Exercise 3.7.6 (Representations of $p$ by forms of discriminant $-23$)

Proof.

(a)

Let $f=(a,b,c)$ a reduced positive definite quadratic form of discriminant $d=-23$. By Theorem 3.19, $0<a\le \sqrt{-d∕3}=\sqrt{23∕3}$, thus $a=1$ or $a=2$.

If $a=1$, then $-1<b\le 1$, and $b^2-4c=-23$, thus $b$ is odd, so $b=1$, and $c=(b^2+23)∕4=6$, which gives $f=f_0=(1,1,6)$.

If $a=2$, then $-2<b\le 2$, and $b$ is odd, thus $b=\pm 1$, and $c=(b^2+23)∕8=3$, so $f=f_1=(2,1,3)$ or $f=f_2=(2,-1,3)$. These three reduced forms are primitive, so

$$H(-23)=h(-23)=3.$$

By Theorem 3.25, these three forms are not equivalent. Since every form $f$ of discriminant $-23$ is equivalent to a reduced form, $f$ is primitive, and equivalent to exactly one of the forms $f_0,f_1,f_2$.

(b)

Let $p$ be a prime number such that $\left(\frac{-23}{p}\right)=-1$. Then $p\ne 2,p\ne 23$, and $d=-23\equiv 1(\mathit{mod}4)$ satisfies $p\nmid d$ and $\left(\frac{d}{p}\right)\ne 1$. Then Corollary 3.14 shows that $p$ is not represented by any form of discriminant $-23$, so is not represented by $f_0,f_1,f_2$.

(c)

The prime $2$ is not represented by $f_0$, and has two representations apiece by $f_1$ and $f_2$. We suppose now that $p$ is an odd prime.

Assume now that $\left(\frac{-23}{p}\right)=1$. By Corollary 3.14, there is a binary quadratic form $f$ of discriminant $-23$ that represents $p$. Since $f$ is properly equivalent to $f_0$, $f_1$ or $f_2$, $p$ is represented by one of these three forms.

By the last formula of section 3.7, applied to $n=p$,

$$\underset{f\in ℱ}{\sum }{r}_f(p)=w{N}_d(p),$$

where $ℱ=\{f_0,f_1,f_2\}$, and by Theorem 3.26, $w=w(f_0)=w(f_1)=w(f_2)=2$.

By definition,

$$\begin{array}{llll}\hfill N_d(p)& =\mathrm{Card}\{u\in [[0,2p[[\mid u^2\equiv -23(\mathit{mod}4p)\}& \hfill & \\ \hfill & =\frac{1}{2}\mathrm{Card}\{u\in [[0,4p[[\mid u^2\equiv -23(\mathit{mod}4p)\}& \hfill & \end{array}$$

Since $\left(\frac{-23}{p}\right)=1$, the congruence $x^2\equiv -23(\mathit{mod}p)$ has two solutions, and the congruence $x^2\equiv 23\equiv 1(\mathit{mod}4)$ has also $2$ solutions, so the congruence $x^2\equiv -23(\mathit{mod}4p)$ has $4$ solutions, so $N_d(p)=4$, therefore

$$\underset{i=0}{\overset{2}{\sum }}{r}_{f_i}(p)=4.$$

Since every representation of the prime $p$ is proper, we can write this formula in the form

$$R_{f_0}(p)+R_{f_1}(p)+R_{f_2}(p)=4,$$

which means that $p$ has a total of $4$ representations by the forms $f_0,f_1,f_2$.

Since $f_1(x,-y)=f_2(x,y)$, there are as many representations of $p$ by $f_1$ and $f_2$, so

$$R_{f_1}(p)=R_{f_2}(p).$$

Put $a=R_{f_0}(p),b=R_{f_1}(p)=R_{f_2}(p)$. Then $a+2b=4$. The only solutions $(a,b)\in {\mathbb{N}}^2$ of $a+2b$ are $(4,0),(2,1)$ and $(0,2)$. Since $f_1(x,y)=f_1(-x,-y)$, $R_{f_1}(p)=1$ is impossible, so the only possibilities are

$$\begin{array}{llll}\hfill R_{f_0}(p)=4,& R_{f_1}(p)=R_{f_2}(p)=0,& \hfill & \\ \hfill & \text{or}& \hfill & \\ \hfill R_{f_0}(p)=0,& R_{f_1}(p)=R_{f_2}(p)=2.& \hfill & \end{array}$$

(d)

Consider the example $p=139$. Following the hint, we search a solution of the congruence $h^2\equiv -23(\mathit{mod}4p)$. The Tonelli-Shanks algorithm gives a square root of $-23$ modulo $p$: $$33^2\equiv -23(\mathit{mod}139).$$

Moreover

$$1^2\equiv -23(\mathit{mod}4).$$

A solution of the system of congruence

$$\{\begin{array}{cc}h\equiv 33\hfill & (\mathit{mod}139)\hfill \\ h\equiv 1\hfill & (\mathit{mod}4)\hfill \end{array}$$

is $h=33$ (since $33\equiv 1(\mathit{mod}4)$), thus

$$33^2\equiv -23(\mathit{mod}4\cdot 139).$$

( Check: $33^2+23=1112=2\cdot (4\cdot 139)$.) Now we take $k$ such that the form $g(x,y)=139x^2+\mathit{hxy}+k{y}^2$ satisfies $d=\mathrm{discr}(g)=33^2-4\cdot 129k=-23$, which gives $k=2$, so

$$g(x,y)=139x^2-33\mathit{xy}+2y^2.$$

Now we reduce $g$ with the usual algorithm (see the program given in the notes of Problem 3.6.4.) We obtain

$$\begin{array}{llll}\hfill & g=(139,-33,2),& \hfill & \\ \hfill & g\cdot S=(2,33,139)& \hfill & \\ \hfill & g\cdot (S{T}^8)=(2,1,3)=f_1& \hfill & \end{array}$$

Thus

$$g\cdot A=f_1,$$

where $A=S{T}^{-8}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 8\hfill \end{array}\right)$.

Verification:

$$\begin{array}{llll}\hfill g(y,-x+8y)& =139y^2-33y(-x+8y)+2{(-x+8y)}^2& \hfill & \\ \hfill & =2x^2+\mathit{xy}+3y^2& \hfill & \\ \hfill & =f_1(x,y).& \hfill & \end{array}$$

So $g$ and $f_1$ are properly equivalent. Since $p=139=g(1,0)$ is represented by $g$, we obtain that $p$ is represented by the equivalent form $f_1$. By part (c), we obtain that $p$ has two representations apiece by $f_1$ and $f_2$ and no representation by $f_0$.

Explicitly, $p=g(1,0)=(f_1\cdot A^{-1})(1,0)=f_1(8,1)=2\cdot 8^2+1\cdot 8+3\cdot 1^2$, so

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