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Suppose $x,y,z,w \in \mathbb{F}_{q^2}$, where $q=p^k$ for some prime $p$. Consider the system of equations $$ \left\{ \begin{array}{l} xy + zw = 0; \\ xy^q + yx^q + zw^q + wz^q = 0. \end{array} \right. $$

Is it possible to say something about the number of solutions (excluding the trivial one) over $\mathbb{F}_{q^2}$? I am not that familiar with the finite field arithmetics, so sorry if this is something known.

If we substitute $xy$ from the first equation to the second, we get $$ (xy)(x^{q-1}+y^{q-1}-z^{q-1}-w^{q-1}) = 0, $$ but this does not make it easier for me. Thanks in advance for your help and suggestions!

UPD: By performing some computations in Mathematica, I have found that for $q=2,3,5,7$ the number of solutions takes respectively values $64,513,6625,35329$ which suggests that the number of solutions in $\mathbb{F}_{q^2}$ must be $q^3(2q^2+q-2)$, however I still don't know how to prove this in general.

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To just get started, consider the fact that if $(a,b,c,d)$ is a solution, then so is $(\lambda a, \lambda b, \lambda c, \lambda d)$ for every $\lambda \in \mathbb{F}_{q^{2}}$. Use this to consider two cases: first, $x = 0$ (this is rather easy; I will leave this part to you). Second, if $x \neq 0$, then you can assume $x = 1$. At the end, you can multiply the number of solutions with $x = 1$ by $q^{2}-1$ to get the total number of solutions with $x \neq 0$.

Addressing the second case, taking $x = 1$ we then have $y = -zw$ in order for the first equation to hold. Then the second equation becomes $$-z^{q}w^{q} - zw +zw^{q}+z^{q}w=0.$$ This simplifies to $$(z-z^{q})(w^{q}-w) = 0,$$ implying that either $z = z^{q}$ or $w = w^{q}$. This happens when $z \in \mathbb{F}_{q}$ or $w \in \mathbb{F}_{q}$, respectively, so there are $2q^{3}-q^{2}$ such solutions.

Remember this only counts solutions with $x = 1$, so we have to multiply these by $(q^{2}-1)$ to get a total of $2q^{3}(q^{2}-1)$ solutions with $x \neq 0$. Then there are $2q^{4}-q^{2}$ further solution with $x=0$. This gives a total of $$(2q^{3}-q^{2})(q^{2}-1)+2q^{4}-q^{2} = q^{3}(2q^{2}+q-2)$$ solutions.

For a more technical description of this problem, you can think of your vectors $(x,y,z,w)$ as being points in a projective $3$-space $PG(3,q)$; this means that two vectors represent the same point iff they are a scalar multiple of each other (0 vector does not represent a point). There are $q^3+q^2+q+1$ points. Under this model you are looking for the intersection between a hyperbolic quadric (contains $(q^{2}+1)^{2}$ points) and a Hermitian surface (contains $(q^{2}+1)(q^{3}+1)$ points). Each point in this intersection then corresponds to $q^{2}-1$ nonzero vectors that satisfy both equations.

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  • $\begingroup$ Can you suggest some literature? Actually, I am trying to consider the four different possibilities: 1) $xy = xy^q + x^qy = 0, zw = zw^q + z^qw$; 2) $xy = 0, xy^q + x^qy = a, zw = 0, zw^q + z^qw = -a$; 3) $xy = b, xy^q + x^qy = 0, zw = -b, zw^q + z^qw = 0$; 4) $xy = b, xy^q + x^qy = a, zw = -b, zw^q + z^qw = -a$. The last case is the hardest one, so I am having huge problems with it. $\endgroup$ – kissanpentu May 27 '16 at 23:50
  • $\begingroup$ Thank you very much for such a brilliant answer! However, could you still recommend some references for the counting techniques in projective spaces as I am trying to find an inductive technique to count the points in $2k$-dimensional space which satisfy both hermitian form and the quadratic form of hyperbolic type (by "satisfy" I mean being isotropic)? $\endgroup$ – kissanpentu May 28 '16 at 12:46
  • $\begingroup$ And tiny correction: when $x=0$ there are actually $2q^2-1$ solutions for the remaining system (not $2q^2-2$, since we only need to count the case $z=0,w=0$ once), so there are $2q^4-q^2$ solutions in total for this case. But anyway, thanks, you have saved the day for me :) $\endgroup$ – kissanpentu May 28 '16 at 13:38
  • $\begingroup$ @kissanpentu Maybe a good reference, it's extensive though. Feel free to skip ahead to what interests you. math.ucdenver.edu/~spayne/classnotes/topics.pdf $\endgroup$ – Morgan Rodgers May 29 '16 at 14:53

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