Full proof of the generalized Markoff solution-count conjecture

Let

$$Q(X,Y,Z) =a_{11}X^2+a_{22}Y^2+a_{33}Z^2 +a_{12}XY+a_{13}XZ+a_{23}YZ,$$

and suppose that $p$ is prime, $q=p^r$ with $r\geq1$, and $p\nmid a_{11}a_{22}a_{33}d$. Define

$$\begin{aligned} G&= \begin{pmatrix} 2a_{11}&a_{12}&a_{13}\\ a_{12}&2a_{22}&a_{23}\\ a_{13}&a_{23}&2a_{33} \end{pmatrix}, &D&=\det G,\\ A_{11}&=4a_{22}a_{33}-a_{23}^{2}, &A_{22}&=4a_{11}a_{33}-a_{13}^{2}, &A_{33}&=4a_{11}a_{22}-a_{12}^{2}. \end{aligned}$$

We prove all four cases of Xiao-Jie Zhu's Conjecture 10.1 for every prime power $q$. Theorem 1.6 of that source proves the prime-field case $r=1$; the argument below supplies the conjectured extension to every $r$.

1. Projection from the singular origin

Write

$$\begin{aligned} \mathcal C &=\bigl\{[X:Y:Z]\in\mathbb P^2(\mathbb F_q):Q(X,Y,Z)=0\bigr\},\\ \mathcal B &=\bigl\{[X:Y:Z]\in\mathcal C:XYZ=0\bigr\},\\ N_q &=\#\bigl\{(x,y,z)\in\mathbb F_q^3:Q(x,y,z)=dxyz\bigr\}. \end{aligned}$$

Every nonzero affine point lies on a unique line through the origin with projective direction $P=[X:Y:Z]$. Its nonzero points have the form $t(X,Y,Z)$, with $t\in\mathbb F_q^\times$, and satisfy

$$t^2Q(X,Y,Z)=dt^3XYZ.$$

If $XYZ\neq0$, this line contributes exactly one nonzero solution when $Q(P)\neq0$, and contributes none when $Q(P)=0$. There are $(q-1)^2$ projective directions with $XYZ\neq0$, of which $\#\mathcal C-\#\mathcal B$ satisfy $Q(P)=0$.

If $XYZ=0$, the line contributes all $q-1$ nonzero points precisely when $P\in\mathcal B$, and none otherwise. Including the origin gives the characteristic-independent identity

$$\boxed{ N_q=1+(q-1)^2-\#\mathcal C+q\,\#\mathcal B. } \tag{1}$$

Thus the entire problem reduces to counting points on a projective conic and on its three coordinate sections.

2. Odd characteristic

Suppose that $p>2$, and let $\chi_q$ be the quadratic character of $\mathbb F_q$, extended by $\chi_q(0)=0$. Set

$$\chi_i=\chi_q(-A_{ii}), \qquad S=\chi_1+\chi_2+\chi_3.$$

None of the three coordinate vertices lies on $\mathcal C$, since each $a_{ii}$ is nonzero. The coordinate sides therefore contribute disjointly to $\mathcal B$. For example, on $Z=0$ the defining equation is

$$a_{11}X^2+a_{12}XY+a_{22}Y^2=0,$$

which has exactly $1+\chi_q(a_{12}^2-4a_{11}a_{22})=1+\chi_3$ projective roots. The other two sides give the analogous expressions, so

$$\boxed{\#\mathcal B=3+S.} \tag{2}$$

Nonsingular conic

If $p\nmid D$, the conic $\mathcal C$ is nonsingular. Every nonsingular projective plane conic over a finite field has exactly $q+1$ rational points: diagonalization and the intersection of two subsets of $\mathbb F_q$ having $(q+1)/2$ elements give a rational point, and projection from that point parametrizes the conic by $\mathbb P^1(\mathbb F_q)$. Hence

$$\#\mathcal C=q+1.$$

Substitution into (1) and (2) gives

$$\boxed{ N_q=q^2+q(\chi_1+\chi_2+\chi_3)+1. } \tag{3}$$

Moreover, $p\nmid D$ automatically implies $p>2$: modulo $2$, the matrix $G$ is alternating of odd order and therefore singular.

Degenerate conic

Suppose now that $p\mid D$. Since all diagonal coefficients of $Q$ are nonzero, the rank of $G$ is either $1$ or $2$.

If $\operatorname{rank}G=1$, then $Q$ is a nonzero scalar multiple of the square of a linear form. Consequently,

$$A_{11}=A_{22}=A_{33}=0, \qquad \#\mathcal C=q+1, \qquad N_q=q^2+1. \tag{4}$$

Suppose instead that $\operatorname{rank}G=2$. Its adjugate has rank one and is symmetric, so

$$\operatorname{adj}(G)=c\,vv^{\mathsf T} \tag{5}$$

for some $c\in\mathbb F_q^\times$ and nonzero $v\in\mathbb F_q^3$. In particular, at least one diagonal cofactor $A_{jj}=cv_j^2$ is nonzero, and every nonzero diagonal cofactor has the same quadratic character:

$$\chi_q(-A_{ii})=\chi_q(-c) \qquad\text{whenever }A_{ii}\neq0. \tag{6}$$

In fact at least two diagonal cofactors are nonzero. Otherwise $v$ would be supported on a single coordinate $j$, while $G\operatorname{adj}(G)=0$ implies $Gv=0$; its $j$th coordinate would give $2a_{jj}v_j=0$, contradicting $p\nmid 2a_{jj}$.

