# On a certain equation in finite fields

I am interested in the following question. Let $$q$$ be a prime power and let $$\mathbb{F}_q$$ be the finite field of cardinality $$q$$. Suppose $$q>61$$. Is it true that, for every $$b\in \mathbb{F}_q$$ and for every $$c\in \mathbb{F}_q$$ with $$c\ne 0$$, there exists $$x,y,z\in\mathbb{F}_q$$ such that \begin{align*} x^2+y^2+z^2&=b\\ xyz&=c. \end{align*} I have computational evidence towards this and in fact the request $$q>61$$ is suggested by the computational data.

Generically the intersection of the surfaces described by these two equations is an algebraic curve of genus $$4$$. Once one has made sure that this curve is absolutely irreducible, by Weil there are $$\mathbb F_q$$-points provided that $$q$$ is big enough. Weil assumes a smooth curve, so it can be a pain to handle singularities. However, there are explicit bounds, like Theorem 5.4.1 in the third edition of Fried and Jarden's Field Arithmetic: If the absolutely irreducible affine curve has degree $$d$$, then the number of $$\mathbb F_q$$-points is at least $$q+1-(d-1)(d-2)\sqrt{q}-d$$. In your case $$d=6$$. This lower bound is positive once $$q\ge419$$. So one has to check the smaller cases directly, or has to resort to Joe Silverman's suggestion from the comments.

• Are you claiming there is always a solution for $q \ge 43$? But e.g. in the case $q = 61$ there is no solution when $b = 0$ and $c = 4$. Mar 31 at 16:44
• For what it's worth, I think that the projective curve $$x^2+y^2+z^2=bw^2,\quad xyz=cw^3$$ in $\mathbb P^3$ is nonsingular provided $bc\ne0$ and $b^3\ne27c^2$. (Although I didn't double check my calculation.) So in those cases one should be able to use Weil, plus estimate the number of points at infinity, i.e., with $w=0$. Mar 31 at 16:53
• @RobertIsrael The $d$ is $6$ and not $4$, I fixed the mistake. Mar 31 at 17:20

Following Joe Silverman's suggestion, we study the projective system of equations $$x^2+y^2+z^2 -b w^2=0 , xyz- cw^3=0$$ with Jacobian matrix $$\begin{pmatrix} 2x & 2y & 2z & -2bw \\ yz & xz & xy & - 3 c w^2 \end{pmatrix}.$$ The singular locus is contained in the locus where this has rank $$\leq 1$$, i.e. the vanishing set of the $$2 \times 2$$ minors. One of the minors is $$2 x^2 z - 2 y^2 z$$ so in odd characteristic we get $$z= 0$$ or $$x= \pm y$$, and symmetrically $$x=0$$ or $$y = \pm z$$ and $$y=0$$ or $$x=\pm z$$, which means either two of $$x,y,z$$ are $$0$$ or they're all equal to $$\pm$$ each other.

If two are $$0$$ then $$xyz=0$$ so since $$c\neq 0$$ we must have $$w=0$$, which contradicts $$x^2+y^2+z^2 -b w^2=0$$. After rescaling, we may assume $$(x,y,z)= (1,1,1), (1,1,-1), (1,-1,1)$$, or $$(1,-1,-1)$$. Set $$\epsilon=xyz=\pm 1$$. This gives $$3=bw^2$$ and $$\epsilon=c w^3$$, so $$w\neq 0$$ and $$b^3 = 27 / w^6 = 27 \epsilon^2/ w^6= 27 c^2$$.

So in the case of odd characteristic and $$b^3 \neq 27 c^2$$ we have no projective singularities, i.e. a smooth complete intersection of a quadric and a cubic, which has genus $$4$$, and thus has at least $$q +1- 8 \sqrt{6}$$ rational points. The points at $$\infty$$ are the projective solutions of the equations $$x^2+y^2+z^2= 0 = xyz$$, which are points where one of the three coordinates is $$0$$ and the other two have ratio a square root of $$-1$$. The number of these is $$6$$ if $$q \equiv 1 \bmod 4$$ and $$0$$ otherwise, so in total the number of points is at least $$q - 8 \sqrt{q}-5$$ which is positive for $$q>73$$, so there is always a solution in the smooth case for $$q=73$$.

In the case $$b^3 = 27 c^2$$, there are four singularities of the form $$(\pm 1: \pm 1 : \pm 1 : \pm b /(3c))$$. One must understand whether these singularities cause the curve to split into two distinct geometrically irreducible components.

To do this, note that $$x,y,z$$ are the three roots of the polynomial $$x^3 - ax + (a^2-b)x /2 -c$$ for some $$a$$. If $$b^3= 27c^2$$, the discriminant factors as $$( a- 9 c/b)^2 (a + 3c/b)^2 (a ^2+ 12 c/b + 54 c^2/b^2)$$ and the single roots of the quadratic factor correspond to reflections in the monodromy group, whereas the point $$a- 9 c/b$$ corresponds to the polynomial $$(a - 3c/b)^3$$ which has one root and so since the discriminant only vanishes to second order must have monodromy of order $$3$$. Thus the monodromy group of the cover is $$S_3$$ and so the curve is geometrically irreducible.

Since the arithmetic genus is still $$4$$, the geometric genus is $$0$$, and each of the singularities can have at most two branches, so the Weil lower bound is $$q+1-4-6= q-9$$ in this case.

So modulo difficulties in characteristic $$2$$, and, when $$b=0$$, characteristic $$3$$ and numerical checks for $$q$$ between $$61$$ and $$73$$, your claim is verified

• Apropos "difficulties in characteristic 2", in this case we may replace the first equation by $x+y+z=e$ (where $b=e^2$), so the resulting curve is an elliptic one. Mar 31 at 18:15