Let $q$ be an odd prime power and consider $\mathbb{F}_q^*$
I am working on something that has led me to the following question:
How many pairs $(a,b) \in \mathbb{F}_q^* \times \mathbb{F}_q^*$ are there such that $ab$ is a square (in $\mathbb{F}_q^*$)?
I know the number of elements $a \in \mathbb{F}_q^*$ such that $a$ is a square, is $\frac{1}{2}(q-1)$, noting that $q$ is odd. And by some reverse engineering of what I'm working on, I have a feeling the answer to my question is $\frac{1}{2}(q-1)^2$, but I don't know how to show this.
Any help appreciated, it may be obvious but I'm a bit brain-tired right now. Thanks!
For any fixed $a \in \mathbb F_q^*$, the map $\mathbb F_q^* \to \mathbb F_q^*: x \mapsto ax$ is a bijection, thus the number of elements $b \in \mathbb F_q^*$ such that $ab$ is square is the same as the number of squares in $\mathbb F_q^*$.
I think the rest is clear.