Exercise 5.5.7 (Value of $N(ax^2 + by^2 +cz^2 \equiv 0 \pmod p)$ if $abc \equiv 0 \pmod p$.)

Proof. Let ${𝔽}_p=\mathbb{Z}∕\mathit{p\mathbb{Z}}$ the field with $p$ elements.

We write $\alpha =\bar{a},\beta =\bar{b},\gamma =\bar{c}$ the classes of $a,b,c$ in $\mathbb{Z}∕\mathit{p\mathbb{Z}}$, and

We define

so that

• If $p\mid a,p\mid b,p\mid c$, then $\alpha =\beta =\gamma =0$, and $f(x,y,z)=0$. Therefore $S_p={𝔽}_p^3$, and

  $$N(p)=p^3.$$

• We suppose that $p\mid a,p\mid b$ and $p\nmid c$. Then $f(x,y,z)=\gamma z^2$, where $\gamma \ne 0$. Therefore $f(x,y,z)=0⟺z=0$, thus $S_p={𝔽}_p\times {𝔽}_p\times \{0\}$. This gives

  $$N(p)=p^2.$$

  More generally, we prove similarly that if $p$ divides exactly two of the coefficients $a,b,c$ then $N(p)=p^2$.

• Suppose now that $p\mid a,p\nmid b,p\nmid c$. Then $f(x,y,z)=\beta y^2+\gamma z^2$, where $\beta \ne 0,\gamma \ne 0$, and

  $$\begin{array}{llll}\hfill f(x,y,z)=0& ⟺\beta y^2=-\gamma z^2& \hfill & \\ \hfill & ⟺(y,z)=(0,0)\text{ or }{(z{y}^{-1})}^2=-\beta {\gamma }^{-1}.& \hfill & \\ \hfill & & \hfill \end{array}$$

  • If $-\mathit{bc}$ is not a square modulo $p$, then $-\beta {\gamma }^{-1}=(-\mathit{\beta \gamma }){\gamma }^{-2}$ is not a square in ${𝔽}_p$. Therefore $f(x,y,z)=0⟺y=z=0$, so $S_p={𝔽}_p\times \{0\}\times \{0\}$, thus

    $$N(p)=p.$$

  • If $-\mathit{bc}$ is a square modulo $p$, then $-\beta {\gamma }^{-1}={\delta }^2$, for some $\delta \in {𝔽}_p^{\text{∗}}$. Then

    $$\begin{array}{llll}\hfill f(x,y,z)=0& ⟺(y,z)=(0,0)\text{ or }{(z{y}^{-1})}^2={\delta }^2& \hfill & \\ \hfill & ⟺z=\mathit{\delta y}\text{ or }z=-\mathit{\delta y}.& \hfill & \end{array}$$

    Let $A_{\delta }$ denote

    $$\begin{array}{llll}\hfill A_{\delta }& =\{(x,y,z)\in {𝔽}_p^3\mid z=\mathit{\delta y}\}& \hfill & \\ \hfill & =\{(x,y,\mathit{\delta y})\mid x\in {𝔽}_p,y\in {𝔽}_p\}.& \hfill & \end{array}$$

    Then $S_p=A_{\delta }\cup A_{-\delta }$. Moreover the map $\phi :{𝔽}_p^2\to A_{\delta }$ defined by $\phi (x,y)=(x,y,\mathit{\delta y})$ is bijective, thus $\mathrm{Card}(A_{\delta })=p^2$

    • If $p\ne 2$, since $\delta \ne 0$, $\mathit{\delta y}=-\mathit{\delta y}⟺2\mathit{\delta y}=0⟺y=0$, thus $A_{\delta }\cap A_{-\delta }={𝔽}_p\times \{0\}\times \{0\}$, so $\mathrm{Card}(A_{\delta }\cap A_{-\delta })=p$, and

      $$\begin{array}{llll}\hfill N(p)& =\mathrm{Card}(A_{\delta }\cup A_{-\delta })& \hfill & \\ \hfill & =\mathrm{Card}(A_{\delta })+\mathrm{Card}(A_{-\delta })-\mathrm{Card}(A_{\delta }\cap A_{-\delta })& \hfill & \\ \hfill & =2p^2-p.& \hfill & \end{array}$$

    • If $p=2$, then $\beta =\gamma =\delta =1$, and $f(x,y,z)=0⟺y^2+z^2=0⟺z=y$, thus $S_p=\{0,0,0),(0,1,1),(1,0,0),(1,1,1)\}$, so

      $$N(p)=4.$$

    This shows that if $p$ divides exactly one of the coefficients $a,b,c$ then $N(p)=p$ or $N(p)=2p^2-p$, except that $N(2)=4$.