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.


2 Answers 2


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.

  • $\begingroup$ 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 $c = 4$. $\endgroup$ Mar 31 at 16:44
  • $\begingroup$ 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 $w=0$. $\endgroup$ Mar 31 at 16:53
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    $\begingroup$ @RobertIsrael The $d$ is $6$ and not $4$, I fixed the mistake. $\endgroup$ 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

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    $\begingroup$ 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. $\endgroup$ Mar 31 at 18:15

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