Fix an index $j$ with $A_{jj}\neq0$. The restriction of $Q$ to the coordinate line $X_j=0$ is a nondegenerate binary quadratic form with discriminant $-A_{jj}$. Its projective zero set therefore consists of $1+\chi_j$ points. The radical of $G$ is spanned by $v$ and does not lie on this coordinate line, because $v_j\neq0$. Accordingly, $\mathcal C$ is the union of the lines joining this radical point to the $1+\chi_j$ points of the binary section. Therefore

$$\#\mathcal C =1+q(1+\chi_j) =q+1+q\chi_j. \tag{7}$$

Equations (1), (2), and (7) yield

$$\boxed{ N_q=q^2+q\left(\chi_1+\chi_2+\chi_3-\chi_j\right)+1. } \tag{8}$$

If at least two diagonal cofactors vanish modulo $p$, the preceding observation excludes rank two. Thus $G$ has rank one, all three diagonal cofactors vanish, and (4) gives

$$\boxed{N_q=q^2+1.} \tag{9}$$

If at most one diagonal cofactor vanishes, (8) is exactly the third conjectured formula. Its value is independent of the admissible index $j$ by (6).

Finally, because all coefficients come from the prime subfield, for every integer $b$ one has

$$\chi_q(b) =\left(\frac{b}{p}\right)^{r} =\left(\frac{b}{p^r}\right)_{K}. \tag{10}$$

Thus (3), (9), and (8) agree exactly with the Kronecker-symbol expressions in the first three parts of Conjecture 10.1.

3. Characteristic two

Now let $p=2$ and $q=2^r$. The source assumes integer coefficients with $2\nmid a_{11}a_{22}a_{33}d$, so reduction to $\mathbb F_q$ gives

$$a_{11}=a_{22}=a_{33}=d=1.$$

Write

$$e=a_{12}\bmod2, \qquad f=a_{13}\bmod2, \qquad g=a_{23}\bmod2, \qquad k=e+f+g\in\{0,1,2,3\}, \qquad \varepsilon=(-1)^r.$$

Here $k$ is the ordinary integer count of nonzero mixed coefficients. On a coordinate side, a zero mixed coefficient produces the equation

$$U^2+V^2=(U+V)^2=0,$$

with one projective root. A nonzero mixed coefficient produces

$$U^2+UV+V^2=0.$$

This has two projective roots precisely when $\mathbb F_q$ contains $\mathbb F_4$, equivalently when $r$ is even, and has no projective roots when $r$ is odd. Its number of roots is therefore $1+\varepsilon$. Since the coordinate vertices remain disjoint from $\mathcal C$, we obtain

$$\boxed{\#\mathcal B=3+k\varepsilon.} \tag{11}$$

The conic itself has four elementary forms, up to permutation of coordinates.

If $k=0$, then

$$Q(X,Y,Z)=(X+Y+Z)^2, \qquad \#\mathcal C=q+1. \tag{12}$$

If $k=1$, take the mixed term to be $XY$. Then

$$Q(X,Y,Z)=X^2+Y^2+Z^2+XY.$$

Its partial derivatives are $Y$, $X$, and $0$; their unique common projective zero is $[0:0:1]$, which is not on the conic. Thus the conic is nonsingular. The rational point $[1:0:1]$ and projection from that point give

$$\#\mathcal C=q+1. \tag{13}$$

If $k=2$, take the mixed terms to be $XY$ and $XZ$. With $U=Y+Z$,

$$Q(X,Y,Z)=X^2+XU+U^2. \tag{14}$$

If $k=3$, set $U=X+Y$ and $V=X+Z$. Then

$$Q(X,Y,Z)=U^2+UV+V^2. \tag{15}$$

In either (14) or (15), the conic is the union of two distinct rational lines when $r$ is even and has only its rational vertex when $r$ is odd. Consequently,

$$\#\mathcal C=q+1+q\varepsilon \qquad\text{for }k\in\{2,3\}. \tag{16}$$

Combining (1), (11), and (12)–(16), we find

$$\boxed{ N_q= \begin{cases} q^2+1,&k=0,\\ q^2+(-1)^r q+1,&k\in\{1,2\},\\ q^2+(-1)^r 2q+1,&k=3. \end{cases} } \tag{17}$$

These are precisely the three characteristic-two alternatives in part (4) of Conjecture 10.1. Together, (3), (9), (8), and (17) prove the complete conjecture over every finite field $\mathbb F_{p^r}$.