Consider the split-octonions $\mathbb{O}$ with coefficients in $\mathbb{F}_q$. Suppose $a \in \mathbb{F}_q$ and $b \in \mathbb{F}_q^*$. I want to find the amount of elements $x \in \mathbb{O}$ such that $\overline{x} x = ac^{-1}$. As far as I understand, this reduces to finding the number of solutions of $$ x_0^2 + x_1^2 + ... + x_7^2 = ac^{-1} $$ for each choice of $a$ and $c$ over $\mathbb{F}_q$, but I don't know how to count these. Moreover, I even know that the answer is $q^7-q^3$ (I am reading a paper, where this is stated but never proved), but I cannot understand why. Could you please help?
Here is some kind of an insight: there are $q$ choices for $a$ and $q-1$ choices for $c$. Thus, I need to prove that the given equation has $q^2+q^3+q^4+q^5$ solutions over $\mathbb{F}_q$.
UPD: The paper I am talking about is http://arxiv.org/abs/1310.5886. Basically, I am dealing with the proof of Theorem $1$ on page $10$.
Live and learn. In Ireland and Rosen, A Classical Introduction to Modern Number Theory (Second edition), on page 102 we find Proposition 8.6.1. Evidently $p$ is an odd prime and we are in the field $\mathbb Z / p \mathbb Z.$ If $r$ is even, the number of solutions of $$ x_1^2 + \cdots + x_r^2 = 1 $$ is $$ p^{r-1} - (-1)^{(r/2)((p-1)/2)} p^{(r/2)-1}, $$ so there is your $p^7 - p^3$ for one important case.
I see no difference if the number represented is a quadratic residue $\pmod p$ rather than $1.$ Not entirely sure how many representations of $0$ there are, so there is more work to be done to count representations of non-residues. Good start, though.
Borrowed Arithmetic of Finite Fields by Charles Small. In a finite field $\mathbb F$ with odd number $q$ elements, $q$ a prime power, and $n$ divisible by $4,$ the number of solutions to $$ x_1^2 + \cdots x_n^2 = b $$ is $$ q^{n-1} - q^{(n-2)/2} $$ for nonzero $b,$ while the number of solutions to $$ x_1^2 + \cdots x_n^2 = 0 $$ is $$ q^{n-1} + q^{n/2} - q^{(n-2)/2} $$ This is Theorem 4.5 on page 91, with necessary definitions on pages 86 and 88.