For which elements $t$ in a finite field $\mathbb{F}_{p^n}$ is $t^2 - 4$ a square?

Question

That is, how to characterize the elements $t \in \mathbb{F}_{p^n}$ for which there exists $x \in \mathbb{F}_{p^n}$ such that $t^2 - 4 = x^2$?

Answer 1

If $p = 2$, then any $t$ will do. Hence, there are $2^n$ possible values of $t$ in this case.

Now assume that $p>2$. If $x^2=t^2-4$, then $(t-x)(t+x)=t^2-x^2=4$. Hence, $t-x$ can be any nonzero element $2u \in \mathbb{F}_{p^n}$ and $t+x=2u^{-1}$. Therefore, $t=u+u^{-1}$ for some nonzero $u \in \mathbb{F}_{p^n}$. Conversely, if $t=u+u^{-1}$ for some $u\in\mathbb{F}_{p^n}\setminus\{0\}$, then $t^2-4=\left(u-u^{-1}\right)^2$. There are $\frac{p^n+1}{2}$ possible values of $t$.