Exercise 3.5.8 (Least positive integer represented by $44x^2 - 97 xy + 35 y^2$)

Proof.

(a)

The discriminant $d$ of $f$ is $d=97^2-4\cdot 44\cdot 35=3249=57^2$, so $d$ is a positive perfect square.

First we factor $f(x,y)$.

$$\begin{array}{llll}\hfill 4\cdot 44\cdot f(x,y)& =4\cdot 44^2{x}^2-4\cdot 44\cdot 97\mathit{xy}+4\cdot 44\cdot 35y^2& \hfill & \\ \hfill & ={(88x-97y)}^2-(97^2-4\cdot 44\cdot 35)y^2& \hfill & \\ \hfill & ={(88x-97y)}^2-{(57y)}^2& \hfill & \\ \hfill & =(88x-97y-57y)(88x-97y+57y)& \hfill & \\ \hfill & =(88x-154y)(88x-40y)& \hfill & \\ \hfill & =22(4x-7y)\cdot 8(11x-5y).& \hfill & \end{array}$$

Therefore

$$f(x,y)=(4x-7y)(11x-5y).$$

Following the steps of Problem 7, we note that $f(7,4)=0$, where $7\wedge 4=1$. Then $-7+2\cdot 4=1$ is a Bézout’s relation between $7$ and $4$, which shows that

$$A=\left(\begin{array}{cc}\hfill 7\hfill & \hfill -2\hfill \\ \hfill 4\hfill & \hfill -1\hfill \end{array}\right)\in \Gamma .$$

Put $h=f\cdot A$, so that $g$ is properly equivalent to $f$, and

$$\begin{array}{llll}\hfill h(x,y)& =f(7x-2y,4x-y)& \hfill & \\ \hfill & =[4(7x-2y)-7(4x-y)][11(7x-2y)-5(4x-y)]& \hfill & \\ \hfill & =-y(57x-17y).& \hfill & \end{array}$$

To obtain a reduced form as in Problem 7, we put $q=h\cdot S$, where

$$S=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right),$$

then $q=(17,57,0)$, so

$$q(x,y)=x(17x+57y).$$

So $q(x,y)=x(17x+57y)$ is equivalent to $f$ and satisfies the conditions $0\le a=17<|b|=57$, but this is not the waited answer. To prove that the two forms $q$ and $g$ are equivalent, we search a matrix $B\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\in \Gamma$ such that $g=q\cdot B$. This is equivalent to the system of equations

$$\{\begin{array}{cc}a(17a+57c)\hfill & =47,\hfill \\ 34\mathit{ab}+57\mathit{bc}+57\mathit{ad}\hfill & =-57,\hfill \\ b(17b+57d)\hfill & =0.\hfill \end{array}$$

This system has a solution, given by $a=47,b=-57,c=-14,d=17$, and $\mathit{ad}-\mathit{bc}=1$, so

$$B=\left(\begin{array}{cc}\hfill 47\hfill & \hfill -57\hfill \\ \hfill -14\hfill & \hfill 17\hfill \end{array}\right)\in \Gamma .$$

To check this result, we observe that

$$\begin{array}{llll}\hfill (q\cdot B)(x,y)& =q(47x-57y,-14x+17y)& \hfill & \\ \hfill & =(47x-57y)(17(47x-57y)+57(-14x+17y))& \hfill & \\ \hfill & =(47x-57y)x& \hfill & \\ \hfill & =g(x,y).& \hfill & \end{array}$$

This shows that $g$ and $q$ are properly equivalent, and so $f$ and $g$ are properly equivalent. (The forms $q$ and $g$ are equivalent and reduced in the sense of Problem 7, but are distinct.)

Note: If $M=\mathit{ASB}$ then $f\cdot M=g$, and

$$M=\left(\begin{array}{cc}\hfill -4\hfill & \hfill 5\hfill \\ \hfill -9\hfill & \hfill 11\hfill \end{array}\right).$$

Indeed,

$$\begin{array}{llll}\hfill f(-4x+5y,-9x+11y)& =[4(-4x+5y)-7(-9x+11y),(11(-4x+5y)-5(-9x+11y)]& \hfill & \\ \hfill & =(47x-57y)x.& \hfill & \end{array}$$

These two lines are sufficient to prove the sentence, but I don’t know by which method NZM obtained this matrix $M=S{T}^{-2}S{T}^{4}S{T}^{-1}{S}^2.$

(b)

Since $f$ and $g$ are equivalent, an integer $n$ is represented by $f(x,y)$ if and only if $n$ is represented by $g(x,y)=x(47x-57y)$. So $n=g(x_0,y_0)=x_0(47x_0-57y_0)$ where $x_0,y_0$ are integers. Then $n=\mathit{ab}$, where $a=x_0,b=47x_0-57y_0\equiv 47a(\mathit{mod}57)$.

Conversely, if $n=\mathit{ab}$, where $b\equiv 47a(\mathit{mod}57)$, then $b=47a-57k$ for some integer $k$, thus $n=g(a,k)$ is represented by $g$.

This shows that $n$ is represented by $f$ if and only if $n$ can be written in the form $n=\mathit{ab}$ where $b\equiv 47a(\mathit{mod}57)$.

(c)

For instance $17$ is represented by $f$ since $f\sim q=(17,57,0)$, and $17=q(0,1)$. The condition is satisfied since $17=17(47\cdot 17-57\cdot 14)=\mathit{ab}$, where $a=17,b=1=47\cdot 17-57\cdot 14\equiv 47a(\mathit{mod}57)$.

We show that no positive integer $n$ less that $17$ can be represented by $f$. By part (b), $n=\mathit{ab}$, where $b\equiv 47a(\mathit{mod}57)$, and so $n=a(47a-57k)$.

Since $n=a(47a-57k)=(-a)(47(-a)+57k)$, we can suppose that $a>0$, and since $0<n=\mathit{ab}<17$, we know that $0<a<17,0<b<17$. Hence $b$ is the remainder in the division of $47a$ by $57$.

The following array

$a$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
$b$	47	37	27	17	7	54	44	34	24	14	4	51	41	31	21	11
$\mathit{ab}$	47	74	81	68	35	324	308	272	216	140	44	612	533	434	315	176

shows that $n=\mathit{ab}>17$ if $1\le a\le 16$ (but $a=17$ gives $b=1$ and $\mathit{ab}=17$).

The least positive integer $n$ represented by $f$ is $17$. □