Exercise 5, § 12

Let K be a finite field of q elements where q is not a power of 2, and let $a_1, a_2, b$ be three elements of K such that $a_1a_2 \neq 0$. Show that the number $\nu$ of solutions $(x_1, x_2) \in K^2$ of the equation $a_1x_1^2 + a_2x_2^2 = b$ is given by the following formulae : 1" if $b = 0$ and $-a_1a_2$ is not a square in K, $\nu = 1$; 2" if $b \neq 0$ and $-a_1a_2$ is not a square in K, $\nu = q + 1$; 3" if $b = 0$ and $-a_1a_2$ is a square in K, $\nu = 2q - 1$; 4" if $b \neq 0$ and $-a_1a_2$ is a square in K, $\nu = q - 1$. (When $-a_1a_2$ is not a square, adjoin to K a root of $X^2 + a_1a_2$ and use the fact that the multiplicative group of a finite field is cyclic.)