Let $\Bbb F_q$ be a finite field with q elements, with q odd. For a quadratic polynomial $Q(T)=T^2+a \in \Bbb F_q[T]$, show that if $a \neq 0$, then: $$|\{(x,y)\in \Bbb F_q^2 : y^2 = Q(x) \}|=q-1$$
If $a$ is a quadratic residue, then we have $q-1$ solutions of the form: $(0,b),(b,0)$ when $b$ is a quadratic residue. How do I show that those are the only solutions? Also, what if $a$ is not a quadratic residue?
Let's look at an example, say $\mathbf{F}_5$ and $Q(x)=x^2+2$. Now $2=1\cdot2=2\cdot1=3\cdot4=4\cdot3$. We can convert each of these products to a solution, and there are $5-1=4$ of them. For example, $2=4\cdot3=(1+3)(1-3)=1^2-3^2$ so that $(3,1)$ is a solution. Similarly $2=1\cdot2=(4+2)(4-2)=4^2-2^2$ so that $(2,4)$ is a solution.
This can be generalized to any finite field of odd characteristic and any $a$. Note that odd characteristic is required when writing each product as the difference of two squares.