How many different solutions does $x^2_{1}+2x^2_{2}+3x^3_{3}=4x^2_{4} $ have? How many different solutions does 
$$x^2_{1}+2x^2_{2}+3x^3_{3}=4x^2_{4}$$
have over the field with $13$ elements $\mathbb F_{13}$?
I don't know how to approach to this problem. Need help
 A: Let's start by slightly rearranging:
$$x_1^2 + 2x_2^2 - 4x_3^2 = -3x_4^3$$
to get the cubic term on the right. 
Now fix a value of $x_4$ and write $c = -3x^4$. If we count the number of solutions to 
$$x_1^2 + 2x_2^2 - 4x_3^2 = c$$
for each value of $x_4$, and sum them, we get the total count.
It is easily seen that for each value $c$ there is some solution: all elements of $\mathbb F_{13}$ (so certainly all possible values of $c$) are of the form $x^2$ or $2x^2$, since $2$ is not a square.
If we now homogenize the equation to
$$x_1^2 + 2x_2^2 - 4x_3^2 = cw^2$$
for $c \ne 0$ we get the equation for a nonsingular projective quadric surface with a rational point, and a rational curve given by
$$x_1^2 + 2x_2^2 - 4x_3^2 = 0,\ w = 0$$
at infinity.
This leaves us with $|\mathbb A^2(\mathbb F_{13})| = 13^2$ points on the original affine quadric independent of the value of $c \ne 0$. 
When $x_4 = c = 0$ the equation is homogeneous already, and defines a nonsingular rational projective curve with a rational point, giving $13 + 1 = 14$ points on the projective curve. Each of these corresponds to 12 solutions to the original equation (by multiplication with an element of $\mathbb F_{13}^\ast$), and finally with the solution $(0,0,0)$ we have 169 solutions again.
This gives a total of $13\cdot 169 = 2197$ points.
EDIT As Jyrki Lahtonen pointed out, this is not quite correct. It can be salvaged along the same lines though. 
All non-tangent lines through the rational point intersect the quadric in exactly one other point, giving a bijection between points not in the tangent plane and $\mathbb A^2(\mathbb F_{13})$. I had erroneously assumed that the points in the tangent plane formed a rational curve, but in fact it is degenerate. 
For different $c$ the number of solutions is different. The case of $c = 0$ is discussed in the preceding. By linear changes of coordinates all cases where $c\ne 0$ is a square are equivalent, as are the cases where $c$ is not a square. It is immediately seen that both cases occur equally often (6 each).
The two possiblilities are two intersecting (projective) lines, or the single point, giving either 1 or 27 additional points. Subtracting the 14 points at infinity gives a total of $169 \pm 13$, and that should depend on $c$ being a square or not. I'll get back later to fill in the detail, but I got to run now.
Note that applying the results in the links by Jyrki Lahtonen the answer follows as well.
A: You ought to be able to do this systematically.
The cubes modulo $13$ are $0,\pm 1, \pm 5$ - since $13-1=12$ there are just $12/3=4$ non-zero cubes, each with three cube roots. We also have $0^3=0$.
Note that $-1$ is a quadratic residue modulo $13$ so we can solve $y_4^2=-4x_4^2$ for any $x_4$.
We can rearrange the equation to be $$y_1^2+2y_2^2+y_4^2=3y_3^3=0, \pm 3,\pm 2$$
The squares modulo $13$ are $0,\pm 1, \pm 3,\pm 4$ so you have $$\{0,\pm 1,\pm 3,\pm 4\}+\{0,\pm 2,\pm 6, \pm 5\}+\{0,\pm 1,\pm 3,\pm 4\}=\{0,\pm 2\ , \pm 3\}$$
Each non-zero square has two square roots. Each non-zero cube has three cube roots. If $y_4\neq 0$ there are two possible values for $x_4$.
So we have $$0+0+0=0$$$$0+0\pm 3=\pm 3$$ (this gives twelve solutions when signs are chosen and roots are taken) $$0\pm 2+0=\pm 2$$$$0\pm 2\pm 1=\pm 3$$$$0\pm 6\mp3=\pm 3$$$$0\pm 6\mp 4=\pm 2$$
etc
You can reduce the task by exploiting the symmetry between $y_1$ and $y_4$.
Also starting with the cubes gives just three possibilities to consider, so it may probably more efficient to write
$$\{0,\pm 2,\pm 3\}+\{0,\pm 1,\pm 3,\pm 4\}+\{0,\pm 1,\pm 3,\pm 4\}+\{0,\pm 2,\pm 5, \pm 6\}=0$$ and look for solutions $$\pm a\pm b\pm c\pm d=0$$ with $b\le c$ and count a factor of two whenever $b\lt c$
With this you get $$(0,0,0,0); (0\pm 1,\mp 1,0); (0,\pm 1,\pm 1\mp 2);(0,\pm 1, \pm 4,\mp 5); (0,\pm 3, \pm 3, \mp 6); (0,\pm 3,\pm 4, \pm 6);(0,\pm 4,\pm 4, \pm 5)$$ and then other solutions for other choices of $a$.
Listing takes some care, but there are only $30$ choices from the first three brackets which have $b\le c$
A: Let $Q(x)$ be a form of degree $d$ on $F^n$, and let $k$ be a positive integer. We map bijectively the set $\{(x,y) \in F^{n+1}\ | Q(x) = y^{kd+1}\}$ onto $\{(x,y) \in F^{n+1} \ | Q(x) = y\}$ as follows: if $y= 0$ then $(x,0) \mapsto (x,0)$. If $y\ne 0$ then $(x,y) \mapsto (\frac{x}{y^k}, y)$. We conclude that the set has $q^n$ elements.
