Consider the finite field $\mathbb{F}_{q}$, where $q=p^{a}$, $p$ is an odd prime. The question is that in how many ways can one choose three numbers $\lambda_{1}, \lambda_{2}, \lambda_{3}$, such that all three are distinct and $\lambda_{1}\lambda_{2}\lambda_{3}=1$. I tried it in the following way: We can choose $\lambda_{1}$ first in $q-1$ ways. After once we choose $\lambda_{2}\neq \lambda_{1}$, $\lambda_{3}=\lambda_{2}^{-1}\lambda_{1}^{-1}$. But we have to make sure that all are distinct. So, $\lambda_{2}^{-1}\lambda_{1}^{-1}=\lambda_{1} \implies \lambda_{2}=\lambda_{1}^{-2}$. So we have to choose $\lambda_{2}\neq \lambda_{1}^{-2}$. Now also, $\lambda_{2}^{-1}\lambda_{1}^{-1}=\lambda_{2} \implies \lambda_{1}=\lambda_{2}^{-2}$.
This is where I get stuck at and it is not clear to me how to compute further.
Any kind of help will be really appreciated. Thanks in advance!!