(a+c+\sqrt{{(a+c)}^2+b^2}),& \hfill & \\ \hfill \underset{0\le \varphi \le 2\pi }{\min <!--\; FUNCTION\; APPLICATION\; -->}f(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)& =\frac{r^2}{2}(a+c-\sqrt{{(a+c)}^2+b^2}).& \hfill & \end{array}$$

Moreover

$$\begin{array}{llll}\hfill {(a-c)}^2+b^2& =a^2+c^2-2\mathit{ac}+b^2& \hfill & \\ \hfill & =a^2+c^2+2\mathit{ac}+(b^2-4\mathit{ac})& \hfill & \\ \hfill & ={(a+c)}^2+d,& \hfill & \end{array}$$

where $d$ is the discriminant of the form $f$. So

$$\begin{array}{llll}\hfill \underset{0\le \varphi \le 2\pi }{\max <!--\; FUNCTION\; APPLICATION\; -->}f(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)& =\frac{r^2}{2}(a+c+\sqrt{{(a+c)}^2+d}),& \hfill & \\ \hfill \underset{0\le \varphi \le 2\pi }{\min <!--\; FUNCTION\; APPLICATION\; -->}f(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)& =\frac{r^2}{2}(a+c-\sqrt{{(a+c)}^2+d}).& \hfill & \end{array}$$

(Here ${(a+c)}^2+d={(a-c)}^2+b^2\ge 0$.)

(c)

Put $$C_1=\frac{a+c-\sqrt{{(a+c)}^2+d}}{2},C_2=\frac{a+c+\sqrt{{(a+c)}^2+d}}{2}.$$

We must verify that $C_1>0,C_2>0$ for a definite positive quadratic form $f$. In this case $d<0$, and $a>0,c>0$ (Theorem 3.11). Moreover, since $a+c>0$,

$$\begin{array}{llll}\hfill C_1>0& ⟺a+c<\sqrt{{(a+c)}^2+d}& \hfill & \\ \hfill & ⟺{(a+c)}^2<{(a+c)}^2+d& \hfill & \\ \hfill & ⟺d<0,& \hfill & \end{array}$$

and

$$\begin{array}{lll}\hfill C_2=\frac{a+c+\sqrt{{(a+c)}^2+d}}{2}\ge \frac{a+c}{2}>0.& & \hfill \end{array}$$

By part (b), for all real $r$, $𝜃$,

$$C_1{r}^2\le f(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)\le C_2{r}^2.$$

Here $x=r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,y=r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃$, thus $r^2=x^2+y^2$. Therefore, for all real numbers $x,y$ (a fortiori for all integers $x,y$)

$$C_1(x^2+y^2)\le f(x,y)\le C_2(x^2+y^2),$$

where $C_1,C_2$ are positive constants depending only on the coefficients of $f$.

(d)

Let $f$ be a definite positive quadratic form, and let $n$ be an integer. If $(x,y)$ is a representation of $n$ for the form $f(x,y)$, so that $f(x,y)=n$, then by part (c) $$C_1(x^2+y^2)\le n,$$

where $C_1>0$. Therefore

$$\begin{array}{lll}\hfill |x|\le \sqrt{x^2+y^2}\le \sqrt{\frac{n}{C_1}},& & \hfill \\ \hfill |y|\le \sqrt{x^2+y^2}\le \sqrt{\frac{n}{C_1}},& & \hfill \end{array}$$

thus

$$\begin{array}{lll}\hfill -A\le x\le A,& & \hfill \\ \hfill -A\le y\le A.& & \hfill \end{array}$$

where $A=\lfloor \sqrt{\frac{n}{C_1}}\rfloor$.

Hence there are at most ${(2A+1)}^2$ representations.

If $f$ is a positive definite quadratic form then an integer $n$ has at most a finite number of representations by $f$.