Exercise 6, § 12

Let K be a finite field of q elements where q is not a power of 2. a) Let $a_1, a_2, \ldots, a_n, b$ be elements of K such that $a_1a_2 ... a_n \neq 0$. Show that the number $\nu$ of solutions $(x_1, ..., x_n) \in K^{2m}$ of the equation $a_1x_1^2 + \cdots + a_{2m}x_{2m}^2 = b$ is given by the following formulae : 1" if $b = 0$ and $(-1)^ma_1a_2 ... a_n$ is not a square in K, $\nu = q^{2m-1} - q^m + q^{m-1}$; 2" if $b \neq 0$ and $(-1)^ma_1a_2 ... a_n$ is not a square in K, $\nu = q^{2m-1} + q^{m-1}$; 3" if $b = 0$ and $(-1)^ma_1a_2 ... a_n$ is a square in K, $\nu = q^{2m-1} + q^m - q^m$; 4" if $b \neq 0$ and $(-1)^ma_1a_2 ... a_n$ is a square in K, $\nu = q^{2m-1} - q^{m-1}$. (Argue by induction on m, using Ex. 5.) b) Let $a_1, a_2, \ldots, a_{2m+1}, b$ be elements of K such that $a_1a_2 \ldots a_{2m+1} \neq 0$. Show that the number $\nu$ of solutions $(x_1, x_2, ..., x_{2m-1}) \in K^{2m+1}$ of the equation

is given by the following formulae : 1" if $b = 0$, $\nu = q^{2m}$; 2" if $b \neq 0$ and $(-1)^ma_1a_2 ... a_{2m-1}b$ is not a square in K, $\nu = q^{2m} - q^m$; 3" if $b \neq 0$ and $(-1)^ma_1a_2 ... a_{2m+1}b$ is a square in K, $\nu = q^{2m} + q^m$. (Reduce to the case a.).