# Number of solutions of a degree 4 polynomial equation over a finite field

Suppose that $$q$$ is a prime power and $$\xi, \eta\in \mathbb{F}_q$$ are nonzero. A computer calculation for $$q<70$$ suggests that the number $$N$$ of $$4$$-tuples $$(a,b,c,d)\in\mathbb{F}_q^{4}$$ satisfying $$(ac-\xi bd)^2-(a^2-\xi b^2+1)(c^2-\xi d^2-\eta)=0$$ is $$q^3-q$$.

Question. Is there some theory, or nice method, for computing $$N$$?

It may help, when $$\xi$$ is a non-square, to observe that the norm of $$a+b\sqrt{\xi}\in\mathbb{F}_{q^2}$$ is $$a^2-\xi b^2$$. The non-homogeneous polynomial above seems hard to me, but I am not an expert in such matters. I am not surprised that $$N$$ is a cubic in $$q$$.

• Why are you interested in that polynomial in particular? – Daniel Hast Jun 15 at 13:06
• @Danial Hast. This polynomial arose from a geometric problem regarding quadratic forms. The size of its fibres seem to be piecewise polynomial. Indeed, this is the first of a series of polynomials that arise essentially from the determinant of a Gram matrix. – Glasby Jun 15 at 14:16

I will concentrate on the case of odd $$q$$ and $$\xi,-\eta$$ being squares, but the solution should be extendable to the remaining cases.

It is convenient to use the language of characters sums in order to compute $$N$$. To reduce your point counting problem to a character sum problem, the following observation is useful: $$\#\{ x \in \mathbb{F}_q : ax^2+bx+c=0\} = 1 + \chi(b^2-4ac),$$ where $$\chi$$ is the unique non-trivial quadratic character of $$\mathbb{F}_q^{\times}$$, extended to give $$0$$ on $$0$$, and $$a \neq 0$$. This observation is proved by completing the square.

We will need the following well-known formula: $$(\sim) \, \sum_{x \in \mathbb{F}_q} \chi(x^2+t) = -1$$ for $$t\neq 0$$. This formula encodes the fact that the number of points on the genus-0 curve $$y^2=x^2+t$$ is $$q-1$$. In your case we are definitely lucky, as the computation of $$N$$ reduces to point counting on genus-0 curves, while for higher genus these point counts are not polynomial in general.

1. (Cosmetic step) Replacing $$(b,d)$$ by $$(b/\sqrt{\xi},d/\sqrt{\xi})$$ we see that we may assume that $$\xi = 1$$. Further replacing $$(c,d)$$ by $$(c\sqrt{-\eta},d\sqrt{-\eta})$$ we see that we may assume that $$-\eta = 1$$ as well.

2. Let us expand your defining hypersurface and express it as a quadratic polynomial in $$a$$:

$$(*) \, a^2(d^2-1) + a(-2bcd) + b^2c^2+b^2-c^2+d^2-1 = 0.$$

1. The case $$d^2 =1$$ simplifies to $$a(2bcd) = b^2c^2+b^2-c^2$$. If $$bc \neq 0$$, this determines $$a$$ uniquely. If $$bc=0$$, we must have $$b=c=0$$, and $$a$$ can be arbitrary. This case yields $$2((q-1)^2 + q)$$ solutions.

2. Let us assume $$d^2 \neq 1$$. The discriminant of $$(*)$$ factorizes as $$4(c^2+1-d^2)(b^2+d^2-1)$$, which is a lucky coincidence, and possibly the heart the matter. Hence we see that given $$b,c,d$$ contribute to $$(*)$$ $$1+\chi( (c^2+1-d^2)(b^2 +d^2-1) )$$ solutions. Summarizing, we have $$(**)\, N=2(q^2-q+1) + q^2(q-2) + \sum_{d^2 \neq 1, \, b,c,d \in \mathbb{F}_q} \chi( (c^2+1-d^2)(b^2 +d^2-1) ).$$

3. By $$(\sim)$$, $$\sum_{b \in \mathbb{F}_q} \chi( (c^2+1-d^2)(b^2 +d^2-1) ) = \chi(c^2+1-d^2) \sum_{b \in \mathbb{F}_q} \chi(b^2 +d^2-1 ) = -\chi(c^2+1-d^2)$$ when $$d^2 \neq 1$$, and so $$\sum_{d^2 \neq 1, \, b,c,d \in \mathbb{F}_q} \chi( (c^2+1-d^2)(b^2 +d^2-1) )= -\sum_{d^2 \neq 1, \, c,d \in \mathbb{F}_q} \chi(c^2+1-d^2),$$ and applying $$(\sim)$$ once more this becomes $$-\sum_{d^2 \neq 1, \, c,d \in \mathbb{F}_q} \chi(c^2+1-d^2) = \sum_{d^2 \neq 1} 1 = q-2.$$ Plugging this character sum evaluation in $$(**)$$, $$N=q^3-q$$ is obtained, confirming your empirical observation.

For general $$\xi$$ and $$\eta$$, a very similar argument will work, because the discriminant of your defining equation (considered as a quadratic polynomial in $$a$$) still factorizes nicely, specifically it is $$4(c^2 - \eta-d^2\xi) (-b^2\xi\eta + d^2 \xi + \eta).$$

Even $$q$$ is easier. The defining equation can now be written as $$(a+1)^2 (\xi d^2 +\eta) = \xi b^2 (c^2+\eta) + c^2.$$ Recall that in $$\mathbb{F}_q$$ with even $$q$$, $$x \mapsto x^2$$ is a field automorphism.

• If $$\xi d^2+\eta \neq 0$$ (happens for all but a unique $$d$$), $$a+1$$ is uniquely determined by $$b,c$$, giving $$(q-1)q^2$$ solutions.
• If $$\xi d^2+\eta = 0$$, $$d$$ is determined uniquely, $$a$$ is arbitrary and it remains to count solutions $$(b,c)$$ to $$c^2(\xi b^2+1) = \eta \xi b^2$$. Specifying $$b$$ determines $$c$$ uniquely, unless $$\xi b^2 + 1=0$$ (happens for a unique $$b$$), in which case there are no solutions. So this case contributes $$q(q-1)$$ solutions.

All in all, $$N = (q-1)q^2+q(q-1) = q^3-q$$ for even $$q$$ as well.