Number of solutions of $x^2 + y^2 + z^2 = 0$ over finite fields.

Question

I want to prove that the number of elements of the set $\{(x,y,z)\in \mathbb{F}_p^3: x^2 + y^2 + z^2 = 0\}$ is $p^2$.

I know that the number of elements of the set is a multiple of $p$ using the Chevalley-Warning theorem, but I don't know how to continue.

Answer 1

Exercise $19$ in Chapter $8$ of "A Classical Introduction to Modern Number Theory" by Kenneth Ireland, Michael Rosen is the following result: If $m$ is odd, then the number of solutions to $$ x_1^2+x_2^2+\cdots +x_m^2=0 $$ equals $p^{m-1}$. A proof is given in the book via Gauss and Jacobi sums in Theorem $5$ of Chapter $8$.

Answer 2

There is also a very direct argument for a ternary form: here, ...

For $p=2$, $x^2+y^2+z^2=(x+y+z)^2$, so the number of zeros is the number of points in a hyperplane, $p^2$.

For $p=1\mod 4$ there is a $j=\sqrt{-1}$ in $\mathbb F_p$. Change variables to $u=x+jy$ and $v=x-jy$, so $x=(u+v)/2$ and $y=(u-v)/2j$. To solve the indicated equation is to solve $uv=-z^2$. For $z=0$ there are $2p-1$ solutions. For each of the $p-1$ values of $z\not=0$ there are $p-1$ solutions. Altogether, there are $(2p-1)+(p-1)^2=p^2$ solutions.

For $p=3\mod 4$, $x^2+y^2$ is the norm of $x+jy$ where $j^2=-1$ and $j$ lies in the unique quadratic extension of $\mathbb F_p$. The norm map is $\alpha \to \alpha\cdot \alpha^p=\alpha^{1+p}$, and is (thus) surjective. On non-zero elements it is a group homomorphism with kernel of order $p+1$. For $z=0$, there is the unique solution $(0,0,0)$. For each of the $p-1$ $z\not=0$, surjectivity gives $p+1$ solutions. Altogether, this is $1+(p-1)(p+1)=p^2$.

Answer 3

The case $p=2$ is dealt with by explcit counting. So assume $p>2$.

Let $m$ be the number of solutions. $\mathbb F_p^\times$ acts on the set of solution by coordinate-wise multiplication, where we have $(cx,cy,cz)=(x,y,z)$ iff $c=1$ or $x=y=z=0$, that is: all orbits except that of the trivial solution have length $p-1$. We conclude that $p-1\mid m-1$, hence together with $p\mid m$ $$m\equiv p\pmod{p^2-p}$$ by the Chinese Remainder Theorem. Moreover, if $(x,y,z)$ is a nontrivial solution and wlog $x\ne 0$, then at least one of $y,z$ is nonzero and $(-x,y,z)$ is a solution not in the orbit of $(x,y,z)$. Hence there must be at least two nontrivial orbits, i.e., $m\ge 2(p-1)+1>p$.

For $x\in\mathbb F_p$ let $S(x)=\{\,(y,z)\in\mathbb F_p^2:x^2+y^2+z^2=0\,\} $ and let $f(x)=|S(x)|$. Similarly, let $g(x,y)$ the number of $z$ with $x^2+y^2+z^2=0$. Then $g(x,y)\in\{0,1,2\}$ and $$\tag1f(x)=\left|\{\,y:g(x,y)=1\,\}\right|+2\left|\{\,y:g(x,y)=2\,\}\right|.$$ Note that $g(x,y)=1$ occurs only if $x^2+y^2=0$. From $(1)$ we immediately get $f(x)\le 2p$. But $f(x)=2p$ can only occur if $g(x,y)=2$ for all $y$, including $y=0$. But if $x^2+0^2+z^2=0$, then $g(x,z)=1$. Hence $$\tag2f(x)\le 2p-1.$$ The group $\mathbb Z/4\mathbb Z$ acts $S(x)$ per $(x,y,z)\mapsto(x,z,-y)$. A fixpoint or an orbit of length $2$ can only occur if $y=z=0$. Hence for $x\ne 0$, we have $4\mid f(x)$ and so $$f(x)\le 2p-2\quad\text{if }x\ne 0.$$ We conclude that $$m\le (2p-1)+(p-1)(2p-2) <2p^2-p$$ as desired.

From $p<m<2p^2-p$ and $m\equiv p\pmod{p^2-p}$ the claim $m=p^2$ follows.