Exercise 5.5.8* ($N(ax^2 + by^2 + cy^2 \equiv 0 \pmod p) = p^2$ if $abc \not \equiv 0 \pmod p$.)

Proof.

If $a\in \mathbb{Z}$, and $\alpha =\bar{a}\in \mathbb{Z}∕\mathit{p\mathbb{Z}}$, we write $\left(\frac{\alpha }{p}\right)=\left(\frac{a}{p}\right)$. This makes sense because $\left(\frac{a}{p}\right)$ depends only of the class of $a$ modulo $p$.

Let $N(x^2=u)$ denote the number of solutions of $x^2=u$, where $u\in {𝔽}_p$. Then (see remark on p.132)

By Problem 3.3.16, if $p\ne 2$, for every $\alpha ,\beta \in {𝔽}_p$ such that $\alpha \ne 0$,

in particular, for $\alpha =1,\beta =0$ (or see Problem 3.3.5),

By Problem 3.3.17, if $p\ne 2$, for every $\alpha \in {𝔽}_p$,

As in problem 7, $\alpha =\bar{a},\beta =\bar{b},\gamma =\bar{c}$ are the classes of $a,b,c$ in $\mathbb{Z}∕\mathit{p\mathbb{Z}}$,

and

so that

By hypothesis $\alpha \ne 0,\beta \ne 0,\gamma \ne 0$.

We suppose first that $p$ is an odd prime.

For every solution of $f(x,y,z)=0$, there is a unique triple $(u,v,w)\in {𝔽}_p^3$ such that $\mathit{\alpha u}+\mathit{\beta v}+\mathit{\gamma w}=0$, where $x^2=u,y^2=v,z^2=w$. Therefore

For simplicity, we write $s=-{\gamma }^{-1}\alpha ,t=-{\gamma }^{-1}\beta$, so that $s\ne 0,t\ne 0$, and

Therefore

By expanding the inner product, we obtain

We compute these eight sums.

• Since $|{𝔽}_p^2|=p^2$,

  $$\underset{(u,v)\in {𝔽}_p^2}{\sum }1=p^2.$$

• By (3),

  $$\underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{u}{p}\right)=\underset{v\in {𝔽}_p}{\sum }\underset{u\in {𝔽}_p}{\sum }\left(\frac{u}{p}\right)=0.$$

• Similarly

  $$\underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{v}{p}\right)=0.$$

• By (2), since $s\ne 0,t\ne 0$,

  $$\begin{array}{llll}\hfill \underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{\mathit{su}+\mathit{tv}}{p}\right)& =\left(\frac{t}{p}\right)\underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{s{t}^{-1}u+v}{p}\right)=0.& \hfill & \end{array}$$

• By (3),

  $$\underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{u}{p}\right)\left(\frac{v}{p}\right)=\underset{v\in {𝔽}_p}{\sum }\left(\frac{v}{p}\right)\underset{u\in {𝔽}_p}{\sum }\left(\frac{u}{p}\right)=0.$$

• By (4), since $t{s}^{-1}\ne 0$,

  $$\underset{u\in {𝔽}_p}{\sum }\left(\frac{u}{p}\right)\left(\frac{u(u+t{s}^{-1}v)}{p}\right)=\{\begin{array}{cc}-1\hfill & \text{if }v\ne 0\hfill \\ p-1\hfill & \text{if }v=0.\hfill \end{array}$$

  thus

  $$\begin{array}{llll}\hfill \underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{u}{p}\right)\left(\frac{\mathit{su}+\mathit{tv}}{p}\right)& =\left(\frac{s}{p}\right)\underset{v\in {𝔽}_p}{\sum }\underset{u\in {𝔽}_p}{\sum }\left(\frac{u(u+t{s}^{-1}v)}{p}\right)& \hfill & \\ \hfill & =\left(\frac{s}{p}\right)[(p-1)(-1)+(p-1)]& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

• Similarly,

  $$\underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{v}{p}\right)\left(\frac{\mathit{su}+\mathit{tv}}{p}\right)=0.$$

• Finally,

  $$\begin{array}{llll}\hfill \underset{(u,v)\in {𝔽}_p^2}{\sum }\left(\frac{u}{p}\right)\left(\frac{v}{p}\right)\left(\frac{\mathit{su}+\mathit{tv}}{p}\right)& =\left(\frac{s}{p}\right)\underset{v\in {𝔽}_p}{\sum }\left(\frac{v}{p}\right)\underset{u\in {𝔽}_p}{\sum }\left(\frac{u(u+t{s}^{-1}v)}{p}\right)& \hfill & \\ \hfill & =-\left(\frac{s}{p}\right)\underset{v\in {𝔽}_p^{\text{∗}}}{\sum }\left(\frac{v}{p}\right)& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

This shows that if $p\ne 2$, $N(p)=p^2.$

If $p=2$, then $\alpha =\beta =\gamma =1$, so $f(x,y,z)=0⟺x^2+y^2+z^2=0$ has $4=2^2$ solutions $(0,0,0,)(0,1,1),(1,0,1),(1,1,0)$. In all cases

If $p$ divides none of the numbers $a,b,c$, the congruence $a{x}^2+b{y}^2+c{y}^2\equiv 0(\mathit{mod}p)$ has $p^2$ solutions. □