I am stuck on the following problem. Let $GF(q)$ be the finite field of order $q$, where $q$ is an odd prime power. How can I show by elementary methods that the number of points $(x,y,z)$ satisfying
$x^2+y^2+z^2=d$,
where $-d \neq 0$ is a non-square in $GF(q)$, is equal to $q^2-q$?
How elementary? Do you want to ignore all geometric/theory of quadratic forms stuff?
If you want to be completely straightforward, consider the quadratic form $Q:(x,y,z) \mapsto x^{2}+y^{2}+z^{2}$. You want show to that of the $q^{3}$ vectors in $\mathbb{F}_{q}^{3}$, $q^2$ have $-Q(v)=0$, $\frac{1}{2}(q^{3}-q)$ have $-Q(v)$ a nonzero square, and $\frac{1}{2}q(q-1)^{2}$ have $-Q(v)$ nonsquare. Then you want to show that for each of the $\frac{1}{2}(q-1)$ choices of a nonsquare $d$, the same number of vectors $v$ have $-Q(v) = d$.
The easiest way is to work in the projective plane $\mathrm{PG}(2,q)$, where the points are taken to be the normalized vectors (those of the form $(1,y,z)$, $(0,1,z)$, or $(0,0,1)$, there are $q^2+q+1$ of these), with quadratic form $Q(v) = v^{T}v$. The zeroes of this quadratic form give a conic in the plane, having $q+1$ points. Multiplying one of these special vectors by a scalar doesn't effect whether the quadratic form is $0$/square/nonsquare, so this corresponds to $(q+1)(q-1)+1 = q^2$ total vectors giving $0$.
You want to show that, of the remaining $q^{2}$ points, if the point is represented by a normalized vector $v$ with $-Q(v)$ a nonsquare, then there are precisely two scalars (call them $\pm \lambda)$ with $-Q(\lambda v) = d$. Then you want to show that there are precisely $\frac{1}{2}(q^{2}-q)$ such points. This is usually done through geometric arguments, looking at interior and exterior points of the conic (interior points are on no tangent of the conic, exterior points are on two tangents of the conic). Once you have counted the number of interior/exterior points, you can show that they are characterized by whether $-Q(v)$ is a nonsquare/nonzero square.